JEE Main 2023 Mathematics Question Paper Jan 29 Shift 2- Download PDF

We verify the equivalence of \( B \Rightarrow ((\sim A) \lor B) \) by using truth tables.


Step 1: Analyze the given expression

The statement is \( B \Rightarrow ((\sim A) \lor B) \). By the definition of implication: \[ P \Rightarrow Q \equiv (\sim P) \lor Q, \]
the statement can be rewritten as: \[ (\sim B) \lor ((\sim A) \lor B). \]


Step 2: Construct the truth table for \( B \Rightarrow ((\sim A) \lor B) \)
\[ \begin{array}{c|c|c|c|c} A & B & \sim A & (\sim A) \lor B & B \Rightarrow ((\sim A) \lor B)
\hline T & T & F & T & T
T & F & F & F & T
F & T & T & T & T
F & F & T & T & T
\end{array} \]

Explanation:

Column \( \sim A \): Negation of \( A \).
Column \( (\sim A) \lor B \): Logical OR of \( \sim A \) and \( B \).
Column \( B \Rightarrow ((\sim A) \lor B) \): True if \( B \) implies \( (\sim A) \lor B \).



Step 3: Verify equivalence with the options


% Option
(A) Option (1): \( B \Rightarrow (A \Rightarrow B) \)

Simplify \( A \Rightarrow B \equiv (\sim A) \lor B \). Thus:
\[ B \Rightarrow ((\sim A) \lor B). \]
The truth table matches exactly with \( B \Rightarrow ((\sim A) \lor B) \).

(B)Option (3): \( A \Rightarrow ((\sim A) \Rightarrow B) \)

Simplify \( (\sim A) \Rightarrow B \equiv A \lor B \). Thus:
\[ A \Rightarrow ((\sim A) \lor B) \equiv (\sim A) \lor ((\sim A) \lor B), \]
which matches the truth table of \( B \Rightarrow ((\sim A) \lor B) \).

(C)Option (4): \( B \Rightarrow ((\sim A) \Rightarrow B) \)

Simplify \( (\sim A) \Rightarrow B \equiv A \lor B \). Thus:
\[ B \Rightarrow (A \lor B), \]
which also matches the truth table of \( B \Rightarrow ((\sim A) \lor B) \). Quick Tip: When checking logical equivalences, construct truth tables and compare them to confirm equivalence.

The parametric equations of the lines are:
\begin{align*
\vec{r_1 &= \hat{i - 8\hat{j + 4\hat{k + \lambda(2\hat{i - 7\hat{j + 5\hat{k),

\vec{r_2 &= \hat{i + 2\hat{j + 6\hat{k + \mu(2\hat{i + \hat{j - 3\hat{k).
\end{align*


Step 1: Define the points and direction vectors

\( \vec{a_1} = \hat{i} - 8\hat{j} + 4\hat{k} \) (point on line 1).
\( \vec{a_2} = \hat{i} + 2\hat{j} + 6\hat{k} \) (point on line 2).
\( \vec{b_1} = 2\hat{i} - 7\hat{j} + 5\hat{k} \) (direction vector of line 1).
\( \vec{b_2} = 2\hat{i} + \hat{j} - 3\hat{k} \) (direction vector of line 2).



Step 2: Find the cross product \( \vec{b_1} \times \vec{b_2} \) \[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -7 & 5
2 & 1 & -3 \end{vmatrix} = (16\hat{i} + 16\hat{j} + 16\hat{k}) = 16(\hat{i} + \hat{j} + \hat{k}). \]


Step 3: Find \( |\vec{b_1} \times \vec{b_2}| \) \[ |\vec{b_1} \times \vec{b_2}| = 16\sqrt{3}. \]


Step 4: Compute \( \vec{a_2} - \vec{a_1} \) \[ \vec{a_2} - \vec{a_1} = 10\hat{j} + 2\hat{k}. \]


Step 5: Find the scalar projection of \( \vec{a_2} - \vec{a_1} \) on \( \vec{b_1} \times \vec{b_2} \) \[ Shortest distance = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}. \]

Substitute: \[ (\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 16\sqrt{104}. \]
\[ Shortest distance = \frac{\sqrt{104}}{\sqrt{3}} = 4\sqrt{3}. \] Quick Tip: The shortest distance between skew lines is calculated using the cross product of their direction vectors.

Step 1: Simplify \( \vec{r} \times \vec{b} + \vec{b} \times \vec{c} = 0 \)
\[ \vec{r} \times \vec{b} = -\vec{b} \times \vec{c}. \]

This implies \( \vec{r} \) can be written as: \[ \vec{r} = \vec{c} + \lambda \vec{b}, \quad for some scalar \lambda. \]


Step 2: Use \( \vec{r} \cdot \vec{a} = 0 \)

Substitute \( \vec{r} = \vec{c} + \lambda \vec{b} \) into \( \vec{r} \cdot \vec{a} = 0 \): \[ (\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0. \]

Simplify: \[ \vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0. \]


Step 3: Compute dot products
\[ \vec{c} \cdot \vec{a} = (7)(1) + (-3)(0) + (4)(2) = 15, \] \[ \vec{b} \cdot \vec{a} = (1)(1) + (1)(0) + (1)(2) = 3. \]

Substitute: \[ 15 + 3\lambda = 0 \implies \lambda = -5. \]


Step 4: Find \( \vec{r} \)
\[ \vec{r} = \vec{c} + \lambda \vec{b} = (7\hat{i} - 3\hat{j} + 4\hat{k}) - 5(\hat{i} + \hat{j} + \hat{k}), \] \[ \vec{r} = 2\hat{i} - 8\hat{j} - \hat{k}. \]


Step 5: Compute \( \vec{r} \cdot \vec{c} \)
\[ \vec{r} \cdot \vec{c} = (2)(7) + (-8)(-3) + (-1)(4), \] \[ \vec{r} \cdot \vec{c} = 14 + 24 - 4 = 34. \] Quick Tip: To solve vector problems with constraints, express vectors in parametric form and solve for scalars systematically.

(A)Let \( P(W_1) = \lambda \). Then \( P(W_2) = \frac{\lambda}{2} \), \( P(W_3) = \frac{\lambda}{4} \), and in general:
\[ P(W_n) = \frac{\lambda}{2^{n-1}}. \]

(B)The total probability of the sample space \( S \) is 1:
\[ \sum_{n=1}^\infty P(W_n) = \lambda \left(1 + \frac{1}{2} + \frac{1}{4} + \cdots \right). \]
This is a geometric series with sum:
\[ \lambda \sum_{n=0}^\infty \left(\frac{1}{2}\right)^n = \lambda \cdot \frac{1}{1 - \frac{1}{2}} = \lambda \cdot 2. \]
Setting \( \lambda \cdot 2 = 1 \), we get \( \lambda = \frac{1}{2} \).

(C)For \( A = \{2k + 3l; k, l \in \mathbb{N}\} \), \( A \) includes numbers such as \( 5, 7, 8, 9, 10, \ldots \). Thus, \( B = \{W_5, W_7, W_8, W_9, \ldots\} \).

(D)The probability \( P(B) \) is:
\[ P(B) = \sum_{n \in A} P(W_n). \]
Substituting \( P(W_n) = \frac{1}{2^n} \), we calculate:
\[ P(B) = 1 - \left(P(W_1) + P(W_2) + P(W_3) + P(W_4) + P(W_6)\right). \]

(E)Compute individual probabilities:
\[ P(W_1) = \frac{1}{2}, \, P(W_2) = \frac{1}{4}, \, P(W_3) = \frac{1}{8}, \, P(W_4) = \frac{1}{16}, \, P(W_6) = \frac{1}{64}. \]
Summing these:
\[ P(W_1) + P(W_2) + P(W_3) + P(W_4) + P(W_6) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} = \frac{36}{64}. \]

(F) Thus:
\[ P(B) = 1 - \frac{36}{64} = \frac{36}{64}. \] Quick Tip: For probabilities involving subsets, calculate the total probability of excluded events and subtract it from 1.

(A)The given integral is:
\[ \int_{1}^2 \frac{t^4 + 1}{t^6 + 1} \, dt. \]

(B)Factorize \( t^6 + 1 \):
\[ t^6 + 1 = (t^2 + 1)(t^4 - t^2 + 1). \]
Rewrite the integrand:
\[ \frac{t^4 + 1}{t^6 + 1} = \frac{t^4 + 1}{(t^2 + 1)(t^4 - t^2 + 1)}. \]

(C)Use partial fractions:
\[ \frac{t^4 + 1}{(t^2 + 1)(t^4 - t^2 + 1)} = \frac{A}{t^2 + 1} + \frac{Bt + C}{t^4 - t^2 + 1}. \]
Multiply through and equate terms to solve for \( A, B, C \). Solving gives:
\[ A = 1, \, B = 0, \, C = \frac{1}{3}. \]

(D)Substituting, the integral becomes:
\[ \int_{1}^2 \frac{1}{t^2 + 1} \, dt + \frac{1}{3} \int_{1}^2 \frac{t}{t^4 - t^2 + 1} \, dt. \]

(E)Solve the first term:
\[ \int_{1}^2 \frac{1}{t^2 + 1} \, dt = \tan^{-1}(2) - \tan^{-1}(1). \]
Since \( \tan^{-1}(1) = \frac{\pi}{4} \), this simplifies to:
\[ \tan^{-1}(2) - \frac{\pi}{4}. \]

(F) For the second term, perform substitution \( u = t^2 - 1 \), \( du = 2t \, dt \):
\[ \frac{1}{3} \int_{1}^2 \frac{t}{t^4 - t^2 + 1} \, dt = \frac{1}{3} \int \frac{1}{u^2 + 1} \, du = \frac{1}{3} \tan^{-1}(u). \]
Back-substitute and evaluate:
\[ \frac{1}{3} \left[\tan^{-1}(8) - \tan^{-1}(0)\right] = \frac{1}{3} \tan^{-1}(8). \]


(G) Combine the results:
\[ \tan^{-1}(2) - \frac{\pi}{4} + \frac{1}{3} \tan^{-1}(8). \]
Simplify:
\[ \tan^{-1}(2) + \frac{1}{3} \tan^{-1}(8) - \frac{\pi}{3}. \] Quick Tip: Factorize the denominator and use trigonometric substitutions for integrals involving rational functions.

In the expansion of \[ (1 + x)^{99} = C_0 + C_1 x + C_2 x^2 + \dots + C_{99} x^{99} \]
we define \( K \) as: \[ K = C_1 + C_3 + \dots + C_{99} = 2^{98} \]

To find the middle term in the expansion of \[ \left( 2 + \frac{1}{\sqrt{2}} \right)^{200} \]
we consider the term:
\[ T_{\frac{200}{2} + 1} = C_{100}^{200} (2)^{100} \left( \frac{1}{\sqrt{2}} \right)^{100} \]
\[ = C_{100}^{200} \cdot 2^{50} \]

Thus, we get:
\[ \frac{200}{200} \cdot \frac{C_{99} \times 2^{98}}{C_{100}^{200} \times 2^{50}} = \frac{100}{101} \times 2^{48} \]

So,
\[ \frac{25}{101} \times 2^{50} = \frac{m}{n} 2^{\ell} \]

Since \( m \) and \( n \) are odd, we conclude that:
\[ (\ell, n) = (50, 101) \quad Ans. \]

- The problem involves the binomial expansion of \((1 + x)^{99}\) and the summation of alternating binomial coefficients.

- The middle term in the expansion of \(\left( 2 + \frac{1}{\sqrt{2}} \right)^{200}\) is found using the binomial theorem.

- Simplifications using properties of binomial coefficients and powers of 2 lead to the final result.

- The values of \( m \) and \( n \) are determined to be odd, helping us find the required answer. Quick Tip: For binomial expansions, use symmetry and simplify middle terms systematically. For large factorials, simplify ratios directly using combinatorial identities.

(A)From \( f'(x) = g''(x) + 6x \), integrate to find \( f(x) \):
\[ f'(x) = g'(x) + 3x^2 + C_1. \]

(B)Using \( f'(1) = 9 \):
\[ 9 = g'(1) + 3(1) + C_1 \implies C_1 = 2. \]

(C)Integrate again to find \( f(x) \):
\[ f(x) = g(x) + x^3 + 2x + C_2. \]

(D)Using \( f(2) = 3g(2) = 12 \):
\[ 12 = g(2) + 8 + 4 + C_2 \implies C_2 = -4. \]

(E)Thus:
\[ f(x) = g(x) + x^3 + 2x - 4. \]

(F) Compute \( h(x) = f(x) - g(x) = x^3 + 2x - 4 \). For \( -1 < x < 2 \), the range of \( h(x) \) is:
\[ -7 < h(x) < 4 \implies |h(x)| < 8. \]
However, \( |h(x)| \geq 7 \), so option (2) is not true. Quick Tip: For inequalities involving differentiable functions, analyze monotonicity using derivatives to find range bounds.

(A)The given matrix is invertible if its determinant is non-zero:
\[ \det\left(\begin{bmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t)
e^t & e^{-t}(2\sin t + \cos t) & e^{-t}(\sin t - 2\cos t)
e^t & e^{-t}\cos t & e^{-t}\sin t \end{bmatrix}\right) \neq 0. \]

(B)Perform row operations to simplify the determinant. Subtract the first row from the second and third rows:
\[ R_2 \to R_2 - R_1, \quad R_3 \to R_3 - R_1. \]
After simplification, the determinant becomes:
\[ \det\begin{bmatrix} e^t & e^{-t}(\sin t - 2\cos t) & e^{-t}(-2\sin t - \cos t)
0 & e^{-t} & 0
0 & 0 & e^{-t} \end{bmatrix}. \]

(C)Expanding along the first row:
\[ \det = e^t \cdot \det\begin{bmatrix} e^{-t} & 0
0 & e^{-t} \end{bmatrix}. \]
Compute the determinant of the \( 2 \times 2 \) submatrix:
\[ \det = e^t \cdot (e^{-t} \cdot e^{-t}) = e^t \cdot e^{-2t} = e^{-t}. \]

(D)For invertibility, \( e^{-t} \neq 0 \). Since \( e^{-t} \neq 0 \) for all \( t \in \mathbb{R} \), the matrix is invertible for all \( t \in \mathbb{R} \). Quick Tip: For matrix invertibility, simplify determinants using row/column operations and check when the determinant is non-zero.

(A)The region \( A \) is bounded by \( y = |\cos x - \sin x| \) and \( y = \sin x \) over the interval \( 0 \leq x \leq \frac{\pi}{2} \).

(B) Split the region into two parts based on the behavior of \( |\cos x - \sin x| \):
\[ |\cos x - \sin x| = \begin{cases} \cos x - \sin x, & if \cos x \geq \sin x,
\sin x - \cos x, & if \cos x < \sin x. \end{cases} \]

(C)Evaluate the area in two parts:
\[ Area = \int_{0}^{\frac{\pi}{4}} (\sin x - (\cos x - \sin x)) dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - (\sin x - \cos x)) dx. \]

(D)Simplify the integrands:
\[ Area = \int_{0}^{\frac{\pi}{4}} (2\sin x - \cos x) dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\cos x) dx. \]

(E)Solve each integral:
\begin{align*
\int (2\sin x - \cos x) dx &= -2\cos x - \sin x,

\int \cos x dx &= \sin x.
\end{align*

(F) Substitute limits:
\begin{align*
First integral: & \left[-2\cos x - \sin x\right]_0^{\frac{\pi{4 = (-2\cos\frac{\pi{4 - \sin\frac{\pi{4) - (-2\cos 0 - \sin 0),

& = (-2\frac{\sqrt{2{2 - \frac{\sqrt{2{2) - (-2 - 0) = -\sqrt{2 - \sqrt{2 + 2,

& = 2 - 2\sqrt{2.
\end{align*
\begin{align*
\text{Second integral: & \left[\sin x\right]_{\frac{\pi{4^{\frac{\pi{2 = \sin\frac{\pi{2 - \sin\frac{\pi{4,

& = 1 - \frac{\sqrt{2{2.
\end{align*

Combine results:
\[ \text{Area = (2 - 2\sqrt{2}) + \left(1 - \frac{\sqrt{2}}{2}\right) = \sqrt{5} - 2\sqrt{2} + 1. \] Quick Tip: Break regions with absolute values into separate intervals and evaluate integrals piecewise.

(A)Start with the given equation:
\[ \cos^2(2x) - 2\sin x - 2\cos(2x) = \lambda. \]

(B)Use the trigonometric identity \( \cos^2(2x) = 1 - \sin^2(2x) \):
\[ 1 - \sin^2(2x) - 2\sin x - 2\cos(2x) = \lambda. \]

(C)Expand \( \cos(2x) = 1 - 2\sin^2(x) \):
\[ 1 - \sin^2(2x) - 2\sin x - 2(1 - 2\sin^2(x)) = \lambda. \]

(D)Simplify the equation:
\[ 1 - \sin^2(2x) - 2\sin x - 2 + 4\sin^2(x) = \lambda. \]
\[ -1 - \sin^2(2x) - 2\sin x + 4\sin^2(x) = \lambda. \]

(E)Replace \( \sin^2(2x) = 4\sin^2(x)\cos^2(x) = 4\sin^2(x)(1 - \sin^2(x)) \):
\[ -1 - 4\sin^2(x)(1 - \sin^2(x)) - 2\sin x + 4\sin^2(x) = \lambda. \]

(F) Simplify:
\[ -1 - 4\sin^2(x) + 4\sin^4(x) - 2\sin x + 4\sin^2(x) = \lambda. \]
\[ -1 + 4\sin^4(x) - 2\sin x = \lambda. \]


(G) Analyze the range of the function \( f(x) = 4\sin^4(x) - 2\sin x - 1 \). The critical points occur when \( f'(x) = 0 \):
\[ f'(x) = 16\sin^3(x)\cos(x) - 2\cos(x). \]
Factorize:
\[ f'(x) = \cos(x)(16\sin^3(x) - 2). \]

Solve \( f'(x) = 0 \):
\[ \cos(x) = 0 \quad or \quad 16\sin^3(x) - 2 = 0 \implies \sin^3(x) = \frac{1}{8}. \]

Evaluate \( f(x) \) at critical points and endpoints \( \sin(x) = -1 \) to \( \sin(x) = 1 \):
\[ f(x) \in [-1, -\frac{1}{2}]. \]

Thus, the range of \( \lambda \) is:
\[ [-1, -\frac{1}{2}]. \] Quick Tip: To find ranges of trigonometric functions, rewrite using identities and analyze using calculus.

(A)Arrange the letters of OUGHT in alphabetical order: G, H, O, T, U.

(B) Count words starting with each letter before "T":
\[ G \to 4!, \quad H \to 4!, \quad O \to 4!. \]
(C) Count words starting with "T" but second letter as:
\[ TG \to 3!, \quad TH \to 3!, \quad TOG \to 2!, \quad TOH \to 2!. \]
(D) Add the counts:
\[ 4! + 4! + 4! + 3! + 3! + 2! + 2! + 1! = 89. \] Quick Tip: To find the serial number of a word in dictionary order, systematically count permutations based on preceding letters.

(A)The coordinates of \( C \) using the section formula:
\[ C = \left(\frac{a + 4}{3}, \frac{2b - 2}{3}, \frac{2 - 2}{3}\right). \]

(B)Substituting \( C \) into the plane equation \( 2x - y + z = 4 \):
\[ 2\left(\frac{a + 4}{3}\right) - \left(\frac{2b - 2}{3}\right) + \left(\frac{2}{3}\right) = 4. \]
Simplify:
\[ \frac{2a + 8 + 2b - 2 + 2}{3} = 4 \implies 2a + 2b = 4 \implies a + b = 2. \quad (1) \]

(C)The distance of \( C \) from the origin:
\[ \left(\frac{a + 4}{3}\right)^2 + \left(\frac{2b - 2}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = 5. \]

Solve using \( a + b = 2 \) and simplify:
\[ (b + 6)^2 + (2b - 2)^2 = 41 \implies 5b^2 + 4b - 1 = 0. \]

(D)Roots are \( b = -1 \) or \( b = \frac{1}{5} \). Using \( ab < 0 \), \( (a, b) = (1, -1) \).

(E)Coordinates of \( C \):
\[ C = \left(\frac{5}{3}, \frac{-4}{3}, \frac{2}{3}\right). \]
Coordinates of \( P \):
\[ P = (2, -1, -3). \]

(F) Compute \( CP^2 \):
\[ CP^2 = \left(\frac{5}{3} - 2\right)^2 + \left(\frac{-4}{3} + 1\right)^2 + \left(\frac{2}{3} + 3\right)^2. \]
Simplify:
\[ CP^2 = \frac{17}{3}. \] Quick Tip: To solve geometry problems involving planes and line segments, use the section formula and solve constraints systematically.

(A)Compute \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
4 & 3 & 0
3 & -4 & 5 \end{vmatrix} = 15\hat{i} - 20\hat{j} - 25\hat{k}. \]

(B)Let \( \vec{c} = x\hat{i} + y\hat{j} + z\hat{k} \). Using the condition \( (\vec{a} \times \vec{b}) \cdot \vec{c} + 25 = 0 \):
\[ 15x - 20y - 25z + 25 = 0 \implies 3x - 4y - 5z = -5. \quad (1) \]

(C)Using \( \vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4 \):
\[ x + y + z = 4. \quad (2) \]

(D)Using the projection condition \( \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1 \):
\[ \frac{4x + 3y}{5} = 1 \implies 4x + 3y = 5. \quad (3) \]

(E)Solve equations (1), (2), and (3) to find \( \vec{c} = 2\hat{i} - \hat{j} + 3\hat{k} \).

(F) Compute the projection of \( \vec{c} \) on \( \vec{b} \):
\[ Projection = \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|}. \]
Simplify:
\[ Projection = \frac{25}{\sqrt{50}} = \frac{5}{\sqrt{2}}. \] Quick Tip: To solve vector problems, combine dot and cross product conditions systematically and solve equations step-by-step.

(A)Let the parametric equations of the first line be:
\[ x = 1 + \lambda, \quad y = 2 + 2\lambda, \quad z = -3 + \lambda. \]
For the second line:
\[ x = a + 2\mu, \quad y = -2 + 3\mu, \quad z = -3 + \mu. \]

(B)For intersection, equate coordinates:
\[ 1 + \lambda = a + 2\mu, \quad 2 + 2\lambda = -2 + 3\mu, \quad -3 + \lambda = -3 + \mu. \]

(C)From the third equation:
\[ \lambda = \mu. \]

(D)Substitute \( \lambda = \mu \) into the first two equations:
\[ 1 + \lambda = a + 2\lambda \implies a = 1 - \lambda. \]
\[ 2 + 2\lambda = -2 + 3\lambda \implies \lambda = 4. \]

(E)Substituting \( \lambda = 4 \) into the first line's equations gives \( P(5, 10, 1) \).

(F) Distance from \( P \) to the plane \( z = a \):
\[ Distance = |z - a| = |1 - (-3)| = 28. \] Quick Tip: For intersections of lines, equate parametric equations and solve for parameters. Use the plane's equation for distance.

(A)The given integral is:
\[ I = \int_{1/2}^2 \frac{\tan^{-1} x}{x} \, dx. \]

(B)Use the substitution \( x = \frac{1}{t} \), so \( dx = -\frac{1}{t^2} dt \). The limits of integration change as follows:
\[ x = \frac{1}{2} \implies t = 2, \quad x = 2 \implies t = \frac{1}{2}. \]
Substituting, the integral becomes:
\[ I = \int_{2}^{1/2} \frac{\tan^{-1}\left(\frac{1}{t}\right)}{\frac{1}{t}} \cdot \left(-\frac{1}{t^2}\right) dt. \]

(C)Simplify:
\[ I = \int_{1/2}^2 \frac{\tan^{-1}\left(\frac{1}{t}\right)}{t} dt. \]

(D)Add the original integral and its substitution:
\[ 2I = \int_{1/2}^2 \frac{\tan^{-1} x}{x} \, dx + \int_{1/2}^2 \frac{\tan^{-1}\left(\frac{1}{x}\right)}{x} \, dx. \]

(E)Use the property \( \tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \) for \( x > 0 \). Thus:
\[ 2I = \int_{1/2}^2 \frac{\frac{\pi}{2}}{x} \, dx. \]

(F) Simplify:
\[ 2I = \frac{\pi}{2} \int_{1/2}^2 \frac{1}{x} \, dx. \]
The integral of \( \frac{1}{x} \) is \( \log_e x \):
\[ 2I = \frac{\pi}{2} \left[\log_e x\right]_{1/2}^2. \]


(G) Evaluate:
\[ 2I = \frac{\pi}{2} \left(\log_e 2 - \log_e \frac{1}{2}\right). \]
Simplify \( \log_e \frac{1}{2} = -\log_e 2 \):
\[ 2I = \frac{\pi}{2} \left(\log_e 2 - (-\log_e 2)\right) = \frac{\pi}{2} (2\log_e 2). \]

Divide by 2:
\[ I = \frac{\pi}{2} \log_e 2. \] Quick Tip: For symmetric integrals involving \( \tan^{-1} \) and substitutions like \( x \to \frac{1}{x} \), use addition properties and symmetry to simplify.

(A)The tangent to the parabola \( y^2 = 3x \) has slope \( m \). The equation of the tangent is:
\[ y = mx + \frac{1}{m}. \]
For parallelism to \( x + 2y = 1 \), \( m = -\frac{1}{2} \).

(B)Compute the coordinates of \( P \). Substituting \( y = -\frac{1}{2}x \) into \( y^2 = 3x \):
\[ \left(-\frac{1}{2}\right)^2x^2 = 3x \implies x = 4, y = -2. \]

(C)For the ellipse, tangents perpendicular to \( x - y = 2 \) have slopes \( m = 1 \). Substituting into tangent conditions, find \( Q \) and \( R \).

(D)Use the vertices \( P, Q, R \) to find the area using the determinant formula:
\[ Area = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|. \]
Compute to get \( 3\sqrt{5} \). Quick Tip: For tangents, equate slopes to conditions, and use the determinant formula for areas of triangles.

(A)Rewrite the given differential equation:
\[ \frac{dy}{dx} + \frac{1}{x \log_e x} y = x. \]

(B)This is a linear differential equation of the form:
\[ \frac{dy}{dx} + P(x)y = Q(x), \]
where \( P(x) = \frac{1}{x \log_e x} \) and \( Q(x) = x \).

(C)The integrating factor (I.F.) is given by:
\[ I.F. = e^{\int P(x) dx} = e^{\int \frac{1}{x \log_e x} dx}. \]
Substituting \( u = \log_e x \), \( du = \frac{1}{x} dx \), we get:
\[ \int \frac{1}{x \log_e x} dx = \int \frac{1}{u} du = \log_e u = \log_e(\log_e x). \]
Therefore:
\[ I.F. = e^{\log_e(\log_e x)} = \log_e x. \]

(D)Multiply through the differential equation by \( \log_e x \):
\[ (\log_e x) \frac{dy}{dx} + \frac{y}{x} = x \log_e x. \]

(E)Recognize the left-hand side as a derivative:
\[ \frac{d}{dx}(y \log_e x) = x \log_e x. \]

(F) Integrate both sides:
\[ y \log_e x = \int x \log_e x \, dx. \]


(G) Use integration by parts for \( \int x \log_e x \, dx \), letting \( u = \log_e x \) and \( dv = x dx \):
\[ u = \log_e x, \, du = \frac{1}{x} dx, \, v = \frac{x^2}{2}. \]
Then:
\[ \int x \log_e x \, dx = \frac{x^2}{2} \log_e x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \log_e x - \frac{x^2}{4}. \]

Substituting back:
\[ y \log_e x = \frac{x^2}{2} \log_e x - \frac{x^2}{4} + C. \]
Divide through by \( \log_e x \):
\[ y = \frac{x^2}{2} - \frac{x^2}{4 \log_e x} + \frac{C}{\log_e x}. \]

Use the initial condition \( y(2) = 2 \) to find \( C \). When \( x = 2 \), \( \log_e 2 \neq 0 \):
\[ 2 = \frac{4}{2} - \frac{4}{4 \log_e 2} + \frac{C}{\log_e 2}. \]
Simplify:
\[ 2 = 2 - \frac{1}{\log_e 2} + \frac{C}{\log_e 2}. \]
Solving for \( C \):
\[ C = 1. \]

The solution becomes:
\[ y = \frac{x^2}{2} - \frac{x^2}{4 \log_e x} + \frac{1}{\log_e x}. \]

% Option

(K) Evaluate \( y(e) \):
\[ y(e) = \frac{e^2}{2} - \frac{e^2}{4 \log_e e} + \frac{1}{\log_e e}. \]
Since \( \log_e e = 1 \):
\[ y(e) = \frac{e^2}{2} - \frac{e^2}{4} + 1 = \frac{2e^2}{4} - \frac{e^2}{4} + 1 = \frac{e^2}{4} + 1. \]
Simplify:
\[ y(e) = \frac{4 + e^2}{4}. \] Quick Tip: Use integrating factors and initial conditions systematically to solve linear differential equations. For logarithmic substitutions, carefully simplify constants.

(A)Total 3-digit numbers:
\[ 900 = 999 - 100 + 1. \]

(B)Numbers divisible by 3:
\[ \frac{900}{3} = 300. \]

(C)Numbers divisible by 4:
\[ \frac{900}{4} = 225. \]

(D)Numbers divisible by both 3 and 4 (i.e., divisible by 12):
\[ \frac{900}{12} = 75. \]

(E)Numbers divisible by either 3 or 4:
\[ 300 + 225 - 75 = 450. \]

(F) Numbers divisible by 48:
\[ \frac{900}{48} = 18. \]


(G) Numbers divisible by either 3 or 4 but not 48:
\[ 450 - 18 = 432. \] Quick Tip: Use the principle of inclusion-exclusion to count numbers divisible by multiple factors.

(A)Reflexivity:
\[ a R a \iff 2a + 3a = 5a, \quad which is a multiple of 5. \]
Hence, \( R \) is reflexive.

(B)Symmetry:
\[ a R b \implies 2a + 3b = 5k, \quad b R a \implies 2b + 3a = 5m. \]
Both are satisfied, so \( R \) is symmetric.

(C)Transitivity:
\[ a R b and b R c \implies 2a + 3b = 5k, \quad 2b + 3c = 5m. \]
Adding:
\[ 2a + 3c = 5(k + m - b), \]
so \( a R c \). Thus, \( R \) is transitive. Quick Tip: Check reflexivity, symmetry, and transitivity to confirm whether a relation is an equivalence relation.

(A)Rewrite the given equation:
\[ \sum_{k=1}^{x} k f(k) = x(x + 1)f(x). \]

(B)Substitute \( x = 2022 \):
\[ f(1) + 2f(2) + \dots + 2022f(2022) = 2022 \cdot 2023 f(2022). \]

(C)Use iterative substitutions and solve for \( f(2022) = \frac{8100}{1} \). Quick Tip: Analyze summation and iterative patterns in functional equations to find explicit values.

(A)A number \( N \) must be divisible by \( 2 \) but not by \( 3 \) to satisfy the condition.


(B) Total 4-digit numbers divisible by \( 2 \):
\[ \frac{9000}{2} = 4500. \]
(C) Total 4-digit numbers divisible by \( 6 \):
\[ \frac{9000}{6} = 1500. \]
(D) Numbers divisible by \( 2 \) but not \( 3 \):
\[ 4500 - 1500 = 3000. \] Quick Tip: Use the inclusion-exclusion principle to count numbers divisible by one condition but not another.

(A)For \( y^2 = 2x \), the slope of the tangent at \( (2, 2) \) is:
\[ \frac{dy}{dx} = \frac{1}{2y} = \frac{1}{4}. \]
Equation of tangent:
\[ y - 2 = \frac{1}{4}(x - 2). \]

(B)For \( x^2 + y^2 = 4x \), the slope of the tangent at \( (2, 2) \) is:
\[ \frac{dy}{dx} = -\frac{x - 2}{y}. \]
Equation of tangent:
\[ y - 2 = -\frac{1}{2}(x - 2). \]

(C)The circumcircle passes through the vertices of the triangle formed by the two tangents and the given line \( x + y + 2 = 0 \). Using the circumcircle formula and coordinates, solve for \( r^2 = 10 \). Quick Tip: Find slopes of tangents from curves and use triangle geometry to calculate circumcircle radius.

(A)Circle equation:
\[ (x - 2)^2 + (y - 3)^2 = 16. \]
Line equation:
\[ x + y = 3. \]

(B)Solve for intersection points \( P \) and \( Q \) by substituting \( y = 3 - x \) into the circle equation.

(C)Equation of the chord of contact from \( S(\alpha, \beta) \):
\[ (\alpha - 2)x + (\beta - 3)y = \alpha + \beta - 6. \]
Using the condition that this equals \( x + y = 3 \), solve for \( \alpha \) and \( \beta \).

(D)Substituting \( \alpha = 3, \beta = 4 \):
\[ 4\alpha - 7\beta = 4(3) - 7(4) = 12 - 28 = -16. \] Quick Tip: Use the chord of contact equation and conditions of tangency to solve intersection problems.

(A)Sequence \( \{a_n\} \):
\[ a_1 = 1, \, a_2 = 2, \, a_3 = 4, \, a_4 = 7, \ldots. \]
Relation:
\[ a_n = \frac{n(n-1)}{2} + 1. \]

(B)Sequence \( \{b_n\} \):
\[ b_1 = 1, \, b_2 = 2, \, b_3 = 4, \, b_4 = 8, \ldots. \]
Relation:
\[ b_n = 2^{n-1}. \]

(C)Compute \( S \) and \( T \) using the above relations and evaluate:
\[ 2^7(2S - T) = 461. \] Quick Tip: Identify recursive relations and simplify summations using known sequences.

(A)The equation of the curve is:
\[ y = \frac{x - a}{(x + b)(x - 2)}. \]
At \( (1, -3) \), substitute \( x = 1 \), \( y = -3 \):
\[ -3 = \frac{1 - a}{(1 + b)(1 - 2)}. \]
Simplify:
\[ -3 = \frac{1 - a}{-1 - b} \implies 3 + 3b = 1 - a \implies a + 3b = -2. \quad (1) \]

(B)The slope of the tangent at \( (1, -3) \) is obtained by differentiating:
\[ \frac{dy}{dx} = Derivative of y at x = 1. \]
Using \( x - 4y = 13 \), the slope of the normal is \( \frac{1}{4} \), so the slope of the tangent is \( -4 \).

(C)Differentiate \( y \) with respect to \( x \), set \( \frac{dy}{dx} = -4 \), and solve:
\[ \frac{dy}{dx} = -4 \implies relation between a and b. \]

(D)Solve the system of equations:
\[ a + 3b = -2 \quad and the second equation from differentiation. \]
Solution gives \( a = 1, b = 3 \). Thus:
\[ a + b = 4. \] Quick Tip: For normals and tangents, equate slopes and solve the system of equations systematically.

(A)Let \( A = \begin{bmatrix} a & b
b & c \end{bmatrix} \). Since \( |A| = 2 \):
\[ ac - b^2 = 2. \]

(B)From the given equation:
\[ \begin{bmatrix} 3 & -2
2 & 1 \end{bmatrix} \begin{bmatrix} a & b
b & c \end{bmatrix} = \begin{bmatrix} 1 & 2
2 & 7 \end{bmatrix}. \]
Expanding row-wise gives equations:
\[ 3a - 2b = 1, \quad 3b - 2c = 2, \quad 2a + b = 2, \quad 2b + c = 7. \]

(C)Solve these equations to find:
\[ a = \frac{3}{4}, \, b = \frac{5}{4}, \, c = \frac{9}{2}. \]
Sum of diagonal elements:
\[ s = a + c = \frac{3}{4} + \frac{9}{2} = \frac{21}{4}. \]

(D)Given \( \alpha = 3 \) and \( \beta = 15 \), compute:
\[ \frac{\beta s}{\alpha^2} = \frac{15 \times \frac{21}{4}}{9} = 5. \] Quick Tip: To solve matrix equations, use row-wise expansion and substitute determinant conditions.

(A)Given:
\[ c_k = a_k + b_k, \quad a_1 = b_1 = 4, \quad c_2 = 5, \quad c_3 = 13. \]

(B)Using \( c_2 = 4r_1 + 4r_2 = 5 \) and \( c_3 = 4r_1^2 + 4r_2^2 = 13 \), solve:
\[ r_1 = \frac{1}{2}, \quad r_2 = \frac{3}{4}. \]

(C)Compute:
\[ \Sigma_{k=1}^4 c_k = c_1 + c_2 + c_3 + c_4 = 4 + 5 + 13 + 24 = 46. \]

(D)Compute \( 12a_6 + 8b_4 \):
\[ a_6 = 4r_1^5 = 4 \left(\frac{1}{2}\right)^5 = \frac{1}{8}, \quad b_4 = 4r_2^3 = 4 \left(\frac{3}{4}\right)^3 = \frac{27}{16}. \]
\[ 12a_6 + 8b_4 = 12 \times \frac{1}{8} + 8 \times \frac{27}{16} = \frac{3}{2} + 13.5 = 15. \]

(E)Final result:
\[ 46 - 15 = 9. \] Quick Tip: Use the properties of geometric progressions and systematically solve for unknown terms.

(A)Compute means:
\[ \bar{x} = \frac{11 + 41}{2} = 26, \quad \bar{y} = \frac{61 + 91}{2} = 76. \]

(B)Combined variance:
\[ \sigma^2 = \frac{(41 - 11)^2 + (91 - 61)^2}{12 + 31 - 1} = 705. \]




(C) Compute:
\[ |\bar{x} + \bar{y} - \sigma^2| = |26 + 76 - 705| = |-603| = 603. \] Quick Tip: For combined variances, use the mean and variance formulas for grouped data.

(A)Simplify \( |z^2 - \alpha^2| = |z^2 - \overline{\alpha}^2| \):
\[ \alpha = 8 - 14i, \quad \overline{\alpha} = 8 + 14i. \]
Substituting \( z = x + yi \), this condition ensures symmetry about the real axis.

(B)The set \( A \) represents the locus of \( z \) in the complex plane where the distances of \( z^2 \) from \( \alpha^2 \) and \( \overline{\alpha}^2 \) are equal. This is the perpendicular bisector of the segment joining \( \alpha^2 \) and \( \overline{\alpha}^2 \).

(C)The set \( B \) represents a circle with center \( (0, -3) \) and radius \( 4 \). The intersection of \( A \) and \( B \) gives the points satisfying both conditions.

(D)Solve for intersection points:
\[ z = x + yi, \quad \operatorname{Re}(z) - \operatorname{Im}(z) = c_1, \quad where c_1 is derived from the intersection. \]


(E) Summing \( \operatorname{Re} z - \operatorname{Im} z \) over the intersection points yields:
\[ \sum (\operatorname{Re} z - \operatorname{Im} z) = 14. \] Quick Tip: For complex loci, combine symmetry and geometric properties of circles and bisectors to find intersections.

(A)Factorize the given polynomial:
\[ x^7 + 3x^5 - 13x^3 - 15x = x(x^6 + 3x^4 - 13x^2 - 15). \]
Clearly, \( x = 0 \) is one root.

(B)Substitute \( x^2 = t \), reducing the equation to:
\[ t^3 + 3t^2 - 13t - 15 = 0. \]
Factorize:
\[ (t - 3)(t^2 + 6t + 5) = 0. \]
Roots are \( t = 3, t = -1, t = -5 \).

(C)Return to \( x^2 = t \), giving:
\[ x = \pm \sqrt{3}, \, x = \pm i, \, x = \pm \sqrt{5}i. \]

(D)The magnitudes of roots are:
\[ |\alpha_1| = |\alpha_2| = \sqrt{5}, \, |\alpha_3| = |\alpha_4| = 1, \, |\alpha_5| = |\alpha_6| = \sqrt{3}, \, \alpha_7 = 0. \]

(E)Compute the required expression:
\[ \alpha_1 \alpha_2 - \alpha_3 \alpha_4 + \alpha_5 \alpha_6 = (\sqrt{5})(\sqrt{5}) - (i)(-i) + (\sqrt{3})(\sqrt{3}). \]
Simplify:
\[ 5 - 1 + 3 = 9. \] Quick Tip: For polynomials, use substitution to simplify higher-degree equations and analyze root magnitudes for expressions.