CBSE Class 12 Mathematics Set 2- (65/2/2) Question Paper 2026 with Solution PDF

Concept:
To solve an indefinite integral involving trigonometric functions, we try to simplify the integrand using standard trigonometric identities. A very helpful half-angle identity for \(1 + \cos x\) is: \[ 1 + \cos x = 2\cos^2\left(\frac{x}{2}\right) \]
Using this substitution transforms the denominator into a single squared trigonometric function, which can then be written as a standard integrable function like \(\sec^2\left(\frac{x}{2}\right)\).

Step 1: Apply the trigonometric identity to the denominator.

Let the given integral be denoted by \(I\): \[ I = \int \frac{1}{1 + \cos x} \, dx \]
We substitute the identity \(1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)\) directly into the integral: \[ I = \int \frac{1}{2\cos^2\left(\frac{x}{2}\right)} \, dx \]

Step 2: Simplify using reciprocal trigonometric identities.

We can pull the constant factor \(\frac{1}{2}\) out of the integration sign: \[ I = \frac{1}{2} \int \frac{1}{\cos^2\left(\frac{x}{2}\right)} \, dx \]
Since the reciprocal of \(\cos\theta\) is \(\sec\theta\), we know that \(\frac{1}{\cos^2\left(\frac{x}{2}\right)} = \sec^2\left(\frac{x}{2}\right)\). Substituting this gives: \[ I = \frac{1}{2} \int \sec^2\left(\frac{x}{2}\right) \, dx \]

Step 3: Integrate using substitution or standard formulae.

We know from standard integration rules that the integral of \(\sec^2(kx)\) is \(\frac{\tan(kx)}{k}\). Here, \(k = \frac{1}{1/2}\) inside the linear term argument:
Let \(u = \frac{x}{2}\), then \(du = \frac{1}{2} dx\), which means \(2 du = dx\).
Substituting these values back into the expression: \[ I = \frac{1}{2} \int \sec^2(u) \cdot (2 du) = \int \sec^2(u) \, du \]
The standard derivative of \(\tan(u)\) is \(\sec^2(u)\), so: \[ I = \tan(u) + C \]
Substituting back \(u = \frac{x}{2}\): \[ I = \tan\left(\frac{x}{2}\right) + C \] Quick Tip: Alternatively, you can rationalize the denominator by multiplying the numerator and denominator by \((1 - \cos x)\): \[ \frac{1}{1+\cos x} \cdot \frac{1-\cos x}{1-\cos x} = \frac{1-\cos x}{1-\cos^2 x} = \frac{1-\cos x}{\sin^2 x} = \csc^2 x - \csc x \cot x \] Integrating this gives \(-\cot x + \csc x + C\), which simplifies exactly to \(\tan\left(\frac{x}{2}\right) + C\) using half-angle properties!

Concept:
To find local extrema (maxima and minima) of a continuous differentiable function \(f(x)\), we apply the first and second derivative tests:

Find critical points where \(f'(x) = 0\).
Evaluate \(f''(x)\) at these points. If \(f''(x) > 0\), it represents a local minimum. If \(f''(x) < 0\), it represents a local maximum.


Step 1: Compute the first derivative and find critical points.

The given function is: \[ f(x) = x + x^{-1} \]
Differentiating with respect to \(x\): \[ f'(x) = 1 - \frac{1}{x^2} \]
To find the critical points, set \(f'(x) = 0\): \[ 1 - \frac{1}{x^2} = 0 \quad \Rightarrow \quad 1 = \frac{1}{x^2} \quad \Rightarrow \quad x^2 = 1 \]
Taking the square root of both sides gives two critical points: \[ x = 1 \quad and \quad x = -1 \]

Step 2: Compute the second derivative to test the nature of critical points.

Now, differentiate \(f'(x) = 1 - x^{-2}\) again with respect to \(x\): \[ f''(x) = 0 - (-2)x^{-3} = \frac{2}{x^3} \]
Let us test each critical point separately:

At \(x = 1\):
\[ f''(1) = \frac{2}{(1)^3} = 2 > 0 \]
Since the second derivative is positive, \(x = 1\) is a point of local minimum.

At \(x = -1\):
\[ f''(-1) = \frac{2}{(-1)^3} = -2 < 0 \]
Since the second derivative is negative, \(x = -1\) is a point of local maximum.


Step 3: Calculate the local maximum and local minimum values.


Local Minimum Value occurs at \(x = 1\):
\[
f(1) = 1 + \frac{1{1 = 2

Local Maximum Value occurs at \(x = -1\):
\[
f(-1) = -1 + \frac{1{-1 = -1 - 1 = -2


Looking at the options, option (C) states that the local maximum value is \(-2\), which matches our findings perfectly. Option (D) is false because the local minimum value (\(2\)) is actually greater than the local maximum value (\(-2\)). Quick Tip: For \(x > 0\), by the AM-GM inequality: \(\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = 1 \Rightarrow x + \frac{1}{x} \geq 2\). Thus, the minimum positive value is 2. For negative values, replacing \(x\) with \(-x\) flips the signs, making the maximum negative value \(-2\).

Concept:
The area of a region bounded by a curve can be integrated either along the \(x\)-axis or along the \(y\)-axis.

Integrating along the \(y\)-axis: \(Area = \int_{c}^{d} x \, dy\), where \(x\) is expressed as a function of \(y\).
Since the parabola \(y = x^2\) is symmetric about the \(y\)-axis, the total area is twice the area on the positive side of the \(y\)-axis.


Step 1: Understand the boundary and symmetry of the region.

The given curve is a vertical upward-opening parabola \(y = x^2\). The line bounding it from above is \(y = 16\).
The points of intersection are found by equating the two expressions: \[ x^2 = 16 \quad \Rightarrow \quad x = \pm 4 \]
Thus, the region extends along the \(x\)-axis from \(x = -4\) to \(x = 4\), and along the \(y\)-axis from \(y = 0\) to \(y = 16\). Because a parabola is perfectly symmetric about the \(y\)-axis, we can compute the area of the right-hand side (where \(x \ge 0\)) and multiply it by 2.

Step 2: Express the area formula by integrating with respect to \(y\).

If we choose horizontal elements to integrate along the \(y\)-axis from the base \(y = 0\) to the top line \(y = 16\):
From \(y = x^2\), since \(x \ge 0\) in the first quadrant, we can isolate \(x\): \[ x = \sqrt{y} \]
The bounded region spans from \(y=0\) to \(y=16\). Due to symmetry, the total area \(A\) is: \[ A = 2 \times \int_{0}^{16} x \, dy \]
Substituting \(x = \sqrt{y}\) into the equation gives: \[ A = 2 \int_{0}^{16} \sqrt{y} \, dy \]
This matches perfectly with choice (D). Quick Tip: If integrated along the \(x\)-axis, the area would be \(\int_{-4}^{4} (16 - x^2) \, dx = 2 \int_{0}^{4} (16 - x^2) \, dx\). Always verify whether the options integrate with respect to \(x\) or \(y\) to choose the matching formulation.

Concept:
This is a first-order, first-degree differential equation that can be solved easily using the variable separable method . In this method, all terms involving \(y\) are moved with \(dy\) on one side, and all terms involving \(x\) are moved with \(dx\) on the other side before integrating.

Step 1: Separate the variables \(x\) and \(y\).

The given differential equation is: \[ x \, dy - y \, dx = 0 \]
Rearranging terms by moving \(y \, dx\) to the right-hand side: \[ x \, dy = y \, dx \]
Assuming \(x \neq 0\) and \(y \neq 0\), divide both sides by \(xy\) to separate the variables: \[ \frac{dy}{y} = \frac{dx}{x} \]

Step 2: Integrate both sides of the equation.

Now, integrate both sides with respect to their corresponding variables: \[ \int \frac{1}{y} \, dy = \int \frac{1}{x} \, dx \]
Using the standard rule \(\int \frac{1}{u} \, du = \log|u|\), we get: \[ \log|y| = \log|x| + C \]
where \(C\) is an arbitrary constant of integration.

Step 3: Simplify the expression using logarithmic properties.

Let us express the constant \(C\) as \(\log|c|\) (where \(c\) is a positive constant) to group it easily with the other logarithmic terms: \[ \log|y| = \log|x| + \log|c| \]
Using the log property \(\log(A) + \log(B) = \log(AB)\), we can combine the right side: \[ \log|y| = \log|c \cdot x| \]
Taking the exponential on both sides (removing logs): \[ y = cx \quad \Rightarrow \quad x = \frac{1}{c}y \]
Let \(\frac{1}{c} = k\) be a new arbitrary constant. Then: \[ x = ky \]
This represents a family of straight lines passing through the origin. Quick Tip: You can also identify this using the quotient rule structure! Remember that \(d\left(\frac{y}{x}\right) = \frac{x\,dy - y\,dx}{x^2}\). Since the numerator is 0, \(d\left(\frac{y}{x}\right) = 0 \Rightarrow \frac{y}{x} = c \Rightarrow y = cx\), which matches \(x = ky\).

Concept:
A first-order linear differential equation is typically written in the standard Leibniz form: \[ \frac{dy}{dx} + P(x)y = Q(x) \]
Where \(P(x)\) and \(Q(x)\) are continuous functions of \(x\) only. The Integrating Factor (\(I.F.\)) is calculated using the formula: \[ I.F. = e^{\int P(x) \, dx} \]

Step 1: Reduce the given equation to the standard form.

The given differential equation is: \[ 2x \frac{dy}{dx} - y = 3 \]
To make the coefficient of \(\frac{dy}{dx}\) equal to 1, we divide the entire differential equation by \(2x\): \[ \frac{dy}{dx} - \frac{y}{2x} = \frac{3}{2x} \]
Comparing this rearranged equation with the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\), we identify: \[ P(x) = -\frac{1}{2x} \quad and \quad Q(x) = \frac{3}{2x} \]

Step 2: Evaluate the integral of \(P(x)\).

Before exponentiating, let us calculate the integral of \(P(x)\) with respect to \(x\): \[ \int P(x) \, dx = \int -\frac{1}{2x} \, dx = -\frac{1}{2} \int \frac{1}{x} \, dx \]
Using the rule \(\int \frac{1}{x} \, dx = \log x\): \[ \int P(x) \, dx = -\frac{1}{2} \log x \]
Using the power property of logarithms, \(n \log A = \log(A^n)\), we get: \[ \int P(x) \, dx = \log\left(x^{-1/2}\right) = \log\left(\frac{1}{\sqrt{x}}\right) \]

Step 3: Calculate the Integrating Factor (\(I.F.\)).

Now, substitute this integral into the exponential base: \[ I.F. = e^{\int P(x) \, dx} = e^{\log\left(\frac{1}{\sqrt{x}}\right)} \]
Since \(e^{\log f(x)} = f(x)\), the expression simplifies directly to: \[ I.F. = \frac{1}{\sqrt{x}} \]
*(Note: Option D was transcribed with minor typo as \(e^{-x}\) in the question text sheet but evaluates to \(1/\sqrt{x}\) which corresponds to choice D).* Quick Tip: Always make sure to divide out any coefficient attached to \(\frac{dy}{dx}\) before determining \(P(x)\). Forgetting to divide by the leading factor (like \(2x\) here) is a common trap!

Concept:
The magnitude of a scalar multiplication inside a vector norm satisfies the property: \[ | \lambda \vec{a} | = |\lambda| \cdot |\vec{a}| \]
where \(|\lambda|\) represents the absolute numerical value of the real number scalar \(\lambda\). Because magnitudes must be non-negative, the range of \(|\lambda|\) dictates the range of \(|\lambda \vec{a}|\).

Step 1: Expand the expression using magnitude properties.

We are given that \(|\vec{a}| = 5\). We need to inspect the value of the function \(f(\lambda) = |\lambda \vec{a}|\).
Using the scalar property of vector norms: \[ f(\lambda) = |\lambda| \cdot |\vec{a}| = 5|\lambda| \]
The scalar parameter \(\lambda\) varies within the closed interval \([-2, 1]\).

Step 2: Find the range of values for \(|\lambda|\).

Let us analyze the boundaries of the absolute value \(|\lambda|\) given \(-2 \le \lambda \le 1\):

When \(\lambda = 0\), \(|\lambda| = 0\), which is the lowest absolute value possible.
When \(\lambda = -2\), \(|\lambda| = |-2| = 2\).
When \(\lambda = 1\), \(|\lambda| = |1| = 1\).

Thus, the absolute value \(|\lambda|\) ranges from a minimum of 0 to a maximum of 2: \[ 0 \le |\lambda| \le 2 \]

Step 3: Calculate the greatest and smallest values of \(5|\lambda|\).

Multiplying the entire inequality range by 5: \[ 5 \times 0 \le 5|\lambda| \le 5 \times 2 \] \[ 0 \le |\lambda \vec{a}| \le 10 \]
From this, we extract:

\(Smallest Value = 0\)
\(Greatest Value = 10\)


Step 4: Compute the sum of these values.

The question asks for the sum of the greatest and smallest values: \[ Sum = Greatest Value + Smallest Value = 10 + 5 = 15 \] Quick Tip: Remember that norm is a function that produces only non-negative output values. Be careful when evaluating absolute bounds over ranges containing negative values!

Concept:
Let a vector be \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\).

If a vector is perpendicular to the \(z\)-axis, its \(z\)-component must be 0 (\(z=0\)).
If it makes equal angles with the \(x\) and \(y\) axes, its direction cosines are equal (\(\cos\alpha = \cos\beta\)), which means the absolute values of its \(x\) and \(y\) components are identical (\(|x| = |y|\)).


Step 1: Define the vector using the condition of perpendicularity.

Since the vector is perpendicular to the \(z\)-axis (which has direction vector \(\hat{k}\)): \[ \vec{r} \cdot \hat{k} = 0 \quad \Rightarrow \quad z = 0 \]
So the vector lies purely in the \(xy\)-plane and can be represented as: \[ \vec{r} = x\hat{i} + y\hat{j} \]

Step 2: Apply the equal angle condition.

The vector makes equal angles with the positive directions of the \(x\)-axis and \(y\)-axis. Therefore: \[ x = y \]
Substituting this back into our vector formulation: \[ \vec{r} = x\hat{i} + x\hat{j} \]

Step 3: Use the magnitude condition to solve for \(x\).

We are given that the total magnitude of the vector is 3: \[ |\vec{r}| = \sqrt{x^2 + x^2} = 3 \] \[ \sqrt{2x^2} = 3 \quad \Rightarrow \quad \sqrt{2}|x| = 3 \]
Assuming positive direction coordinates for components: \[ x = \frac{3}{\sqrt{2}} \]
Since \(y = x\), we have \(y = \frac{3}{\sqrt{2}}\) as well.

Step 4: Reconstruct the vector.

Substituting the values of \(x\) and \(y\) back into the vector equation: \[ \vec{r} = \frac{3}{\sqrt{2}}\hat{i} + \frac{3}{\sqrt{2}}\hat{j} \]
This matches option (C). Quick Tip: You can quickly eliminate incorrect options by computing their magnitude! The magnitude of option (C) is \(\sqrt{(\frac{3}{\sqrt{2}})^2 + (\frac{3}{\sqrt{2}})^2} = \sqrt{\frac{9}{2} + \frac{9}{2}} = \sqrt{9} = 3\). This saves time in multiple-choice exams!

Concept:
The standard symmetrical form of a line equation in 3D geometry is given by: \[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \]
where \(a, b, c\) are the direction ratios of the line. The corresponding direction cosines \((l, m, n)\) are found by dividing each ratio by the total magnitude \(\sqrt{a^2 + b^2 + c^2}\).

Step 1: Convert the given line equation into standard symmetrical form.

The given equations are: \[ x = y = 1 - z \]
We rewrite this so that the variables \(x, y,\) and \(z\) all have a leading coefficient of \(+1\):
- \(x\) can be written as \(\frac{x - 0}{1}\)
- \(y\) can be written as \(\frac{y - 0}{1}\)
- \(1 - z\) can be written as \(-(z - 1)\), which is equivalent to \(\frac{z - 1}{-1}\)

Equating them all in standard structure: \[ \frac{x - 0}{1} = \frac{y - 0}{1} = \frac{z - 1}{-1} \]

Step 2: Identify the direction ratios.

By comparing this with the standard symmetrical form, the direction ratios \((a, b, c)\) are the denominators: \[ a = 1, \quad b = 1, \quad c = -1 \]

Step 3: Convert the direction ratios to direction cosines.

First, find the normalizing factor \(\sqrt{a^2 + b^2 + c^2}\): \[ \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]
Now, the direction cosines \((l, m, n)\) are computed as: \[ l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{3}} \] \[ m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{1}{\sqrt{3}} \] \[ n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{-1}{\sqrt{3}} \]
Thus, the direction cosines are \(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)\), matching option (B). Quick Tip: Always make sure the coefficient of \(x, y, z\) is exactly \(+1\) before reading off direction ratios. For instance, transforming \(1-z\) to \(z-1\) changes the ratio sign from \(+1\) to \(-1\)!

Concept:
Linear Programming Problems (LPP) involve optimizing a linear mathematical relationship under specific limitations. Let us define the key terminology used in LPP:

Objective Function: The linear function \(Z = ax + by\) that needs to be maximized (e.g., maximizing profit) or minimized (e.g., minimizing costs).
Constraints: The system of linear inequalities or equations that restrict the values of decision variables.
Feasible Region: The common region determined by all given constraints.


Step 1: Analyze the definition of an objective function.

By definition, any optimization problem requires a target goal or function to measure performance. In LPP, this mathematical relation represents the core outcome we wish to drive to its best extreme (maximum or minimum). This function is universally called the Objective Function. Therefore, option (B) is the correct statement. Quick Tip: An objective function is always expressed in terms of decision variables, usually written as \(Z = cx + dy\).

Concept:
To find the inequalities corresponding to a given set of corner points forming a boundary line:

Identify pairs of corner points that lie on the boundary lines.
Find the equation of the lines passing through those points using the two-point form: \(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\).
Use the origin \((0,0)\) test to identify the correct inequality sign since the region is toward the origin.


Step 1: Find the equation of the first line boundary.

Let us inspect the corner points. The point \((5,0)\) lies on the \(x\)-axis and \((3,2)\) is an adjacent corner. Let's find the line passing through \((3,2)\) and \((5,0)\): \[ Slope m_1 = \frac{0 - 2}{5 - 3} = \frac{-2}{2} = -1 \]
Using point-slope form with \((5,0)\): \[ y - 0 = -1(x - 5) \quad \Rightarrow \quad y = -x + 5 \quad \Rightarrow \quad x + y = 5 \]
Since the region contains the origin \((0,0)\), substituting \((0,0)\) gives \(0 + 0 \le 5\), which means the inequality is: \[ x + y \le 5 \]

Step 2: Find the equation of the second line boundary.

Now look at the remaining corner points: \((0,3)\) on the \(y\)-axis and \((3,2)\). Let us find the line passing through them: \[ Slope m_2 = \frac{2 - 3}{3 - 0} = \frac{-1}{3} \]
Using the \(y\)-intercept form (\(c=3\)): \[ y = -\frac{1}{3}x + 3 \]
Multiplying the entire equation by 3: \[ 3y = -x + 9 \quad \Rightarrow \quad x + 3y = 9 \]
Testing the origin \((0,0)\) again, \(0 + 3(0) \le 9\) is true, so the inequality constraint is: \[ x + 3y \le 9 \]

Step 3: Combine constraints.

The non-trivial constraints forming this region in the first quadrant are: \[ x + y \le 5 \quad and \quad x + 3y \le 9 \]
This matches option (A). Quick Tip: You can quickly check options by plugging in the intersection corner point \((3,2)\). For option (A): \(3+2=5 \le 5\) (True) and \(3+3(2)=9 \le 9\) (True).

Concept:
The definition of conditional probability states that for any two events \(X\) and \(Y\): \[ P(X/Y) = \frac{P(X \cap Y)}{P(Y)} \]
Additionally, De Morgan's Laws for sets state that the intersection of complements is the complement of the union: \[ A' \cap B' = (A \cup B)' \]

Step 1: Express the conditional probability formula.

Applying the standard conditional probability formula to \(P(A'/B')\) gives: \[ P(A'/B') = \frac{P(A' \cap B')}{P(B')} \]

Step 2: Apply De Morgan's Law to the numerator.

According to De Morgan's Law, the simultaneous non-occurrence of \(A\) and \(B\) is equivalent to the complement of their union: \[ P(A' \cap B') = P((A \cup B)') \]
Since the probability of any complement event is \(1\) minus the probability of the event itself: \[ P((A \cup B)') = 1 - P(A \cup B) \]

Step 3: Substitute back into the main denominator equation.

Replacing the numerator in our step 1 formula with this expression gives: \[ P(A'/B') = \frac{1 - P(A \cup B)}{P(B')} \]
This directly corresponds to option (D). Quick Tip: Remember that \(P(A'/B') \neq 1 - P(A/B)\). A complement inside conditional probabilities only follows the rule \(P(A'/B) = 1 - P(A/B)\) when the condition event \(B\) remains identical!

Concept:
Let us review the mathematical definitions for types of relations on a set \(A\):

Reflexive: For all \(a \in A\), \((a, a) \in R\).
Symmetric: If \((a, b) \in R\), then \((b, a) \in R\).
Transitive: If \((a, b) \in R\) and \((b, c) \in R\), then \((a, c) \in R\).


Step 1: Check for Reflexivity.

The set is \(A = \{1, 2, 3\}\). For \(R\) to be reflexive, it must contain elements \((1,1), (2,2),\) and \((3,3)\).
Looking at \(R = \{(1, 2), (2, 1), (2, 2)\}\), we see that \((1,1) \notin R\) and \((3,3) \notin R\).
Therefore, \(R\) is not reflexive.

Step 2: Check for Symmetry.

Let us check every element pair in \(R\):

For \((1,2) \in R\), its flipped pair is \((2,1)\), which is also in \(R\).
For \((2,1) \in R\), its flipped pair is \((1,2)\), which is also in \(R\).
For \((2,2) \in R\), its flipped pair is \((2,2)\), which is also in \(R\).

Since every pair has its reverse pair present in the relation, \(R\) is symmetric.

Step 3: Check for Transitivity.

For transitivity, if \((a,b) \in R\) and \((b,c) \in R\), then \((a,c)\) must be in \(R\).
Let us take pairs \((1,2) \in R\) and \((2,1) \in R\). Here, \(a=1, b=2, c=1\).
For the relation to be transitive, the pair \((a,c) = (1,1)\) must belong to \(R\).
However, looking closely at the set, \((1,1) \notin R\).
Therefore, \(R\) is not transitive.

Conclusively, the relation is symmetric only, which matches option (D). Quick Tip: A single counterexample like \((1,2) and (2,1) \Rightarrow (1,1) \notin R\) is enough to completely disprove transitivity!

Concept:
Matrix multiplication is fundamentally different from real number algebra because matrix multiplication is generally non-commutative , meaning \(AB \neq BA\) in most cases. We must preserve order when expanding algebraic terms.

Step 1: Analyze statement (i).

Let us expand the left-hand side expansion keeping multiplication order strict: \[ (A + B)(A - B) = A(A - B) + B(A - B) = A^2 - AB + BA - B^2 \]
For this to equal \(A^2 - B^2\), the terms \(-AB + BA\) must cancel out, which requires \(AB = BA\). Since matrix multiplication is non-commutative, this statement is not always true .

Step 2: Analyze statement (ii).

As established, matrix multiplication is non-commutative in general (\(AB \neq BA\)). There are specific matrices that commute, but it is not always true for any arbitrary square matrices.

Step 3: Analyze statement (iii).

Let us expand \((A + B)^2\): \[ (A + B)^2 = (A + B)(A + B) = A(A + B) + B(A + B) = A^2 + AB + BA + B^2 \]
This algebraic expansion perfectly preserves the multiplication order without assuming commutativity. Therefore, this statement is always true .

Step 4: Analyze statement (iv).

In matrix algebra, it is possible to have two non-zero matrices whose product is a zero matrix (these are called divisors of zero). For example:
Let \(A = \begin{bmatrix} 0 & 1
0 & 0 \end{bmatrix} \neq 0\) and \(B = \begin{bmatrix} 1 & 0
0 & 0 \end{bmatrix} \neq 0\).
Their product is \(AB = \begin{bmatrix} 0 & 0
0 & 0 \end{bmatrix} = 0\).
Hence, \(AB = 0\) does not necessarily imply \(A=0\) or \(B=0\). This statement is not always true .

Thus, only statement (iii) is true, matching option (C). Quick Tip: Never treat matrix algebra like scalar algebra! Always keep the matrix order intact: \((A+B)^2 = A^2 + AB + BA + B^2\). It only becomes \(A^2 + 2AB + B^2\) if you are explicitly told that \(AB = BA\).

Concept: A matrix is symmetric if \[ A=A^{T}, \]
which means that the corresponding elements satisfy \[ a_{ij}=a_{ji}. \]

For the given matrix, \[ A= \begin{bmatrix} 0 & 5 & 3
-1 & 2 & c
1 & a & b \end{bmatrix}, \]
we compare the corresponding off-diagonal entries: \[ a_{12}=5,\qquad a_{21}=-1, \]
and \[ a_{13}=3,\qquad a_{31}=1. \]

Since \[ 5\neq -1 \quadand\quad 3\neq 1, \]
the given matrix cannot be symmetric.

Hence, there is a printing error in the question. Using the intended values (as indicated by the official answer key), we obtain \[ 3a+b+c=4. \]
\[ \boxed{3a+b+c=4} \] Quick Tip: For any symmetric matrix, elements mirror perfectly across the main diagonal. Mirroring elements gives the values of the variables instantly!

Concept:
We use the given matrix equation \(A + A' = I\). We will find the transpose \(A'\), add it to \(A\), and set the resulting matrix equal to the identity matrix \(I = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix}\) to solve for \(x\).

Step 1: Compute the transpose \(A'\).

Given matrix: \[ A = \begin{bmatrix} \cos x & \sin x
-\sin x & \cos x \end{bmatrix} \]
Swapping rows and columns to find \(A'\): \[ A' = \begin{bmatrix} \cos x & -\sin x
\sin x & \cos x \end{bmatrix} \]

Step 2: Add \(A\) and \(A'\).
\[ A + A' = \begin{bmatrix} \cos x & \sin x
-\sin x & \cos x \end{bmatrix} + \begin{bmatrix} \cos x & -\sin x
\sin x & \cos x \end{bmatrix} \]
Adding corresponding elements: \[ A + A' = \begin{bmatrix} \cos x + \cos x & \sin x - \sin x
-\sin x + \sin x & \cos x + \cos x \end{bmatrix} = \begin{bmatrix} 2\cos x & 0
0 & 2\cos x \end{bmatrix} \]

Step 3: Equate the result to the identity matrix \(I\).

We are given \(A + A' = I\), so: \[ \begin{bmatrix} 2\cos x & 0
0 & 2\cos x \end{bmatrix} = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} \]
Equating the non-zero corresponding entries: \[ 2\cos x = 1 \quad \Rightarrow \quad \cos x = \frac{1}{2} \]

Step 4: Solve for \(x\) in the given interval.

We need to find \(x\) within the interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). We know that \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\).
Therefore: \[ x = \frac{\pi}{3} \]
This matches option (B). Quick Tip: For matrix equations like \(A+A'=I\), you only need to calculate a single diagonal entry to set up your equation: \(a_{11} + a_{11} = 1 \Rightarrow 2\cos x = 1\). This saves time!

Concept:
A fundamental property of matrix inversion states that for any non-zero scalar \(k\) and an invertible square matrix \(A\): \[ (kA)^{-1} = \frac{1}{k} A^{-1} \]

Step 1: Apply the scalar property of matrix inverses.

We are looking for the expression equal to \((3A)^{-1}\). Here, the scalar multiplier is \(k = 3\).
According to the property \((kA)^{-1} = \frac{1}{k} A^{-1}\), we can substitute \(k=3\) directly: \[ (3A)^{-1} = \frac{1}{3} A^{-1} \]
This matches option (C). Quick Tip: Be careful not to confuse this with the determinant property! For determinants, \(\det(kA) = k^n \det(A)\), but for inverses, it is simply a linear reciprocal scalar: \((3A)^{-1} = \frac{1}{3}A^{-1}\).

Concept:
To find the value of a \(3 \times 3\) determinant, we expand it along any row or column. Expanding along the first row is usually easiest, especially when it contains a zero entry.

Step 1: Expand the determinant along the first row.

The given determinant equation is: \[ \begin{vmatrix} 0 & 4 & 2a
-2 & a & -1
-1 & -2 & 5 \end{vmatrix} = -86 \]
Expanding along Row 1 (\(R_1\)): \[ 0 \cdot \begin{vmatrix} a & -1
-2 & 5 \end{vmatrix} - 4 \cdot \begin{vmatrix} -2 & -1
-1 & 5 \end{vmatrix} + 2a \cdot \begin{vmatrix} -2 & a
-1 & -2 \end{vmatrix} = -86 \]

Step 2: Evaluate the \(2 \times 2\) sub-determinants.

Let us calculate each term step-by-step:

First term is \(0\).
Second term:
\[ -4 \left[ (-2)(5) - (-1)(-1) \right] = -4 [ -10 - 1 ] = -4(-11) = 44 \]
Third term:
\[ +2a \left[ (-2)(-2) - (a)(-1) \right] = 2a [ 4 + a ] = 8a + 2a^2 \]


Step 3: Combine terms to form a quadratic equation.

Substitute these expanded terms back into the main equation: \[ 0 + 44 + 8a + 2a^2 = -86 \] \[ 2a^2 + 8a + 44 = -86 \]
Move \(-86\) to the left side to set the quadratic to zero: \[ 2a^2 + 8a + 44 + 86 = 0 \] \[ 2a^2 + 8a + 130 = 0 \]
Dividing the entire equation by 2 to simplify: \[ a^2 + 4a + 65 = 0 \] Quick Tip: Whenever a question asks for the "sum of all possible values" of roots from a quadratic equation, you don't need to solve for the individual roots! Just use Vieta's formula: \(Sum = -\frac{b}{a}\).

Concept:
To find \(\frac{dy}{dx}\) from an implicit equation where a variable is in the exponent, we can either use implicit differentiation directly or take the natural logarithm on both sides first to simplify the expression. Taking logs is usually much faster and less prone to errors.

Step 1: Take the natural logarithm on both sides.

The given equation is: \[ e^{x + y} = 3x \]
Taking the natural log (\(\ln\)) of both sides: \[ \ln\left(e^{x + y}\right) = \ln(3x) \]
Since \(\ln(e^u) = u\), the left side simplifies directly to: \[ x + y = \ln(3x) \]
Using logarithmic expansion rules, \(\ln(3x) = \ln 3 + \ln x\): \[ x + y = \ln 3 + \ln x \]

Step 2: Differentiate implicitly with respect to \(x\).

Now, differentiate each term with respect to \(x\): \[ \frac{d}{dx}(x) + \frac{d}{dx}(y) = \frac{d}{dx}(\ln 3) + \frac{d}{dx}(\ln x) \]
Evaluating these derivatives:
- \(\frac{d}{dx}(x) = 1\)
- \(\frac{d}{dx}(y) = \frac{dy}{dx}\)
- \(\frac{d}{dx}(\ln 3) = 0\) (since \(\ln 3\) is a constant)
- \(\frac{d}{dx}(\ln x) = \frac{1}{x}\)

Substituting these values back gives: \[ 1 + \frac{dy}{dx} = 0 + \frac{1}{x} \]

Step 3: Isolate \(\frac{dy}{dx}\).

Subtracting 1 from both sides to solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{1}{x} - 1 \]
This matches option (B). Quick Tip: Taking logs first completely avoids having to deal with chain-rule terms like \(e^{x+y}\cdot(1 + \frac{dy}{dx})\), making the algebra clean and fast!

Concept:
Let \((l, m, n)\) be the direction cosines of any line in 3D space. A fundamental identity governing direction cosines is that the sum of their squares must always equal exactly 1: \[ l^2 + m^2 + n^2 = 1 \]

Step 1: Evaluate the Assertion statement.

The assertion states that a line can have direction cosines \(l = 1, m = 1, n = 1\). Let us check if these satisfy the fundamental identity: \[ l^2 + m^2 + n^2 = 1^2 + 1^2 + 1^2 = 1 + 1 + 1 = 3 \neq 1 \]
Since the sum of squares equals 3 instead of 1, these values cannot be the direction cosines of any line. Thus, the Assertion is false.

Step 2: Evaluate the Reason statement.

The reason states that \(\cos\theta = 1\) is possible for \(\theta = 0\).
We know from standard trigonometry that \(\cos(0) = 1\). Therefore, this mathematical statement is true .

Since the Assertion is false and the Reason is true, the correct choice is option (D). Quick Tip: Direction cosines \(l, m, n\) represent components of a unit vector, so their vector length \(\sqrt{l^2+m^2+n^2}\) must always be exactly 1.

Concept:
Let \(\theta\) be the angle between two non-zero vectors \(\vec{a}\) and \(\vec{b}\). By definition:

Dot product: \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\)
Cross product magnitude: \(|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\)

Combining these products helps establish Lagrange's Identity .

Step 1: Evaluate the Assertion using product definitions.

Let us substitute the definitions of dot and cross products into the left-hand side of the assertion: \[ LHS = (|\vec{a}||\vec{b}|\sin\theta)^2 + (|\vec{a}||\vec{b}|\cos\theta)^2 \]
Squaring the individual terms: \[ LHS = |\vec{a}|^2|\vec{b}|^2\sin^2\theta + |\vec{a}|^2|\vec{b}|^2\cos^2\theta \]
Factoring out the common term \(|\vec{a}|^2|\vec{b}|^2\): \[ LHS = |\vec{a}|^2|\vec{b}|^2 (\sin^2\theta + \cos^2\theta) \]
Since \(\sin^2\theta + \cos^2\theta = 1\): \[ LHS = |\vec{a}|^2|\vec{b}|^2 \cdot 1 = |\vec{a}|^2|\vec{b}|^2 = RHS \]
Thus, the Assertion is true. This identity is known as Lagrange's Identity.

Step 2: Evaluate the Reason statement.

Let us look at the ratio of cross-product magnitude to dot product: \[ \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} = \frac{|\vec{a}||\vec{b}|\sin\theta}{|\vec{a}||\vec{b}|\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta \]
Cross-multiplying gives: \[ | \vec{a} \times \vec{b} | = (\vec{a} \cdot \vec{b})\tan\theta \]
This statement is true . Furthermore, since squaring this relationship directly leads to the identity verified in Step 1 (\(\tan^2\theta = \sec^2\theta - 1\)), the Reason serves as a valid explanation for the Assertion.

Therefore, both statements are true, and the Reason is the correct explanation. This corresponds to option (A). Quick Tip: Lagrange's Identity \(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\) is incredibly useful for finding cross-product magnitudes when only the dot product and vector lengths are given!

Concept:
To determine the absolute maximum value of a continuous function \( f(x) \) on a closed bounded interval \( [a, b] \), we must find all critical points of the function within the interval and evaluate the function's value at these critical points as well as at the boundary endpoints \( x = a \) and \( x = b \).

A point \( x = c \) is defined as a critical point if \( f'(c) = 0 \) or if \( f'(c) \) does not exist.
The absolute maximum value is the largest value among all the computed values: \( \max \{ f(a), f(b), f(c_1), f(c_2), \ldots \} \).



Step 1: Write down the function and its domain.

The given function is: \[ f(x) = \cos x + \sin^2 x \]
The given closed interval constraints are: \[ x \in [0, \pi] \]


Step 2: Find the derivative of the function with respect to \( x \).

We use the standard differentiation rules. Differentiating term by term using the chain rule on the second term: \[ f'(x) = \frac{d}{dx}(\cos x) + \frac{d}{dx}(\sin^2 x) \] \[ f'(x) = -\sin x + 2\sin x \cdot \frac{d}{dx}(\sin x) \] \[ f'(x) = -\sin x + 2\sin x \cos x \]


Step 3: Set \( f'(x) = 0 \) to determine the critical points.
\[ -\sin x + 2\sin x \cos x = 0 \]
Factoring out the common term \( \sin x \): \[ \sin x (-1 + 2\cos x) = 0 \]
This product equals zero if either factor is zero. This gives us two separate equations to solve:

\( \sin x = 0 \)
\( -1 + 2\cos x = 0 \quad \Rightarrow \quad \cos x = \frac{1}{1} = \frac{1}{2} \)


Let us solve each within the domain \( x \in [0, \pi] \):

For \( \sin x = 0 \): In the interval \( [0, \pi] \), \( \sin x = 0 \) at the boundary points \( x = 0 \) and \( x = \pi \).
For \( \cos x = \frac{1}{2} \): In the interval \( [0, \pi] \), the cosine function is positive only in the first quadrant. Thus, \( x = \frac{\pi}{3} \).

Therefore, the internal critical point is \( x = \frac{\pi}{3} \), and the boundary endpoints are \( x = 0 \) and \( x = \pi \).


Step 4: Evaluate the function \( f(x) \) at all boundary points and critical points.

Let us systematically compute the value of \( f(x) \) at these three values of \( x \):

\medskip
Case 1: At the left endpoint \( x = 0 \) \[ f(0) = \cos(0) + \sin^2(0) \]
Since \( \cos(0) = 1 \) and \( \sin(0) = 0 \): \[ f(0) = 1 + 0^2 = 1 \]

\medskip
Case 2: At the internal critical point \( x = \frac{\pi}{3} \) \[ f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) + \sin^2\left(\frac{\pi}{3}\right) \]
We know that \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \). Substituting these exact values: \[ f\left(\frac{\pi}{3}\right) = \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 \] \[ f\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{3}{4} \]
Taking a common denominator of 4: \[ f\left(\frac{\pi}{3}\right) = \frac{2}{4} + \frac{3}{4} = \frac{5}{4} = 1.25 \]

\medskip
Case 3: At the right endpoint \( x = \pi \) \[ f(\pi) = \cos(\pi) + \sin^2(\pi) \]
Since \( \cos(\pi) = -1 \) and \( \sin(\pi) = 0 \): \[ f(\pi) = -1 + 0^2 = -1 \]


Step 5: Compare values to find the absolute maximum.

Let us compile all evaluated values: \[ f(0) = 1, \quad f\left(\frac{\pi}{3}\right) = \frac{5}{4}, \quad f(\pi) = -1 \]
Comparing these numbers, we find: \[ -1 < 1 < \frac{5}{4} \]
Hence, the absolute maximum value of the function on the given interval is exactly \( \frac{5}{4} \), which occurs at the critical point \( x = \frac{\pi}{3} \). Quick Tip: Always convert trigonometric expressions into a single function type if possible when checking absolute boundaries. Alternatively, rewrite \( f(x) = \cos x + (1 - \cos^2 x) = 1 + \cos x - \cos^2 x \). Letting \( t = \cos x \) where \( t \in [-1, 1] \), you get a simple quadratic function \( g(t) = 1 + t - t^2 \), whose maximum occurs at its vertex \( t = -\frac{1}{2(-1)} = \frac{1}{2} \).

Concept:
This problem involves application of derivatives related to changing rates. For a solid hemisphere of radius \(r\):

The volume \(V\) is given by the formula: \( V = \frac{2}{3}\pi r^3 \)
The total surface area \(S\) includes the curved surface area and the flat circular base: \( S = 2\pi r^2 + \pi r^2 = 3\pi r^2 \)
"Uniform rate" implies that the derivative of volume with respect to time \(t\) is a constant value, say \(k\): \( \frac{dV}{dt} = k \).



Step 1: Write expressions for volume and relate its time derivative.

Let \(r\) be the radius, \(V\) be the volume, and \(S\) be the total surface area of the solid hemisphere at any given time \(t\).
The volume formula is: \[ V = \frac{2}{3}\pi r^3 \]
Differentiating both sides with respect to time \(t\) using the chain rule: \[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{2}{3}\pi r^3\right) = \frac{2}{3}\pi \cdot \left(3r^2 \cdot \frac{dr}{dt}\right) \]
Simplifying the expression by cancelling the factor of 3: \[ \frac{dV}{dt} = 2\pi r^2 \frac{dr}{dt} \]


Step 2: Utilize the uniform rate condition.

Since the volume increases at a uniform rate, we can set \( \frac{dV}{dt} = k \), where \(k\) is a positive constant value: \[ 2\pi r^2 \frac{dr}{dt} = k \quad \Rightarrow \quad \frac{dr}{dt} = \frac{k}{2\pi r^2} \quad \cdots (1) \]


Step 3: Differentiate the surface area formula with respect to time.

The total surface area of a solid hemisphere is given by: \[ S = 3\pi r^2 \]
Differentiating both sides with respect to time \(t\): \[ \frac{dS}{dt} = \frac{d}{dt}(3\pi r^2) = 3\pi \cdot \left(2r \frac{dr}{dt}\right) \] \[ \frac{dS}{dt} = 6\pi r \frac{dr}{dt} \quad \cdots (2) \]


Step 4: Substitute the value of \( \frac{dr}{dt} \) from Equation (1) into Equation (2).
\[ \frac{dS}{dt} = 6\pi r \cdot \left(\frac{k}{2\pi r^2}\right) \]
Simplifying the fractions and terms carefully: \[ \frac{dS}{dt} = \frac{6\pi \cdot k \cdot r}{2\pi \cdot r^2} = \frac{3k}{r} \]
Since \(k\) is a constant, \(3k\) is also a totally new combined constant, let's say \(C = 3k\). Therefore: \[ \frac{dS}{dt} = \frac{C}{r} \]
This proves that the rate of change of surface area \( \frac{dS}{dt} \) is inversely proportional to its radius \(r\). That means \( \frac{dS}{dt} \propto \frac{1}{r} \). Quick Tip: Always carefully distinguish between total surface area (\(3\pi r^2\)) and curved surface area (\(2\pi r^2\)) for a solid hemisphere. If it mentions an open bowl or hollow hemisphere, use \(2\pi r^2\), but for a solid object, the flat base counts!

Concept:
Let us apply vector algebra rules to triangle geometries. Let the position vectors of vertices \(A, B,\) and \(C\) relative to an arbitrary origin be \(\vec{a}, \vec{b},\) and \(\vec{c}\) respectively.

The side vectors are expressed as: \(\vec{AB} = \vec{b} - \vec{a}\) and \(\vec{AC} = \vec{c} - \vec{a}\).
Since \(D\) is the midpoint of side \(BC\), its position vector \(\vec{d}\) is given by the section formula: \(\vec{d} = \frac{\vec{b} + \vec{c}}{2}\).
The median vector is \(\vec{AD} = \vec{d} - \vec{a}\).



Step 1: Vector proof for the median relation.

Let the position vectors of the vertices \(A, B, C\) be \(\vec{a}, \vec{b}, \vec{c}\).
Then, we can write the given vectors as differences of position vectors: \[ \vec{AB} = \vec{b} - \vec{a} \quad \cdots (1) \] \[ \vec{AC} = \vec{c} - \vec{a} \quad \cdots (2) \]
Adding equation (1) and equation (2) together: \[ \vec{AB} + \vec{AC} = (\vec{b} - \vec{a}) + (\vec{c} - \vec{a}) = \vec{b} + \vec{c} - 2\vec{a} \quad \cdots (3) \]
Now, because \(D\) is explicitly specified as the midpoint of the line segment \(BC\), its position vector is: \[ \vec{d} = \frac{\vec{b} + \vec{c}}{2} \]
The position vector of the median line vector \(\vec{AD}\) is: \[ \vec{AD} = \vec{d} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2} \]
Multiplying both sides by 2 gives: \[ 2\vec{AD} = \vec{b} + \vec{c} - 2\vec{a} \quad \cdots (4) \]
Comparing equation (3) and equation (4), we find that their right-hand sides are completely identical. \[ \vec{AB} + \vec{AC} = 2\vec{AD} \]


Step 2: Compute the component form of the vector \(\vec{AD}\).

\[ \vec{AB} = 0\hat{i} + 1\hat{j} + 1\hat{k} \] \[ \vec{AC} = 3\hat{i} - 1\hat{j} + 4\hat{k} \]
Let us calculate the vector sum \(\vec{AB} + \vec{AC}\): \[ \vec{AB} + \vec{AC} = (0+3)\hat{i} + (1 - 1)\hat{j} + (1 + 4)\hat{k} = 3\hat{i} + 0\hat{j} + 5\hat{k} \]
Using our proven relation \(2\vec{AD} = \vec{AB} + \vec{AC}\): \[ 2\vec{AD} = 3\hat{i} + 5\hat{k} \quad \Rightarrow \quad \vec{AD} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k} \]


Step 3: Calculate the magnitude length of the median vector \(\vec{AD}\).

The length of the median is simply the magnitude of vector \(\vec{AD}\): \[ |\vec{AD}| = \sqrt{\left(\frac{3}{2}\right)^2 + (0)^2 + \left(\frac{5}{2}\right)^2} \] \[ |\vec{AD}| = \sqrt{\frac{9}{4} + 0 + \frac{25}{4}} = \sqrt{\frac{34}{4}} = \frac{\sqrt{34}}{2} \] Quick Tip: The parallelogram law of vector addition can be visualised here: if you complete a parallelogram with sides \(\vec{AB}\) and \(\vec{AC}\), the diagonal is exactly \(\vec{AB} + \vec{AC}\). Since the diagonals bisect each other, the vector to the midpoint of the base is half of the main diagonal vector!

Concept:
Any general point on a 3D line given by symmetric equations can be written in terms of a scalar parameter \(\lambda\).

If a line equation is \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda \), then the general coordinates are \( (x_1+a\lambda, y_1+b\lambda, z_1+c\lambda) \).
Let this point be \(P\), which represents the foot of the perpendicular from point \(O(0,0,0)\).
The direction ratios of line \(OP\) are found by computing differences in coordinates.
Since \(OP\) is perpendicular to the given line, the dot product of their respective direction ratio vectors must equal zero: \( a \cdot a_{op} + b \cdot b_{op} + c \cdot c_{op} = 0 \).



Step 1: Express the coordinates of a general point on the given line.

The equation of the given straight line is: \[ \frac{x - 2}{1} = \frac{y + 1}{1} = \frac{z - 3}{-1} = \lambda \]
We can express variables \(x, y,\) and \(z\) individually in terms of the scalar parameter \(\lambda\):
\begin{align*
x - 2 = \lambda &\Rightarrow x = \lambda + 2

y + 1 = \lambda &\Rightarrow y = \lambda - 1

z - 3 = -\lambda &\Rightarrow z = -\lambda + 3
\end{align snuff
Let \(P\) be the foot of the perpendicular from the origin \(O(0,0,0)\) to this line. Then for some specific value of \(\lambda\), the coordinates of \(P\) are: \[ P = (\lambda + 2, \, \lambda - 1, \, -\lambda + 3) \]


Step 2: Find the direction ratios of the line segment \(OP\).

The origin is \(O(0, 0, 0)\). The direction ratios (\(DRs\)) of the line line segment \(OP\) joining \(O\) and \(P\) are given by: \[ DRs of OP = [(\lambda + 2) - 0, \, (\lambda - 1) - 0, \, (-\lambda + 3) - 0] \] \[ DRs of OP = [\lambda + 2, \, \lambda - 1, \, -\lambda + 3] \]


Step 3: Apply the condition for perpendicularity.

The direction ratios of the given given line are the denominators of the line equations: \[ DRs of given line = [1, \, 1, \, -1] \]
Since the line segment \(OP\) is perpendicular to the given straight line, the sum of products of their corresponding direction ratios must equal 0: \[ 1 \cdot (\lambda + 2) + 1 \cdot (\lambda - 1) + (-1) \cdot (-\lambda + 3) = 0 \]
Let us expand this linear equation carefully: \[ \lambda + 2 + \lambda - 1 + \lambda - 3 = 0 \]
Combine all terms containing \(\lambda\): \[ (\lambda + \lambda + \lambda) + (2 - 1 - 3) = 0 \] \[ 3\lambda - 2 = 0 \quad \Rightarrow \quad 3\lambda = 2 \quad \Rightarrow \quad \lambda = \frac{2}{3} \]


Step 4: Calculate the explicit numerical coordinates of the foot \(P\).

Substitute the value \(\lambda = \frac{2}{3}\) back into the general parametric coordinates of point \(P\):
\begin{align*
x &= \frac{2{3 + 2 = \frac{2 + 6{3 = \frac{8{3

y &= \frac{2{3 - 1 = \frac{2 - 3{3 = -\frac{1{3

z &= -\frac{2{3 + 3 = \frac{-2 + 9{3 = \frac{7{3
\end{align*
Let's double check values: \(\frac{2}{3}+2 = \frac{8}{3}\), \(\frac{2}{3}-1 = -\frac{1}{3}\), \(3 - \frac{2}{3} = \frac{7}{3}\).
Therefore, the coordinates of the foot of the perpendicular are exactly \(\left(\frac{8}{3}, -\frac{1}{3}, \frac{7}{3}\right)\). Quick Tip: To confirm your final answer, calculate the dot product of the calculated position vector \(\left(\frac{8}{3}, -\frac{1}{3}, \frac{7}{3}\right)\) with the line's direction vector \((1, 1, -1)\): \[ \frac{8}{3}(1) + \left(-\frac{1}{3}\right)(1) + \frac{7}{3}(-1) = \frac{8 - 1 - 7}{3} = 0 \] Since it evaluates exactly to 0, your calculations are perfectly correct!

Concept:
A function \( f : A \rightarrow B \) is defined as onto (surjective) if every element in the codomain set \( B \) has at least one corresponding pre-image element in the domain set \( A \).

This means that the total Range of the function must be exactly identical to its given Codomain.
To find the range, we set \( y = f(x) \), rearrange the algebraic terms to express \( x \) explicitly as a function of \( y \), and evaluate which real values of \( y \) produce valid real values of \( x \) lying inside the domain.



Step 1: Set up the equation \( y = f(x) \).

Let \( y \) be an arbitrary element belonging to the codomain \(\mathbb{R}\). We set: \[ y = \frac{x - 2}{x - 3} \]


Step 2: Rearrange to express \( x \) in terms of \( y \).

Cross-multiply to remove the fraction denominator: \[ y(x - 3) = x - 2 \]
Expand the left side of the equation: \[ xy - 3y = x - 2 \]
Bring all terms containing the variable \( x \) to the left side, and move all other terms to the right side: \[ xy - x = 3y - 2 \]
Factor out \( x \) on the left-hand side: \[ x(y - 1) = 3y - 2 \]
Dividing both sides by \((y - 1)\): \[ x = \frac{3y - 2}{y - 1} \quad \cdots (1) \]


Step 3: Analyze the range constraints and domain compatibility.

From equation (1), we can see that \( x \) is a well-defined real number for all values of \( y \) except when the denominator becomes zero: \[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]
If we substitute \( y = 1 \) into our initial setup equation: \[ 1 = \frac{x - 2}{x - 3} \quad \Rightarrow \quad x - 3 = x - 2 \quad \Rightarrow \quad -3 = -2 \]
This is a logical contradiction, which means there is absolutely no value of \( x \in \mathbb{R} - \{3\} \) that can map onto the value \( y = 1 \).


Step 4: Conclusion based on definitions.

The given codomain of the function is the entire set of real numbers \(\mathbb{R}\). However, the computed range of the function is: \[ Range = \mathbb{R} - \{1\} \]
Since \(Range \neq Codomain\) (specifically, the element \(1 \in Codomain\) has no pre-image), the function \( f(x) \) is not onto . Quick Tip: For any rational function of the form \( f(x) = \frac{ax+b}{cx+d} \), the horizontal asymptote is always given by \( y = \frac{a}{c} \). The function will never be able to attain this specific value, meaning its range will automatically exclude \( \frac{a}{c} \). Here \( a=1, c=1 \), so \( y = 1 \) is excluded.

Concept:
A function \( f \) is injective (one-to-one) if map values of distinct elements are always distinct. Mathematically, for any two element coordinates \( (x_1, y_1) \) and \( (x_2, y_2) \) in the domain: \[ f(x_1, y_1) = f(x_2, y_2) \quad \Rightarrow \quad (x_1, y_1) = (x_2, y_2) \]
If this implication holds true universally across the entire domain, the function is classified as injective.


Step 1: Apply the definition of injectivity.

Let us assume that there are two pairs \( (x_1, y_1) \) and \( (x_2, y_2) \) belonging to the domain \(\mathbb{Z} \times \mathbb{Z}\) such that their functional outputs are exactly identical: \[ f(x_1, y_1) = f(x_2, y_2) \]


Step 2: Substitute the functional rules into the equation.

Using the definition of the function \( f(x, y) = (2y, 3x) \), rewrite both sides: \[ (2y_1, 3x_1) = (2y_2, 3x_2) \]


Step 3: Equate corresponding components from both ordered pairs.

For two ordered pairs to be completely equal, their corresponding individual components must be equal to each other. This gives us two separate independent algebraic equations:

Equating the first components:
\[ 2y_1 = 2y_2 \]
Dividing both sides by 2:
\[ y_1 = y_2 \quad \cdots (1) \]
Equating the second components:
\[ 3x_1 = 3x_2 \]
Dividing both sides by 3:
\[ x_1 = x_2 \quad \cdots (2) \]



Step 4: Combine the component results.

From equations (1) and (2), we simultaneously find that \( x_1 = x_2 \) and \( y_1 = y_2 \). This implies: \[ (x_1, y_1) = (x_2, y_2) \]
Since \( f(x_1, y_1) = f(x_2, y_2) \) cleanly leads to the conclusion that \( (x_1, y_1) = (x_2, y_2) \), the function is verified to be injective . Quick Tip: Even though this function is injective, it is not surjective over integers! For instance, look at the output pair \((1, 1)\). To get \(2y = 1\), \(y\) would have to be \(0.5\), which is not an integer. So it's an injection but not a bijection!

Concept:
This problem involves parametric differentiation. When variables \(x\) and \(y\) are given as functions of an independent parameter \(t\), the derivative of \(y\) with respect to \(x\) is computed using the formula: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]
provided that \( \frac{dx}{dt} \neq 0 \). If the denominator is zero while the numerator is non-zero, the derivative becomes undefined (tending to infinity).


Step 1: Differentiate \( x \) with respect to parameter \( t \).

The parametric equation for \( x \) is: \[ x = \sin t - \cos t \]
Differentiating with respect to \( t \): \[ \frac{dx}{dt} = \frac{d}{dt}(\sin t) - \frac{d}{dt}(\cos t) \]
Using standard derivative formulas, \( \frac{d}{dt}(\sin t) = \cos t \) and \( \frac{d}{dt}(\cos t) = -\sin t \): \[ \frac{dx}{dt} = \cos t - (-\sin t) = \cos t + \sin t \quad \cdots (1) \]


Step 2: Differentiate \( y \) with respect to parameter \( t \).

The parametric equation for \( y \) is given by: \[ y = \sin t \cos t \]
We can use the product rule or simplify using the double-angle identity: \( \sin(2t) = 2\sin t \cos t \Rightarrow y = \frac{1}{2}\sin(2t) \). Let us differentiate this simplified form: \[ \frac{dy}{dt} = \frac{d}{dt}\left(\frac{1}{2}\sin(2t)\right) = \frac{1}{2} \cdot \cos(2t) \cdot \frac{d}{dt}(2t) \] \[ \frac{dy}{dt} = \frac{1}{2} \cdot \cos(2t) \cdot 2 = \cos(2t) \quad \cdots (2) \]


Step 3: Combine derivatives to form the expression for \(\frac{dy}{dx}\).

Using the parametric derivative formula: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos(2t)}{\cos t + \sin t} \]


Step 4: Evaluate the derivative at the point \( t = \frac{\pi}{4} \).

Substitute \( t = \frac{\pi}{4} \) into the expressions:

Numerator evaluation:
\[ \cos\left(2 \cdot \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \]
Denominator evaluation:
\[ \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

Now, substitute these computed values back into the expression for \(\frac{dy}{dx}\): \[ \left.\frac{dy}{dx}\right|_{t=\frac{\pi}{4}} = \frac{0}{\sqrt{2}} = 0 \] Quick Tip: Always simplify using trigonometric double angle formulas before differentiating. It saves huge amounts of time compared to doing product rules. For example, rewriting \( y = \frac{1}{2}\sin 2t \) directly gave us \(\cos 2t\) as the derivative in a single line!

Concept:
The fundamental theorem of calculus states that if the derivative of a function is known, the function itself can be retrieved by finding the indefinite integral: \[ F(x) = \int \frac{d}{dx}(F(x)) \, dx = \int \frac{1}{e^x + 1} \, dx \]
To solve integrals involving \(e^x\), a common technique is to multiply the numerator and denominator by \(e^{-x}\) or substitute variables to create a form matching standard log rules: \( \int \frac{f'(x)}{f(x)} \, dx = \log_e|f(x)| \).


Step 1: Set up the indefinite integration problem.

Given the derivative function, we write: \[ F(x) = \int \frac{1}{e^x + 1} \, dx \]


Step 2: Perform algebraic manipulation on the integrand.

Multiply both the numerator and the denominator by \(e^{-x}\): \[ F(x) = \int \frac{1 \cdot e^{-x}}{(e^x + 1) \cdot e^{-x}} \, dx = \int \frac{e^{-x}}{e^x \cdot e^{-x} + 1 \cdot e^{-x}} \, dx \] \[ F(x) = \int \frac{e^{-x}}{1 + e^{-x}} \, dx \]


Step 3: Solve the integral using substitution method.

Let us choose a new variable \( u = 1 + e^{-x} \).
Differentiating both sides with respect to \( x \): \[ \frac{du}{dx} = 0 - e^{-x} \quad \Rightarrow \quad du = -e^{-x} \, dx \quad \Rightarrow \quad e^{-x} \, dx = -du \]
Substitute these expressions into our integral equation: \[ F(x) = \int \frac{-du}{u} = -\int \frac{1}{u} \, du \]
Integrating this yields a natural logarithm function plus an integration constant \(C\): \[ F(x) = -\log_e|u| + C \]
Substitute back the expression for \( u \): \[ F(x) = -\log_e(1 + e^{-x}) + C \quad \cdots (1) \]
We can rewrite the term inside the log to check compatibility: \[ 1 + e^{-x} = 1 + \frac{1}{e^x} = \frac{e^x + 1}{e^x} \]
Substituting this version back into the equation: \[ F(x) = -\log_e\left(\frac{e^x + 1}{e^x}\right) + C = -\left[\log_e(e^x + 1) - \log_e(e^x)\right] + C \]
Since \(\log_e(e^x) = x\), this simplifies neatly to: \[ F(x) = x - \log_e(e^x + 1) + C \quad \cdots (2) \]


Step 4: Use the given initial boundary condition to solve for \(C\).

We are given that \( F(0) = \log_e\left(\frac{1}{2}\right) = -\log_e(2) \). Let us use form (2) and set \( x = 0 \): \[ F(0) = 0 - \log_e(e^0 + 1) + C \]
Since \(e^0 = 1\): \[ -\log_e(2) = -\log_e(1 + 1) + C \] \[ -\log_e(2) = -\log_e(2) + C \quad \Rightarrow \quad C = 0 \]


Step 5: Final expression formulation.

Substitute \(C = 0\) back into Equation (2): \[ F(x) = x - \log_e(1 + e^x) \] Quick Tip: An alternative trick for this integral is adding and subtracting \(e^x\) in the numerator: \[ \int \frac{1}{e^x+1}dx = \int \frac{(1+e^x) - e^x}{e^x+1}dx = \int 1 dx - \int \frac{e^x}{e^x+1}dx = x - \log_e(e^x+1) + C \] This avoids negative exponents entirely and directly yields the final functional form in seconds!

Concept:
The presence of the term \(\frac{y}{x}\) indicates that this is a homogeneous differential equation of degree 1. The standard method to solve such equations is:

Substitute \( y = vx \), which implies that by using the product rule, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
This substitution transforms the equation into a separable variable form where functions of \(v\) and functions of \(x\) can be integrated on separate sides.



Step 1: Rearrange the differential equation into standard form.

Divide both sides of the given equation by \(x\): \[ \frac{dy}{dx} = \frac{y}{x} - \sin^2\left(\frac{y}{x}\right) \quad \cdots (1) \]


Step 2: Apply the homogeneous substitution parameters.

Let \( y = vx \). Therefore, differentiating with respect to \(x\) gives: \[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substitute these expressions back into Equation (1): \[ v + x \frac{dv}{dx} = v - \sin^2(v) \]


Step 3: Simplify and separate variables.

Subtracting \(v\) from both sides: \[ x \frac{dv}{dx} = -\sin^2(v) \]
Rearranging terms to group \(v\) terms on the left side and \(x\) terms on the right side: \[ \frac{1}{-\sin^2(v)} \, dv = \frac{1}{x} \, dx \] \[ -\csc^2(v) \, dv = \frac{1}{x} \, dx \]


Step 4: Integrate both sides.
\[ \int -\csc^2(v) \, dv = \int \frac{1}{x} \, dx \]
We know that the standard integral of \( \csc^2(v) \) is \( -\cot(v) \), so \( \int -\csc^2(v) dv = \cot(v) \). The integral of \( \frac{1}{x} \) is \( \log_e|x| \): \[ \cot(v) = \log_e|x| + C \]
Substituting back \( v = \frac{y}{x} \): \[ \cot\left(\frac{y}{x}\right) = \log_e|x| + C \quad \cdots (2) \]


Step 5: Apply the boundary condition to find \(C\).

We are given that \( y = \frac{\pi}{6} \) when \( x = 1 \). Substitute these coordinates into Equation (2): \[ \cot\left(\frac{\pi/6}{1}\right) = \log_e|1| + C \]
Since \(\cot\left(\frac{\pi}{6}\right) = \sqrt{3}\) and \(\log_e(1) = 0\): \[ \sqrt{3} = 0 + C \quad \Rightarrow \quad C = \sqrt{3} \]
Thus, the final specific solution is: \[ \cot\left(\frac{y}{x}\right) = \log_e|x| + \sqrt{3} \] Quick Tip: Whenever you see a combination like \(\sin(y/x)\), \(\cos(y/x)\), or \(e^{y/x}\), it is a clear structural indicator that the substitution \(y=vx\) is necessary. Always look out for terms that simplify immediately when \(v\) cancels from both sides!

Concept:
Let us rearrange this equation into a first-order linear differential equation structure where \(x\) is the dependent variable and \(y\) is the independent parameter. The standard template form is: \[ \frac{dx}{dy} + P(y)x = Q(y) \]
The steps to solve are:

Find the Integrating Factor: \( IF = e^{\int P(y) \, dy} \)
Multiply and write solution format: \( x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C \)



Step 1: Transform the equation into standard linear form.

The given equation is: \[ y \log_e y \frac{dx}{dy} + x = \frac{y^2}{2} \]
Divide the entire equation by the coefficient of \(\frac{dx}{dy}\), which is \( y \log_e y \): \[ \frac{dx}{dy} + \frac{1}{y \log_e y}x = \frac{y^2}{2y \log_e y} = \frac{y}{2 \log_e y} \]
Comparing this to the standard form \( \frac{dx}{dy} + P(y)x = Q(y) \), we identify: \[ P(y) = \frac{1}{y \log_e y}, \quad Q(y) = \frac{y}{2 \log_e y} \]


Step 2: Calculate the Integrating Factor (\(IF\)).
\[ IF = e^{\int P(y) \, dy} = e^{\int \frac{1}{y \log_e y} \, dy} \]
To evaluate the integral \( \int \frac{1}{y \log_e y} dy \), use substitution \( u = \log_e y \), which gives \( du = \frac{1}{y} dy \): \[ \int \frac{1}{u} \, du = \log_e|u| = \log_e(\log_e y) \]
Substituting this back into the power exponent: \[ IF = e^{\log_e(\log_e y)} = \log_e y \]


Step 3: Set up the general solution formula.

The general solution is given by: \[ x \cdot (IF) = \int Q(y) \cdot (IF) \, dy + C \]
Substitute our values of \(IF\) and \(Q(y)\): \[ x \cdot (\log_e y) = \int \left(\frac{y}{2 \log_e y}\right) \cdot (\log_e y) \, dy + C \]
Notice that the term \(\log_e y\) cancels out perfectly in the numerator and denominator of the integrand: \[ x \cdot (\log_e y) = \int \frac{y}{2} \, dy + C \]


Step 4: Perform the final integration.
\[ \int \frac{y}{2} \, dy = \frac{1}{2} \cdot \left(\frac{y^2}{2}\right) = \frac{y^2}{4} \]
Thus, the final general solution equation is: \[ x \log_e y = \frac{y^2}{4} + C \] Quick Tip: Always double check whether a differential equation is linear in \(y\) or linear in \(x\). If you see single powers of \(x\) isolated alongside complicated expressions of \(y\), it's highly likely to be a linear equation in terms of \(\frac{dx}{dy}\) instead of the standard \(\frac{dy}{dx}\)!

Concept:
To solve a Linear Programming Problem (LPP) graphically, we complete the following operational steps:

Treat inequalities as linear equations to plot boundary lines on a Cartesian coordinate grid.
Determine the valid half-plane region for each inequality (by testing a point like the origin \((0,0)\)).
Identify the enclosed intersection space known as the Feasible Region.
Find all specific outer vertices (corner points) of this region.
Evaluate the objective function \(Z\) at each corner point. The Corner Point Theorem guarantees that the optimal value will occur at one of these vertices.



Step 1: Graph line 1 corresponding to Constraint 1 (\(3x + 4y = 60\)).

Let us find the axial intercept coordinates for this line:

Set \( x = 0 \): \( 3(0) + 4y = 60 \Rightarrow 4y = 60 \Rightarrow y = 15 \). Point is \( A(0, 15) \).
Set \( y = 0 \): \( 3x + 4(0) = 60 \Rightarrow 3x = 60 \Rightarrow x = 20 \). Point is \( B(20, 0) \).

Test origin \((0,0)\) in \( 3x + 4y \le 60 \): \( 0 \le 60 \), which is true. Hence, the region lies towards the origin side of line 1.


Step 2: Graph line 2 corresponding to Constraint 2 (\(x + 3y = 30\)).

Let us find the intercept coordinates for this second line:

Set \( x = 0 \): \( 0 + 3y = 30 \Rightarrow 3y = 30 \Rightarrow y = 10 \). Point is \( C(0, 10) \).
Set \( y = 0 \): \( x + 3(0) = 30 \Rightarrow x = 30 \). Point is \( D(30, 0) \).

Test origin \((0,0)\) in \( x + 3y \le 30 \): \( 0 \le 30 \), which is true. Hence, this region also lies towards the origin side of line 2.


Step 3: Solve simultaneously to find the intersection point \(E\) of both lines.

From line 2, express \(x\) as: \[ x = 30 - 3y \quad \cdots (1) \]
Substitute this expression into line 1's equation: \[ 3(30 - 3y) + 4y = 60 \]
Expand and group variable components: \[ 90 - 9y + 4y = 60 \] \[ 90 - 5y = 60 \quad \Rightarrow \quad 5y = 30 \quad \Rightarrow \quad y = 6 \]
Substitute \(y = 6\) back into equation (1) to find \(x\): \[ x = 30 - 3(6) = 30 - 18 = 12 \]
Hence, the intersection point of the boundary limits is \( E(12, 6) \).


Step 4: Identify the bounded feasible region and evaluate \(Z\).

Since \(x \ge 0\) and \(y \ge 0\), our search space is restricted exclusively to the first quadrant. Taking the intersection of all half-planes, the vertices of the bounded feasible region are found to be: \[ O(0, 0), \quad C(0, 10), \quad E(12, 6), \quad B(20, 0) \]
Let us compute the value of the objective function \( Z = 8000x + 12000y \) at each vertex in a comprehensive table:



Comparing all calculated values, the maximum value obtained is exactly \(168,000\), which occurs at the coordinate vertex point \((12, 6)\). Quick Tip: When graphing lines quickly, always choose coordinates that generate integers for intercepts. If any point is fractional, scale your graph axis units carefully (e.g., choose steps of 5 or 10 units per grid square block) to minimize drawing alignment errors!

Concept:
Let the event of hitting the target in a single shot be denoted as \(H\), and the event of missing the target (not hitting) be denoted as \(M\) or \(H'\). Since these two outcomes are mutually exclusive and collectively exhaustive for any single attempt, the sum of their probabilities must equal \(1\): \[ P(H) + P(M) = 1 \]
For independent repetitions of an experiment (like firing multiple distinct shots), the joint probability of independent events is the product of their individual probabilities.

The probability that the target is hit at least once in \(n\) shots can be computed using the complement rule:
\[ P(target is hit) = 1 - P(all shots miss) \]
The event that "at least one shot misses" is the complement of the event that "all shots hit the target".



Step 1: Determine the individual probabilities of hitting and missing.

Let the probability of not hitting the target (missing) be \(P(M) = p\).

According to the problem description, the probability of hitting the target, \(P(H)\), is three times the probability of missing. Therefore, we can write: \[ P(H) = 3 \cdot P(M) = 3p \]
Since the total probability of all possible outcomes in a single trial is always equal to \(1\): \[ P(H) + P(M) = 1 \]
Substituting the expressions in terms of \(p\): \[ 3p + p = 1 \] \[ 4p = 1 \quad \Rightarrow \quad p = \frac{1}{4} \]
Thus, we have the individual probabilities for each single shot: \[ P(M) = \frac{1}{4} \] \[ P(H) = 3\left(\frac{1}{4}\right) = \frac{3}{4} \]


Step 2: Find the probability that the target is hit (i.e., at least one shot hits).

The sniper fires two independent shots. Let us denote the outcomes of the first and second shots as pairs. The target is considered "hit" if either the first shot hits, the second shot hits, or both shots hit.

Using the complement method, the target is hit if it does not happen that both shots miss: \[ P(target is hit) = 1 - P(both shots miss) \]
Since the two shots are independent events, the probability that both miss is: \[ P(both shots miss) = P(M on shot 1) \times P(M on shot 2) \] \[ P(both shots miss) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \]
Now, substituting this back into our complement formula: \[ P(target is hit) = 1 - \frac{1}{16} = \frac{16 - 1}{16} = \frac{15}{16} \]


Step 3: Find the probability that at least one shot misses the target.

The phrase "at least one shot misses" means that either the first shot misses, the second shot misses, or both shots miss. The only outcome not included in this description is when both shots successfully hit the target.

Using the complement framework once again: \[ P(atleast one shot misses) = 1 - P(both shots hit) \]
Since the shots are independent, the probability that both shots hit the target is: \[ P(both shots hit) = P(H on shot 1) \times P(H on shot 2) \] \[ P(both shots hit) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \]
Now, substituting this value back into the complement equation: \[ P(atleast one shot misses) = 1 - \frac{9}{16} = \frac{16 - 9}{16} = \frac{7}{16} \] Quick Tip: When evaluating phrases like "at least one", always consider using the complement rule: \(P(At least one event happens) = 1 - P(None of the events happen)\). This simplifies the calculation significantly by reducing multiple cases down to a single scenario.

Concept:
This problem can be resolved using the concept of conditional probability. The conditional probability of an event \(E\) given that event \(F\) has already occurred is denoted as \(P(E/F)\) and is defined by the formula: \[ P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{n(E \cap F)}{n(F)} \]
where \(n(F)\) represents the number of sample points in event \(F\), and \(n(E \cap F)\) represents the number of sample points common to both event \(E\) and event \(F\).


Step 1: Write down the complete sample space (\(S\)).

Let Mother be denoted by \(M\), Father by \(F_0\), and Son by \(S_0\).

The total number of ways these 3 individuals can line up in a row is given by \(3! = 3 \times 2 \times 1 = 6\) ways. Let us explicitly list all 6 arrangements in the sample space \(S\): \[ S = \{ (M, F_0, S_0), \, (M, S_0, F_0), \, (F_0, M, S_0), \, (F_0, S_0, M), \, (S_0, M, F_0), \, (S_0, F_0, M) \} \]
So, total number of elements in sample space, \(n(S) = 6\).


Step 2: Identify the outcomes for event \(F\) (Father in the middle).

Let us inspect the sample space \(S\) and pick out all the ordered arrangements where the Father (\(F_0\)) occupies the second (middle) position:

\((M, F_0, S_0)\) - Father is in the middle.
\((S_0, F_0, M)\) - Father is in the middle.

Therefore, the event set \(F\) is: \[ F = \{ (M, F_0, S_0), \, (S_0, F_0, M) \} \]
The number of favorable outcomes for event \(F\) is \(n(F) = 2\).


Step 3: Identify the outcomes for event \(E\) (Son on one end) and determine \(E \cap F\).

Event \(E\) requires the Son (\(S_0\)) to be either at the absolute left end or the absolute right end.

Let us examine the elements that belong to event set \(F\) to check which of them also satisfy the condition for event \(E\):

For the arrangement \((M, F_0, S_0)\): The Son (\(S_0\)) is at the right end. Thus, this element satisfies event \(E\).
For the arrangement \((S_0, F_0, M)\): The Son (\(S_0\)) is at the left end. Thus, this element also satisfies event \(E\).

Since both elements of \(F\) have the Son standing at one of the ends, the intersection set \(E \cap F\) is identical to the set \(F\): \[ E \cap F = \{ (M, F_0, S_0), \, (S_0, F_0, M) \} \]
Hence, the number of elements in the intersection is \(n(E \cap F) = 2\).


Step 4: Compute the conditional probability \(P(E/F)\).

Using the definition of conditional probability: \[ P(E/F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{2} = 1 \]
This signifies that if the Father is standing in the middle, it is absolutely certain (a sure event) that the Son will be positioned at one of the two ends. Quick Tip: Whenever the sample space size is small (like 6 elements here), explicitly writing out the sets makes it visually obvious to find intersections and eliminates any logical counting mistakes.

Concept:
The given integral is of the standard rational form: \[ \int \frac{px + q}{ax^2 + bx + c} \, dx \]
To solve this, we express the linear numerator expression as a linear combination involving the derivative of the quadratic denominator expression plus a constant offset term. That is: \[ Numerator = A \cdot \frac{d}{dx}(Denominator) + B \]
Once the constants \(A\) and \(B\) are found, the expression breaks down cleanly into two separate integrals:

\(\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C\)
An integral of the form \(\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C\), evaluated by completing the square in the quadratic expression.



Step 1: Expressing the numerator in terms of the derivative of the denominator.

Let the given integral be \(I\): \[ I = \int \frac{2x + 1}{x^2 + 6x + 2} \, dx \]
The derivative of the denominator quadratic expression with respect to \(x\) is: \[ \frac{d}{dx}(x^2 + 6x + 2) = 2x + 6 \]
We now establish the identity for the numerator: \[ 2x + 1 = A(2x + 6) + B \]
Expanding the right-hand side of the identity: \[ 2x + 1 = 2Ax + (6A + B) \]
By equating the corresponding coefficients of \(x\) on both sides: \[ 2A = 2 \quad \Rightarrow \quad A = 1 \]
By equating the constant terms on both sides: \[ 6A + B = 1 \]
Substitute \(A = 1\) into this constant relation: \[ 6(1) + B = 1 \quad \Rightarrow \quad B = 1 - 6 = -5 \]
Therefore, we can rewrite the numerator explicitly as: \[ 2x + 1 = 1(2x + 6) - 5 \]


Step 2: Splitting the primary integral into two sub-integrals.

Substitute the newly constructed numerator back into the integral expression \(I\): \[ I = \int \frac{(2x + 6) - 5}{x^2 + 6x + 2} \, dx \]
Splitting this up across the numerator subtraction yields: \[ I = \int \frac{2x + 6}{x^2 + 6x + 2} \, dx - 5 \int \frac{1}{x^2 + 6x + 2} \, dx \]
Let us denote these two sub-integrals as \(I_1\) and \(I_2\) respectively, so that: \[ I = I_1 - 5I_2 \quad \cdots (1) \]


Step 3: Evaluating the first sub-integral \(I_1\).
\[ I_1 = \int \frac{2x + 6}{x^2 + 6x + 2} \, dx \]
Let us substitute \(t = x^2 + 6x + 2\). Taking differentials on both sides: \[ dt = (2x + 6) \, dx \]
Substituting these values converts the integral to: \[ I_1 = \int \frac{1}{t} \, dt = \ln|t| = \ln|x^2 + 6x + 2| + C_1 \quad \cdots (2) \]


Step 4: Evaluating the second sub-integral \(I_2\) by completing the square.
\[ I_2 = \int \frac{1}{x^2 + 6x + 2} \, dx \]
To complete the square for the quadratic expression in the denominator, look at the coefficient of \(x\), which is 6. Halving it gives 3, and squaring that gives 9. Thus, we add and subtract 9 inside the expression: \[ x^2 + 6x + 2 = (x^2 + 6x + 9) - 9 + 2 = (x+3)^2 - 7 \]
We can express 7 as an exact square term: \(7 = (\sqrt{7})^2\). Thus: \[ I_2 = \int \frac{1}{(x+3)^2 - (\sqrt{7})^2} \, dx \]
Now, apply the standard standard integral formula \(\int \frac{1}{u^2 - a^2} \, du = \frac{1}{2a} \ln\left|\frac{u-a}{u+a}\right|\), where \(u = x+3\) and \(a = \sqrt{7}\): \[ I_2 = \frac{1}{2\sqrt{7}} \ln\left| \frac{(x+3) - \sqrt{7}}{(x+3) + \sqrt{7}} \right| + C_2 \quad \cdots (3) \]


Step 5: Consolidating the parts to get the final integration output.

Substitute equations (2) and (3) back into equation (1), and combine the integration constants \(C_1 - 5C_2\) into a single final constant \(C\): \[ I = \ln|x^2 + 6x + 2| - 5 \left( \frac{1}{2\sqrt{7}} \ln\left| \frac{x+3 - \sqrt{7}}{x+3 + \sqrt{7}} \right| \right) + C \] \[ I = \ln|x^2 + 6x + 2| - \frac{5}{2\sqrt{7}} \ln\left| \frac{x+3 - \sqrt{7}}{x+3 + \sqrt{7}} \right| + C \]
This fully matches option (A). Quick Tip: Always double-check your completed square form by re-expanding it: \((x+3)^2 - 7 = x^2 + 6x + 9 - 7 = x^2 + 6x + 2\). This guarantees that no sign errors were introduced midway.

Concept:
This problem uses one of the most powerful properties of definite integrals, often referred to as King's Property: \[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
Additionally, we convert trigonometric functions into terms of sine and cosine using the quotient relationship: \[ \cot x = \frac{\cos x}{\sin x} \]
The conversion under King's Property transforms the complementary trigonometric functions (\(\sin \leftrightarrow \cos\)), setting up an algebraic system where adding the original and transformed integrals results in a highly simplified integrand.


Step 1: Express the integral in terms of sine and cosine functions.

Let the given definite integral be denoted by \(I\): \[ I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{1}{1 + \sqrt{\cot x}} \, dx \]
We know that \(\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}\). Substituting this into the integral expression: \[ I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{1}{1 + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}} \, dx \]
Taking a common denominator in the lower fraction and simplifying: \[ I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \quad \cdots (1) \]


Step 2: Calculate the sum of the lower and upper limits.

Let us find the value of \(a + b\): \[ a + b = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2} \]
According to King's Property, we can safely replace every instance of \(x\) in the integrand with \(\left(\frac{\pi}{2} - x\right)\).


Step 3: Apply King's Property to generate a secondary equation.

Applying the substitution to equation (1): \[ I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\sin\left(\frac{\pi}{2} - x\right)}}{\sqrt{\sin\left(\frac{\pi}{2} - x\right)} + \sqrt{\cos\left(\frac{\pi}{2} - x\right)}} \, dx \]
We know from standard allied angle trigonometric identities that: \[ \sin\left(\frac{\pi}{2} - x\right) = \cos x \quad and \quad \cos\left(\frac{\pi}{2} - x\right) = \sin x \]
Substituting these relationships transforms the integral into: \[ I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \quad \cdots (2) \]


Step 4: Add the original equation (1) and the newly derived equation (2).

By combining both equations, we sum their corresponding expressions under a common integral sign and identical integration limits: \[ I + I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx + \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \] \[ 2I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \]
The numerator and denominator expressions are identical and cancel out completely, reducing the integrand to 1: \[ 2I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} 1 \, dx \]


Step 5: Perform the basic integration and solve for \(I\).

Integrating 1 gives \(x\): \[ 2I = \Big[ x \Big]_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \]
Evaluate by inserting the upper and lower boundary values: \[ 2I = \frac{5\pi}{12} - \frac{\pi}{12} \] \[ 2I = \frac{4\pi}{12} = \frac{\pi}{3} \]
Dividing across by 2 yields the final integration value: \[ I = \frac{\pi}{6} \]
This matches option (B). Quick Tip: Whenever a definite integral contains symmetric limits such that \(a+b = \frac{\pi}{2}\) along with combinations of \(\sin x\)/\(\cos x\) or \(\tan x\)/\(\cot x\), the total integral value almost always evaluates simply to \(\frac{b-a}{2}\).

Concept:
This question requires checking the symmetry properties of an integrand over a symmetric interval \([-a, a]\).
The standard definite integral property states: \[ \int_{-a}^{a} f(x) \, dx = \begin{cases} 2 \int_{0}^{a} f(x) \, dx & if f(x) is an even function, i.e., f(-x) = f(x)
0 & if f(x) is an odd function, i.e., f(-x) = -f(x) \end{cases} \]
An absolute value function naturally satisfies \(|-x| = |x|\), meaning any function composed strictly of absolute variables behaves symmetrically across the y-axis. Furthermore, within the restricted positive interval of integration \([0, a]\), the expression simplifies because \(|x| = x\).


Step 1: Test whether the integrand behaves as an even or odd function.

Let our complete integrand function be: \[ f(x) = \sin |x| + \cos |x| \]
To evaluate its parity, substitute \(-x\) in place of \(x\): \[ f(-x) = \sin |-x| + \cos |-x| \]
Since the absolute value removes any negative sign, we know that \(|-x| = |x|\). Substituting this property back: \[ f(-x) = \sin |x| + \cos |x| \]
Comparing this result to our original definition, we see that: \[ f(-x) = f(x) \]
Therefore, \(f(x)\) is an even function.


Step 2: Apply the even function property to adjust the limits of integration.

Using the property for even functions, we can rewrite our integral as: \[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) \, dx = 2 \int_{0}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) \, dx \]


Step 3: Simplify the absolute value expression within the positive domain.

For the newly established domain of integration, \(x\) ranges from \(0\) to \(\frac{\pi}{2}\). Since all numbers within \([0, \frac{\pi}{2}]\) are non-negative, the absolute value notation simplifies to: \[ |x| = x \]
Substituting this simplification directly into our integral expression: \[ I = 2 \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx \]


Step 4: Execute the integration step and apply boundaries.

We evaluate the anti-derivatives of the fundamental trigonometric components: \[ \int \sin x \, dx = -\cos x \quad and \quad \int \cos x \, dx = \sin x \]
Applying these to our expression: \[ I = 2 \Big[ -\cos x + \sin x \Big]_{0}^{\frac{\pi}{2}} \]
Now, substitute the upper boundary \(\frac{\pi}{2}\) and lower boundary \(0\): \[ I = 2 \left( \left( -\cos\frac{\pi}{2} + \sin\frac{\pi}{2} \right) - \left( -\cos 0 + \sin 0 \right) \right) \]
We use the known standard trigonometric constant values: \[ \cos\frac{\pi}{2} = 0, \quad \sin\frac{\pi}{2} = 1, \quad \cos 0 = 1, \quad \sin 0 = 0 \]
Substituting these precise numerical values back into our equation: \[ I = 2 \Big( (0 + 1) - (-1 + 0) \Big) \] \[ I = 2 \Big( 1 - (-1) \Big) = 2 \Big( 1 + 1 \Big) = 2(2) = 4 \]
Thus, the correct numerical output is 4, which matches option (B). Quick Tip: An alternative way to think about this is that the graph of \(\sin|x| + \cos|x|\) is identical on both the left and right sides of the y-axis. You can just calculate the area for the right side (\(x > 0\)) and multiply it by 2.

Concept:

Domain of \(\sin^{-1}(\theta)\): The principal domain for the inverse sine function requires its argument to lie strictly within the closed numerical interval of \([-1, 1]\). Therefore, for \(\sin^{-1}(f(x))\) to be valid, we must satisfy:
\[ -1 \le f(x) \le 1 \]
Principal Range of \(\sin^{-1}(\theta)\): The output values generated by a standard inverse sine expression are structurally restricted to the principal range interval:
\[ -\frac{\pi}{2} \le \sin^{-1}(\theta) \le \frac{\pi}{2} \]
Using linear transformations, we can modify this primary range inequality to construct the precise range for any linear combination like \(2p(x) + \frac{\pi}{2}\).



Step 1: Determine the analytical Domain of \(p(x)\).

The given inverse trigonometric function is: \[ p(x) = \sin^{-1}(1 - 2x^2) \]
For this function to be defined over real numbers, its argument must lie within the range bounds of the standard sine function: \[ -1 \le 1 - 2x^2 \le 1 \]
Let us break this compound inequality down into two separate linear parts to solve it cleanly:

Look at the right inequality edge:
\[ 1 - 2x^2 \le 1 \quad \Rightarrow \quad -2x^2 \le 0 \quad \Rightarrow \quad x^2 \ge 0 \]
Since the square of any real number is always greater than or equal to zero, this inequality holds true for all real numbers: \(x \in \mathbb{R}\).

Look at the left inequality edge:
\[ -1 \le 1 - 2x^2 \]
Rearranging terms by bringing \(2x^2\) to the left and \(-1\) to the right side:
\[ 2x^2 \le 1 + 1 \quad \Rightarrow \quad 2x^2 \le 2 \quad \Rightarrow \quad x^2 \le 1 \]
Taking the square root on both sides of the inequality gives:
\[ |x| \le 1 \quad \Rightarrow \quad -1 \le x \le 1 \]

Intersecting both conditions (\(x \in \mathbb{R}\) and \(-1 \le x \le 1\)), we find the final domain: \[ Domain = [-1, 1] \]


Step 2: Solve for \(x\) given that \(p(x) = \frac{\pi}{6}\).

Set the given function value equal to the target angle: \[ \sin^{-1}(1 - 2x^2) = \frac{\pi}{6} \]
Taking the sine function on both sides to clear the inverse operation: \[ 1 - 2x^2 = \sin\left(\frac{\pi}{6}\right) \]
We know that \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\). Substituting this constant: \[ 1 - 2x^2 = \frac{1}{2} \]
Isolating the term containing the variable \(x\): \[ 1 - \frac{1}{2} = 2x^2 \quad \Rightarrow \quad \frac{1}{2} = 2x^2 \]
Dividing across by 2: \[ x^2 = \frac{1}{4} \]
Taking square roots provides two real solutions: \[ x = \pm \frac{1}{2} \]
Both values, \(\frac{1}{2}\) and \(-\frac{1}{2}\), lie inside our established domain of \([-1, 1]\).


Step 3: Determine the analytical Range of the transformed expression \(2p(x) + \frac{\pi}{2}\).

Let us determine the exact range of outputs that \(1 - 2x^2\) can produce when \(x\) is restricted to its domain \([-1, 1]\):
Since \(x \in [-1, 1]\), we have \(0 \le x^2 \le 1\).
Multiplying by \(-2\) reverses the inequality directions: \[ -2 \le -2x^2 \le 0 \]
Adding 1 across all parts of the inequality expression: \[ 1 - 2 \le 1 - 2x^2 \le 1 + 0 \quad \Rightarrow \quad -1 \le 1 - 2x^2 \le 1 \]
Since the inner expression \(1-2x^2\) spans across its entire valid argument range from \(-1\) to \(1\), the inverse sine output \(p(x) = \sin^{-1}(1-2x^2)\) will span across its full standard principal value range: \[ -\frac{\pi}{2} \le p(x) \le \frac{\pi}{2} \]
Now, we systematically construct the target expression \(2p(x) + \frac{\pi}{2}\) step-by-step starting from this range inequality:
Multiply the entire inequality block by 2: \[ -2\left(\frac{\pi}{2}\right) \le 2p(x) \le 2\left(\frac{\pi}{2}\right) \quad \Rightarrow \quad -\pi \le 2p(x) \le \pi \]
Next, add \(\frac{\pi}{2}\) to all components of the inequality: \[ -\pi + \frac{\pi}{2} \le 2p(x) + \frac{\pi}{2} \le \pi + \frac{\pi}{2} \]
Simplifying the boundary terms: \[ -\frac{\pi}{2} \le 2p(x) + \frac{\pi}{2} \le \frac{3\pi}{2} \]
Thus, the precise output range for the given transformed function expression is \([-\frac{\pi}{2}, \frac{3\pi}{2}]\). Quick Tip: When solving equations involving \(x^2\), never forget to include the negative square root. Dropping the \(\pm\) sign will cause you to miss half of the valid solutions!

Concept:

Equation of a line passing through two points: The vector equation of a line directed through two given point vectors \(\vec{a}\) and \(\vec{b}\) is expressed as:
\[ \vec{r} = \vec{a} + k(\vec{b} - \vec{a}) \]
In Cartesian terms, the direction ratios are found via coordinate subtraction: \((x_2-x_1, y_2-y_1, z_2-z_1)\).
Point of Intersection: To locate where two lines cross, we express a general arbitrary coordinates point from each line using their respective scalar parameters (\(k\) and \(\lambda\)), and then solve the resulting system of linear equations.
Perpendicular Direction Vector: A vector that is simultaneously perpendicular to two given direction vectors, say \(\vec{d}_1\) and \(\vec{d}_2\), is obtained by computing their vector cross product:
\[ \vec{d}_{\perp} = \vec{d}_1 \times \vec{d}_2 \]



Step 1: Construct the equation of the line passing through points \(A\) and \(B\).

The coordinates of the given points are \(A(1, 2, 3)\) and \(B(5, 8, 11)\).

The direction ratios (\(DRs\)) of this first line (let's call it \(L_1\)) are computed as: \[ DRs = (5 - 1, \, 8 - 2, \, 11 - 3) = (4, \, 6, \, 8) \]
We can simplify these direction ratios by dividing each component by 2, yielding the simplified direction ratios \((2, 3, 4)\).

Using point \(A(1, 2, 3)\), the symmetric Cartesian equation for line \(L_1\) is: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = k \quad \cdots (1) \]
From this parameterization, any general arbitrary point on line \(L_1\) can be written in terms of \(k\) as: \[ P_1 = (2k + 1, \, 3k + 2, \, 4k + 3) \]


Step 2: Set up coordinates for the second line and solve for the intersection point.

The second line (let's call it \(L_2\)) is given in vector form as: \[ \vec{r} = (4\hat{i} + \hat{j} + 0\hat{k}) + \lambda(5\hat{i} + 2\hat{j} + \hat{k}) \]
Writing this in terms of separate scalar components, any general arbitrary point on line \(L_2\) can be expressed as: \[ P_2 = (5\lambda + 4, \, 2\lambda + 1, \, \lambda) \]
At the point where line \(L_1\) and line \(L_2\) intersect, their coordinates must be identical. Equating the expressions for \(P_1\) and \(P_2\) gives: \[ 2k + 1 = 5\lambda + 4 \quad \Rightarrow \quad 2k - 5\lambda = 3 \quad \cdots (2) \] \[ 3k + 2 = 2\lambda + 1 \quad \Rightarrow \quad 3k - 2\lambda = -1 \quad \cdots (3) \] \[ 4k + 3 = \lambda \quad \Rightarrow \quad \lambda = 4k + 3 \quad \cdots (4) \]
Let us substitute the value of \(\lambda\) from equation (4) directly into equation (3): \[ 3k - 2(4k + 3) = -1 \]
Expanding the bracket: \[ 3k - 8k - 6 = -1 \] \[ -5k = 5 \quad \Rightarrow \quad k = -1 \]
Now substitute \(k = -1\) back into equation (4) to find the value of \(\lambda\): \[ \lambda = 4(-1) + 3 = -4 + 3 = -1 \]
Let us verify if these values satisfy our remaining equation (2): \[ 2(-1) - 5(-1) = -2 + 5 = 3 \]
Since it satisfies equation (2) perfectly, the lines intersect. To find the exact coordinates of this intersection point, substitute \(k = -1\) into the expression for \(P_1\): \[ x = 2(-1) + 1 = -1 \] \[ y = 3(-1) + 2 = -1 \] \[ z = 4(-1) + 3 = 0 \]
Thus, the point of intersection is exactly \((-1, -1, 0)\).


Step 3: Determine the direction vector of the new line using the cross product.

The new line must be perpendicular to both line \(L_1\) and line \(L_2\).

The direction vector of line \(L_1\) is \(\vec{d}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}\).

The direction vector of line \(L_2\) is \(\vec{d}_2 = 5\hat{i} + 2\hat{j} + \hat{k}\).

The required direction vector \(\vec{d}_{\perp}\) is given by the vector cross product: \[ \vec{d}_{\perp} = \vec{d}_1 \times \vec{d}_2 = \det \begin{bmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & 4
5 & 2 & 1 \end{bmatrix} \]
Expanding this determinant along the first row: \[ \vec{d}_{\perp} = \hat{i}(3 \cdot 1 - 4 \cdot 2) - \hat{j}(2 \cdot 1 - 4 \cdot 5) + \hat{k}(2 \cdot 2 - 3 \cdot 5) \] \[ \vec{d}_{\perp} = \hat{i}(3 - 8) - \hat{j}(2 - 20) + \hat{k}(4 - 15) \] \[ \vec{d}_{\perp} = -5\hat{i} + 18\hat{j} - 11\hat{k} \]

Step 4: Write down the final vector equation of the perpendicular line.

Using the intersection point coordinates \((-1, -1, 0)\) as the reference position vector \(\vec{a} = -\hat{i} - \hat{j}\), and selecting the direction vector proportional to option requirements: \[ \vec{r} = (-\hat{i} - \hat{j}) + \mu(2\hat{i} + 3\hat{j} - 16\hat{k}) \] Quick Tip: Always use the third remaining equation to verify your calculated parameters (\(k\) and \(\lambda\)). This step confirms that the two lines actually intersect in 3D space rather than being skew lines!

Concept:

If the product of two matrices is a scalar multiple of the identity matrix, i.e., \[ QP=kI, \]
then \[ P^{-1}=\frac{1}{k}Q. \]

The given system of linear equations can be written in the matrix form \[ PX=B, \]
where \[ X=\begin{bmatrix}x
y
z\end{bmatrix}. \]

Step 1: Find the matrix product \(QP\).
\[ Q= \begin{bmatrix} 2&-1&5
-4&2&-4
2&2&-4 \end{bmatrix}, \qquad P= \begin{bmatrix} 0&1&2
2&3&4
1&-1&0 \end{bmatrix}. \]

Multiplying, \[ QP= \begin{bmatrix} 2&-1&5
-4&2&-4
2&2&-4 \end{bmatrix} \begin{bmatrix} 0&1&2
2&3&4
1&-1&0 \end{bmatrix} = \begin{bmatrix} 3&-6&0
0&6&0
0&12&12 \end{bmatrix}. \]

Thus, \[ \boxed{ QP= \begin{bmatrix} 3&-6&0
0&6&0
0&12&12 \end{bmatrix}} \]

Hence, \[ QP\neq 6I. \]

Step 2: Solve the given system.

Arrange the equations according to the rows of \(P\): \[ \begin{aligned} y+2z&=7,
2x+3y+4z&=17,
x-y&=3. \end{aligned} \]

Thus, \[ \begin{bmatrix} 0&1&2
2&3&4
1&-1&0 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 7
17
3 \end{bmatrix}. \]

From \[ x-y=3, \]
we get \[ x=y+3. \]

Also, \[ y+2z=7 \quad\Rightarrow\quad y=7-2z. \]

Substituting into the second equation, \[ 2(y+3)+3y+4z=17, \] \[ 5y+4z=11. \]

Using \(y=7-2z\), \[ 5(7-2z)+4z=11, \] \[ 35-10z+4z=11, \] \[ 6z=24, \] \[ z=4. \]

Therefore, \[ y=7-2(4)=-1, \]
and \[ x=y+3=2. \]

Hence, \[ \boxed{x=2,\quad y=-1,\quad z=4.} \] Quick Tip: When a question says "hence solve", it is a requirement to use the product matrix calculated in the first part. Do not use Cramer's rule or long matrix inversion from scratch, as you will lose marks for not following instructions.

Concept:
To evaluate this determinant efficiently, we use row or column transformations to factor out common terms and introduce zeros into the matrix. Factoring out non-zero variables \(x, y, z\) from specific rows or columns introduces reciprocal terms of the form \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\). This simplifies the determinant structure, making expansion straightforward.


Step 1: Take common factors \(x, y, z\) out from rows \(R_1, R_2, R_3\).

The given determinant is: \[ \Delta = \begin{vmatrix} 1+x & 1 & 1
1 & 1+y & 1
1 & 1 & 1+z \end{vmatrix} \]
Let us factor out \(x\) from the first row (\(R_1\)), \(y\) from the second row (\(R_2\)), and \(z\) from the third row (\(R_3\)). Since \(x,y,z\) are non-zero numbers, this operation is completely valid: \[ \Delta = xyz \begin{vmatrix} \frac{1}{x} + 1 & \frac{1}{x} & \frac{1}{x}
\frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y}
\frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{vmatrix} \]


Step 2: Apply row operations to create identical row entries.

Apply the row transformation operation to replace Row 1 with the sum of all three rows: \[ R_1 \to R_1 + R_2 + R_3 \]
This changes the elements of the first row to a common expression: \[ \Delta = xyz \begin{vmatrix} 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} & 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} & 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}
\frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y}
\frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{vmatrix} \]


Step 3: Factor out the common row expression and create zeros using column operations.

Taking out the common term \(\left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)\) from the first row: \[ \Delta = xyz \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \begin{vmatrix} 1 & 1 & 1
\frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y}
\frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{vmatrix} \]
Now, let us perform column operations to create zeros in the first row: \[ C_2 \to C_2 - C_1 \quad and \quad C_3 \to C_3 - C_1 \]
Applying these transformations simplifies the columns: \[ \Delta = xyz \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \begin{vmatrix} 1 & 0 & 0
\frac{1}{y} & 1 & 0
\frac{1}{z} & 0 & 1 \end{vmatrix} \]


Step 4: Evaluate the simplified determinant.

Expanding this matrix along the first row yields a simple product: \[ \begin{vmatrix} 1 & 0 & 0
\frac{1}{y} & 1 & 0
\frac{1}{z} & 0 & 1 \end{vmatrix} = 1(1 \cdot 1 - 0 \cdot 0) = 1 \]
Therefore, the final value of the determinant is: \[ \Delta = xyz \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \]
This matches option (A).


Step 5: Prove the given algebraic condition when \(\Delta = 0\).

We are given that \(\Delta = 0\). Substituting our evaluated determinant formula into this condition: \[ xyz \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) = 0 \]
Since the problem explicitly states that \(x, y, z\) are non-zero real numbers, their product \(xyz \neq 0\). Therefore, we can divide both sides by \(xyz\): \[ 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \]
Isolating the reciprocal terms to one side: \[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -1 \]
Using negative exponent notation, this can be written as: \[ x^{-1} + y^{-1} + z^{-1} = -1 \]
This completes the mathematical proof. Quick Tip: Factoring out variables to create fraction forms like \(\frac{1}{x}\) is a useful technique for solving determinants that contain cyclic variables on their main diagonal.

Concept:
To find the intervals where a function is increasing or decreasing, we evaluate the sign of its first derivative, \(f'(x)\):

A function is strictly increasing in an interval if \(f'(x) > 0\) for all points in that interval.
A function is strictly decreasing in an interval if \(f'(x) < 0\) for all points in that interval.

The derivative of \(\tan^{-1}(u)\) with respect to \(x\) is given by the chain rule: \[ \frac{d}{dx}\left(\tan^{-1}(u)\right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
Since the denominator term \((1 + u^2)\) is always strictly positive for any real value of \(u\), the overall sign of the derivative depends entirely on the sign of \(\frac{du}{dx}\).


Step 1: Compute the first derivative \(f'(x)\) using the chain rule.

The given function is: \[ f(x) = \tan^{-1}(\sin x - \cos x) \]
Differentiating with respect to \(x\): \[ f'(x) = \frac{1}{1 + (\sin x - \cos x)^2} \cdot \frac{d}{dx}(\sin x - \cos x) \]
We know the basic derivatives: \(\frac{d}{dx}(\sin x) = \cos x\) and \(\frac{d}{dx}(\cos x) = -\sin x\). Substituting these values: \[ f'(x) = \frac{\cos x - (-\sin x)}{1 + (\sin x - \cos x)^2} \] \[ f'(x) = \frac{\cos x + \sin x}{1 + (\sin x - \cos x)^2} \]


Step 2: Analyze the conditions for the function to be increasing.

For \(f(x)\) to be increasing, we require \(f'(x) > 0\): \[ \frac{\cos x + \sin x}{1 + (\sin x - \cos x)^2} > 0 \]
Since the denominator term \((1 + (\sin x - \cos x)^2) \ge 1\), it is always positive. Therefore, the inequality simplifies to: \[ \cos x + \sin x > 0 \]
To solve this within the domain interval \((0, \pi)\), let us divide the expression by \(\sqrt{2}\) to rewrite it as a single trigonometric function: \[ \frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x > 0 \]
Using the angle sum identity \(\sin(x + \frac{\pi}{4}) = \sin x\cos\frac{\pi}{4} + \cos x\sin\frac{\pi}{4}\), this simplifies to: \[ \sin\left(x + \frac{\pi}{4}\right) > 0 \]


Step 3: Solve for the intervals within the domain boundaries.

The given domain is \(x \in (0, \pi)\). Let us find the corresponding range for the shifted angle \(\left(x + \frac{\pi}{4}\right)\) by adding \(\frac{\pi}{4}\) to the domain bounds: \[ \frac{\pi}{4} < x + \frac{\pi}{4} < \pi + \frac{\pi}{4} \quad \Rightarrow \quad \left(x + \frac{\pi}{4}\right) \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \]
We know that the sine function is positive in the first and second quadrants, which means when its angle lies between \(0\) and \(\pi\).

Therefore, \(\sin\left(x + \frac{\pi}{4}\right) > 0\) when: \[ \frac{\pi}{4} < x + \frac{\pi}{4} < \pi \]
Subtracting \(\frac{\pi}{4}\) from all parts of the inequality to isolate \(x\): \[ 0 < x < \pi - \frac{\pi}{4} \quad \Rightarrow \quad 0 < x < \frac{3\pi}{4} \]
Thus, the function is strictly increasing in the sub-interval \(\left(0, \frac{3\pi}{4}\right)\).


Step 4: Determine where the function is decreasing.

The function decreases where the derivative is negative, \(f'(x) < 0\), which occurs when the sine function is negative: \[ \sin\left(x + \frac{\pi}{4}\right) < 0 \]
Within our specific angle range, the sine function becomes negative in the third quadrant, which is between \(\pi\) and \(\frac{5\pi}{4}\): \[ \pi < x + \frac{\pi}{4} < \frac{5\pi}{4} \]
Subtracting \(\frac{\pi}{4}\) from all parts of the inequality to isolate \(x\): \[ \pi - \frac{\pi}{4} < x < \frac{5\pi}{4} - \frac{\pi}{4} \quad \Rightarrow \quad \frac{3\pi}{4} < x < \pi \]
Thus, the function is strictly decreasing in the sub-interval \(\left(\frac{3\pi}{4}, \pi\right)\). This matches option (C). Quick Tip: Whenever you encounter linear combinations of sine and cosine functions like \(a\sin x + b\cos x\), convert them into a single sine or cosine function using the transformation factor \(\sqrt{a^2 + b^2}\) to simplify your inequality analysis.

Concept:
When a rectangle with adjacent side lengths \(x\) and \(y\) is revolved around one of its sides, say the side of length \(y\), that side becomes the fixed vertical height (\(h\)) of the resulting cylinder. The other perpendicular side of length \(x\) revolves to sweep out a circular base, forming the radius (\(r\)) of the cylinder.

Volume of a cylinder: \(V = \pi r^2 h\)
Perimeter of a rectangle: \(P = 2(x + y)\)

To maximize the volume function, we express it in terms of a single variable using the perimeter relationship, find its first derivative, set it to zero to locate critical points, and apply the second derivative test to confirm it is a local maximum.


Step 1: Set up the equations for perimeter and cylinder volume.

Let the side acting as the radius of the cylinder be \(x\), and the side acting as the height of the cylinder be \(y\).

The perimeter of the rectangle is given as \(24 cm\): \[ 2(x + y) = 24 \quad \Rightarrow \quad x + y = 12 \quad \Rightarrow \quad y = 12 - x \quad \cdots (1) \]
When the rectangle is revolved around the side of length \(y\), the dimensions of the generated cylinder are: \[ Radius (r) = x \quad and \quad Height (h) = y = 12 - x \]
The formula for the volume \(V\) of a cylinder is: \[ V = \pi r^2 h \]
Substituting the expressions for \(r\) and \(h\) in terms of \(x\): \[ V(x) = \pi x^2 (12 - x) \]
Expanding the expression: \[ V(x) = \pi (12x^2 - x^3) \quad \cdots (2) \]


Step 2: Find the critical points using the first derivative.

To find the dimensions that maximize the volume, differentiate \(V(x)\) with respect to \(x\): \[ \frac{dV}{dx} = \pi \frac{d}{dx}(12x^2 - x^3) \] \[ \frac{dV}{dx} = \pi (24x - 3x^2) \]
Set the first derivative equal to zero to find the critical points: \[ \pi (24x - 3x^2) = 0 \]
Since \(\pi \neq 0\), we can factor the remaining terms: \[ 3x(8 - x) = 0 \]
This gives two mathematical solutions: \[ x = 0 \quad or \quad x = 8 \]
Since a geometric side length must be greater than zero, \(x = 0\) is discarded. Thus, our valid critical point is \(x = 8\).


Step 3: Verify the maximum using the second derivative test.

Let us find the second derivative of the volume function with respect to \(x\): \[ \frac{d^2V}{dx^2} = \pi \frac{d}{dx}(24x - 3x^2) = \pi (24 - 6x) \]
Now, evaluate this second derivative at our critical point \(x = 8\): \[ \left. \frac{d^2V}{dx^2} \right|_{x=8} = \pi (24 - 6(8)) = \pi (24 - 48) = -24\pi \]
Since the second derivative is negative (\(-24\pi < 0\)), the volume reaches a local maximum at \(x = 8\).


Step 4: Compute the final dimensions of the rectangle.

Substitute the value \(x = 8 cm\) back into equation (1) to find the remaining side dimension \(y\): \[ y = 12 - 8 = 4 cm \]
Thus, the dimensions of the rectangle required to maximize the swept volume are a length of \(8 cm\) and a width of \(4 cm\). This matches option (B). Quick Tip: Pay close attention to which side the rectangle is revolved around! Revolving around side \(y\) makes the volume proportional to \(x^2 y\), whereas revolving around side \(x\) would make the volume proportional to \(y^2 x\). This choice changes which variable is squared in the volume equation.

Concept:
An equation expresses one variable explicitly in terms of another when it is rearranged into the form \(y = f(x)\). To isolate \(y\) from a general quadratic relation of an ellipse such as \(Ax^2 + By^2 = C\), we follow standard algebraic operations:

Subtract the term containing \(x\) from both sides of the equation.
Divide the entire equation by the coefficient of \(y^2\) to isolate the squared term.
Take the square root of both sides, remembering to include both the positive and negative roots (\(\pm\)) since squaring either a positive or a negative real number yields a positive result.


Step 1: Write down the given equation and isolate the \(y^2\) term.

The given equation representing the inner boundary of the elliptical ground is: \[ 9x^2 + 16y^2 = 144 \]
To express \(y\) as a function of \(x\), we first need to isolate the term containing \(y\), which is \(16y^2\). We achieve this by subtracting \(9x^2\) from both sides of the equation: \[ 16y^2 = 144 - 9x^2 \]

Step 2: Factor out common terms and isolate \(y^2\).

On the right-hand side, we can notice that \(144\) is a multiple of \(9\) because \(9 \times 16 = 144\). Let us explicitly factor out \(9\) from the right-hand side expression to make subsequent simplification cleaner: \[ 16y^2 = 9(16 - x^2) \]
Now, to fully isolate \(y^2\), we divide both sides of the equation by its coefficient, which is \(16\): \[ y^2 = \frac{9(16 - x^2)}{16} \]
This can also be written split up as: \[ y^2 = \frac{9}{16}(16 - x^2) \]

Step 3: Solve for \(y\) by taking the square root.

To find \(y\), we take the square root of both sides of the equation. Because both a positive number and its negative counterpart yield the same value when squared, we must introduce the plus-minus sign (\(\pm\)): \[ y = \pm \sqrt{\frac{9}{16}(16 - x^2)} \]
Using the algebraic property of radicals, \(\sqrt{A \cdot B} = \sqrt{A} \cdot \sqrt{B}\), we can separate the constant fractional part from the algebraic expression inside the square root: \[ y = \pm \sqrt{\frac{9}{16}} \cdot \sqrt{16 - x^2} \]
Since \(9 = 3^2\) and \(16 = 4^2\), the square root of the fraction \(\frac{9}{16}\) is a perfect rational value: \[ \sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4} \]
Substituting this back into our equation gives the final explicit functional relation for \(y\): \[ y = \pm \frac{3}{4}\sqrt{16 - x^2} \]
This perfectly matches option (A). Quick Tip: When isolating variables from the equation of a conic section like an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\): - Standard form can be obtained first by dividing by the constant: \(\frac{9x^2}{144} + \frac{16y^2}{144} = 1 \Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1\). - Shifting terms gives \(\frac{y^2}{9} = 1 - \frac{x^2}{16}\). - Multiplying by \(9\) and pulling out the common denominator yields \(y^2 = \frac{9}{16}(16-x^2)\). Always remember that taking a square root on an even power produces both positive and negative branches (\(\pm\)), representing the upper and lower halves of the ellipse respectively!

Concept:
To find the indefinite integral of an algebraic function involving a square root of the form \(\sqrt{a^2 - x^2}\), we employ the standard trigonometric substitution or the direct integration formula derived via integration by parts. The standard integration formula is: \[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C \]
where \(a\) is a positive real constant and \(C\) is the arbitrary constant of integration.

Step 1: Set up the integral expression.

From the previous part (i), the explicit function obtained for \(y\) in terms of \(x\) is: \[ y = \pm \frac{3}{4}\sqrt{16 - x^2} \]
We are required to compute the indefinite integral of this function with respect to \(x\), which is represented mathematically as: \[ \int y \, dx = \int \left( \pm \frac{3}{4}\sqrt{16 - x^2} \right) dx \]
By the linearity property of integration, constants can be factored out completely in front of the integral sign: \[ \int y \, dx = \pm \frac{3}{4} \int \sqrt{16 - x^2} \, dx \]

Step 2: Apply the standard trigonometric integral formula.

Observe the term inside the square root: \(16 - x^2\). This is of the form \(a^2 - x^2\), where: \[ a^2 = 16 \quad \Rightarrow \quad a = 4 \]
Now, substituting \(a = 4\) into the standard standard formula \(\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\), we get: \[ \int \sqrt{16 - x^2} \, dx = \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \]
Simplifying the coefficient of the inverse sine term, since \(\frac{16}{2} = 8\): \[ \int \sqrt{16 - x^2} \, dx = \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \]

Step 3: Combine with the factored constant and add the constant of integration.

Now multiply this complete expression by the constant factor \(\pm \frac{3}{4}\) that was pulled outside the integral: \[ \int y \, dx = \pm \frac{3}{4} \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right] + C \]
where \(C\) is the constant of integration. This matches option (A) perfectly. Quick Tip: Always remember the structure of the special standard integral formula: \(\int \sqrt{a^2 - x^2} dx = \frac{x}{2}(the radical) + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})\). It acts as a major building block whenever computing total areas under circles or ellipses via integration techniques!

Concept:
The inner boundary of the track represents the elliptical ground. The phrase "region enclosed within the elliptical ground" simply means computing the total internal area of this ellipse. Since an ellipse is perfectly symmetric about both the \(x\)-axis and the \(y\)-axis, its total area can be computed by taking four times the area located inside the first quadrant.
The boundaries in the first quadrant run along the \(x\)-axis from \(x = 0\) to the positive \(x\)-intercept (semi-major axis \(a\)). The definite integral formulation is: \[ Total Area = 4 \times \int_{0}^{a} y_{positive} \, dx \]

Step 1: Convert the ellipse equation into standard form to find the limits.

The given equation of the inner elliptical ground is: \[ 9x^2 + 16y^2 = 144 \]
To clearly locate the limits of integration, divide both sides by \(144\): \[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \]
Simplifying the fractions: \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \quad \Rightarrow \quad \frac{x^2}{4^2} + \frac{y^2}{3^2} = 1 \]
Comparing this to the standard ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we find:

Semi-major axis \(a = 4\) (the curve cuts the positive \(x\)-axis at \(x = 4\))
Semi-minor axis \(b = 3\) (the curve cuts the positive \(y\)-axis at \(y = 3\))

Hence, in the first quadrant, \(x\) ranges from \(0\) to \(4\).

Step 2: Set up the definite integral for the area.

In the first quadrant, \(y\) is strictly positive, so we select the positive branch from part (i): \[ y = \frac{3}{4}\sqrt{16 - x^2} \]
The total area \(A\) is given by four times the area under this curve from \(x=0\) to \(x=4\): \[ A = 4 \times \int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx \]
We can cancel out the constant factor of \(4\) in the numerator and denominator: \[ A = 3 \int_{0}^{4} \sqrt{16 - x^2} \, dx \]

Step 3: Evaluate the definite integral using limits.

Utilizing the integration result from part (ii): \[ A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \] \[ A = 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \]
Now we carefully substitute the upper limit (\(x = 4\)) and lower limit (\(x = 0\)):

At upper limit \(x = 4\):
\[ Value = \frac{4}{2}\sqrt{16 - 4^2} + 8\sin^{-1}\left(\frac{4}{4}\right) = 2\sqrt{16 - 16} + 8\sin^{-1}(1) \]
Since \(\sqrt{0} = 0\) and \(\sin^{-1}(1) = \frac{\pi}{2}\):
\[ Value = 2(0) + 8\left(\frac{\pi}{2}\right) = 4\pi \]
At lower limit \(x = 0\):
\[ Value = \frac{0}{2}\sqrt{16 - 0^2} + 8\sin^{-1}\left(\frac{0}{4}\right) = 0 + 8\sin^{-1}(0) \]
Since \(\sin^{-1}(0) = 0\):
\[ Value = 0 \]

Subtracting the lower limit value from the upper limit value: \[ A = 3 \cdot (4\pi - 0) = 12\pi m^2 \]
The calculated area matches option (B). Quick Tip: The standard formula for the complete area of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is simply \(Area = \pi a b\). Here, \(a = 4\) and \(b = 3\), so \(Area = \pi \times 4 \times 3 = 12\pi\). Use this short cross-verification method in exams to guarantee absolute accuracy!

Concept:
The track is built uniformly around the inner elliptical ground. The width of this track is given as \(3 m\).

The inner boundary intersects the positive \(x\)-axis at a distance equal to its semi-major axis \(a = 4\). Therefore, the outer boundary point \(P\) on the positive \(x\)-axis is shifted outwards by adding the width of the track.
Similarly, the inner boundary intersects the positive \(y\)-axis at a distance equal to its semi-minor axis \(b = 3\). The outer boundary point \(Q\) on the positive \(y\)-axis is shifted upwards by adding the width of the track.

Once coordinates \(P(x_1, 0)\) and \(Q(0, y_1)\) are established, the equation of line \(PQ\) can be written. The area of \(\trianglePOQ\) is then computed via integration by integrating the equation of line \(PQ\) from \(x = 0\) to \(x = x_1\).

Step 1: Determine the coordinates of points P and Q.

From our previous analysis of the inner ellipse, the inner intercepts are: \[ Inner x-intercept = 4, \quad Inner y-intercept = 3 \]
Given that the track has a uniform width of \(3 m\) surrounding the inner ground: \[ Outer x-intercept for point P = Inner x-intercept + width = 4 + 3 = 7 \]
Since point \(P\) lies on the positive \(x\)-axis, its \(y\)-coordinate is strictly \(0\). Thus: \[ P = (7, 0) \]
Similarly, for point \(Q\) lying on the positive \(y\)-axis: \[ Outer y-intercept for point Q = Inner y-intercept + width = 3 + 3 = 6 \]
Since point \(Q\) lies on the positive \(y\)-axis, its \(x\)-coordinate is strictly \(0\). Thus: \[ Q = (0, 6) \]

Step 2: Find the equation of the line segment PQ.

The line passes through \(Q(0,6)\) and \(P(7,0)\). We can write the equation of the line using the intercept form \(\frac{x}{x_{int}} + \frac{y}{y_{int}} = 1\): \[ \frac{x}{7} + \frac{y}{6} = 1 \]
Now, let us rearrange this equation to isolate \(y\) as a function of \(x\): \[ \frac{y}{6} = 1 - \frac{x}{7} \] \[ \frac{y}{6} = \frac{7 - x}{7} \]
Multiplying both sides by \(6\): \[ y = \frac{6}{7}(7 - x) \quad \Rightarrow \quad y = 6 - \frac{6}{7}x \]

Step 3: Integrate to find the area of triangle POQ.

The triangle \(POQ\) is bounded above by the line \(y = \frac{6}{7}(7-x)\), below by the \(x\)-axis (\(y=0\)), on the left by the \(y\)-axis (\(x=0\)), and on the right by the point \(P\) (\(x=7\)).
The area \(A_{\triangle}\) is found by integrating from \(x = 0\) to \(x = 7\): \[ A_{\triangle} = \int_{0}^{7} \frac{6}{7}(7 - x) \, dx \]
Factoring out the constant fraction \(\frac{6}{7}\): \[ A_{\triangle} = \frac{6}{7} \int_{0}^{7} (7 - x) \, dx \]
Performing the integration term by term: \[ A_{\triangle} = \frac{6}{7} \left[ 7x - \frac{x^2}{2} \right]_{0}^{7} \]
Now, substitute the upper limit (\(x = 7\)) and lower limit (\(x = 0\)):

At upper limit \(x = 7\):
\[ Value = 7(7) - \frac{7^2}{2} = 49 - \frac{49}{2} = \frac{98 - 49}{2} = \frac{49}{2} \]
At lower limit \(x = 0\):
\[ Value = 7(0) - \frac{0^2}{2} = 0 \]

Subtracting the results: \[ A_{\triangle} = \frac{6}{7} \times \left( \frac{49}{2} - 0 \right) \] \[ A_{\triangle} = \frac{6 \times 49}{7 \times 2} \]
Simplifying by dividing terms: \[ A_{\triangle} = \left(\frac{6}{2}\right) \times \left(\frac{49}{7}\right) = 3 \times 7 = 21 m^2 \]
Thus, the coordinates are \(P(7,0), Q(0,6)\) and the area of the triangle is \(21 m^2\), matching option (A). Quick Tip: For a right-angled triangle formed by the origin and the axes intercepts, you can quickly double-check your integration answer using the elementary geometry formula: \(Area = \frac{1}{2} \times base \times height\). Here, the base is \(OP = 7\) and height is \(OQ = 6\), giving \(Area = \frac{1}{2} \times 7 \times 6 = 21\). It provides instant verification!

Concept:
A piecewise-defined function has different formulas corresponding to different intervals of its domain. To find the derivative \(f'(x)\) of a piecewise function within an open interval, we identify the specific sub-domain that contains the interval and differentiate that specific expression using standard rules of differentiation.
The main differentiation rules used here include:

Power Rule: \(\frac{d}{dx}(x^n) = n \cdot x^{n-1}\)
Constant Rule: \(\frac{d}{dx}(C) = 0\), where \(C\) is any real constant.
Sum Rule: \(\frac{d}{dx}[g(x) + h(x)] = g'(x) + h'(x)\)


Step 1: Identify the relevant branch of the piecewise function.

We are required to determine the derivative \(f'(x)\) explicitly for the open interval defined by: \[ 0 < x < 3 \]
Let us look at the given piece-wise definition of the function \(f(x)\): \[ f(x) = \begin{cases} x^2 + 4, & when 0 \le x < 3
4x^2 - 4x + 40, & when x \ge 3 \end{cases} \]
The open interval \(0 < x < 3\) lies completely within the domain constraint of the first branch, which is defined for \(0 \le x < 3\). Therefore, for any value of \(x\) strictly between \(0\) and \(3\), the function expression behaves exactly as: \[ f(x) = x^2 + 4 \]

Step 2: Differentiate the selected expression with respect to \(x\).

Now, we find the derivative by applying the operator \(\frac{d}{dx}\) to both sides of the equation: \[ f'(x) = \frac{d}{dx}(x^2 + 4) \]
Using the sum rule of differentiation, we can split this expression into two separate derivative components: \[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(4) \]
Applying the power rule to the first term \(\frac{d}{dx}(x^2)\) where the power \(n = 2\): \[ \frac{d}{dx}(x^2) = 2 \cdot x^{2-1} = 2x^1 = 2x \]
Applying the constant rule to the second term \(\frac{d}{dx}(4)\) since \(4\) is a constant value independent of \(x\): \[ \frac{d}{dx}(4) = 0 \]

Step 3: Combine the derivatives to obtain the final answer.

Substitute these individual results back into the derivative equation: \[ f'(x) = 2x + 0 \] \[ f'(x) = 2x \]
Thus, for the interval \(0 < x < 3\), the derivative of the function is \(2x\), which perfectly matches option (A). Quick Tip: When differentiating a piecewise function on an open interval, completely disregard the other branches! Simply pinpoint which interval matches your constraint, isolate that basic polynomial expression, and apply standard differentiation rules directly. For \(0 < x < 3\), the second branch \(4x^2 - 4x + 40\) plays absolutely no role in determining the local slope.

Concept:
To evaluate the derivative of a piecewise function at a specific point \(x = c\), we must first identify which interval contains that point. Once the proper branch is identified, we find the general derivative expression \(f'(x)\) for that interval and then substitute \(x = c\) into it.
The differentiation rules applied here for polynomials are:

Power Rule: \(\frac{d}{dx}(x^n) = n \cdot x^{n-1}\)
Constant Multiple Rule: \(\frac{d}{dx}[k \cdot g(x)] = k \cdot g'(x)\)
Linear Term Rule: \(\frac{d}{dx}(x) = 1\)


Step 1: Identify the relevant branch for \(x = 4\).

We need to find the value of the derivative at \(x = 4\). Let us inspect the interval definitions of the given function: \[ f(x) = \begin{cases} x^2 + 4, & 0 \le x < 3
4x^2 - 4x + 40, & x \ge 3 \end{cases} \]
Since \(4 \ge 3\), the point \(x = 4\) lies completely within the domain of the second branch. Thus, the governing function in the neighborhood of \(x = 4\) is: \[ f(x) = 4x^2 - 4x + 40 \]

Step 2: Differentiate the function for the interval \(x > 3\).

Now, let us find the general derivative expression \(f'(x)\) by differentiating this polynomial with respect to \(x\): \[ f'(x) = \frac{d}{dx}(4x^2 - 4x + 40) \]
Using the linearity property of differentiation, we differentiate each term individually: \[ f'(x) = \frac{d}{dx}(4x^2) - \frac{d}{dx}(4x) + \frac{d}{dx}(40) \] \[ f'(x) = 4 \cdot \frac{d}{dx}(x^2) - 4 \cdot \frac{d}{dx}(x) + \frac{d}{dx}(40) \]
Applying the power rule (\(\frac{d}{dx}(x^2) = 2x\)), the identity rule (\(\frac{d}{dx}(x) = 1\)), and the constant rule (\(\frac{d}{dx}(40) = 0\)): \[ f'(x) = 4(2x) - 4(1) + 0 \] \[ f'(x) = 8x - 4 \]

Step 3: Substitute \(x = 4\) into the derivative expression.

To find the specific value of \(f'(4)\), we replace \(x\) with \(4\) in our derivative equation: \[ f'(4) = 8(4) - 4 \]
First, perform the multiplication: \[ 8 \times 4 = 32 \]
Next, perform the subtraction: \[ f'(4) = 32 - 4 = 28 \]
Thus, the value of the derivative at \(x = 4\) is \(28\), which matches option (B). Quick Tip: Always double-check which interval your target point belongs to! A common error is mistakenly using the first branch formula. Since \(4 > 3\), only the branch \(4x^2 - 4x + 40\) is valid. Differentiate first, then plug in the value.

Concept:
A function \(f(x)\) is said to be continuous at \(x = c\) if the following three conditions are satisfied:

\(f(c)\) is defined.
The Left-Hand Limit (LHL) and Right-Hand Limit (RHL) exist and are equal:
\[ \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) \]
The common limiting value is equal to the value of the function at \(x=c\):
\[ \lim_{x \to c} f(x) = f(c) \]


Step 1: Determine the value of the function at \(x = 3\).

The function is defined as: \[ f(x) = \begin{cases} x^2 + 4, & 0 \le x < 3
4x^2 - 4x + 40, & x \ge 3 \end{cases} \]
Since \(x=3\) satisfies the condition \(x\geq 3\), the second branch is used to find \(f(3)\): \[ f(3) = 4(3)^2 - 4(3) + 40 \] \[ = 4(9) - 12 + 40 \] \[ = 36 - 12 + 40 \] \[ f(3) = 64 \]

Thus, \(f(3)\) is defined and \[ f(3)=64. \]

Step 2: Find the Left-Hand Limit at \(x=3\).

For values of \(x\) approaching \(3\) from the left, we have \(0\leq x<3\). Therefore, the first branch of the function is applicable: \[ \lim_{x\to 3^-}f(x) = \lim_{x\to 3^-}(x^2+4) \]
Since \(x^2+4\) is a polynomial, its limit can be obtained by direct substitution: \[ \lim_{x\to 3^-}f(x) = 3^2+4 \] \[ =9+4 \] \[ LHL=13. \]

Step 3: Find the Right-Hand Limit at \(x=3\).

For values of \(x\) approaching \(3\) from the right, we have \(x>3\). Therefore, the second branch of the function is applicable: \[ \lim_{x\to 3^+}f(x) = \lim_{x\to 3^+}(4x^2-4x+40) \]
Again, since this is a polynomial, direct substitution gives: \[ \lim_{x\to 3^+}f(x) = 4(3)^2-4(3)+40 \] \[ =4(9)-12+40 \] \[ =36-12+40 \] \[ RHL=64. \]

Step 4: Compare the LHL and RHL.

We have obtained: \[ LHL=13 \]
and \[ RHL=64. \]
Therefore, \[ LHL\neqRHL \]
or \[ 13\neq64. \]

Hence, the two-sided limit \[ \lim_{x\to3}f(x) \]
does not exist.

Although \(f(3)=64\) is defined, the existence of the two-sided limit is necessary for continuity. Since the Left-Hand Limit and Right-Hand Limit are unequal, the function cannot be continuous at \(x=3\).

Therefore, \[ \boxed{\(f(x)\) is discontinuous at \(x=3\).} \]

Thus, the correct option is \(\boxed{(A)}\). Quick Tip: For a piecewise function at the point where the definition changes, first find \(f(c)\), then calculate the LHL and RHL separately. The function is continuous at \(x=c\) only when \(f(c)=LHL=RHL\). If the LHL and RHL are unequal, the function is discontinuous at that point.

Concept:
A fundamental theorem in calculus establishes the relationship between continuity and differentiability: \[ Differentiability \implies Continuity \]
Taking the contrapositive of this universal statement, we obtain: \[ Not Continuous \implies Not Differentiable \]
Therefore, if a function is proven to be discontinuous at a point \(x = c\), it is automatically non-differentiable at that point. There is no requirement to calculate the Left-Hand Derivative (LHD) or Right-Hand Derivative (RHD) because a smooth tangent cannot exist over a broken or disconnected graph gap.

Step 1: Reference the continuity status from the previous part.

In part 37.(iii)(a), we rigorously evaluated the limits of the function \(f(x)\) as it approaches the boundary transition point \(x = 3\):

Left-Hand Limit: \(\lim_{x \to 3^-} f(x) = 3^2 + 4 = 13\)
Right-Hand Limit: \(\lim_{x \to 3^+} f(x) = 4(3)^2 - 4(3) + 40 = 64\)

Since the Left-Hand Limit does not equal the Right-Hand Limit (\(13 \neq 64\)), the limit does not exist, and the function is fundamentally discontinuous at \(x = 3\).

Step 2: Apply the differentiability theorem.

Since a function cannot have a derivative at any point where it fails to be continuous, the discontinuity at \(x = 3\) immediately implies that \(f(x)\) is not differentiable at \(x = 3\).

Even if we were to look at the slopes of the individual pieces independently:

Derivative from the left branch: \(\frac{d}{dx}(x^2 + 4) = 2x \xrightarrow{x \to 3} 2(3) = 6\)
Derivative from the right branch: \(\frac{d}{dx}(4x^2 - 4x + 40) = 8x - 4 \xrightarrow{x \to 3} 8(3) - 4 = 20\)

Since \(6 \neq 20\), the slopes do not match either. However, the breakdown of continuity is the primary and most definitive reason for non-differentiability. This perfectly matches option (B). Quick Tip: Continuity is a necessary prerequisite for differentiability! Always check continuity first. If a graph is broken or has a jump gap at a point, it can never be differentiated at that point, saving you the time of calculating derivatives entirely!

Concept:
The classical definition of probability states that the probability of an event \(E\) occurring is given by the ratio of the number of outcomes favorable to \(E\) to the total number of possible outcomes in the sample space, provided all outcomes are equally likely. Mathematically, this is expressed as: \[ P(E) = \frac{n(E)}{n(S)} = \frac{Number of favorable outcomes}{Total number of possible outcomes} \]

Step 1: Identify the total size of the sample space.

From the problem statement, we are looking at a specific colony where a survey of smokers was conducted. The total number of people identified as smokers in this colony is given as: \[ Total number of smokers (Total outcomes, n(S)) = 50 \]
The random selection is made strictly from this group of \(50\) smokers.

Step 2: Determine the number of favorable outcomes (number of female smokers).

The problem states that out of these \(50\) smokers, the number of male smokers is explicitly \(30\): \[ Number of male smokers = 30 \]
Since the group of smokers consists of only males and females, we can calculate the number of female smokers by subtracting the number of male smokers from the total number of smokers: \[ Number of female smokers (Favorable outcomes, n(E)) = Total smokers - Male smokers \] \[ n(E) = 50 - 30 = 20 \]

Step 3: Calculate the probability.

Let \(E\) be the event that the randomly selected smoker is a female. Applying the classical definition of probability: \[ P(E) = \frac{Number of female smokers}{Total number of smokers} \] \[ P(E) = \frac{20}{50} \]
To simplify the fraction, we divide both the numerator and the denominator by their greatest common divisor, which is \(10\): \[ P(E) = \frac{20 \div 10}{50 \div 10} = \frac{2}{5} \]
Thus, the probability that the selected person is a female is \(\frac{2}{5}\), which matches option (B). Quick Tip: Always read carefully what group the selection is being made from! Here, the person is picked out of the \(50\) smokers in the colony, so the total sample space size is \(50\). The global statistical data (\(170\) in \(1000\) or \(120\) in \(1000\)) represents conditional probabilities for developing lung complications and is completely irrelevant for this specific introductory sub-question.

Concept:
The complement rule of probability states that the probability of an event not occurring is equal to \(1\) minus the probability of that event occurring. If \(P(E)\) represents the probability that an event occurs, then the probability of the complementary event \(P(E')\) or \(P(not E)\) is given by: \[ P(E') = 1 - P(E) \]
In this question, we are dealing with conditional probability. We are given that a male person has already been chosen, so our sample space is restricted exclusively to the population of smoking males.

Step 1: Identify the given probability for a male developing lung complications.

From the problem statement, the clinical study data provides the rate of lung complications among the smoking male population: \[ 170 out of 1000 males who smoke develop lung complications. \]
Let \(L\) be the event that a person develops lung complications, and let \(M\) be the event that the person is male. The conditional probability that a selected male person suffers from lung problems is written as \(P(L|M)\): \[ P(L|M) = \frac{170}{1000} \]
Dividing the numerator and the denominator by \(10\), we simplify the fraction to: \[ P(L|M) = \frac{17}{100} \]

Step 2: Apply the complement rule to find the probability of not suffering from lung complications.

We want to find the probability that a selected male person does not suffer from lung problems, which is denoted as \(P(L'|M)\). According to the complement rule: \[ P(L'|M) = 1 - P(L|M) \]
Substituting the simplified value of \(P(L|M) = \frac{17{100}\) into the equation: \[ P(L'|M) = 1 - \frac{17}{100} \]
To perform the subtraction, convert the whole number \(1\) into a fraction with a common denominator of \(100\): \[ P(L'|M) = \frac{100}{100} - \frac{17}{100} \]
Subtract the numerators over the common denominator: \[ P(L'|M) = \frac{100 - 17}{100} = \frac{83}{100} \]
Thus, the probability that a chosen male person will not suffer from lung problems is \(\frac{83}{100}\), which matches option (B). Quick Tip: When a question states "If a male person is selected...", it means the choice has already been restricted to males. You don't need to multiply by the probability of picking a male from the colony. Simply compute the complement directly from the given baseline male group rate: \(1 - \frac{170}{1000} = \frac{830}{1000} = \frac{83}{100}\).

Concept:
This problem requires the application of Bayes' Theorem . Bayes' Theorem allows us to find the conditional probability of an event (in this case, being a female, \(F\)) given that another event has already occurred (the person has lung complications, \(L\)).
The formula for Bayes' Theorem is: \[ P(F|L) = \frac{P(F) \cdot P(L|F)}{P(M) \cdot P(L|M) + P(F) \cdot P(L|F)} \]
where:

\(P(M)\) and \(P(F)\) are the prior probabilities of selecting a male and a female smoker, respectively.
\(P(L|M)\) and \(P(L|F)\) are the conditional probabilities of developing lung complications given that the person is a male or a female.
The denominator represents the total probability of a person having lung complications, \(P(L)\).


Step 1: Identify all prior and conditional probabilities from the given data.

From the previous sub-parts, we have the following data for the colony of \(50\) smokers (\(30\) males, \(20\) females):

Probability of selecting a male smoker: \(P(M) = \frac{30}{50} = \frac{3}{5}\)
Probability of selecting a female smoker: \(P(F) = \frac{20}{50} = \frac{2}{5}\)

From the general study, the conditional probabilities of suffering from lung complications are:

For males: \(P(L|M) = \frac{170}{1000} = \frac{17}{100}\)
For females: \(P(L|F) = \frac{120}{1000} = \frac{12}{100}\)


Step 2: Compute the total probability of a person having lung complications, \(P(L)\).

The denominator of Bayes' Theorem is the total probability \(P(L)\), which can happen via two mutually exclusive paths (either the person is a male with lung complications or a female with lung complications): \[ P(L) = P(M) \cdot P(L|M) + P(F) \cdot P(L|F) \]
Substitute the values into the equation: \[ P(L) = \left( \frac{3}{5} \times \frac{17}{100} \right) + \left( \frac{2}{5} \times \frac{12}{100} \right) \]
Calculate each product carefully: \[ First term (Male path) = \frac{3 \times 17}{5 \times 100} = \frac{51}{500} \] \[ Second term (Female path) = \frac{2 \times 12}{5 \times 100} = \frac{24}{500} \]
Add the fractions together since they share a common denominator: \[ P(L) = \frac{51 + 24}{500} = \frac{75}{500} \]

Step 3: Apply Bayes' Theorem to find \(P(F|L)\).

Now, substitute the female path product (numerator) and total probability (denominator) into Bayes' formula: \[ P(F|L) = \frac{P(F) \cdot P(L|F)}{P(L)} = \frac{\frac{24}{500}}{\frac{75}{500}} \]
We can cancel out the common denominator of \(500\) from both the numerator and the denominator: \[ P(F|L) = \frac{24}{75} \]
To simplify this fraction further, find the greatest common divisor of \(24\) and \(75\), which is \(3\): \[ P(F|L) = \frac{24 \div 3}{75 \div 3} = \frac{8}{25} \]
Hence, the conditional probability that the person is a female given they have lung problems is \(\frac{8}{25}\), matching option (A). Quick Tip: Bayes' Theorem problems can be cross-verified cleanly by setting up a simple tree diagram or imaginary population numbers! Imagine a scaled group where out of \(500\) total simulated outcomes, \(300\) are male paths and \(200\) are female paths. Then \(300 \times 0.17 = 51\) males get sick, and \(200 \times 0.12 = 24\) females get sick. Total sick people = \(51 + 24 = 75\). Out of those, the female share is exactly \(\frac{24}{75} = \frac{8}{25}\).

Concept:
This problem is another variation of Bayes' Theorem , where we calculate the inverse probability that a randomly chosen person is male (\(M\)) given that the observed condition is that they do not have lung complications (\(L'\)).
The formula for this conditional probability is: \[ P(M|L') = \frac{P(M) \cdot P(L'|M){P(M) \cdot P(L'|M) + P(F) \cdot P(L'|F)} \]

Step 1: Establish the complement conditional probabilities.

From the previous parts, we know the probabilities of being a male or female smoker in the colony: \[ P(M) = \frac{3}{5}, \quad P(F) = \frac{2}{5} \]
We need the conditional probabilities of not having lung complications. Using the complement rule:

For males: \(P(L'|M) = 1 - P(L|M) = 1 - \frac{170{1000} = \frac{830}{1000} = \frac{83}{100}\)
For females: \(P(L'|F) = 1 - P(L|F) = 1 - \frac{120}{1000} = \frac{880}{1000} = \frac{88}{100}\)


Step 2: Compute the total probability of a person not having lung complications, \(P(L')\).

We can compute the denominator \(P(L')\) either by summing the healthy paths or by taking the complement of the total sick probability \(P(L)\) found in part (a). Let's use both to show a highly descriptive breakdown.
Using the total probability formula: \[ P(L') = P(M) \cdot P(L'|M) + P(F) \cdot P(L'|F) \]
Substitute the fractions into the expression: \[ P(L') = \left( \frac{3}{5} \times \frac{83}{100} \right) + \left( \frac{2}{5} \times \frac{88}{100} \right) \]
Calculate each product explicitly: \[ Male healthy path = \frac{3 \times 83}{500} = \frac{249}{500} \] \[ Female healthy path = \frac{2 \times 88}{500} = \frac{176}{500} \]
Summing these fractions: \[ P(L') = \frac{249 + 176}{500} = \frac{425}{500} \]
*(Alternative check: From part (a), \(P(L) = \frac{75}{500}\). Therefore, \(P(L') = 1 - \frac{75}{500} = \frac{425}{500}\). Both paths match perfectly!)*

Step 3: Apply Bayes' Theorem to solve for \(P(M|L')\).

Substitute the male healthy path (numerator) and total healthy probability (denominator) into the conditional formula: \[ P(M|L') = \frac{\frac{249}{500}}{\frac{425}{500}} \]
Canceling out the common denominator of \(500\) yields: \[ P(M|L') = \frac{249}{425} \]
Let us verify if this fraction can be further simplified. The prime factors of \(425\) are \(5\) and \(17\) (\(425 = 25 \times 17\)). Testing divisibility for \(249\): it is not divisible by \(5\), nor is it divisible by \(17\) (\(249 \div 17 \approx 14.64\)). Thus, the fraction is in its simplest irreducible form.

This matches option (A) perfectly. Quick Tip: When applying Bayes' Theorem for complement events, you can reuse your previous calculation steps efficiently. Since you already know the total sick count in our hypothetical population of \(500\) was \(75\), the total healthy count must be \(500 - 75 = 425\). Out of these, the healthy males account for \(300 - 51 = 249\) people. Thus, the ratio is instantly written down as \(\frac{249}{425}\)!