\(A\) is any matrix, defined by \(A = [a_{ij}]_{3\times 3}\), where \(a_{ij} = |i - j|\), then which of the following statements are true ? A. \(A\) is symmetric matrix. B. \(A\) is skew-symmetric matrix. C. \(A\) is diagonal matrix. D. \(A\) is non singular matrix. Choose the correct answer from the options given below:
(A) A and B only
(B) B and C only
(C) A and D only
(D) B and D only
Correct Answer: (C) A and D only
View SolutionStep 1: Write out the matrix:
Put \(i, j = 1, 2, 3\) in \(a_{ij} = |i-j|\). The diagonal entries are \(|i-i| = 0\). The entries next to the diagonal are 1 and the corner entries are 2.
\[ A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 0 & 1 \\ 2 & 1 & 0 \end{bmatrix} \]
Step 2: Check statement A:
A matrix is symmetric when \(a_{ij} = a_{ji}\). Here \(|i-j| = |j-i|\) always, so \(A^T = A\). So A is TRUE.
Step 3: Check statement B:
A skew-symmetric matrix needs \(a_{ij} = -a_{ji}\). But \(a_{12} = 1\) and \(a_{21} = 1\), not \(-1\). So B is FALSE.
Step 4: Check statement C:
A diagonal matrix has zero everywhere off the diagonal. Here \(a_{12} = 1 \neq 0\). So C is FALSE.
Step 5: Check statement D:
A matrix is non singular when its determinant is not zero. Expand along the first row:
\[ |A| = 0(0-1) - 1(0-2) + 2(1-0) = 0 + 2 + 2 = 4 \]
Since \(|A| = 4 \neq 0\), A is non singular. So D is TRUE.
Final Answer:
Statements A and D are true, and B and C are false. This is option 3.
\[ \boxed{\text{A and D only}} \]
Question 2:
If \(AB = A\) and \(BA = B\), where \(A\) and \(B\) are square matrices of same order, then
(A) \(B^2 = B, A^2 = A\)
(B) \(B^2 \neq B\) and \(A^2 = A\)
(C) \(A^2 \neq A, B^2 = B\)
(D) \(A^2 \neq A, B^2 \neq B\)
Correct Answer: (A) \(B^2 = B, A^2 = A\)
View SolutionStep 1: Find \(A^2\):
Start with \(A^2 = A \cdot A\). Replace the first \(A\) by \(AB\), since \(AB = A\).
\[ A^2 = (AB)A = A(BA) \]
Now use \(BA = B\):
\[ A^2 = AB = A \]
So \(A^2 = A\).
Step 2: Find \(B^2\):
Start with \(B^2 = B \cdot B\). Replace the first \(B\) by \(BA\), since \(BA = B\).
\[ B^2 = (BA)B = B(AB) \]
Now use \(AB = A\):
\[ B^2 = BA = B \]
So \(B^2 = B\).
Step 3: Match with the options:
Both \(A^2 = A\) and \(B^2 = B\) hold. Option 2 says \(B^2 \neq B\), option 3 says \(A^2 \neq A\) and option 4 denies both. All three contradict what we proved, so they are wrong. Only option 1 is correct.
Final Answer:
Both matrices are idempotent. This is option 1.
\[ \boxed{B^2 = B,\ A^2 = A} \]
Question 3:
If \(\Delta = \begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} = (1 + ax^3)^b\), then which of the following statements are TRUE? A. \(a = -1\) and \(b = 2\) B. \(x = 1\) is a multiple root of \(\Delta = 0\) C. \(x = 1\) is a simple root of \(\Delta = 0\) D. \(x = 3\) is a simple root of \(\Delta = 0\) Choose the correct answer from the options given below:
(A) A and D only
(B) A and C only
(C) A, B and D only
(D) A and B only
Correct Answer: (D) A and B only
View SolutionStep 1: Expand the determinant:
Expand along the first row:
\[ \Delta = 1(1 \cdot 1 - x \cdot x^2) - x(x^2 \cdot 1 - x \cdot x) + x^2(x^2 \cdot x^2 - 1 \cdot x) \]
\[ \Delta = (1 - x^3) - x(x^2 - x^2) + x^2(x^4 - x) = 1 - x^3 + x^6 - x^3 \]
\[ \Delta = 1 - 2x^3 + x^6 = (1 - x^3)^2 \]
Step 2: Check statement A:
Compare \((1 - x^3)^2\) with \((1 + ax^3)^b\). This gives \(a = -1\) and \(b = 2\). So A is TRUE.
Step 3: Check statements B and C:
\(\Delta = (1-x^3)^2 = (1-x)^2(1 + x + x^2)^2\). The factor \((1-x)\) appears twice, so \(x = 1\) is a root of multiplicity 2, a multiple root. So B is TRUE and C is FALSE.
Step 4: Check statement D:
Put \(x = 3\): \(\Delta = (1 - 27)^2 = 676 \neq 0\). So 3 is not a root at all. D is FALSE.
Final Answer:
Only A and B are true. This is option 4.
\[ \boxed{\text{A and B only}} \]
Question 4:
If \(A\) is a square matrix of order \(3 \times 3\), and \(|\text{adj } A| = 25\), then the value of \(|2A|\) is
(A) 20
(B) \(\pm 20\)
(C) 40
(D) \(\pm 40\)
Correct Answer: (D) \(\pm 40\)
View SolutionStep 1: Recall the formulas:
For an \(n \times n\) matrix, \(|\text{adj } A| = |A|^{n-1}\) and \(|kA| = k^n |A|\). Here \(n = 3\).
Step 2: Find \(|A|\):
\(|\text{adj } A| = |A|^2 = 25\). So \(|A| = 5\) or \(|A| = -5\). Both values are possible because a square is positive either way.
Step 3: Find \(|2A|\):
\[ |2A| = 2^3 |A| = 8|A| \]
For \(|A| = 5\) this is 40. For \(|A| = -5\) this is \(-40\). So \(|2A| = \pm 40\).
Step 4: Check the other options:
Options 1 and 2 use 20, which would come from \(4|A|\) (the wrong power of 2). Option 3 gives only the positive case and ignores \(|A| = -5\).
If \(y = \sqrt{\dfrac{1-x}{1+x}}\), then the value of \((1 - x^2)\dfrac{dy}{dx} + y\) is
(A) \(\dfrac{1}{1+x}\)
(B) \(\dfrac{-1}{(1+x)^2}\)
(C) 0
(D) 1
Correct Answer: (C) 0
View SolutionStep 1: Rewrite y:
\(y = \sqrt{\dfrac{1-x}{1+x}} = \dfrac{1-x}{\sqrt{1-x^2}}\), obtained by multiplying top and bottom by \(\sqrt{1-x}\). This form is easier to differentiate.
Step 2: Differentiate with the quotient rule:
Let \(u = 1 - x\) and \(v = (1-x^2)^{1/2}\). Then \(u' = -1\) and \(v' = \dfrac{-x}{\sqrt{1-x^2}}\).
\[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{-\sqrt{1-x^2} + \dfrac{x(1-x)}{\sqrt{1-x^2}}}{1-x^2} \]
Put the top over one denominator:
\[ \frac{dy}{dx} = \frac{-(1-x^2) + x(1-x)}{(1-x^2)^{3/2}} = \frac{-1 + x^2 + x - x^2}{(1-x^2)^{3/2}} = \frac{x - 1}{(1-x^2)^{3/2}} \]
Step 5: Check the options:
\(e^x\) is always positive, so the sum can never be \(-1\) or 0. The sum is never below 2, so 1 is not reached either.
Final Answer:
The minimum value is 2, option 4.
\[ \boxed{2} \]
Question 7:
The function \(f(x) = \dfrac{x}{\log x}\) is increasing in the interval (consider \(\log_e x = \log x\))
(A) \((1, \infty)\)
(B) \((e, \infty)\)
(C) \((-\infty, e)\)
(D) \((-\infty, -1)\)
Correct Answer: (B) \((e, \infty)\)
View SolutionStep 1: Find the domain:
\(\log x\) exists only for \(x > 0\), and the function needs \(\log x \neq 0\), so \(x \neq 1\). Hence the domain is \((0,1) \cup (1,\infty)\).
Step 2: Differentiate:
Use the quotient rule:
\[ f'(x) = \frac{\log x \cdot 1 - x \cdot \frac{1}{x}}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} \]
Step 3: Find where f' is positive:
The denominator \((\log x)^2\) is positive on the domain. So \(f'(x) > 0\) when \(\log x - 1 > 0\), that is \(\log x > 1\), that is \(x > e\).
Step 4: Check the options:
Option 1, \((1, \infty)\), includes \((1, e)\) where \(\log x < 1\) and \(f'\) is negative, so it is wrong. Options 3 and 4 include negative numbers, outside the domain. Only option 2 is correct.
Final Answer:
The function increases on \((e, \infty)\), option 2.
\[ \boxed{(e, \infty)} \]
Question 8:
\(\displaystyle\int e^{x}\left(\log x+\frac{1}{x^{2}}\right)dx\) is equal to (consider \(\log_e x = \log x\))
(A) \(e^{x}\log_e x + c\) : where c is an arbitrary constant
(B) \(e^{x}\left(\log_e x+\frac{1}{x}\right)+c\): where c is an arbitrary constant
(C) \(e^{x}\left(\log_e x-\frac{1}{x}\right)+c\): where c is an arbitrary constant
(D) \(e^{x}\cdot\frac{1}{x}+c\) : where c is an arbitrary constant
Correct Answer: (C) \(e^{x}\left(\log_e x-\frac{1}{x}\right)+c\): where c is an arbitrary constant
View SolutionStep 1: Understanding the Concept:
The integrand looks like \(e^{x}\left[f(x)+f^{\prime}(x)\right]\). Such integrals have a neat result. But here the second term is \(\frac{1}{x^{2}}\), so we must first find which function \(f\) gives a derivative that produces this term.
Step 2: Key Formula or Approach:
We know \(\displaystyle\int e^{x}\left[f(x)+f^{\prime}(x)\right]dx = e^{x}f(x)+c\). Since the options are given, we can also differentiate each option and see which one returns the integrand.
Step 3: Try \(f(x)=\log x - \frac{1}{x}\).
Then \(f^{\prime}(x)=\frac{1}{x}+\frac{1}{x^{2}}\). So \(f(x)+f^{\prime}(x)=\log x-\frac{1}{x}+\frac{1}{x}+\frac{1}{x^{2}}=\log x+\frac{1}{x^{2}}\). This matches the integrand exactly.
Step 5: Check the other options.
Option 1: \(\frac{d}{dx}\left(e^{x}\log x\right)=e^{x}\left(\log x+\frac{1}{x}\right)\), which has \(\frac{1}{x}\) and not \(\frac{1}{x^{2}}\). So it is wrong. Option 2: \(\frac{d}{dx}\left[e^{x}\left(\log x+\frac{1}{x}\right)\right]=e^{x}\left(\log x+\frac{1}{x}+\frac{1}{x}-\frac{1}{x^{2}}\right)\), which is not the integrand. So it is wrong. Option 4: \(\frac{d}{dx}\left(\frac{e^{x}}{x}\right)=e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\), which has no \(\log x\). So it is wrong.
Final Answer:
The integral equals \(e^{x}\left(\log x-\frac{1}{x}\right)+c\), which is option 3. \[ \boxed{e^{x}\left(\log x-\frac{1}{x}\right)+c} \]
Question 9:
If \(2f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}-5,\ x\neq 0\), then \(\displaystyle\int_{1}^{2}f(x)dx\) is equal to: (consider \(\log_e x = \log x\))
(A) \(\frac{1}{6}\left[\log 16-13\right]\)
(B) \(\frac{1}{3}\left[\log 8-13\right]\)
(C) \(\frac{1}{3}\left[\log 8-14\right]\)
(D) \(\left[\log 16-14\right]\)
Correct Answer: (A) \(\frac{1}{6}\left[\log 16-13\right]\)
View SolutionStep 1: Understanding the Question:
We are given one equation linking \(f(x)\) and \(f\left(\frac{1}{x}\right)\). We need \(f(x)\) itself before we can integrate it from 1 to 2.
Step 2: Key Formula or Approach:
Replace \(x\) by \(\frac{1}{x}\) to get a second equation. Then solve the two equations for \(f(x)\) like a pair of simultaneous equations.
Step 3: Form the two equations.
Equation (1): \(2f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}-5\). Replace \(x\) by \(\frac{1}{x}\): Equation (2): \(2f\left(\frac{1}{x}\right)+f(x)=x-5\).
Step 4: Solve for \(f(x)\).
Multiply equation (1) by 2: \(4f(x)+2f\left(\frac{1}{x}\right)=\frac{2}{x}-10\). Subtract equation (2): \(3f(x)=\frac{2}{x}-10-x+5=\frac{2}{x}-x-5\). \[ f(x)=\frac{1}{3}\left(\frac{2}{x}-x-5\right) \]
Step 5: Integrate from 1 to 2.
\[ \int_{1}^{2}f(x)dx=\frac{1}{3}\left[2\log x-\frac{x^{2}}{2}-5x\right]_{1}^{2} \] At \(x=2\): \(2\log 2-2-10=2\log 2-12\). At \(x=1\): \(0-\frac{1}{2}-5=-\frac{11}{2}\). Difference: \(2\log 2-12+\frac{11}{2}=2\log 2-\frac{13}{2}\). So the integral is \(\frac{1}{3}\left(2\log 2-\frac{13}{2}\right)=\frac{4\log 2-13}{6}=\frac{\log 16-13}{6}\).
Step 6: Check the other options.
Options 2 and 3 have the factor \(\frac{1}{3}\) with \(\log 8\), which would come from a wrong coefficient of \(\log 2\). Option 4 has no factor \(\frac{1}{6}\). Only option 1 matches our value.
Final Answer:
The value is \(\frac{1}{6}\left[\log 16-13\right]\), option 1. \[ \boxed{\frac{1}{6}\left[\log 16-13\right]} \]
Question 10:
Area of the region bounded by the lines \(|x|+|y|=1,\ x\geq 0,\ y\geq 0\) is
(A) 1 square units
(B) \(\frac{1}{2}\) square units
(C) 2 square units
(D) 4 square units
Correct Answer: (B) \(\frac{1}{2}\) square units
View SolutionStep 1: Understanding the Question:
With \(x\geq 0\) and \(y\geq 0\), the absolute value signs can be dropped. The line becomes \(x+y=1\).
Step 2: Key Formula or Approach:
The region is bounded by the two axes and the line \(x+y=1\). This is a right triangle in the first quadrant. Its area can also be found by integration.
Step 3: Find the vertices.
Put \(y=0\): \(x=1\), so the point is \((1,0)\). Put \(x=0\): \(y=1\), so the point is \((0,1)\). The third vertex is the origin \((0,0)\).
Step 4: Find the area.
Both legs have length 1. \[ \text{Area}=\frac{1}{2}\times 1\times 1=\frac{1}{2} \] Check by integration: \(\int_{0}^{1}(1-x)dx=\left[x-\frac{x^{2}}{2}\right]_{0}^{1}=\frac{1}{2}\).
Step 5: Check the other options.
An area of 1 would be a full unit square, but the triangle is only half of it. Areas 2 and 4 are much larger than the unit square, so they are impossible.
Final Answer:
The area is \(\frac{1}{2}\) square units, option 2. \[ \boxed{\frac{1}{2}} \]
Question 11:
The order and degree of the differential equation \(\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}=\left(\frac{d^{2}y}{dx^{2}}\right)^{\frac{1}{3}}\) are
(A) order =2, degree = 2
(B) order =2, degree=\(\frac{1}{3}\)
(C) order = 3, degree is not defined
(D) order = 2, degree = 3
Correct Answer: (A) order =2, degree = 2
View SolutionStep 1: Understanding the Concept:
The order is the highest derivative present. The degree is the power of that highest derivative after the equation is made free of radicals and fractional powers in the derivatives.
Step 2: Find the order.
The highest derivative is \(\frac{d^{2}y}{dx^{2}}\). So the order is 2.
Step 3: Remove the radical and fractional power.
Square both sides: \(1+\left(\frac{dy}{dx}\right)^{2}=\left(\frac{d^{2}y}{dx^{2}}\right)^{\frac{2}{3}}\). Now cube both sides: \[ \left[1+\left(\frac{dy}{dx}\right)^{2}\right]^{3}=\left(\frac{d^{2}y}{dx^{2}}\right)^{2} \]
Step 4: Find the degree.
Now the equation is a polynomial in the derivatives. The highest derivative \(\frac{d^{2}y}{dx^{2}}\) has power 2. So the degree is 2.
Step 5: Check the other options.
Degree cannot be \(\frac{1}{3}\), because degree is defined only after removing fractional powers. Order 3 is wrong since no third derivative exists. Degree 3 is wrong because after clearing powers the second derivative appears squared, not cubed.
Final Answer:
Order = 2 and degree = 2, option 1. \[ \boxed{\text{order}=2,\ \text{degree}=2} \]
Question 12:
The solution of the differential equation \(\frac{dy}{dx}=\frac{x}{y}\) represents a family of
(A) circles
(B) ellipses
(C) parabolas
(D) hyperbolas
Correct Answer: (D) hyperbolas
View SolutionStep 1: Understanding the Question:
We must solve the differential equation and then identify the curve given by its general solution.
Step 2: Key Formula or Approach:
The equation has variables that can be separated. Put all \(y\) terms with \(dy\) and all \(x\) terms with \(dx\), then integrate.
Step 3: Separate and integrate.
\[ y\,dy=x\,dx \] \[ \frac{y^{2}}{2}=\frac{x^{2}}{2}+k \] Multiply by 2 and let \(c=2k\): \[ y^{2}-x^{2}=c \]
Step 4: Identify the curve.
The equation \(x^{2}-y^{2}=-c\) has the form \(x^{2}-y^{2}=\text{constant}\). Such an equation is a hyperbola (a rectangular hyperbola). For \(c=0\) it degenerates into the pair of lines \(y=\pm x\).
Step 5: Check the other options.
A circle or ellipse has a plus sign between \(x^{2}\) and \(y^{2}\), but we have a minus sign. A parabola has only one squared variable. So these three are wrong.
Final Answer:
The solution is a family of hyperbolas, option 4. \[ \boxed{\text{hyperbolas}} \]
Question 13:
The solution of initial value problem: \(\frac{dy}{dx}=e^{3x+4y};\ y(0)=-\frac{1}{4}\) is
(A) \(e^{-4y}+e^{3x}=3e\)
(B) \(3e^{-4y}+4e^{3x}=3e+4\)
(C) \(-4e^{-4y}+3e^{3x}=3e+4\)
(D) \(e^{-4y}-3e^{3x}=0\)
Correct Answer: (B) \(3e^{-4y}+4e^{3x}=3e+4\)
View SolutionStep 1: Understanding the Question:
We have a differential equation with a starting condition. We must find the particular solution that passes through \(x=0,\ y=-\frac{1}{4}\).
Step 2: Key Formula or Approach:
Write \(e^{3x+4y}=e^{3x}\cdot e^{4y}\). Then separate the variables and integrate. Use the condition to find the constant.
Step 3: Separate the variables.
\[ \frac{dy}{dx}=e^{3x}e^{4y} \] \[ e^{-4y}dy=e^{3x}dx \]
Step 5: Use the initial condition.
Put \(x=0\) and \(y=-\frac{1}{4}\). Then \(e^{-4y}=e^{1}=e\) and \(e^{3x}=1\). \[ -\frac{e}{4}=\frac{1}{3}+C \implies C=-\frac{e}{4}-\frac{1}{3} \]
Step 7: Check the other options.
Test the point \((0,-\frac{1}{4})\) in each option. Option 1 gives \(e+1\), not \(3e\). Option 3 gives \(-4e+3\), not \(3e+4\). Option 4 gives \(e-3\), not 0. Only option 2 gives \(3e+4\).
Final Answer:
The solution is \(3e^{-4y}+4e^{3x}=3e+4\), option 2. \[ \boxed{3e^{-4y}+4e^{3x}=3e+4} \]
Question 14:
Three coins are tossed, then Match the LIST-I with LIST-II
LIST-I
LIST-II
A.
Probability of occurrence of 3 heads or 3 tails
I.
\(\frac{1}{8}\)
B.
Probability of occurrence of at least two heads
II.
\(\frac{7}{8}\)
C.
Probability of occurrence of at most two heads
III.
\(\frac{1}{2}\)
D.
Probability of occurrence of no head
IV.
\(\frac{1}{4}\)
Choose the correct answer from the options given below:
(A) A-I, B-II, C-III, D-IV
(B) A-IV, B-III, C-I, D-II
(C) A-III, B-IV, C-II, D-I
(D) A-IV, B-III, C-II, D-I
Correct Answer: (D) A-IV, B-III, C-II, D-I
View SolutionStep 1: Understanding the Question:
Three coins give \(2^{3}=8\) equally likely outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. We find each probability and match it with List-II.
Step 2: Key Formula or Approach:
\(P(E)=\frac{\text{favourable outcomes}}{8}\). Count the favourable outcomes for each case.
Step 3: Case A.
3 heads or 3 tails: outcomes HHH and TTT, which is 2. So \(P=\frac{2}{8}=\frac{1}{4}\). This is IV.
Step 4: Case B.
At least two heads means 2 or 3 heads: HHT, HTH, THH, HHH, which is 4. So \(P=\frac{4}{8}=\frac{1}{2}\). This is III.
Step 5: Case C.
At most two heads means every outcome except HHH. That is 7 outcomes. So \(P=\frac{7}{8}\). This is II.
Step 6: Case D.
No head means TTT only, which is 1 outcome. So \(P=\frac{1}{8}\). This is I.
Step 7: Check the options.
The matching is A-IV, B-III, C-II, D-I. Option 1 pairs A with I, which is wrong. Option 2 pairs C with I, which is wrong. Option 3 pairs A with III, which is wrong. Option 4 is fully right.
Final Answer:
The correct matching is A-IV, B-III, C-II, D-I, option 4. \[ \boxed{\text{A-IV, B-III, C-II, D-I}} \]
Question 15:
Match the LIST-I with LIST-II
LIST-I Inequality
LIST-II Solution Space
A.
\(2x+3y \geq 6\)
I.
half plane contains the origin, also points on the line \(2x+3y=6\)
B.
\(2x+3y \leq 6\)
II.
half plane contains the origin but not points on the line \(2x+3y=6\)
C.
\(2x+3y > 6\)
III.
half plane not contains the origin but contains points on the line \(2x+3y=6\)
D.
\(2x+3y < 6\)
IV.
half plane neither contains the origin nor contains points on the line \(2x+3y=6\)
Choose the correct answer from the options given below:
(A) A-II, B-III, C-I, D-IV
(B) A-III, B-II, C-I, D-IV
(C) A-III, B-I, C-IV, D-II
(D) A-II, B-IV, C-I, D-III
Correct Answer: (C) A-III, B-I, C-IV, D-II
View SolutionStep 1: Understanding the Concept:
A linear inequality in two variables shades one side (a half plane) of its boundary line. If the sign is \(\geq\) or \(\leq\), the boundary line is part of the solution. If the sign is \(>\) or \(<\), the line is left out. To decide which side is shaded, test the origin \((0,0)\) in the inequality.
Step 2: Key Formula or Approach:
The boundary line is \(2x+3y=6\). At the origin, the left side is \(2(0)+3(0)=0\). So we only need to check whether \(0 \geq 6\), \(0 \leq 6\), \(0 > 6\) or \(0 < 6\) is true.
Step 3: Check A: \(2x+3y \geq 6\).
At the origin, \(0 \geq 6\) is false. So the origin is not in the half plane. The sign includes equality, so points on the line are included. This matches III.
Step 4: Check B: \(2x+3y \leq 6\).
At the origin, \(0 \leq 6\) is true. So the origin is in the half plane. The line is included because of the equality sign. This matches I.
Step 5: Check C: \(2x+3y > 6\).
At the origin, \(0 > 6\) is false, so the origin is not in the half plane. The sign is strict, so the line is excluded. This matches IV.
Step 6: Check D: \(2x+3y < 6\).
At the origin, \(0 < 6\) is true, so the origin is in the half plane. The sign is strict, so the line is excluded. This matches II.
Step 7: Compare with the options.
We got A-III, B-I, C-IV, D-II. Option 1 has A-II, so it is wrong. Option 2 has B-II, so it is wrong. Option 4 has A-II, so it is wrong. Only option 3 matches.
Final Answer:
The correct matching is A-III, B-I, C-IV, D-II, which is option 3.
\[ \boxed{\text{Option 3: A-III, B-I, C-IV, D-II}} \]
Question 16:
Relation \(R\) in the set \(A = \{1,2,3,4,\ldots,16\}\) defined as \(R = \{(x,y): 3x = y,\ x,y \in A\}\), then which of the following are correct ? A. domain of \(R = \{1, 2, 3, 4, 5\}\). B. range of \(R = \{3, 6, 9, 12, 15\}\). C. \(R\) is reflexive. D. \(R\) is not symmetric. Choose the correct answer from the options given below:
(A) A and B only
(B) B and D only
(C) A,B and C only
(D) A,B and D only
Correct Answer: (D) A,B and D only
View SolutionStep 1: Understanding the Concept:
A relation \(R\) on a set is a set of ordered pairs. The domain is the set of first entries and the range is the set of second entries. Reflexive means \((x,x) \in R\) for every \(x\). Symmetric means \((x,y) \in R\) gives \((y,x) \in R\).
Step 2: List the pairs of R.
We need \(y=3x\) with both \(x\) and \(y\) in \(\{1,\ldots,16\}\). So \(3x \leq 16\), which gives \(x \leq 5\). \[ R=\{(1,3),(2,6),(3,9),(4,12),(5,15)\} \]
Step 3: Check statement A.
The first entries are \(1,2,3,4,5\). So the domain is \(\{1,2,3,4,5\}\). Statement A is TRUE.
Step 4: Check statement B.
The second entries are \(3,6,9,12,15\). So the range is \(\{3,6,9,12,15\}\). Statement B is TRUE.
Step 5: Check statement C.
For \(R\) to be reflexive, we need \((1,1)\) in \(R\). But \(3 \times 1 = 3 \neq 1\). So \((1,1) \notin R\). Statement C is FALSE.
Step 6: Check statement D.
We have \((1,3) \in R\). For symmetry, we need \((3,1) \in R\), that is \(3 \times 3 = 1\), which is false. So \(R\) is not symmetric. Statement D is TRUE.
Step 7: Match with the options.
The true statements are A, B and D. That is option 4. Option 1 misses D. Option 2 misses A. Option 3 includes the false statement C.
Final Answer:
Statements A, B and D are correct.
\[ \boxed{\text{Option 4: A, B and D only}} \]
Question 17:
A function \(f: N \to N\), (\(N\): set of natural numbers) defined by \(f(x) = x^2 + x + 1\) is
(A) one-one and onto
(B) one-one but not onto
(C) many-one and onto
(D) many-one but not onto
Correct Answer: (B) one-one but not onto
View SolutionStep 1: Understanding the Concept:
A function is one-one if different inputs always give different outputs. It is onto if every element of the codomain is an output. Here the domain and codomain are both \(N=\{1,2,3,\ldots\}\).
Step 2: Test one-one.
Suppose \(f(a)=f(b)\). Then \(a^2+a+1=b^2+b+1\), so \(a^2-b^2+a-b=0\). \[ (a-b)(a+b+1)=0 \] Since \(a,b\) are natural numbers, \(a+b+1 \geq 3\), so it is never zero. Hence \(a=b\). The function is one-one.
Step 3: Test onto.
Compute a few values: \(f(1)=3\), \(f(2)=7\), \(f(3)=13\). The smallest output is 3. So the numbers 1 and 2 in the codomain are never reached. The function is not onto.
Step 4: Check the options.
Option 1 says onto, which is false. Option 3 and option 4 say many-one, which is false. Option 2, one-one but not onto, is correct.
Final Answer:
The function is one-one but not onto.
\[ \boxed{\text{Option 2: one-one but not onto}} \]
Choose the correct answer from the options given below:
(A) A-IV, B-I, C-III, D-II
(B) A-II, B-III, C-I, D-IV
(C) A-IV, B-III, C-I, D-II
(D) A-II, B-IV, C-III, D-I
Correct Answer: (A) A-IV, B-I, C-III, D-II
View SolutionStep 1: Understanding the Concept:
Each inverse trigonometric function has a fixed principal value branch, chosen so that the function is one-one and onto. These ranges are standard facts from the NCERT chapter on inverse trigonometric functions.
Step 2: Check A: tan inverse x.
\(\tan^{-1}x\) takes all real \(x\) and gives angles strictly between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). The end points are not reached. So the range is the open interval in IV.
Step 3: Check B: sec inverse x.
\(\sec^{-1}x\) is defined for \(|x| \geq 1\). Its range is \([0,\pi]\) but \(\frac{\pi}{2}\) is left out, because \(\sec\frac{\pi}{2}\) is undefined. This is I.
Step 4: Check C: cosec inverse x.
\(\operatorname{cosec}^{-1}x\) is defined for \(|x| \geq 1\). Its range is \([-\frac{\pi}{2},\frac{\pi}{2}]\) without 0, because \(\operatorname{cosec}0\) is undefined. This is III.
Step 5: Check D: sin inverse x.
\(\sin^{-1}x\) is defined for \(-1 \leq x \leq 1\). Its range is the closed interval \([-\frac{\pi}{2},\frac{\pi}{2}]\). This is II.
Step 6: Compare with the options.
We have A-IV, B-I, C-III, D-II. This is exactly option 1. The other options pair A with II, or B with III, or D with I, and these are all wrong.
Final Answer:
The matching is A-IV, B-I, C-III, D-II.
\[ \boxed{\text{Option 1}} \]
Question 19:
If \(A\) is square matrix such that \(A^2 = A\), then \((2I + A)^4 - 65A\) is equal to (Where \(I\) is identity matrix)
(A) \(8I\)
(B) \(16I\)
(C) \(16I + A\)
(D) \(8I + A\)
Correct Answer: (B) \(16I\)
View SolutionStep 1: Understanding the Concept:
A matrix with \(A^2=A\) is called idempotent. Then every power \(A^n\) equals \(A\). Also \(I\) commutes with every matrix, so we can expand \((2I+A)^n\) like an ordinary binomial.
Step 2: Square the bracket first.
\[ (2I+A)^2 = 4I^2 + 4IA + A^2 = 4I + 4A + A \] Here we used \(I^2=I\), \(IA=A\) and \(A^2=A\). So \[ (2I+A)^2 = 4I+5A \]
Step 3: Square again to get the fourth power.
\[ (2I+A)^4 = (4I+5A)^2 = 16I + 40A + 25A^2 \] Using \(A^2=A\), this becomes \[ 16I + 40A + 25A = 16I + 65A \]
Step 5: Check the options.
Option 1, \(8I\), is too small. Options 3 and 4 keep an extra \(A\), but the \(A\) terms cancel fully. So only option 2 fits. A quick test with \(A=I\) gives \(3^4-65=16\), which equals \(16I\) for the identity, so it matches.
Final Answer:
The value is \(16I\).
\[ \boxed{16I} \]
Question 20:
If \(A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\), such that \(A^2 - 4A + \lambda I = 0\), where \(I\) is identity matrix of order 2, then '\(\lambda\)' is equal to
(A) \(-1\)
(B) \(1\)
(C) \(-2\)
(D) \(2\)
Correct Answer: (B) \(1\)
View SolutionStep 1: Understanding the Concept:
We compute \(A^2\) by matrix multiplication, then put it into the equation and compare entries. The equation must hold for every entry of the matrix.
Step 4: Form the left side.
\[ A^2-4A = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I \] So the equation becomes \(-I+\lambda I = 0\), that is \((\lambda-1)I=0\).
Step 5: Solve for lambda.
Since \(I\) is not the zero matrix, \(\lambda - 1 = 0\), so \(\lambda = 1\).
Step 6: Check the options.
The values \(-1\), \(-2\) and \(2\) would leave a non zero matrix. For example \(\lambda=2\) gives \(I \neq 0\). Only \(\lambda=1\) works, which is option 2.
Final Answer:
The value of \(\lambda\) is 1.
\[ \boxed{\lambda = 1} \]
Question 21:
If \(A = [a_{ij}]_{3\times 3}\), \(a_{ij} = \begin{cases} 0, & i \neq j \\ i+j, & i = j \end{cases}\), then which of the following statements are correct ? A. \(A\) is scalar matrix B. \(A\) is diagonal matrix C. \(A\) is unit matrix D. \(A\) is symmetric matrix Choose the correct answer from the options given below:
(A) B and C only
(B) A, B and C only
(C) B and D only
(D) C and D only
Correct Answer: (C) B and D only
View SolutionStep 1: Understanding the Concept:
A diagonal matrix has zero entries off the main diagonal. A scalar matrix is a diagonal matrix with equal diagonal entries. A unit matrix has all diagonal entries equal to 1. A symmetric matrix satisfies \(a_{ij}=a_{ji}\).
Step 2: Build the matrix.
Off the diagonal, \(a_{ij}=0\). On the diagonal, \(a_{11}=1+1=2\), \(a_{22}=2+2=4\), \(a_{33}=3+3=6\). \[ A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix} \]
Step 3: Check statement A.
The diagonal entries 2, 4, 6 are not all equal. So \(A\) is not a scalar matrix. A is FALSE.
Step 4: Check statement B.
All entries off the main diagonal are zero. So \(A\) is a diagonal matrix. B is TRUE.
Step 5: Check statement C.
A unit matrix needs 1 on every diagonal place. Here the diagonal is 2, 4, 6. C is FALSE.
Step 6: Check statement D.
Every off diagonal entry is 0, so \(a_{ij}=a_{ji}\) for all \(i,j\). That means \(A^T=A\). D is TRUE.
Step 7: Match with the options.
The true statements are B and D. That is option 3.
Final Answer:
Only statements B and D are correct.
\[ \boxed{\text{Option 3: B and D only}} \]
Question 22:
If \(A(x_1,y_1)\), \(B(x_2,y_2)\) and \(C(x_3,y_3)\) are the vertices of an equilateral triangle whose each side is equal to \(k\), then the value of \(\begin{vmatrix} x_1 & y_1 & 2 \\ x_2 & y_2 & 2 \\ x_3 & y_3 & 2 \end{vmatrix}^2\) is
(A) \(3k^2\)
(B) \(4k^2\)
(C) \(3k^4\)
(D) \(4k^4\)
Correct Answer: (C) \(3k^4\)
View SolutionStep 1: Recall the area formula.
The area of a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\) is half the absolute value of the determinant with a column of 1s.
\[ \Delta = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right| \]
Step 2: Pull out the common factor.
Every entry of the last column is 2. Taking 2 out of that column gives a factor of 2 outside the determinant.
\[ \begin{vmatrix} x_1 & y_1 & 2 \\ x_2 & y_2 & 2 \\ x_3 & y_3 & 2 \end{vmatrix} = 2\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]
Step 3: Use the area of an equilateral triangle.
For side \(k\), the area is \(\Delta = \frac{\sqrt{3}}{4}k^2\). From Step 1, the determinant with 1s equals \(\pm 2\Delta = \pm\frac{\sqrt{3}}{2}k^2\).
Step 4: Find the required value.
The given determinant is \(2 \times \left(\pm\frac{\sqrt{3}}{2}k^2\right) = \pm\sqrt{3}k^2\). Squaring removes the sign.
\[ \left(\sqrt{3}k^2\right)^2 = 3k^4 \]
Step 5: Check the options.
Option 1 and option 2 have only \(k^2\), which is wrong because squaring a term with \(k^2\) must give \(k^4\). Option 4 would need the determinant to be \(2k^2\), but it is \(\sqrt{3}k^2\). So only option 3 fits.
Final Answer:
The value is \(3k^4\), which is option 3.
\[ \boxed{3k^4} \]
Question 23:
If \(x, y, z\) are all different from zero and \(\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z \end{vmatrix} = 0\), then the value of \(x^{-1} + y^{-1} + z^{-1}\) is
(A) \(xyz\)
(B) \(-1\)
(C) \(1\)
(D) \(0\)
Correct Answer: (B) \(-1\)
View SolutionStep 1: Expand the determinant.
Expand along the first row.
\[ (1+x)\big[(1+y)(1+z)-1\big] - 1\big[(1+z)-1\big] + 1\big[1-(1+y)\big] \]
Simplify each bracket: the first is \(y+z+yz\), the second is \(z\), the third is \(-y\).
Step 2: Collect the terms.
\[ (1+x)(y+z+yz) - z - y = xy + xz + xyz + yz \]
So the determinant equals \(xyz + xy + yz + zx\).
Step 3: Use the condition.
The determinant is 0, so \(xyz + xy + yz + zx = 0\). Since \(x, y, z \neq 0\), divide by \(xyz\).
\[ 1 + \frac{1}{z} + \frac{1}{x} + \frac{1}{y} = 0 \]
Step 4: Read off the answer.
Hence \(x^{-1}+y^{-1}+z^{-1} = -1\).
Step 5: Check the other options.
Option 1 (\(xyz\)) is not fixed by the condition. Options 3 and 4 would need the sum of reciprocals to be 1 or 0, but the equation gives \(-1\).
Final Answer:
The sum of reciprocals is -1, option 2.
\[ \boxed{-1} \]
Question 24:
If \(A=\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}\), then Match the LIST-I with LIST-II
LIST-I
LIST-II
A. \(|\text{adj }A|\)
I. 81
B. \(|A|\)
II. \(-27\)
C. \(|A\ \text{adj }A|\)
III. 9
D. \(|\text{adj (adj }A)|\)
IV. \(-3\)
Choose the correct answer from the options given below:
(A) A-II, B-III, C-IV, D-I
(B) A-I, B-IV, C-II, D-III
(C) A-IV, B-III, C-II, D-I
(D) A-III, B-IV, C-II, D-I
Correct Answer: (D) A-III, B-IV, C-II, D-I
View SolutionStep 1: Find |A|.
The matrix is lower triangular, so its determinant is the product of the diagonal entries.
\[ |A| = 1 \times 3 \times (-1) = -3 \]
So B matches IV.
Step 2: Recall the adjoint formulas.
For a square matrix of order \(n\): \(|\text{adj }A| = |A|^{n-1}\), and \(A\ (\text{adj }A) = |A| I\). Here \(n=3\).
Step 3: Find |adj A|.
\[ |\text{adj }A| = |A|^{2} = (-3)^2 = 9 \]
So A matches III.
Step 4: Find |A adj A|.
\(A\ \text{adj }A = |A| I = -3I\). The determinant of \(-3I\) (order 3) is \((-3)^3 = -27\). So C matches II.
Step 5: Find |adj(adj A)|.
Use the formula \(|\text{adj}(\text{adj }A)| = |A|^{(n-1)^2}\). With \(n=3\) the power is 4.
\[ (-3)^4 = 81 \]
So D matches I.
Step 6: Compare with the options.
We have A-III, B-IV, C-II, D-I. This is option 4. Option 1 puts A with II, which is \(-27\) and wrong. Option 2 puts A with I (81), wrong. Option 3 puts A with IV, wrong.
Final Answer:
The matching is A-III, B-IV, C-II, D-I, option 4.
\[ \boxed{\text{A-III, B-IV, C-II, D-I}} \]
Question 25:
Function \(f(x)=\begin{cases} k\left(x^2-6x+13\right) & : x<1 \\ |x-3| & : x\ge 1 \end{cases}\) is differentiable at \(x=1\), then the value of \(k\) is
(A) \(4\)
(B) \(\frac{1}{4}\)
(C) \(\frac{-1}{4}\)
(D) \(-4\)
Correct Answer: (B) \(\frac{1}{4}\)
View SolutionStep 1: State the condition.
A function is differentiable at a point only if it is continuous there and the left and right derivatives are equal.
Step 2: Apply continuity at x = 1.
Left limit: \(k(1-6+13) = 8k\). Right value: \(|1-3| = 2\).
\[ 8k = 2 \Rightarrow k = \frac14 \]
Step 3: Check the derivatives.
For \(x<1\), \(f'(x) = k(2x-6)\), so at \(x=1\) the left derivative is \(-4k\). For \(x\ge 1\) near 1, \(x-3<0\), so \(f(x)=3-x\) and \(f'(x)=-1\).
\[ -4k = -1 \Rightarrow k = \frac14 \]
Step 4: Compare.
Both conditions give the same \(k=\frac14\), so the function is differentiable at 1.
Step 5: Check the options.
Option 1 (\(4\)) and option 4 (\(-4\)) give \(8k=\pm32\), not 2. Option 3 (\(-\frac14\)) gives \(8k=-2\), not 2. Only option 2 works.
Final Answer:
The value of \(k\) is \(\frac14\), option 2.
\[ \boxed{\frac{1}{4}} \]
Question 26:
If \(y=x^x,\ x>0\) then \(y^{\prime\prime}(2)-2y^{\prime}(2)\) is equal to
(A) \(8\log_e 2-2\)
(B) \(4(\log_e 2)^2+2\)
(C) \(4(\log_e 2)^2-2\)
(D) \(4(\log_e 2)+2\)
Correct Answer: (C) \(4(\log_e 2)^2-2\)
View SolutionStep 1: Take logarithms.
Since \(y=x^x\), we get \(\ln y = x\ln x\). Differentiate both sides.
\[ \frac{y'}{y} = \ln x + 1 \Rightarrow y' = x^x(1+\ln x) \]
Step 2: Find the second derivative.
Use the product rule on \(y' = y(1+\ln x)\).
\[ y'' = y'(1+\ln x) + y\cdot\frac1x = x^x(1+\ln x)^2 + x^{x-1} \]
Step 3: Put x = 2.
Let \(L=\ln 2\). Then \(y=2^2=4\), so
\[ y'(2) = 4(1+L), \qquad y''(2) = 4(1+L)^2 + 2 \]
Step 5: Check the options.
The result is \(4(\log_e 2)^2-2\). This is option 3. Option 1 has no square term. Option 2 has \(+2\) instead of \(-2\). Option 4 has a plain \(\log_e 2\), which cancelled out.
Final Answer:
The value is \(4(\log_e 2)^2-2\), option 3.
\[ \boxed{4(\log_e 2)^2-2} \]
Question 27:
If \(x=\sec\theta+\cos\theta,\ y=\sec^n\theta+\cos^n\theta\), then \(\frac{dy}{dx}\) is equal to
Step 4: Express in x and y.
Now \(x^2 = \sec^2\theta+2+\cos^2\theta\), so \(x^2-4 = (\sec\theta-\cos\theta)^2\). In the same way \(y^2-4 = (\sec^n\theta-\cos^n\theta)^2\), because \(\sec^n\theta\cos^n\theta=1\).
\[ \frac{dy}{dx} = n\sqrt{\frac{y^2-4}{x^2-4}} \]
Step 5: Check the options.
Options 1 and 3 have \(+4\), which do not match the squares above. Option 2 has an extra square and no root. Option 4 matches.
Final Answer:
The derivative is \(n\sqrt{\frac{y^2-4}{x^2-4}}\), option 4.
\[ \boxed{n\sqrt{\dfrac{y^2-4}{x^2-4}}} \]
Question 28:
If \(f(x)=(x-a)^2+(x-b)^2+(x-c)^2\), then \(f(x)\) has minimum at \(x\) equal to:
(A) \(3\sqrt{abc}\)
(B) \(3abc\)
(C) \(\frac{a+b+c}{3}\)
(D) \(\frac{3}{\frac1a+\frac1b+\frac1c}\)
Correct Answer: (C) \(\frac{a+b+c}{3}\)
View SolutionStep 1: Find the critical point.
Differentiate \(f\).
\[ f'(x) = 2(x-a)+2(x-b)+2(x-c) = 2(3x-a-b-c) \]
Set \(f'(x)=0\). This gives \(x=\frac{a+b+c}{3}\).
Step 2: Test for a minimum.
\(f''(x) = 6 > 0\) for every \(x\). So the critical point is a minimum.
Step 3: Check the options.
Option 1 (\(3\sqrt{abc}\)) and option 2 (\(3abc\)) involve products, but \(f'\) only gives sums of \(a,b,c\). Option 4 is the harmonic mean, which does not solve \(f'(x)=0\). Option 3 is the arithmetic mean, and it does.
Final Answer:
The minimum is at \(x=\frac{a+b+c}{3}\), option 3.
\[ \boxed{\frac{a+b+c}{3}} \]
Question 29:
For the Function \(f(x)=\tan x-4x,\ x\in\left(0,\frac{\pi}{2}\right)\), which of the following statements are correct? A. \(f(x)\) is increasing in \(\left(0,\frac{\pi}{3}\right)\) B. \(f(x)\) is increasing in \(\left(\frac{\pi}{3},\frac{\pi}{2}\right)\) C. \(f(x)\) is decreasing in \(\left(0,\frac{\pi}{3}\right)\) D. \(f(x)\) is decreasing in \(\left(\frac{\pi}{3},\frac{\pi}{2}\right)\) Choose the correct answer from the options given below:
(A) A and B only
(B) A and D only
(C) B and C only
(D) B and D only
Correct Answer: (C) B and C only
View SolutionStep 1: Understanding the Concept:
A function is increasing where its derivative is positive and decreasing where its derivative is negative. So we find \(f'(x)\) and study its sign on \((0,\frac{\pi}{2})\).
Step 3: Find the turning point:
Put \(f'(x)=0\). Then \(\sec^2 x=4\), so \(\sec x=2\) (secant is positive here). This gives \(\cos x=\frac{1}{2}\), so \(x=\frac{\pi}{3}\).
Step 4: Sign of f'(x) on \((0,\frac{\pi}{3})\):
In this interval \(\cos x\) is larger than \(\frac{1}{2}\), so \(\sec x<2\) and \(\sec^2 x<4\). Hence \(f'(x)<0\) and \(f\) is decreasing. So C is true and A is false.
Step 5: Sign of f'(x) on \((\frac{\pi}{3},\frac{\pi}{2})\):
Here \(\cos x\) is smaller than \(\frac{1}{2}\), so \(\sec x>2\) and \(\sec^2 x>4\). Hence \(f'(x)>0\) and \(f\) is increasing. So B is true and D is false.
Step 6: Match with the options:
Correct statements are B and C. Option 1 (A and B) has the false statement A. Option 2 (A and D) has two false statements. Option 4 (B and D) has the false statement D. Option 3 (B and C only) is correct.
Final Answer:
\(f\) decreases on \((0,\frac{\pi}{3})\) and increases on \((\frac{\pi}{3},\frac{\pi}{2})\), so B and C are correct, which is option 3.
\[ \boxed{\text{B and C only}} \]
Question 30:
The difference between the greatest and least values of the function \(f(x)=\sin 2x-x\) on \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) is
(A) \(\frac{\pi}{2}\)
(B) \(\frac{-\sqrt{3}}{2}+\frac{\pi}{6}\)
(C) \(\pi\)
(D) \(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\)
Correct Answer: (C) \(\pi\)
View SolutionStep 1: Understanding the Concept:
On a closed interval, the greatest and least values of a continuous function occur either at critical points inside the interval or at the end points. We compare all these values.
Step 2: Find critical points:
\(f'(x)=2\cos 2x-1\). Setting it to zero gives \(\cos 2x=\frac12\). For \(x\in[-\frac{\pi}{2},\frac{\pi}{2}]\) we have \(2x\in[-\pi,\pi]\), so \(2x=\pm\frac{\pi}{3}\) and \(x=\pm\frac{\pi}{6}\).
Step 3: Value at the critical points:
\[ f\left(\frac{\pi}{6}\right)=\sin\frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt3}{2}-\frac{\pi}{6}\approx 0.342 \]
\[ f\left(-\frac{\pi}{6}\right)=-\frac{\sqrt3}{2}+\frac{\pi}{6}\approx -0.342 \]
Step 4: Value at the end points:
\[ f\left(\frac{\pi}{2}\right)=\sin\pi-\frac{\pi}{2}=-\frac{\pi}{2}\approx -1.571 \]
\[ f\left(-\frac{\pi}{2}\right)=\sin(-\pi)+\frac{\pi}{2}=\frac{\pi}{2}\approx 1.571 \]
Step 5: Greatest and least:
The greatest value is \(\frac{\pi}{2}\) (at \(x=-\frac{\pi}{2}\)). The least value is \(-\frac{\pi}{2}\) (at \(x=\frac{\pi}{2}\)).
Step 6: Difference:
\[ \frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi \]
Options 2 and 4 are only the values at the critical points, which are not the extremes. Option 1 is just the maximum. So option 3 is correct.
Final Answer:
The difference between the greatest and least values is \(\pi\), option 3.
\[ \boxed{\pi} \]
Question 31:
Match the LIST-I with LIST-II
LIST-I Indefinite Integral
LIST-II Solution (where \(c\) is an arbitrary constant)
A. \(\int\sqrt{16-x^2}\,dx\)
I. \(\sin^{-1}\left(\frac{x}{4}\right)+c\)
B. \(\int\frac{dx}{\sqrt{16-x^2}}\)
II. \(\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)+c\)
C. \(\int\frac{dx}{\sqrt{x^2-16}}\)
III. \(\frac{1}{8}\log\left|\frac{4+x}{4-x}\right|+c\)
D. \(\int\frac{dx}{16-x^2}\)
IV. \(\log\left|x+\sqrt{x^2-16}\right|+c\)
Choose the correct answer from the options given below:
(A) A-I, B-IV, C-III, D-II
(B) A-II, B-I, C-III, D-IV
(C) A-IV, B-II, C-III, D-I
(D) A-II, B-I, C-IV, D-III
Correct Answer: (D) A-II, B-I, C-IV, D-III
View SolutionStep 1: Understanding the Concept:
These are the standard integrals of the form \(\sqrt{a^2-x^2}\), \(\frac{1}{\sqrt{a^2-x^2}}\), \(\frac{1}{\sqrt{x^2-a^2}}\) and \(\frac{1}{a^2-x^2}\) with \(a=4\). We match each one with its known result.
Step 2: Match A:
\[ \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+c \]
With \(a=4\), \(\frac{a^2}{2}=8\). This is II, so A matches II.
Step 3: Match B:
\[ \int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+c \]
This is I, so B matches I.
Step 4: Match C:
\[ \int\frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+c \]
This is IV, so C matches IV.
Step 5: Match D:
\[ \int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+c \]
With \(a=4\), \(2a=8\). This is III, so D matches III.
Step 6: Pick the option:
We got A-II, B-I, C-IV, D-III. Option 1 has A-I which is wrong. Option 2 has C-III which is wrong. Option 3 has A-IV which is wrong. Only option 4 fits.
Final Answer:
The matching is A-II, B-I, C-IV, D-III, option 4.
\[ \boxed{\text{A-II, B-I, C-IV, D-III}} \]
Correct Answer: (B) 1
View SolutionStep 1: Understanding the Concept:
For a definite integral over \([-a,a]\), the property \(\int_{-a}^{a}f(x)dx=\int_{0}^{a}\left[f(x)+f(-x)\right]dx\) is useful.
Step 2: Set up:
Let \(f(x)=\frac{\cos x}{1+e^x}\). Then \(f(-x)=\frac{\cos x}{1+e^{-x}}=\frac{e^x\cos x}{e^x+1}\), because \(\cos(-x)=\cos x\).
Step 3: Add f(x) and f(-x):
\[ f(x)+f(-x)=\frac{\cos x+e^x\cos x}{1+e^x}=\frac{\cos x(1+e^x)}{1+e^x}=\cos x \]
Step 5: Check the options:
The answer is 1, so 0, -1 and 2 are wrong. Also the integrand is positive on the whole interval, so the answer cannot be 0 or -1.
Final Answer:
The value of the integral is 1, option 2.
\[ \boxed{1} \]
Question 33:
\(\int\frac{dx}{(x+1)\sqrt{x^2-1}}\) is equal to: [\(c\) is an arbitrary constant]
(A) \(\sqrt{\frac{x+1}{x-1}}+c\)
(B) \(\sqrt{\frac{x}{x-1}}+c\)
(C) \(\sqrt{\frac{x-1}{x+1}}+c\)
(D) \(\sqrt{\frac{x}{x+1}}+c\)
Correct Answer: (C) \(\sqrt{\frac{x-1}{x+1}}+c\)
View SolutionStep 1: Understanding the Concept:
We rewrite \(\sqrt{x^2-1}\) as \(\sqrt{(x-1)(x+1)}\) and then use a substitution that simplifies the ratio \(\frac{x-1}{x+1}\).
Step 2: Rewrite:
\[ \frac{1}{(x+1)\sqrt{(x-1)(x+1)}}=\frac{1}{(x+1)^{2}}\cdot\sqrt{\frac{x+1}{x-1}} \]
because \((x+1)\sqrt{x+1}\sqrt{x-1}=(x+1)^2\sqrt{\frac{x-1}{x+1}}\).
Step 3: Substitute:
Let \(t=\frac{x-1}{x+1}=1-\frac{2}{x+1}\). Then \(dt=\frac{2}{(x+1)^2}dx\), so \(\frac{dx}{(x+1)^2}=\frac{dt}{2}\).
Step 6: Check by differentiating:
\(\frac{d}{dx}\sqrt{\frac{x-1}{x+1}}=\frac{1}{2}\sqrt{\frac{x+1}{x-1}}\cdot\frac{2}{(x+1)^2}=\frac{1}{(x+1)^{3/2}(x-1)^{1/2}}\), which equals the integrand. Option 1 is the reciprocal of the correct ratio, so its derivative has the wrong sign. Options 2 and 4 do not give the integrand either.
Final Answer:
The integral equals \(\sqrt{\frac{x-1}{x+1}}+c\), option 3.
\[ \boxed{\sqrt{\frac{x-1}{x+1}}+c} \]
Question 34:
In the given figure, area of the shaded portion is equal to
(A) 3
(B) \(\frac{3\pi}{2}\)
(C) \(3\left(\frac{\pi}{2}-1\right)\)
(D) \(3(\pi-1)\)
Correct Answer: (C) \(3\left(\frac{\pi}{2}-1\right)\)
View SolutionStep 1: Understanding the Figure:
The shaded region lies between the line \(y=3\) on top and the curve \(y=3\cos x\) below. It runs from \(x=0\) to the line \(x=\frac{\pi}{2}\).
Step 2: Set up the area:
Area between two curves is the integral of (upper minus lower):
\[ A=\int_0^{\pi/2}\left(3-3\cos x\right)dx \]
Step 5: Check the options:
Option 1 is 3, which is the area under the curve, the unshaded part. Option 2 is \(\frac{3\pi}{2}\), the whole rectangle of height 3 and width \(\frac{\pi}{2}\), which also includes the unshaded part. Option 4 uses \(\pi\) instead of \(\frac{\pi}{2}\). So option 3 is right.
Final Answer:
The area is \(3\left(\frac{\pi}{2}-1\right)\) square units, option 3.
\[ \boxed{3\left(\frac{\pi}{2}-1\right)} \]
Question 35:
The area of the region bounded by the line \(y = 3x + 2\), the \(x\)-axis and the ordinates \(x = -1\) and \(x = 1\) is:
(A) \(\frac{25}{6}\)
(B) 4
(C) 0
(D) \(\frac{13}{3}\)
Correct Answer: (D) \(\frac{13}{3}\)
View SolutionStep 1: Understand the region.
The line \(y = 3x + 2\) meets the \(x\)-axis where \(3x + 2 = 0\), that is at \(x = -\frac{2}{3}\). This point lies between \(x = -1\) and \(x = 1\).
So the line is below the axis on \(\left[-1, -\frac{2}{3}\right]\) and above the axis on \(\left[-\frac{2}{3}, 1\right]\). Area is always positive, so we must split the integral at this point.
Step 2: Write the area as two integrals.
\[ A = \int_{-1}^{-2/3} -(3x+2)\,dx + \int_{-2/3}^{1} (3x+2)\,dx \]
Step 3: Evaluate the first integral.
An antiderivative of \(3x + 2\) is \(F(x) = \frac{3x^2}{2} + 2x\).
\(F\left(-\frac{2}{3}\right) = \frac{3}{2}\cdot\frac{4}{9} - \frac{4}{3} = \frac{2}{3} - \frac{4}{3} = -\frac{2}{3}\).
\(F(-1) = \frac{3}{2} - 2 = -\frac{1}{2}\).
First area \(= -\left[F\left(-\frac{2}{3}\right) - F(-1)\right] = -\left[-\frac{2}{3} + \frac{1}{2}\right] = \frac{1}{6}\).
Step 4: Evaluate the second integral.
\(F(1) = \frac{3}{2} + 2 = \frac{7}{2}\).
Second area \(= F(1) - F\left(-\frac{2}{3}\right) = \frac{7}{2} + \frac{2}{3} = \frac{25}{6}\).
Step 5: Add and check the options.
Total \(A = \frac{1}{6} + \frac{25}{6} = \frac{26}{6} = \frac{13}{3}\).
Option 1 counts only the part above the axis. Option 2 (4) is the plain integral without splitting, where the two parts partly cancel, so it is not the area. Option 3 (0) is also wrong. Option 4 matches our total.
Final Answer:
The area of the region is \(\frac{13}{3}\) square units, which is option 4.
\[ \boxed{\frac{13}{3}} \]
Question 36:
Match the LIST-I with LIST-II
LIST-I Differential Equations
LIST-II Solutions[ c is an arbitrary constant]
A. \(x\,dy = y\,dx\)
I. \(y = e^{x} + c\)
B. \(\log\left(\frac{dy}{dx}\right) = x\)
II. \(y = c\,e^{x}\)
C. \(\frac{y\,dx - dy}{y} = 0\)
III. \(xy = c\)
D. \(x\,dy + y\,dx = 0\)
IV. \(y = cx\)
Choose the correct answer from the options given below:
(A) A-IV, B-II, C-I, D-III
(B) A-IV, B-I, C-II, D-III
(C) A-III, B-II, C-I, D-IV
(D) A-III, B-I, C-II, D-IV
Correct Answer: (B) A-IV, B-I, C-II, D-III
View SolutionStep 1: Plan.
We solve each differential equation in List-I and match its general solution with an item in List-II.
Step 2: Solve A.
\(x\,dy = y\,dx\) gives \(\frac{dy}{y} = \frac{dx}{x}\). Integrating, \(\ln y = \ln x + \ln c\), so \(y = cx\). This is IV.
Step 3: Solve B.
\(\log\left(\frac{dy}{dx}\right) = x\) means \(\frac{dy}{dx} = e^{x}\). Integrating, \(y = e^{x} + c\). This is I.
Step 4: Solve C.
\(\frac{y\,dx - dy}{y} = 0\) means \(y\,dx = dy\), so \(\frac{dy}{y} = dx\). Integrating, \(\ln y = x + \ln c\), so \(y = c\,e^{x}\). This is II.
Step 5: Solve D.
\(x\,dy + y\,dx = 0\) is the same as \(d(xy) = 0\). So \(xy = c\). This is III.
Step 6: Compare the options.
We found A-IV, B-I, C-II, D-III. Option 1 has B and C wrongly swapped, so it is wrong. Option 3 has A-III and D-IV, which is wrong. Option 4 also has A-III and D-IV, so it is wrong. Only option 2 matches.
Final Answer:
The correct matching is A-IV, B-I, C-II, D-III, which is option 2.
Question 37:
The general solution of the differential equation of the type \(\frac{dy}{dx} + py = Q\) is: (where \(P\) and \(Q\) are functions of \(x\) only or constant)
(A) \(y\,e^{\int p\,dy} = \int \left(Q\,e^{\int p\,dy}\right) dy + c\) : (where \(c\) is an arbitrary constant)
(B) \(y\,e^{\int p\,dx} = \int \left(Q\,e^{\int p\,dx}\right) dx + c\) : (where \(c\) is an arbitrary constant)
(C) \(x\,e^{\int p\,dy} = \int \left(Q\,e^{\int p\,dy}\right) dy + c\) : (where \(c\) is an arbitrary constant)
(D) \(x\,e^{\int p\,dx} = \int \left(Q\,e^{\int p\,dx}\right) dx + c\) (where \(c\) is an arbitrary constant)
Correct Answer: (B) \(y\,e^{\int p\,dx} = \int \left(Q\,e^{\int p\,dx}\right) dx + c\) : (where \(c\) is an arbitrary constant)
View SolutionStep 1: Recognise the form.
The equation \(\frac{dy}{dx} + Py = Q\) is a first order linear equation. Here \(P\) and \(Q\) depend on \(x\) only, so \(x\) is the independent variable and \(y\) is the dependent one.
Step 2: Find the integrating factor.
The integrating factor is \(\text{I.F.} = e^{\int P\,dx}\). Multiplying the equation by it makes the left side a perfect derivative: \[ \frac{d}{dx}\left(y\,e^{\int P\,dx}\right) = Q\,e^{\int P\,dx} \]
Step 3: Integrate.
Integrating both sides with respect to \(x\): \[ y\,e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\,dx + c \]
Step 4: Check the options.
Options 1 and 3 integrate with respect to \(y\), which is wrong since \(P\) and \(Q\) depend on \(x\). Options 3 and 4 have \(x\) on the left, which is the form for \(\frac{dx}{dy} + Px = Q\) with \(P, Q\) functions of \(y\). Option 2 matches our result.
Final Answer:
The general solution is option 2.
Question 38:
Match the LIST-I with LIST-II
LIST-I
LIST-II
A. If \(|\vec{a}| = \sqrt{26}\), \(|\vec{b}| = 7\) and \(|\vec{a} \times \vec{b}| = 35\), then \(|\vec{a} \cdot \vec{b}|\) is equal to
I. 4
B. If \(2\hat{i} - 3\hat{j} + 4\hat{k}\) and \(a\hat{i} - 6\hat{j} + 8\hat{k}\) are collinear, then 'a' is equal to
II. 2
C. If \(2\hat{i} + \hat{j} + \hat{k}\) and \(2\hat{i} - 4\hat{j} + \lambda\hat{k}\) are perpendicular, then \(\lambda\) is equal to
III. 7
D. \((\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i}\) is equal to
IV. 0
Choose the correct answer from the options given below:
(A) A-III, B-II, C-IV, D-I
(B) A-III, B-I, C-IV, D-II
(C) A-II, B-I, C-III, D-IV
(D) A-I, B-IV, C-II, D-III
Correct Answer: (B) A-III, B-I, C-IV, D-II
View SolutionStep 1: Plan.
We evaluate each item of List-I using vector identities and then match it with List-II.
Step 2: Item A.
We use \(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2\).
\(35^2 + (\vec{a} \cdot \vec{b})^2 = 26 \times 49\), so \((\vec{a} \cdot \vec{b})^2 = 1274 - 1225 = 49\).
Thus \(|\vec{a} \cdot \vec{b}| = 7\). A matches III.
Step 3: Item B.
Collinear vectors have proportional components: \(\frac{a}{2} = \frac{-6}{-3} = \frac{8}{4} = 2\). So \(a = 4\). B matches I.
Step 4: Item C.
Perpendicular vectors have zero dot product: \(2(2) + 1(-4) + 1(\lambda) = 0\), so \(4 - 4 + \lambda = 0\) and \(\lambda = 0\). C matches IV.
Step 5: Item D.
\(\hat{i} \times \hat{j} = \hat{k}\), so \((\hat{i} \times \hat{j}) \cdot \hat{k} = 1\). Also \(\hat{j} \times \hat{k} = \hat{i}\), so \((\hat{j} \times \hat{k}) \cdot \hat{i} = 1\). The sum is 2. D matches II.
Step 6: Check the options.
The matching is A-III, B-I, C-IV, D-II. Option 1 has B-II and D-I, which is wrong. Option 3 has A-II, wrong. Option 4 has A-I, wrong. Only option 2 is right.
Final Answer:
The correct matching is A-III, B-I, C-IV, D-II, which is option 2.
Question 39:
If \(\vec{a} = x\hat{i} + x\hat{j} + y\hat{k}\), \(\vec{b} = \hat{i} + \hat{k}\) and \(\vec{c} = y\hat{i} + y\hat{j} + z\hat{k}\) are three vectors such that \(\vec{a} \times \vec{b} \perp \vec{c}\), then
(A) \(y^2 = xz\)
(B) \(x^2 = yz\)
(C) \(z^2 = xy\)
(D) \(xyz = 1\)
Correct Answer: (A) \(y^2 = xz\)
View SolutionStep 1: Idea.
Two vectors are perpendicular when their dot product is zero. So we need \((\vec{a} \times \vec{b}) \cdot \vec{c} = 0\).
Step 3: Take the dot product with c.
\[ x(y) + (y - x)(y) - x(z) = xy + y^2 - xy - xz = y^2 - xz \]
Setting this equal to zero gives \(y^2 = xz\).
Step 4: Check the options.
Option 1 is \(y^2 = xz\), which matches. Options 2, 3 and 4 do not follow from the condition, since the terms \(x^2\), \(z^2\) and \(xyz\) never appear in our result.
Final Answer:
The condition is \(y^2 = xz\), which is option 1.
\[ \boxed{y^2 = xz} \]
Question 40:
If \(\vec{a}\) is any vector, then the value of \(|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2\) is equal to
(A) \(2|\vec{a}|^2\)
(B) \(|\vec{a}|^2\)
(C) \(4|\vec{a}|^2\)
(D) \(3|\vec{a}|^2\)
Correct Answer: (A) \(2|\vec{a}|^2\)
View SolutionStep 1: Use the Lagrange identity.
For any two vectors, \(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2\).
Step 2: Apply it to each unit vector.
\(|\vec{a} \times \hat{i}|^2 = |\vec{a}|^2 - a_1^2\), \(|\vec{a} \times \hat{j}|^2 = |\vec{a}|^2 - a_2^2\), \(|\vec{a} \times \hat{k}|^2 = |\vec{a}|^2 - a_3^2\). Here \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and each unit vector has length 1.
Step 4: Check the options.
Option 2 would need the sum to be one copy of \(|\vec{a}|^2\), and option 4 forgets to subtract the dot product terms. Option 3 is too large. Option 1 is right.
Final Answer:
The value is \(2|\vec{a}|^2\), option 1.
\[ \boxed{2|\vec{a}|^2} \]
Question 41:
If \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three unit vectors such that \(\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0\) and angle between \(\vec{b}\) and \(\vec{c}\) is \(\frac{\pi}{6}\), then
Correct Answer: (D) \(\vec{a} = \pm 2\left(\vec{b} \times \vec{c}\right)\)
View SolutionStep 1: Direction of a.
Since \(\vec{a} \cdot \vec{b} = 0\) and \(\vec{a} \cdot \vec{c} = 0\), \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\). A vector perpendicular to both is parallel to \(\vec{b} \times \vec{c}\). So \(\vec{a} = \lambda(\vec{b} \times \vec{c})\) for some number \(\lambda\).
Step 2: Length of b x c.
\(|\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\frac{\pi}{6} = 1 \times 1 \times \frac{1}{2} = \frac{1}{2}\).
Step 3: Find lambda.
\(\vec{a}\) is a unit vector, so \(1 = |\lambda| \times \frac{1}{2}\), which gives \(|\lambda| = 2\). So \(\lambda = \pm 2\).
Step 4: Check the options.
Option 1 and option 2 would need \(|\vec{b} \times \vec{c}| = 1\), which is false here. Option 3 misses the negative direction, because \(\vec{a}\) could point either way. Option 4 gives both directions with the right size.
If \(\alpha, \beta, \gamma\) are angles of inclinations of a line with \(x, y\) and \(z\)-axis respectively, then which of the following are correct? A. \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\) B. \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma = -1\) C. \(\sin^2\alpha + \sin^2\beta + \sin^2\gamma = -2\) D. \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 2\) Choose the correct answer from the options given below:
(A) A and B only
(B) A, B and C only
(C) B and D only
(D) A and D only
Correct Answer: (A) A and B only
View SolutionStep 1: Understanding the Concept:
The direction cosines of a line are \(l=\cos\alpha\), \(m=\cos\beta\) and \(n=\cos\gamma\). Here \(\alpha, \beta, \gamma\) are the angles the line makes with the x, y and z axes.
Step 2: Key Formula or Approach:
For any line in space, the direction cosines satisfy \(l^2+m^2+n^2=1\). This is because \((l,m,n)\) is a unit vector along the line. We also use \(\cos 2\theta = 2\cos^2\theta - 1\) and \(\sin^2\theta = 1-\cos^2\theta\).
Step 3: Check statement (A).
By the direction cosine identity, \(\cos^2\alpha+\cos^2\beta+\cos^2\gamma = 1\). So (A) is TRUE.
Step 4: Check statement (B).
Use \(\cos 2\theta = 2\cos^2\theta-1\) on each angle and add: \[ \cos 2\alpha+\cos 2\beta+\cos 2\gamma = 2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma) - 3 = 2(1)-3 = -1 \] So (B) is TRUE.
Step 5: Check statement (C).
Use \(\sin^2\theta = 1-\cos^2\theta\): \[ \sin^2\alpha+\sin^2\beta+\sin^2\gamma = 3 - 1 = 2 \] The statement says \(-2\). A sum of squares can never be negative, so (C) is FALSE.
Step 6: Check statement (D).
The sum of the squares of the cosines is \(1\), not \(2\). So (D) is FALSE.
Step 7: Match with the options.
Only A and B are correct. This is option 1. Option 2 wrongly includes C. Options 3 and 4 wrongly include D.
Final Answer:
Statements A and B are correct, which is option 1.
\[ \boxed{\text{A and B only (Option 1)}} \]
Question 43:
The lines \(x = py + q,\ z = ry + s\) and \(x = p'y + q',\ z = r'y + s'\) are perpendicular if
(A) \(pp' + rr' = 0\)
(B) \(pp' + rr' + 1 = 0\)
(C) \(\dfrac{p}{p'} = \dfrac{r}{r'}\)
(D) \(pp' = rr'\)
Correct Answer: (B) \(pp' + rr' + 1 = 0\)
View SolutionStep 1: Understanding the Concept:
Two lines in space are perpendicular when their direction ratios have zero dot product. So we first need the direction ratios of each line.
Step 2: Write each line with a parameter.
For the first line, let \(y=t\). Then \(x=pt+q\), \(y=t\), \(z=rt+s\). The direction ratios are \((p, 1, r)\). In the same way, the second line has direction ratios \((p', 1, r')\).
Step 3: Apply the perpendicular condition.
The dot product of the direction ratios must be zero: \[ p\cdot p' + 1\cdot 1 + r\cdot r' = 0 \] \[ pp' + rr' + 1 = 0 \]
Step 4: Check the other options.
Option 1 leaves out the middle term \(1\times 1\), because it forgets that the y-component is \(1\), not \(0\). Option 3 is a proportionality relation, which is a parallel type condition, and it also ignores the y-ratio. Option 4 has no basis in the dot product.
Final Answer:
The lines are perpendicular when \(pp'+rr'+1=0\), which is option 2.
\[ \boxed{pp' + rr' + 1 = 0} \]
Question 44:
Two lines \(L_1\) and \(L_2\) are given as \(L_1: \vec{r} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(2\hat{i}+3\hat{j}+4\hat{k})\) and \(L_2: \vec{r} = (2\hat{i}+4\hat{j}+5\hat{k}) + \mu(4\hat{i}+6\hat{j}+8\hat{k})\), where \(\lambda\) and \(\mu\) are real parameters. Then which of the following statements are correct ? A. Shortest distance between line \(L_1\) and \(L_2\) is \(\sqrt{\dfrac{5}{29}}\). B. line \(L_1\) and \(L_2\) are parallel. C. line \(L_1\) and \(L_2\) are concurrent lines. D. line \(L_1\) and \(L_2\) are perpendicular. Choose the correct answer from the options given below:
(A) B and C only
(B) C and D only
(C) A and C only
(D) A and B only
Correct Answer: (D) A and B only
View SolutionStep 1: Understanding the Concept:
Two lines \(\vec r=\vec a_1+\lambda\vec b_1\) and \(\vec r=\vec a_2+\mu\vec b_2\) can be parallel, intersecting (concurrent) or skew. We test this using the direction vectors and the vector between the two given points.
Step 3: Check statement (B).
Since \(\vec b_2=2\vec b_1\), the direction vectors are proportional. So the lines are parallel. (B) is TRUE.
Step 4: Check statement (D).
The dot product is \(\vec b_1\cdot\vec b_2 = 8+18+32 = 58 \neq 0\). So the lines are not perpendicular. (D) is FALSE.
Step 5: Check statement (A).
For parallel lines the distance is \(\dfrac{|(\vec a_2-\vec a_1)\times \vec b_1|}{|\vec b_1|}\). \(\vec a_2-\vec a_1=(1,2,2)\). \[ (1,2,2)\times(2,3,4) = (2\cdot4-2\cdot3,\ 2\cdot2-1\cdot4,\ 1\cdot3-2\cdot2) = (2,0,-1) \] Its magnitude is \(\sqrt{4+0+1}=\sqrt5\). Also \(|\vec b_1|=\sqrt{4+9+16}=\sqrt{29}\). \[ d=\frac{\sqrt5}{\sqrt{29}}=\sqrt{\frac{5}{29}} \] So (A) is TRUE.
Step 6: Check statement (C).
Concurrent lines meet at a point. Our lines are parallel and the distance between them is not zero, so they never meet. (C) is FALSE.
Step 7: Match with the options.
A and B are correct. That is option 4. Options 1, 2 and 3 all include C, which is false.
Final Answer:
Statements A and B are correct, which is option 4.
\[ \boxed{\text{A and B only (Option 4)}} \]
Question 45:
The objective function of a LPP is given by \(z = ax + by\), where \(a\) and \(b\) are constants. If minimum of \(z\) occurs at two points \((50, 30)\) and \((20, 40)\), then
(A) \(3b = a\)
(B) \(3b + a = 0\)
(C) \(3a + b = 0\)
(D) \(3a - b = 0\)
Correct Answer: (D) \(3a - b = 0\)
View SolutionStep 1: Understanding the Concept:
In a linear programming problem, the objective function \(z=ax+by\) can reach its minimum at two corner points only when both points give the same value of \(z\). Then every point on the segment joining them gives that same minimum.
Step 2: Key Formula or Approach:
Put the two points into \(z\) and set the two values equal.
Step 3: Substitute the points.
At \((50,30)\): \(z=50a+30b\). At \((20,40)\): \(z=20a+40b\).
Step 4: Equate and simplify.
\[ 50a+30b=20a+40b \] \[ 30a = 10b \] \[ 3a = b \quad\Rightarrow\quad 3a-b=0 \]
Step 5: Check the options.
Option 1 says \(a=3b\), which is the reverse ratio. Option 2 says \(a=-3b\), which has the wrong sign. Option 3 says \(b=-3a\), also the wrong sign. Only \(3a-b=0\) matches.
Final Answer:
The condition is \(3a-b=0\), which is option 4.
\[ \boxed{3a - b = 0} \]
Question 46:
In the linear programming problem: Minimize \(z = x + y\), subject to the constraints: \(x + 3y \leq 60,\ x + y \geq 10,\ x \leq y,\ x \geq 0,\ y \geq 0\) Which of the following region in graph given below represents the feasible region of the above LPP?
(A) A
(B) B
(C) C
(D) D
Correct Answer: (B) B
View SolutionStep 1: Understanding the Concept:
The feasible region is the set of points that satisfy all the constraints at once. We test one point from each labelled region against every constraint.
Step 2: Key Formula or Approach:
The constraints are \(x+3y\leq 60\), \(x+y\geq 10\), \(x\leq y\), \(x\geq0\), \(y\geq0\). The line \(y=x\) passes through the origin. The condition \(x\leq y\) keeps points on or above it.
Step 3: Test region A.
A lies near the x-axis, below the line \(y=x\). Take the point \((20,5)\). Here \(x\leq y\) reads \(20\leq 5\), which is false. So A is not feasible.
Step 4: Test region B.
B is the hatched region near the y-axis, above the line \(x+y=10\), below \(x+3y=60\) and above \(y=x\). Take \((2,12)\): \(2+36=38\leq 60\) is true, \(2+12=14\geq 10\) is true, and \(2\leq 12\) is true. So B satisfies every constraint.
Step 5: Test region C.
C lies above the line \(x+3y=60\). Take \((10,20)\): \(10+60=70\leq 60\) is false. So C is not feasible.
Step 6: Test region D.
D lies below the line \(y=x\), to the right. Take \((30,10)\): \(x\leq y\) reads \(30\leq10\), which is false. So D is not feasible.
Step 7: Corner points of B.
The corners are \((0,10)\), \((5,5)\), \((15,15)\) and \((0,20)\). These match the shape of region B in the graph.
Final Answer:
Only region B satisfies all the constraints, which is option 2.
\[ \boxed{\text{Region B (Option 2)}} \]
Question 47:
A speaks truth in 80% cases and B in 90% cases. The probability that they contradict each other in stating the same fact, is:
(A) \(\dfrac{17}{50}\)
(B) \(\dfrac{13}{50}\)
(C) \(\dfrac{19}{50}\)
(D) \(\dfrac{11}{50}\)
Correct Answer: (B) \(\dfrac{13}{50}\)
View SolutionStep 1: Understanding the Concept:
A and B contradict each other when exactly one of them speaks the truth. That can happen in two separate ways, and A and B act independently.
Step 2: Key Formula or Approach:
\(P(\text{A truth})=0.8=\dfrac45\), so \(P(\text{A lies})=\dfrac15\). \(P(\text{B truth})=0.9=\dfrac9{10}\), so \(P(\text{B lies})=\dfrac1{10}\).
Step 3: Case 1: A tells the truth and B lies.
\[ \frac45\times\frac1{10}=\frac{4}{50} \]
Step 4: Case 2: A lies and B tells the truth.
\[ \frac15\times\frac9{10}=\frac{9}{50} \]
Step 5: Add the two cases.
The cases cannot happen together, so we add: \[ \frac4{50}+\frac9{50}=\frac{13}{50} \]
Step 6: Check the options.
\(17/50\), \(19/50\) and \(11/50\) do not match the sum of the two cases. Only \(13/50\) is right.
Final Answer:
The probability of contradiction is \(\dfrac{13}{50}\), option 2.
\[ \boxed{\dfrac{13}{50}} \]
Question 48:
If A and B are two events such that \(P(A)=\frac{1}{2}\), \(P(B)=\frac{1}{3}\) and \(P(A \cap B)=\frac{1}{4}\), then \(P\left(\frac{A'}{B}\right)=\)
(A) \(\frac{3}{4}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{5}{8}\)
Correct Answer: (B) \(\frac{1}{4}\)
View SolutionStep 1: Understand what is asked:
We need the probability that A does not happen, given that B has happened. This is a conditional probability, written \(P(A'|B)\).
A' means the complement of A, that is, the event that A does not occur.
Step 2: Recall the formula:
For a conditional probability we use \[ P(A'|B)=\frac{P(A' \cap B)}{P(B)} \]
The event B can be split into two parts that do not overlap: the part inside A and the part outside A. So \[ P(B)=P(A \cap B)+P(A' \cap B) \]
This gives \(P(A' \cap B)=P(B)-P(A \cap B)\).
Step 3: Find the numerator:
Put in the given values: \[ P(A' \cap B)=\frac{1}{3}-\frac{1}{4}=\frac{4-3}{12}=\frac{1}{12} \]
Step 4: Divide by P(B):
\[ P(A'|B)=\frac{1/12}{1/3}=\frac{1}{12}\times 3=\frac{1}{4} \]
Step 5: Check the options:
Option 1 (\(\frac{3}{4}\)) is actually \(P(A|B)=\frac{1/4}{1/3}\), which is the complement of our answer, so it is wrong.
Option 3 (\(\frac{1}{2}\)) is just \(P(A)\), and option 4 (\(\frac{5}{8}\)) does not come from any correct step. Only option 2 matches.
Final Answer:
\(P(A'|B)=\frac{1}{4}\), which is option 2.
\[ \boxed{\frac{1}{4}} \]
Question 49:
The probability of the simultaneous occurrence of two events A and B is \(p\). If the probability that exactly one of A and B occurs is \(q\), then A. \(P(\overline{A})+P(\overline{B})=2+2q-p\) B. \(P(\overline{A})+P(\overline{B})=2-2p-q\) C. \(P(\overline{A}\cap\overline{B})=1-p-q\) D. \(P((A\cap B)|(A\cup B))=\frac{p}{p+q}\)
Choose the correct answer from the options given below:
(A) A, C and D only
(B) B, C and D only
(C) C and D only
(D) B only
Correct Answer: (B) B, C and D only
View SolutionStep 1: Write down what is given:
\(P(A\cap B)=p\). The event 'exactly one of A and B occurs' is \((A\cap B')\cup(A'\cap B)\), and its probability is \(q\).
So \[ q=P(A)+P(B)-2p \] because \(P(A\cap B')=P(A)-p\) and \(P(A'\cap B)=P(B)-p\).
From this, \(P(A)+P(B)=q+2p\).
Step 2: Check statement A:
\[ P(\overline{A})+P(\overline{B})=2-\big(P(A)+P(B)\big)=2-(q+2p)=2-2p-q \]
Statement A says \(2+2q-p\), which is not equal to this in general. So A is FALSE.
Step 3: Check statement B:
From the line above, \(P(\overline{A})+P(\overline{B})=2-2p-q\). This is exactly what statement B says. So B is TRUE.
Step 4: Check statement C:
By De Morgan's law, \(\overline{A}\cap\overline{B}=\overline{A\cup B}\).
Now \(P(A\cup B)=P(A)+P(B)-p=(q+2p)-p=p+q\).
So \(P(\overline{A}\cap\overline{B})=1-(p+q)=1-p-q\). Statement C is TRUE.
Step 5: Check statement D:
\[ P\big((A\cap B)|(A\cup B)\big)=\frac{P\big((A\cap B)\cap(A\cup B)\big)}{P(A\cup B)} \]
Since \(A\cap B\) lies inside \(A\cup B\), the numerator is just \(P(A\cap B)=p\). The denominator is \(p+q\). So the value is \(\frac{p}{p+q}\). Statement D is TRUE.
Step 6: Match with the options:
B, C and D are true and A is false. Option 1 wrongly includes A. Option 3 leaves out B. Option 4 has only B. So option 2 is correct.
Final Answer:
B, C and D are correct. This is option 2.
\[ \boxed{\text{B, C and D only}} \]
Question 50:
A letter is known to have come for 'RAVINDRA' or 'DEVINDRAN'. On the envelop just two consecutive letters 'RA' are visible. The probability that the letter has come for 'RAVINDRA' is:
(A) \(\frac{16}{23}\)
(B) \(\frac{7}{11}\)
(C) \(\frac{4}{11}\)
(D) \(\frac{7}{23}\)
Correct Answer: (A) \(\frac{16}{23}\)
View SolutionStep 1: Name the events:
Let \(E_1\) be the event that the letter is for 'RAVINDRA' and \(E_2\) the event that it is for 'DEVINDRAN'. The letter came for one of the two names with no other information, so \(P(E_1)=P(E_2)=\frac{1}{2}\).
Let \(E\) be the event that two consecutive visible letters on the envelope are 'RA'. We need \(P(E_1|E)\), which is a job for Bayes' theorem.
Step 2: Count the pairs of consecutive letters:
'RAVINDRA' has 8 letters R-A-V-I-N-D-R-A, so it has 7 pairs of consecutive letters: RA, AV, VI, IN, ND, DR, RA. Two of these pairs are 'RA'. So \(P(E|E_1)=\frac{2}{7}\).
'DEVINDRAN' has 9 letters D-E-V-I-N-D-R-A-N, so it has 8 pairs: DE, EV, VI, IN, ND, DR, RA, AN. Only one pair is 'RA'. So \(P(E|E_2)=\frac{1}{8}\).
Step 4: Simplify:
The common factor \(\frac{1}{2}\) cancels: \[ \frac{\frac{2}{7}}{\frac{2}{7}+\frac{1}{8}}=\frac{\frac{16}{56}}{\frac{16}{56}+\frac{7}{56}}=\frac{16}{23} \]
Step 5: Why the other options fail:
\(\frac{7}{23}\) is the probability of the other name, 'DEVINDRAN'. It is the complement of our answer. Options \(\frac{7}{11}\) and \(\frac{4}{11}\) come from wrongly counting pairs. Only option 1 fits.
Final Answer:
The probability is \(\frac{16}{23}\), which is option 1.
\[ \boxed{\frac{16}{23}} \]
Question 51:
The effective rate that is equivalent to a nominal rate of 6% per annum compounded semi-annually is
(A) 6%
(B) 6.09%
(C) 0.069%
(D) 9.06%
Correct Answer: (B) 6.09%
View SolutionStep 1: Understand the terms:
The nominal rate is the yearly rate as quoted, here 6% per year. It is compounded semi-annually, so interest is added twice a year.
The effective rate is the single yearly rate which gives the same final amount as this compounding.
Step 2: Recall the formula:
If the nominal rate is \(r\) per year and interest is compounded \(m\) times a year, then \[ \text{Effective rate}=\left(1+\frac{r}{m}\right)^{m}-1 \]
Here \(r=0.06\) and \(m=2\).
Step 3: Calculate:
The rate for each half-year is \(\frac{0.06}{2}=0.03\).
\[ (1.03)^2-1=1.0609-1=0.0609 \]
As a percentage, this is 6.09%.
Step 4: Check the options:
Option 1 (6%) would be right only for yearly compounding. Option 3 (0.069%) has a wrong decimal shift and is far too small. Option 4 (9.06%) is too large, since compounding twice cannot lift 6% by 3 points. Only 6.09% fits.
Final Answer:
The effective rate is 6.09% per annum, which is option 2.
\[ \boxed{6.09\%} \]
Question 52:
Assume an investment's starting value is Rs. 10,000 and it grows to Rs. 60,000 in 4 years, then the CAGR % of the investment is, [given : \((6)^{1/4}=1.565\)]
(A) 65%
(B) 56.5%
(C) 65.5%
(D) 55.6%
Correct Answer: (B) 56.5%
View SolutionStep 1: Understand CAGR:
CAGR stands for compound annual growth rate. It is the single yearly growth rate at which the starting value would grow to the final value over the given years.
Step 2: Recall the formula:
\[ \text{CAGR}=\left(\frac{\text{Final value}}{\text{Initial value}}\right)^{1/n}-1 \]
where \(n\) is the number of years.
Step 3: Put in the numbers:
Here the initial value is 10,000, the final value is 60,000 and \(n=4\).
\[ \frac{60000}{10000}=6 \]
\[ \text{CAGR}=6^{1/4}-1 \]
Step 4: Use the given value:
The question gives \(6^{1/4}=1.565\). So \[ \text{CAGR}=1.565-1=0.565 \]
In percent, this is \(0.565\times100=56.5\%\).
Step 5: Check the options:
Options 1 (65%) and 3 (65.5%) come from mixing up the digits. Option 4 (55.6%) is a digit swap too. Only 56.5% matches 0.565.
Final Answer:
The CAGR is 56.5%, which is option 2.
\[ \boxed{56.5\%} \]
Question 53:
Vishal purchased a machine costing Rs. 50,000. The estimated useful life of the machine is 10 years with a scrap value of Rs. 5,000. The annual depreciation using linear method is,
(A) Rs. 45,000
(B) Rs. 4,500
(C) Rs. 3500
(D) Rs. 2500
Correct Answer: (B) Rs. 4,500
View SolutionStep 1: Understand the method:
In the linear (straight line) method the machine loses the same amount of value every year. Only the part of the cost which will be lost, that is cost minus scrap value, is spread over its life.
Step 4: Check the options:
Option 1 (45,000) is the total depreciable amount, not the yearly one. Options 3 and 4 do not come from any correct step. If we forgot the scrap value we would get \(\frac{50000}{10}=5000\), which is also not listed.
Final Answer:
The annual depreciation is Rs. 4,500, which is option 2.
\[ \boxed{\text{Rs. }4500} \]
Question 54:
Vatsala takes a loan of Rs. 2,00,000 at an interest rate of 10% per annum for 5 years. Her EMI using flat rate method is,
(A) Rs. 20,000
(B) Rs. 1,666.6
(C) Rs. 5,000
(D) Rs. 15,000
Correct Answer: (C) Rs. 5,000
View SolutionStep 1: Understand the flat rate method:
In the flat rate method, interest is always worked out on the full original loan, for the whole time. It does not reduce as the loan is paid off.
EMI means equated monthly instalment, so the loan and interest are repaid in equal monthly parts.
Step 2: Recall the formula:
\[ \text{EMI}=\frac{P+P\cdot r\cdot t}{12t} \]
where \(P\) is the loan, \(r\) the yearly rate and \(t\) the years.
Step 3: Find the total interest:
\[ I=2{,}00{,}000\times0.10\times5=1{,}00{,}000 \]
Step 4: Find the total amount to repay:
\[ 2{,}00{,}000+1{,}00{,}000=3{,}00{,}000 \]
Step 5: Divide by the number of months:
There are \(5\times12=60\) months.
\[ \text{EMI}=\frac{3{,}00{,}000}{60}=5{,}000 \]
Step 6: Check the options:
Rs. 20,000 is the yearly interest, Rs. 1,666.6 is the monthly interest \(\frac{1{,}00{,}000}{60}\) alone, and Rs. 15,000 does not follow from the formula. Only Rs. 5,000 fits.
Final Answer:
The EMI is Rs. 5,000, which is option 3.
\[ \boxed{\text{Rs. }5000} \]
Question 55:
Face value of a bond is
(A) the rate at which a bond yields interest
(B) the price at which the bond is sold to buyers at the time of issue
(C) the annual interest rate paid by the bond issuer to the bond holder
(D) the discount rate which returns the market price of a bond without embedded optionality
Correct Answer: (B) the price at which the bond is sold to buyers at the time of issue
View SolutionStep 1: Understanding the Concept:
A bond is a loan certificate. The buyer lends money to the issuer, and the issuer promises to pay interest and to return the principal on the maturity date. Every bond carries a printed amount on it, called its face value (also called par value).
Step 2: Key Formula or Approach:
The face value is the base amount on which the issuer calculates the interest, and it is the amount the issuer pays back at maturity. It is a fixed amount, not a percentage. Here, the closest description is the value at which the bond is first offered to buyers at issue.
Step 3: Check option (1).
The rate at which a bond yields interest is the coupon rate or the yield. It is a percentage, not a money value. So option (1) is wrong.
Step 4: Check option (2).
A bond is issued and sold at its stated (par) value. This stated value is the face value. So option (2) is the best match and is correct.
Step 5: Check option (3).
The annual interest rate paid by the issuer to the holder is the coupon rate. The coupon rate is applied on the face value, so it is a different thing. Option (3) is wrong.
Step 6: Check option (4).
The discount rate that returns the market price of a bond is the yield to maturity. It changes with the market, but the face value does not change. Option (4) is wrong.
Final Answer:
Face value is the stated value at which the bond is sold to buyers at issue. This is option (2).
Question 56:
At what rate of interest per annum will the present value of a perpetuity of Rs. 500 payable at the end of every 6 months be Rs. 10,000?
(A) 15%
(B) 10%
(C) 20%
(D) 25%
Correct Answer: (B) 10%
View SolutionStep 1: Understanding the Concept:
A perpetuity is a payment that continues forever. Its present value is the payment divided by the interest rate for one payment period. The payment here is made every 6 months, so the rate must be the rate for 6 months.
Step 2: Key Formula or Approach:
\(PV = \frac{C}{i}\), where \(C\) is the payment each period and \(i\) is the interest rate per period.
Step 3: Detailed Explanation:
Here \(C = 500\) and \(PV = 10000\). The period is 6 months.
\[ i = \frac{500}{10000} = 0.05 = 5\% \text{ per half year} \]
There are two half years in one year, so the annual rate is:
\[ 2 \times 5\% = 10\% \text{ per annum} \]
Step 4: Check the other options.
At 15%, the half yearly rate is 7.5% and the present value is 500/0.075, which is about 6,667. At 20%, it is 500/0.10 = 5,000. At 25%, it is 500/0.125 = 4,000. None of these equals 10,000, so they are wrong.
Final Answer:
The annual rate of interest is 10%.
\[ \boxed{10\%} \]
Question 57:
Let \(X\) denote the number of hours a person drive during a randomly selected working day. The probability that \(X\) can take the value \(x_i\) has the following form, where \(k\) is some unknown constant. \[ P(X = x_i) = \begin{cases} 0.2, & \text{if } x_i = 0 \\ kx_i, & \text{if } x_i = 1 \text{ or } 2 \\ k(4 - x_i), & \text{if } x_i = 3 \\ 0, & \text{otherwise} \end{cases} \] Match the LIST-I with LIST-II
LIST-I
LIST-II
A. The value of \(k\)
I. 1
B. The probability that the person drive atleast two hours on a selected working day
II. 0.4
C. The probability that the person drive atmost three hours on a selected working day
III. 0.2
D. The probability that the person drive atmost one hour on a selected working day
IV. 0.6
Choose the correct answer from the options given below:
(A) A-III, B-II, C-I, D-IV
(B) A-III, B-IV, C-I, D-II
(C) A-IV, B-II, C-III, D-I
(D) A-IV, B-III, C-II, D-I
Correct Answer: (B) A-III, B-IV, C-I, D-II
View SolutionStep 1: Understanding the Concept:
A probability distribution lists every possible value of X with its probability. The probabilities of all values must add up to 1. We use this to find the unknown constant k, then we read the other probabilities from the table.
Step 2: List the probabilities.
For \(x_i=0\): \(P=0.2\).
For \(x_i=1\): \(P=k(1)=k\).
For \(x_i=2\): \(P=k(2)=2k\).
For \(x_i=3\): \(P=k(4-3)=k\).
Step 3: Find k (match A).
The total probability is 1: \[ 0.2 + k + 2k + k = 1 \] \[ 4k = 0.8 \] \[ k = 0.2 \] So A matches III (0.2).
Step 4: Match B.
At least two hours means \(X \geq 2\): \[ P(X \geq 2) = P(2) + P(3) = 2k + k = 3k = 0.6 \] So B matches IV (0.6).
Step 5: Match C.
At most three hours means every possible value 0, 1, 2 and 3, because X cannot take any other value. So the probability is the total, which is 1. So C matches I (1).
Step 6: Match D.
At most one hour means X is 0 or 1: \[ P(0) + P(1) = 0.2 + k = 0.2 + 0.2 = 0.4 \] So D matches II (0.4).
Step 7: Choose the option.
We have A-III, B-IV, C-I, D-II. This is option 2. Option 1 has B-II and D-IV, which is swapped. Options 3 and 4 give A-IV, but k is 0.2, not 0.6, so they are wrong.
Final Answer:
The matching is A-III, B-IV, C-I, D-II.
\[ \boxed{\text{Option (2)}} \]
Question 58:
Match the LIST-I with LIST-II
LIST-I
LIST-II
A. In a binomial distribution, if n = 20, q = 0.75 then its mean is
I. 25
B. An unbiased coin is tossed 4 times. The mean of the number of tails is
II. 5
C. In a binomial distribution, mean is 5 and variance is 4, then the number of trials is
III. 0.5
D. In a binomial distribution, if mean is 8 and variance is 4, then the probability of success is
IV. 2
Choose the correct answer from the options given below:
(A) A-III, B-IV, C-I, D-II
(B) A-II, B-I, C-IV, D-III
(C) A-II, B-IV, C-I, D-III
(D) A-II, B-I, C-III, D-IV
Correct Answer: (C) A-II, B-IV, C-I, D-III
View SolutionStep 1: Understanding the Concept:
In a binomial distribution with n trials and success probability p, the mean is np and the variance is npq, where q = 1 - p.
Step 2: Match A.
Given \(n=20\) and \(q=0.75\), so \(p=1-0.75=0.25\).
\[ \text{Mean} = np = 20 \times 0.25 = 5 \] So A matches II.
Step 3: Match B.
An unbiased coin has p = 0.5 for tails. For 4 tosses: \[ np = 4 \times 0.5 = 2 \] So B matches IV.
Step 4: Match C.
Mean np = 5 and variance npq = 4. Divide the variance by the mean: \[ q = \frac{4}{5} = 0.8 \] so \(p = 0.2\). Then \[ n = \frac{5}{0.2} = 25 \] So C matches I.
Step 5: Match D.
Mean np = 8 and variance npq = 4. Divide: \[ q = \frac{4}{8} = 0.5 \] so \(p = 1 - 0.5 = 0.5\). So D matches III.
Step 6: Choose the option.
The pairs are A-II, B-IV, C-I, D-III. This is option 3. Options 1, 2 and 4 each have at least one wrong pair, for example A-III in option 1 (mean 0.5 is not 5) and B-I in options 2 and 4 (25 is not the mean of 4 tosses).
Final Answer:
The matching is A-II, B-IV, C-I, D-III.
\[ \boxed{\text{Option (3)}} \]
Question 59:
If \(X\) is a Poisson variate such that \(4P(X = 2) = 3P(X = 1)\), then the mean of the distribution is equal to:
(A) 1.5
(B) 2
(C) 3
(D) 4
Correct Answer: (A) 1.5
View SolutionStep 1: Understanding the Concept:
For a Poisson variate with mean \(\lambda\), the probability of r events is \(P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!}\). The mean of the distribution is \(\lambda\).
Step 3: Use the given condition.
\[ 4 \cdot \frac{e^{-\lambda}\lambda^2}{2} = 3 e^{-\lambda}\lambda \] Cancel \(e^{-\lambda}\) and one \(\lambda\) (the mean cannot be 0): \[ 2\lambda = 3 \] \[ \lambda = 1.5 \]
Step 4: Check the options.
The mean equals lambda = 1.5, which is option 1. For lambda = 2, the left side gives 4P(2) = 4 e^-2 x 2 = 8 e^-2 while the right side gives 3P(1) = 6 e^-2, so option 2 fails. Options 3 and 4 fail in the same way.
Final Answer:
The mean of the distribution is 1.5.
\[ \boxed{1.5} \]
Question 60:
In a math aptitude test, students scores are found to be normally distributed having mean as 45 and standard deviation 5. What percent of students scored more than the mean score?
(A) 45%
(B) 50%
(C) 5%
(D) 60%
Correct Answer: (B) 50%
View SolutionStep 1: Understanding the Concept:
A normal distribution is symmetric about its mean. The bell shaped curve has the same area on both sides of the mean. The total area under the curve is 1, or 100%.
Step 2: Apply the symmetry.
The mean score is 45. The area to the left of 45 equals the area to the right of 45. So each area is half of 100%, that is 50%.
Step 3: Check with the z score.
For score 45, \(z = \frac{45-45}{5} = 0\). So \(P(X>45)=P(Z>0)=0.5\), which is 50%.
Step 4: Check the other options.
45%, 5% and 60% cannot be right, because the mean is the exact centre of the curve. The standard deviation 5 does not change this answer.
Final Answer:
50% of the students scored more than the mean.
\[ \boxed{50\%} \]
Question 61:
If \(y=e^{4x}(92)^x\); then \(\frac{dy}{dx}\) is equal to (\(\log_e x=\log x\))
(A) \(4x\,e^{4x}\cdot 92^{x-1}\)
(B) \(e^{4x}92^{x-1}[368+x]\)
(C) \(e^{4x}(92)^x[4+\log 92]\)
(D) \(e^{4x}92^x\log 92\)
Correct Answer: (C) \(e^{4x}(92)^x[4+\log 92]\)
View SolutionStep 1: Understand the function.
The function \(y=e^{4x}\cdot 92^{x}\) is a product of two exponential terms. The variable \(x\) sits in the exponent in both, so the ordinary power rule \(\frac{d}{dx}x^n=nx^{n-1}\) does not apply.
Step 2: Recall the formulas.
1. \(\frac{d}{dx}e^{kx}=k\,e^{kx}\).
2. \(\frac{d}{dx}a^{x}=a^{x}\log a\) for a constant \(a>0\).
3. Product rule: \((uv)'=u'v+uv'\).
Step 3: Differentiate the product.
Let \(u=e^{4x}\) and \(v=92^x\). Then \(u'=4e^{4x}\) and \(v'=92^x\log 92\).
\[ \frac{dy}{dx}=4e^{4x}\cdot 92^x+e^{4x}\cdot 92^x\log 92 \]
Step 4: Take out the common factor.
Both terms contain \(e^{4x}92^x\).
\[ \frac{dy}{dx}=e^{4x}(92)^x\,[\,4+\log 92\,] \]
Step 5: Check option 1.
Option 1 treats \(92^x\) like \(x^{92}\) and brings down a factor \(x\). That is the wrong rule, so it is wrong.
Step 6: Check option 2.
Option 2 also uses \(92^{x-1}\), which comes from the power rule again. It is wrong.
Step 7: Check option 4.
Option 4 differentiates only \(92^x\) and forgets the derivative of \(e^{4x}\). The term with 4 is missing, so it is wrong. Option 3 has both terms.
Final Answer:
The derivative is \(e^{4x}(92)^x[4+\log 92]\), which is option 3.
\[ \boxed{e^{4x}(92)^x\,[4+\log 92]} \]
Question 62:
The order and degree of the following differential equation: \(y=px+\sqrt{a^2p^2+b^2}\), where \(p=\frac{dy}{dx}\), are
(A) Order = 2 : degree = 2
(B) Order = 1 : degree = 1
(C) Order = 1 : degree = 2
(D) Order = 2 : degree = 1
Correct Answer: (C) Order = 1 : degree = 2
View SolutionStep 1: Recall the definitions.
The order of a differential equation is the order of the highest derivative in it. The degree is the power of that highest derivative after the equation is written as a polynomial in the derivatives (free of roots and fractions of derivatives).
Step 2: Find the order.
Here \(p=\frac{dy}{dx}\), so only the first derivative appears. No \(\frac{d^2y}{dx^2}\) is present. The order is 1.
Step 3: Remove the square root.
Move \(px\) to the left side and square both sides.
\[ y-px=\sqrt{a^2p^2+b^2} \]
\[ (y-px)^2=a^2p^2+b^2 \]
Step 4: Write it as a polynomial in p.
Expand the left side and collect the powers of \(p\).
\[ y^2-2pxy+p^2x^2=a^2p^2+b^2 \]
\[ (x^2-a^2)p^2-2xy\,p+(y^2-b^2)=0 \]
The highest power of \(p=\frac{dy}{dx}\) is 2, and the coefficient \(x^2-a^2\) is not zero in general. So the degree is 2.
Step 5: Check the options.
Degree 1 (options 2 and 4) would be true only if the square root could be removed without raising a power of \(p\), which is not the case. Order 2 (options 1 and 4) is wrong because no second derivative appears. Only order 1 and degree 2 fits.
Final Answer:
The order is 1 and the degree is 2, which is option 3.
\[ \boxed{\text{Order}=1,\ \text{Degree}=2} \]
Question 63:
The demand function for a certain product is represented by the equation \(p=150+12x-x^2\), where \(x\) is the number of units demanded and \(p\) is the price per unit. The marginal revenue when 5 units are sold is
(A) Rs. 195
(B) Rs. 282
(C) Rs. 91
(D) Rs. 185
Correct Answer: (A) Rs. 195
View SolutionStep 1: Understand the terms.
Revenue \(R\) is the money received from selling \(x\) units, so \(R=p\cdot x\). Marginal revenue (MR) is the rate of change of revenue with the number of units, so \(MR=\frac{dR}{dx}\).
Step 2: Write the revenue function.
Multiply the price by \(x\).
\[ R=x(150+12x-x^2)=150x+12x^2-x^3 \]
Step 4: Put x = 5.
\[ MR=150+24(5)-3(25)=150+120-75=195 \]
Step 5: Check the other options.
Option 2 (282) and option 3 (91) do not come from \(150+24x-3x^2\) at \(x=5\). Option 4 (185) is the price per unit at \(x=5\), not the marginal revenue. Option 1 is correct.
Final Answer:
The marginal revenue at 5 units is Rs. 195, option 1.
\[ \boxed{195} \]
Question 64:
The function \(f(x)=4x^3-18x^2+27x-7\) has:-
(A) Maxima at \(x=\frac{3}{2}\)
(B) Minima at \(x=\frac{3}{2}\)
(C) Point of inflexion at \(x=\frac{3}{2}\)
(D) Maximum value of function is 156
Correct Answer: (C) Point of inflexion at \(x=\frac{3}{2}\)
View SolutionStep 1: Find the critical points.
Differentiate and set the derivative to zero.
\[ f'(x)=12x^2-36x+27=3(4x^2-12x+9)=3(2x-3)^2 \]
So \(f'(x)=0\) only at \(x=\frac32\).
Step 2: Look at the sign of f'.
Since \(3(2x-3)^2\ge 0\) for every \(x\), the derivative does not change sign at \(x=\frac32\). It is positive on both sides. A maximum or minimum needs a sign change, so there is no extremum there.
Step 3: Use the second derivative.
\[ f''(x)=24x-36 \Rightarrow f''\left(\tfrac32\right)=36-36=0 \]
The test fails, so check the third derivative: \(f'''(x)=24\ne 0\). A non-zero third derivative where \(f''=0\) means a point of inflexion.
Step 4: Check the options.
Option 1 (maxima) and option 2 (minima) are wrong because \(f'\) does not change sign. Option 4 is wrong because a cubic with positive leading coefficient goes to \(+\infty\), so it has no maximum value. Option 3 is correct.
Final Answer:
The function has a point of inflexion at \(x=\frac32\), option 3.
\[ \boxed{\text{Point of inflexion at } x=\tfrac32} \]
Question 65:
The function \(f(x)=\frac{2}{x}+5,\ x\ne 0\) is
(A) increasing function for \(x\in\mathbb{R}-\{0\}\)
(B) increasing function for \(x\in(0,\infty)\)
(C) neither increasing nor decreasing function for \(x\in(0,\infty)\)
(D) decreasing function for \(x\in\mathbb{R}-\{0\}\)
Correct Answer: (D) decreasing function for \(x\in\mathbb{R}-\{0\}\)
View SolutionStep 1: Find the derivative.
Write \(f(x)=2x^{-1}+5\). Then
\[ f'(x)=-\frac{2}{x^2} \]
Step 2: Find its sign.
For every \(x\ne 0\), \(x^2>0\), so \(f'(x)=-\frac{2}{x^2}<0\). The derivative is negative at every point of the domain.
Step 3: Apply the rule.
A function with \(f'(x)<0\) at every point of its domain is a decreasing function there. So \(f\) is decreasing for \(x\in\mathbb{R}-\{0\}\), and in particular on \((0,\infty)\) and on \((-\infty,0)\).
Step 4: Check options 1 and 2.
These say increasing, but \(f'<0\), so both are wrong.
Step 5: Check option 3.
It says neither increasing nor decreasing on \((0,\infty)\). But \(f'<0\) there, so the function is decreasing. It is wrong.
Step 6: Choose option 4.
Option 4 states that \(f\) is decreasing on the whole domain, which matches \(f'(x)<0\) for all \(x\ne0\). This is the intended reading of the question.
Final Answer:
The function is decreasing for \(x\in\mathbb{R}-\{0\}\), option 4.
\[ \boxed{\text{decreasing for } x\in\mathbb{R}-\{0\}} \]
Question 66:
\(\int (x+1)e^x\,dx\) is equal to [ \(C\) is an arbitrary constant]
(A) \((x+1)e^x+C\)
(B) \(e^x+C\)
(C) \(xe^x+C\)
(D) \((x+1)+C\)
Correct Answer: (C) \(xe^x+C\)
View SolutionStep 1: Choose the method.
The integrand is a product of a polynomial \(x+1\) and \(e^x\). Integration by parts is the natural tool. The formula is \(\int u\,dv=uv-\int v\,du\).
Step 2: Set u and dv.
Let \(u=x+1\) so \(du=dx\). Let \(dv=e^x dx\) so \(v=e^x\).
Step 4: Simplify.
\[ (x+1)e^x-e^x=xe^x+e^x-e^x=xe^x \]
So the integral is \(xe^x+C\).
Step 5: Check the options.
Option 1 is what we get before subtracting \(\int e^x dx\), so it forgets a term. Differentiating option 2 gives \(e^x\), not \((x+1)e^x\). Option 4 has no exponential and cannot work. Differentiating option 3 gives \(e^x+xe^x=(x+1)e^x\), which is correct.
Final Answer:
The integral is \(xe^x+C\), option 3.
\[ \boxed{xe^x+C} \]
Question 67:
Particular solution of the differential equation \((x-1)\frac{dy}{dx}=2xy\), when \(y(2)=1\) is:- (consider \(\log_e x=\log x\))
(A) \(\log|y|+\log(x-1)^2=2x+4\)
(B) \(\log|y|-\log(x-1)^2=2x-4\)
(C) \(\log|y|+\log(x-1)^2=2x-4\)
(D) \(\log|y|-\log(x-1)^2=2x+4\)
Correct Answer: (B) \(\log|y|-\log(x-1)^2=2x-4\)
View SolutionStep 1: Recognise the type.
The equation can be written with all \(y\) terms on one side and all \(x\) terms on the other. It is a variable separable equation.
Step 2: Separate the variables.
\[ \frac{dy}{y}=\frac{2x}{x-1}\,dx \]
Divide the numerator: \(\frac{2x}{x-1}=\frac{2(x-1)+2}{x-1}=2+\frac{2}{x-1}\).
Step 3: Integrate both sides.
\[ \log|y|=2x+2\log|x-1|+C \]
Since \(2\log|x-1|=\log(x-1)^2\), this is \(\log|y|-\log(x-1)^2=2x+C\).
Step 4: Use y(2) = 1.
Put \(x=2,\ y=1\): \(\log 1-\log 1=4+C\). Both logs are 0, so \(C=-4\).
Step 5: Write the particular solution.
\[ \log|y|-\log(x-1)^2=2x-4 \]
Step 6: Check the options.
Options 1 and 3 have a plus sign between the logs, but our solution has a minus. Option 4 has \(2x+4\), which would need \(C=+4\). Only option 2 matches.
Final Answer:
The particular solution is \(\log|y|-\log(x-1)^2=2x-4\), option 2.
\[ \boxed{\log|y|-\log(x-1)^2=2x-4} \]
Question 68:
Let R be the feasible region for a linear programming problem, and Let \(z=ax+by\) be the objective function. If the feasible region R is bounded then which of the following are true ? A. The maximum or minimum value of objective function may not exist. B. The objective function \(z\) has both a maximum and a minimum value on R. C. Maximum and minimum values lie in the unbounded region D. The objective function \(z=ax+by\) has both a maximum value and a minimum value and each of these values occurs at a corner point of R. Choose the correct answer from the options given below:
(A) A and B only
(B) A and D only
(C) B and D only
(D) C and D only
Correct Answer: (C) B and D only
View SolutionStep 1: Recall the theorem.
In linear programming, if the feasible region \(R\) is bounded, then the objective function \(Z=ax+by\) has both a maximum and a minimum value on \(R\). Each of these occurs at a corner point (vertex) of \(R\).
Step 2: Check statement A.
A says the maximum or minimum may not exist. For a bounded region this is false, because both always exist. So A is FALSE.
Step 3: Check statement B.
B says \(z\) has both a maximum and a minimum on \(R\). This is exactly the theorem. So B is TRUE.
Step 4: Check statement C.
C says the extreme values lie in the unbounded region. But \(R\) is given as bounded, and the extreme values lie at corner points of \(R\). So C is FALSE.
Step 5: Check statement D.
D says both a maximum and a minimum exist and each occurs at a corner point of \(R\). This matches the corner point theorem. So D is TRUE.
Step 6: Match with the options.
The true statements are B and D only, which is option 3.
Final Answer:
Statements B and D are true, so the answer is option 3.
\[ \boxed{\text{B and D only}} \]
Question 69:
Consider a LPP given by Maximize \(Z=38x+19y\) Subject to \(3x+5y\le 15,\ 5x+2y\le 10,\) and \(x,y\ge 0\) The optimum value of \(Z\) is,
(A) 85
(B) 190
(C) 240
(D) 57
Correct Answer: (A) 85
View SolutionStep 1: Set up the feasible region.
The constraints are \(3x+5y\le 15\), \(5x+2y\le 10\), \(x\ge 0\), \(y\ge 0\). The region is bounded and lies in the first quadrant, so the maximum is at a corner point.
Step 2: Find the axis intercepts.
For \(3x+5y=15\): \((5,0)\) and \((0,3)\). For \(5x+2y=10\): \((2,0)\) and \((0,5)\). The binding limits on the axes are \(x\le 2\) (from the second line) and \(y\le 3\) (from the first line).
Step 3: Find the intersection of the two lines.
Solve \(3x+5y=15\) and \(5x+2y=10\). Multiply the first by 2 and the second by 5: \(6x+10y=30\) and \(25x+10y=50\). Subtract to get \(19x=20\).
\[ x=\frac{20}{19},\qquad y=\frac{45}{19} \]
Step 4: List the corner points.
\((0,0)\), \((2,0)\), \(\left(\frac{20}{19},\frac{45}{19}\right)\) and \((0,3)\).
Step 5: Evaluate Z at each corner.
\(Z(0,0)=0\).
\(Z(2,0)=76\).
\(Z(0,3)=57\).
\(Z\left(\frac{20}{19},\frac{45}{19}\right)=38\cdot\frac{20}{19}+19\cdot\frac{45}{19}=40+45=85\).
Step 6: Choose the largest.
The largest value is 85. Option 4 (57) is the value at \((0,3)\), a smaller vertex. Options 2 and 3 are larger than any corner value, so they cannot be reached in this region.
Final Answer:
The maximum value of \(Z\) is 85, option 1.
\[ \boxed{85} \]
Question 70:
Which of the following is/are true about parameter ? A. It is a characteristic of a sample. B. It is a characterstic of a population. C. It is a numerical value that is taken from the entire population, such as the population mean. D. It is the numerical value taken from a sample and calculated from the sample observations alone. Choose the correct answer from the options given below:
(A) A only
(B) B and D only
(C) B and C only
(D) A and D only
Correct Answer: (C) B and C only
View SolutionStep 1: Understanding the Concept:
A population is the full group we want to study. A sample is a smaller part taken from it. A parameter is a fixed number that describes the population. A statistic is a number worked out from the sample.
Step 2: Check statement A.
A characteristic of a sample is called a statistic, not a parameter. So A is FALSE.
Step 3: Check statement B.
A parameter describes a feature of the whole population, for example its mean or its variance. So B is TRUE.
Step 4: Check statement C.
The population mean is calculated using every member of the population. It is the standard example of a parameter. So C is TRUE.
Step 5: Check statement D.
A value calculated from sample observations alone is a statistic, such as the sample mean. So D is FALSE.
Step 6: Match with the options.
Only B and C are true. Option 1 has A, which is false. Option 2 has D, which is false. Option 4 has A and D, both false. Option 3 (B and C only) is correct.
Final Answer:
A parameter is a population characteristic, so B and C are correct. This is option 3.
\[ \boxed{\text{B and C only}} \]
Question 71:
Consider a office having seating capacity of 58 employees and all of them are to be seated in such a way that an employee can sit on any available vacant seat . What will be the degree of freedom?
(A) 58
(B) 57
(C) 59
(D) 58.5
Correct Answer: (B) 57
View SolutionStep 1: Understanding the Concept:
Degrees of freedom are the number of values that can be chosen freely before the rest get fixed by a condition. In this seating problem the condition is that every employee gets exactly one seat.
Step 2: Count the free choices.
The first employee can pick any of the 58 seats. The second can pick from the 57 that remain, and so on. Each choice is free until only one employee and one seat are left.
Step 3: The last employee.
When 57 employees are seated, exactly one seat is left. The 58th employee has no choice at all. So that last placement is not free.
Step 4: Apply the rule.
\[ \text{degrees of freedom} = n - 1 = 58 - 1 = 57 \]
Step 5: Check the other options.
58 would count the last employee, who has no choice. 59 is more than the number of employees. 58.5 is not possible, because degrees of freedom are whole numbers. So only 57 fits.
Final Answer:
The degrees of freedom are 57, which is option 2.
\[ \boxed{57} \]
Question 72:
The following data are from a simple random sample: 3 , 7 , 4 , 12 , 4 , 12. The point estimate of the population mean is,
(A) 42
(B) 7
(C) 24
(D) 6
Correct Answer: (B) 7
View SolutionStep 1: Understanding the Concept:
A point estimate is a single number used to guess a population value. The best point estimate of the population mean is the sample mean \(\bar{x}\).
Step 2: Key Formula or Approach:
\[ \bar{x} = \frac{\sum x}{n} \]
Step 3: Add the values.
\[ 3+7+4+12+4+12 = 42 \]
Step 4: Divide by the sample size.
There are 6 values. So \[ \bar{x} = \frac{42}{6} = 7 \]
Step 5: Check the options.
42 is only the sum of the values, not the mean. 24 and 6 do not match the calculation. 7 is correct, which is option 2.
Final Answer:
The point estimate of the population mean is 7, which is option 2.
\[ \boxed{7} \]
Question 73:
The patterns and behavior of the data in any time series are based on which of the following components. A. Seasonal component B. Secular trend component C. Irregular component D. Cyclical component Choose the correct answer from the options given below:
(A) A and B only
(B) B and C only
(C) B, C and D only
(D) A, B, C and D
Correct Answer: (D) A, B, C and D
View SolutionStep 1: Understanding the Concept:
A time series is a set of values recorded over time. Its movement is explained by four standard components.
Step 2: Check A, the seasonal component.
Seasonal variation repeats every year, for example higher ice cream sales in summer. It is a component. So A is TRUE.
Step 3: Check B, the secular trend.
The secular trend is the long-term rise or fall of the data. It is a component. So B is TRUE.
Step 4: Check C, the irregular component.
Irregular variation comes from sudden, unpredictable events such as floods or strikes. It is a component. So C is TRUE.
Step 5: Check D, the cyclical component.
Cyclical variation is the wave-like movement over several years, such as business cycles. It is a component. So D is TRUE.
Step 6: Choose the option.
All four are components. Options 1, 2 and 3 each leave out at least one, so option 4 is correct.
Final Answer:
The behavior of a time series depends on all four components, which is option 4.
\[ \boxed{\text{A, B, C and D}} \]
Question 74:
The number of wedding cards, in thousands, distributed in a certain city on each day for a week is given as follows:- 21, 35, 19, 27, 32, 22, 12. The three days moving averages will be:
(A) 25, 28, 26, 23, 21
(B) 25, 24, 23, 27, 26
(C) 25, 27, 26, 27, 22
(D) 24, 28, 26, 22, 25
Correct Answer: (C) 25, 27, 26, 27, 22
View SolutionStep 1: Understanding the Concept:
A three-day moving average takes the mean of three consecutive values, then slides forward by one value. With 7 values, we get \(7-3+1=5\) averages.
Step 6: Compare with the options.
We get 25, 27, 26, 27, 22. Option 1 has 28 as its second value, which is wrong. Option 2 has 24 as its second value, which is wrong. Option 4 starts with 24, which is wrong. Option 3 matches.
Final Answer:
The three-day moving averages are 25, 27, 26, 27, 22, which is option 3.
\[ \boxed{25,\ 27,\ 26,\ 27,\ 22} \]
Question 75:
Match the LIST-I with LIST-II
LIST-I Matrix \(A\)
LIST-II \(|\text{adj}\,A|\)
A. \(A=\begin{bmatrix}2&4\\1&3\end{bmatrix}\)
I. 1
B. \(A=\begin{bmatrix}5&2\\7&4\end{bmatrix}\)
II. 7
C. \(A=\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
III. 2
D. \(A=\begin{bmatrix}6&1\\5&2\end{bmatrix}\)
IV. 6
Choose the correct answer from the options given below:
(A) A - I, B - II, C - III, D - IV
(B) A - IV, B - II, C - III, D - I
(C) A - III, B - IV, C - I, D - II
(D) A - III, B - IV, C - II, D - I
Correct Answer: (C) A - III, B - IV, C - I, D - II
View SolutionStep 1: Understanding the Concept:
For an \(n\times n\) matrix, \(|\text{adj}\,A| = |A|^{n-1}\). For a 2 by 2 matrix, \(n-1=1\), so \(|\text{adj}\,A| = |A|\). We only need the determinant of each matrix.
Step 2: Matrix A.
\[ |A| = 2\times 3 - 4\times 1 = 6-4 = 2 \] So A matches III.
Step 3: Matrix B.
\[ |B| = 5\times 4 - 2\times 7 = 20-14 = 6 \] So B matches IV.
Step 4: Matrix C.
This is the identity matrix, so \[ |C| = 1\times 1 - 0\times 0 = 1 \] So C matches I.
Step 5: Matrix D.
\[ |D| = 6\times 2 - 1\times 5 = 12-5 = 7 \] So D matches II.
Step 6: Choose the option.
The matching is A - III, B - IV, C - I, D - II. Option 1 has A - I, which is wrong. Option 2 has A - IV, which is wrong. Option 4 has C - II and D - I, which is wrong. Option 3 is correct.
Final Answer:
The matching is A - III, B - IV, C - I, D - II, which is option 3.
\[ \boxed{\text{A-III, B-IV, C-I, D-II}} \]
Question 76:
For a given square matrix \(A\) of order n, if there exists another square matrix \(B\) of the same order n, such that \(AB = BA = I\), then \(A\) is said to be invertible and \(B\) is called the inverse matrix of \(A\).
Which of the following statements are not TRUE ?
A. Inverse of a matrix, if it exists, is unique. B. For two invertible matrices of same order, say \(A\) and \(B\) , then \((AB)^{-1} = A^{-1}B^{-1}\) C. For an invertible matrix \(A\), \((A^{-1})^{-1} = A\) D. For an invertible matrix \(A\), \((A^{T})^{-1} = A^{T}\)
Choose the correct answer from the options given below:
(A) A and C only
(B) B and D only
(C) A, C and D only
(D) A and D only
Correct Answer: (B) B and D only
View SolutionStep 1: Understanding the Concept:
We must pick the statements about inverse matrices that are FALSE. So we test each statement against the standard properties of the inverse.
Step 2: Check statement (A).
Suppose \(B\) and \(C\) are both inverses of \(A\). Then \(B = BI = B(AC) = (BA)C = IC = C\). So the inverse is unique. So (A) is TRUE.
Step 3: Check statement (B).
The correct rule is the reversal law: \((AB)^{-1} = B^{-1}A^{-1}\). Matrix multiplication is not commutative, so \(A^{-1}B^{-1}\) is in general a different matrix. So (B) is FALSE.
Step 4: Check statement (C).
Since \(AA^{-1} = A^{-1}A = I\), the matrix \(A\) itself acts as the inverse of \(A^{-1}\). So \((A^{-1})^{-1} = A\), and (C) is TRUE.
Step 5: Check statement (D).
The correct rule is \((A^{T})^{-1} = (A^{-1})^{T}\). It equals \(A^{T}\) only in special cases, for example when \(A\) is orthogonal. For \(A = \begin{bmatrix}1 & 2\\3 & 5\end{bmatrix}\) the two sides are different. So (D) is FALSE.
Step 6: Pick the option.
The statements that are not true are B and D, which is option 2. Option 1 lists true statements. Options 3 and 4 wrongly include a true statement.
Final Answer:
The statements that are not true are B and D only.
\[ \boxed{\text{B and D only}} \]
Question 77:
Match the LIST-I with LIST-II
LIST-I
LIST-II
A.
If A is a square matrix of order 3 such that \(|A| = 3\), then \(|adj (adj A)| =\)
I.
16
B.
If A is a non-singular matrix and \(|A| = \frac{1}{16}\), then \(|A^{-1}| =\)
II.
512
C.
If A is a square matrix of order 4 such that \(|A| = 2\), then \(|adj (adj A)| =\)
III.
9
D.
If A is a square matrix of order 2 such that \(|A| = 9\), then \(|adj (adj A)| =\)
IV.
81
Choose the correct answer from the options given below:
(A) A - IV, B - I, C - III, D - II
(B) A - IV, B - I, C - II, D - III
(C) A - I, B - IV, C - III, D - II
(D) A - III, B - I, C - IV, D - II
Correct Answer: (B) A - IV, B - I, C - II, D - III
View SolutionStep 1: Understanding the Concept:
We use the identity \(|adj A| = |A|^{n-1}\) for a matrix of order n. Applying it twice gives \(|adj(adj A)| = |A|^{(n-1)^2}\). For the inverse we use \(|A^{-1}| = 1/|A|\).
Step 2: Match A.
Here \(n = 3\) and \(|A| = 3\). So \(|adj(adj A)| = 3^{(3-1)^2} = 3^4 = 81\). This is IV.
Step 3: Match B.
\(|A^{-1}| = \frac{1}{|A|} = \frac{1}{1/16} = 16\). This is I.
Step 4: Match C.
Here \(n = 4\) and \(|A| = 2\). So \(|adj(adj A)| = 2^{(4-1)^2} = 2^9 = 512\). This is II.
Step 5: Match D.
Here \(n = 2\) and \(|A| = 9\). So \(|adj(adj A)| = 9^{(2-1)^2} = 9^1 = 9\). This is III.
Step 6: Compare the options.
We got A - IV, B - I, C - II, D - III. That is option 2. Option 1 swaps C and D. Option 3 gives A - I, which is wrong because 81 is not 16. Option 4 gives A - III, which is wrong too.
Final Answer:
The correct matching is A - IV, B - I, C - II, D - III.
\[ \boxed{\text{Option 2}} \]
Choose the correct answer from the options given below:
(A) A - III, B - II, C - I, D - IV
(B) A - III, B - II, C - IV, D - I
(C) A - III, B - IV, C - II, D - I
(D) A - III, B - I, C - IV, D - II
Correct Answer: (C) A - III, B - IV, C - II, D - I
View SolutionStep 1: Understanding the Concept:
We match each name of a matrix type with its example by using the basic definitions.
Step 2: Match A.
A row matrix has only one row. \([1\ 2\ 3]\) is such a matrix. So A matches III.
Step 3: Match B.
A scalar matrix is a diagonal matrix with all diagonal entries equal. The matrix with 2, 2, 2 on the diagonal is scalar. So B matches IV.
Step 4: Match C.
A column matrix has only one column. The matrix with entries 1, 2, 3 in a column is such a matrix. So C matches II.
Step 5: Match D.
A diagonal matrix has zeros off the diagonal. The matrix with 1, 2, 3 on the diagonal fits, and its diagonal entries are not all equal. So D matches I.
Step 6: Choose the option.
We get A - III, B - IV, C - II, D - I, which is option 3. The other options mismatch at least one of B, C, D.
Final Answer:
The correct matching is A - III, B - IV, C - II, D - I.
\[ \boxed{\text{Option 3}} \]
Question 79:
The sum of two skew-symmetric matrices of the same order is a
(A) symmetric matrix
(B) zero matrix
(C) skew-symmetric matrix
(D) diagonal matrix
Correct Answer: (C) skew-symmetric matrix
View SolutionStep 1: Understanding the Concept:
A matrix \(P\) is skew-symmetric if \(P^T = -P\).
Step 2: Take two such matrices.
Let \(P\) and \(Q\) be skew-symmetric of the same order. Then \(P^T = -P\) and \(Q^T = -Q\).
Step 3: Transpose the sum.
\[ (P+Q)^T = P^T + Q^T = -P - Q = -(P+Q) \] So \(P+Q\) is skew-symmetric.
Step 4: Check the other options.
The sum is not always symmetric, because a nonzero skew-symmetric matrix is not symmetric. It is not always the zero matrix, since \(P + P = 2P\) is nonzero when \(P\) is nonzero. It need not be diagonal, because skew-symmetric matrices can have nonzero entries off the diagonal.
Final Answer:
The sum is a skew-symmetric matrix.
\[ \boxed{\text{skew-symmetric matrix}} \]
Question 80:
Which of the following properties are not correct for three real numbers a, b and c?
A. If \(a > b\) and \(b > c\), then \(a < c\) B. If \(a > b\), and \(c < 0\), then \(ac < bc\) C. If \(a > b\), and \(c < 0\), then \(a \div c < b \div c\) D. If \(a > b\) and \(c > 0\), then \(a \div c < b \div c\)
Choose the correct answer from the options given below:
(A) A and D only
(B) B, C and D only
(C) A and B only
(D) B and C only
Correct Answer: (A) A and D only
View SolutionStep 1: Understanding the Concept:
We need the statements that are wrong. We use the rules of inequalities: order is transitive, and multiplying or dividing by a negative number reverses the inequality.
Step 2: Check statement (A).
If \(a > b\) and \(b > c\), transitivity gives \(a > c\). The statement says \(a < c\). So (A) is NOT correct.
Step 3: Check statement (B).
Multiplying \(a > b\) by a negative \(c\) reverses the sign, so \(ac < bc\). For example \(3 > 2\) and \(c = -1\) give \(-3 < -2\). So (B) is correct.
Step 4: Check statement (C).
Dividing by a negative \(c\) also reverses the sign, so \(a/c < b/c\). For example \(3/(-1) = -3 < -2 = 2/(-1)\). So (C) is correct.
Step 5: Check statement (D).
Dividing by a positive \(c\) keeps the sign, so \(a/c > b/c\). The statement says \(a/c < b/c\). So (D) is NOT correct.
Step 6: Pick the option.
The incorrect statements are A and D, which is option 1.
Final Answer:
The incorrect properties are A and D only.
\[ \boxed{\text{A and D only}} \]
Question 81:
A tin container contains 50 litres of cow milk. From this container 5 litres cow milk was taken out and replaced with soya milk. This process was repeated further two more times. How much cow milk is there in the container now?
(A) 364.5 litres
(B) 36.45 litres
(C) 3.645 litres
(D) 40 litres
Correct Answer: (B) 36.45 litres
View SolutionStep 1: Understanding the Concept:
Each time we take out 5 litres of the mixture and add soya milk, the cow milk left is multiplied by the same fraction.
Step 2: Find the fraction kept.
Total volume stays 50 litres. Removing 5 litres removes \(\frac{5}{50} = \frac{1}{10}\) of the cow milk. So \(\frac{9}{10}\) of the cow milk stays after each step.
Step 3: Apply three times.
The process is done once and then repeated two more times, so 3 times in all. \[ 50 \times \left(\frac{9}{10}\right)^3 = 50 \times \frac{729}{1000} = 36.45 \]
Step 4: Check the options.
364.5 is more than 50, which is impossible. 3.645 has the decimal point in the wrong place. 40 would be the amount left if only one replacement of 10 litres was done, which does not match this process.
Final Answer:
The cow milk left is 36.45 litres.
\[ \boxed{36.45 \text{ litres}} \]
Question 82:
If \(53 \equiv y \pmod 6\), then the possible values of \(y\) are:-
(A) ......... 64, 58, 52, 46, 40, ........
(B) ......... 70, 76, 81, 87, 93, 98, ........
(C) ......... 65, 59, 53, 47, 41, ........
(D) ......... 70, 76, 82, 88, 94, 97 ........
Correct Answer: (C) ......... 65, 59, 53, 47, 41, ........
View SolutionStep 1: Understanding the Concept:
\(53 \equiv y \pmod 6\) means that \(53 - y\) is divisible by 6. So \(y\) leaves the same remainder as 53 when divided by 6.
Step 2: Find the remainder.
\(53 = 6 \times 8 + 5\). So the remainder is 5, and \(y \equiv 5 \pmod 6\). Hence \(y = 6k + 5\) for an integer \(k\).
Step 3: Check option 1.
64 divided by 6 leaves remainder 4. So option 1 is wrong.
Step 4: Check option 2.
70 leaves remainder 4, and 81 leaves 3. So option 2 is wrong.
Step 5: Check option 3.
65, 59, 53, 47, 41 all leave remainder 5 and differ by 6 each time. So option 3 is right.
Step 6: Check option 4.
70 leaves remainder 4 and 97 leaves 1. So option 4 is wrong.
Final Answer:
The possible values of y are ...65, 59, 53, 47, 41...
\[ \boxed{\text{Option 3}} \]
Question 83:
A man can row a boat in still water at the rate of 17 km/hr. He finds that it takes him thrice as long to row upstream of a river as to row downstream of the same river. The speed of of the stream is,
(A) 15 km/hr
(B) 11.5 km/hr
(C) 8.5 km/hr
(D) 6 km/hr
Correct Answer: (C) 8.5 km/hr
View SolutionStep 1: Understanding the Concept:
When a boat goes downstream, the stream helps it, so the speed is boat speed plus stream speed. When it goes upstream, the stream opposes it, so the speed is boat speed minus stream speed.
Step 2: Set up the symbols.
Let the stream speed be \(s\) km/hr. The boat speed in still water is 17 km/hr.
Downstream speed \(= 17 + s\).
Upstream speed \(= 17 - s\).
Step 3: Use the time condition.
For the same distance, time is inversely proportional to speed. Upstream time is 3 times the downstream time, so the upstream speed is one third of the downstream speed.
\[ 17 - s = \frac{17 + s}{3} \]
Multiply both sides by 3:
\[ 51 - 3s = 17 + s \]
\[ 34 = 4s \]
\[ s = 8.5 \]
Step 4: Check the options.
With \(s = 8.5\), downstream speed is 25.5 km/hr and upstream speed is 8.5 km/hr. The ratio is 3 : 1, which matches the condition.
If \(s = 15\), the upstream speed is only 2 km/hr and the ratio is 16, so option 1 is wrong.
If \(s = 11.5\), the speeds are 28.5 and 5.5, ratio about 5.2, so option 2 is wrong.
If \(s = 6\), the speeds are 23 and 11, ratio about 2.1, so option 4 is wrong.
Final Answer:
The speed of the stream is 8.5 km/hr, which is option 3.
\[ \boxed{8.5 \text{ km/hr}} \]
Question 84:
Two pipes A and B can fill a tank in 12 minutes and 15 minutes respectively while a third pipe C can empty the full tank in 20 minutes. All the three pipes are opened in the beginning but pipe C is closed 6 minutes before the tank is filled. In what time will the tank is full?
(A) 15 minutes
(B) 10 minutes
(C) 7 minutes
(D) 17 minutes
Correct Answer: (C) 7 minutes
View SolutionStep 1: Understanding the Concept:
In pipe problems we find the part of the tank each pipe fills in one minute. Filling pipes give a positive part and an emptying pipe gives a negative part.
Step 2: Find the one minute work.
Pipe A fills \(\frac{1}{12}\), pipe B fills \(\frac{1}{15}\), pipe C empties \(\frac{1}{20}\) of the tank per minute.
All three together: \[ \frac{1}{12} + \frac{1}{15} - \frac{1}{20} = \frac{5 + 4 - 3}{60} = \frac{6}{60} = \frac{1}{10} \]
A and B alone: \[ \frac{1}{12} + \frac{1}{15} = \frac{9}{60} = \frac{3}{20} \]
Step 3: Set up the equation.
Let the total time be \(T\) minutes. All three pipes work for \(T - 6\) minutes and only A and B work for the last 6 minutes.
\[ \frac{T - 6}{10} + 6 \times \frac{3}{20} = 1 \]
\[ \frac{T - 6}{10} + 0.9 = 1 \]
\[ T - 6 = 1 \]
\[ T = 7 \]
Step 4: Check the options.
In 7 minutes, the first minute gives \(\frac{1}{10}\) and the last 6 minutes give \(\frac{18}{20} = 0.9\). The total is 1, so the tank is exactly full. Options 15, 10 and 17 minutes are too long, because even with C open the tank fills in 10 minutes, and closing C only makes it faster.
Final Answer:
The tank is full in 7 minutes, which is option 3.
\[ \boxed{7 \text{ minutes}} \]
Question 85:
In a 700 m race, A reaches the finishing point in 30 seconds & B reaches in 35 seconds. By how much distance A beats B?
(A) 75 m
(B) 50 m
(C) 100 m
(D) 200 m
Correct Answer: (C) 100 m
View SolutionStep 1: Understanding the Concept:
In a race, A beats B by the distance B still has to run when A reaches the finish line. So we find how far B has run in the time A takes.
Step 2: Find the speed of B.
B runs 700 m in 35 seconds.
\[ \text{Speed of B} = \frac{700}{35} = 20 \text{ m/s} \]
Step 3: Find distance run by B when A finishes.
A finishes in 30 seconds. In 30 seconds B runs \(20 \times 30 = 600\) m.
Step 4: Find the winning margin.
B is still \(700 - 600 = 100\) m from the finish. So A beats B by 100 m.
Option 1 (75 m) and option 2 (50 m) are too small, and option 4 (200 m) would need B to run only 500 m in 30 seconds, which does not match the speed of B.
Final Answer:
A beats B by 100 m, which is option 3.
\[ \boxed{100 \text{ m}} \]
Relation \(R\) in the set \(A = \{1,2,3,4,\ldots,16\}\) defined as \(R = \{(x,y): 3x = y,\ x,y \in A\}\), then which of the following are correct ? A. domain of \(R = \{1, 2, 3, 4, 5\}\). B. range of \(R = \{3, 6, 9, 12, 15\}\). C. \(R\) is reflexive. D. \(R\) is not symmetric. Choose the correct answer from the options given below:
A tin container contains 50 litres of cow milk. From this container 5 litres cow milk was taken out and replaced with soya milk. This process was repeated further two more times. How much cow milk is there in the container now?
A man can row a boat in still water at the rate of 17 km/hr. He finds that it takes him thrice as long to row upstream of a river as to row downstream of the same river. The speed of of the stream is,
If
\[
\left|
\begin{matrix}
x+a & y & z \\
x & y+b & z \\
x & y & z+c
\end{matrix}
\right|
=
abc,
\]
where \(a,b,c\neq0\), then find the value of
\[
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}.
\]
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