Q1. If \( \rho \) be a relation in the set of all integers \( \mathbb{Z} \) and \(\rho = \{ (x,y) \mid |x-y| \leq 5, x,y \in \mathbb{Z} \} \) then the relation \(\rho\) is:
(A) Reflexive and Symmetric
(B) Reflexive and Transitive
(C) Transitive and Symmetric
(D) Equivalence
Show Solution
Reflexive: For any \(x\in\mathbb{Z}\), \(|x-x|=0\le 5\). So \((x,x)\in\rho\) for every \(x\). Hence \(\rho\) is reflexive.
Symmetric: If \((x,y)\in\rho\) then \(|x-y|\le5\). But \(|y-x|=|x-y|\le5\), so \((y,x)\in\rho\). Thus \(\rho\) is symmetric.
Transitive? We need: if \((x,y)\in\rho\) and \((y,z)\in\rho\) (i.e. \(|x-y|\le5\) and \(|y-z|\le5\)) does it follow that \(|x-z|\le5\)?
Counterexample: take \(x=0,\; y=5,\; z=10\). Then \(|0-5|=5\le5\) and \(|5-10|=5\le5\), so \((0,5)\) and \((5,10)\) are in \(\rho\). But \(|0-10|=10>5\), so \((0,10)\notin\rho\). Hence \(\rho\) is not transitive.
Because \(\rho\) is reflexive and symmetric but not transitive, it is not an equivalence relation.
Answer: (A) Reflexive and Symmetric
Q2. Let \(\mathbb{R}\) be the set of real numbers and \( f:\mathbb{R}\to\mathbb{R}, g:\mathbb{R}\to\mathbb{R} \) are two mappings such that \( f(x)=|x|-x^2, \ g(x)=2x+3, \ \forall x \in \mathbb{R} \). Then the value of \( (g \circ f)(-3) \) is:
(A) 9
(B) -9
(C) 6
(D) -6
Show Solution
Solution.
\( f(-3)=|{-3}|-(-3)^2=3-9=-6 \).
\[\therefore (g \circ f)(-3)=g(f(-3))= g(-6)=2(-6)+3=-12+3=-9 \].
Answer: (B) \(-9\)
Q3. Statement (Q): \( f : \mathbb{R} \to \mathbb{R} \) is a function defined as \( f(x) = \lfloor x \rfloor \), greatest integer function. \( f(x) \) is not onto function.
Reason (R): A function \( F : X \to Y \) is one-one if \( F(a) = F(b) \implies a = b \).
Alternatives:
(A) (Q) and (R) both are true, and (R) is a correct explanation of (Q)
(B) (Q) and (R) both are true, but (R) is not a correct explanation of (Q)
(C) (Q) is true but (R) is false
(D) (Q) is false but (R) is true
Show Solution
Solution.
Check (Q): The greatest integer function \( f(x)=\lfloor x\rfloor \) maps every real \(x\) to an integer. So the range of \(f\) is \( \mathbb{Z} \), while the codomain given is \( \mathbb{R} \). Since \( \mathbb{Z}\subsetneq\mathbb{R} \), not every real number is attained. Hence \(f\) is not onto. Therefore (Q) is true.
Check (R): The statement \(F(a)=F(b)\implies a=b\) is the definition of an injective (one-one) function. So (R) is a correct statement about one-one functions and is true.
Does (R) explain (Q)? (Q) concerns surjectivity (onto), while (R) defines injectivity (one-one). These are different properties; the definition of injectivity does not explain why \(f\) fails to be onto. Thus (R) is not a correct explanation of (Q).
Answer: (B) (Q) and (R) both are true, but (R) is not a correct explanation of (Q).
Q4. Let \(\mathbb{R}\) be the set of real numbers and \( f:\mathbb{R}\to\mathbb{R} \) be given by \( f(x)=2x-3 \). Then the value of \( f^{-1}(0) \) is:
(A) -3
(B) \(\tfrac{3}{2}\)
(C) 3
(D) \(\pm 3\)
Show Solution
Solution.
By definition, \( f^{-1}(0) \) is the value of \(x\) such that \( f(x)=0 \).
\( \therefore f(x)=2x-3=0 \).
\( \implies 2x-3=0 \implies 2x=3 \implies x=\tfrac{3}{2}. \)
Therefore \( f^{-1}(0)=\tfrac{3}{2} \).
Answer: (B) \(\tfrac{3}{2}\)
Q5. The principal value of \( \cot^{-1}\!\left(-\tfrac{1}{\sqrt{3}}\right) \) is:
(A) \(\tfrac{2\pi}{3}\)
(B) \(-\tfrac{\pi}{3}\)
(C) \(\tfrac{\pi}{3}\)
(D) \(\tfrac{\pi}{6}\)
Show Solution
Solution.
For \( \cot^{-1}x \), the principal value lies in \( (0,\pi) \).
\(\cot^{-1}\!\left(-\tfrac{1}{\sqrt{3}}\right)\)
\(= \cot^{-1}\!\left(-\cot\tfrac{\pi}{3}\right)\)
\(= \pi - \cot^{-1}\!\left(\cot\tfrac{\pi}{3}\right)\; \bigl[\because \cot^{-1}(-\cot x) = \pi - \cot^{-1}(\cot x)\bigr]\)
\(= \pi - \tfrac{\pi}{3}\)
\(= \tfrac{2\pi}{3}\)
Answer: (A) \(\tfrac{2\pi}{3}\)
Q6. If \( \sin^{-1}x + \sin^{-1}y = \tfrac{2\pi}{3} \), then the value of \( \cos^{-1}x + \cos^{-1}y \) is:
(A) \(\tfrac{\pi}{3}\)
(B) \(\tfrac{\pi}{6}\)
(C) \(\tfrac{2\pi}{3}\)
(D) \(\tfrac{5\pi}{3}\)
Show Solution
Solution.
We will use the identity \( \cos^{-1}x = \tfrac{\pi}{2} - \sin^{-1}x \).
Therefore
\({\cos ^{ - 1}}x + {\cos ^{ - 1}}y\)
\( = \left( {\tfrac{\pi }{2} - {{\sin }^{ - 1}}x} \right) + \left( {\tfrac{\pi }{2} - {{\sin }^{ - 1}}y} \right)\)
\( = \pi - ({\sin ^{ - 1}}x + {\sin ^{ - 1}}y)\)
Given \( \sin^{-1}x+\sin^{-1}y = \tfrac{2\pi}{3} \), so
\( \cos^{-1}x + \cos^{-1}y = \pi - \tfrac{2\pi}{3} = \tfrac{\pi}{3}. \)
Answer: (A) \(\tfrac{\pi}{3}\)
Q7. The value of \( 2\tan^{-1}\!\sqrt{x} - \cos^{-1}\!\left(\tfrac{1-x}{1+x}\right) \) is:
(A) 0
(B) \(\tfrac{1}{4}\)
(C) \(\tfrac{1}{3}\)
(D) \(\tfrac{1}{2}\)
Show Solution
Solution.
Note that
\( 2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)\)
\(\therefore 2{\tan ^{ - 1}}\sqrt x = {\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right)\)
Therefore
\(2{\tan ^{ - 1}}\sqrt x - {\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right)\)
\( = {\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) - {\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right)\)
\( = 0\)
Answer: (A) 0
Q8. If \( A = [a_{ij}] \) is a \(2 \times 2\) matrix, where \( a_{ij} = \tfrac{1}{2}(i+2j)^2 \), then \( A \) is:
(A) \(\begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}\)
(B) \(\begin{bmatrix} 9 & \frac{25}{2} \\ 8 & 18 \end{bmatrix}\)
(C) \(\begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 9 \end{bmatrix}\)
(D) \(\begin{bmatrix} \frac{9}{2} & \frac{15}{2} \\ 4 & 18 \end{bmatrix}\)
Show Solution
Solution.
Given, \({a_{ij}} = \tfrac{1}{2}{(i + 2j)^2}\)
Therefore, for \(i = 1,j = 1\)
\({a_{11}} = \tfrac{1}{2}{(1 + 2 \times 1)^2} = \frac{9}{2}\)
Similarly,
\({a_{12}} = \tfrac{1}{2}{(1 + 2 \times 2)^2} = \frac{{25}}{2}\)
\({a_{21}} = \tfrac{1}{2}{(2 + 2 \times 1)^2} = 8\)
\({a_{22}} = \tfrac{1}{2}{(2 + 2 \times 2)^2} = 18\)
Hence,
\(A = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}} \\ {{a_{21}}}&{{a_{22}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\frac{9}{2}}&{\frac{{25}}{2}} \\ 8&{18} \end{array}} \right)\)
Answer: (A) \(\begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}\)
Q9. If \( A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \) and \( f(x) = x^2 - 2x - 5 \), then \( f(A) \) is equal to:
(A) \(\begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}\)
(B) \(\begin{bmatrix} -3 & 0 \\ 0 & -3 \end{bmatrix}\)
(C) \(\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}\)
(D) \(\begin{bmatrix} -3 & 0 \\ 0 & -2 \end{bmatrix}\)
Show Solution
Solution.
Given, \(A = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 2&1 \end{array}} \right]{\text{ and }}f\left( x \right) = {x^2} - 2x - 5\)
Therefore,
\(f\left( A \right) = {A^2} - 2A - 5{I_2}\)
\( \Rightarrow f\left( A \right) = {\left[ {\begin{array}{*{20}{c}} 1&2 \\ 2&1 \end{array}} \right]^2} - 2\left[ {\begin{array}{*{20}{c}} 1&2 \\ 2&1 \end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]\)
\( \Rightarrow f\left( A \right) = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 2&1 \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} 1&2 \\ 2&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&4 \\ 4&2 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 5&0 \\ 0&5 \end{array}} \right]\)
\( \Rightarrow f\left( A \right) = \left[ {\begin{array}{*{20}{c}} {1 + 4}&{2 + 2} \\ {2 + 2}&{4 + 1} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&4 \\ 4&2 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 5&0 \\ 0&5 \end{array}} \right]\)
\( \Rightarrow f\left( A \right) = \left[ {\begin{array}{*{20}{c}} 5&4 \\ 4&5 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 2&4 \\ 4&2 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 5&0 \\ 0&5 \end{array}} \right]\)
\( \Rightarrow f\left( A \right) = \left[ {\begin{array}{*{20}{c}} {5 - 2 - 5}&{4 - 4 - 0} \\ {4 - 4 - 0}&{5 - 2 - 5} \end{array}} \right]\)
Hence,
\(f\left( A \right) = \left[ {\begin{array}{*{20}{c}} { - 2}&0 \\ 0&{ - 2} \end{array}} \right]\)
Answer: (A) \(\begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix}\)
Q10. Let \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 1 \\ -2 & 0 \end{bmatrix} \). Match the matrix on the left column with the matrix on the right column.
| Left Column | Right Column |
|---|---|
| (i) \( A + A^T \) | (a) \(\begin{bmatrix} -2 & -1 \\ -1 & 0 \end{bmatrix}\) |
| (ii) \( (A+B)^T \) | (b) \(\begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix}\) |
| (iii) \( (AB)^T \) | (c) \(\begin{bmatrix} 0 & 1 \\ 3 & 4 \end{bmatrix}\) |
| (iv) \( B + B^T \) | (d) \(\begin{bmatrix} -5 & -11 \\ 1 & 3 \end{bmatrix}\) |
Options:
(A) (i)-(a), (ii)-(c), (iii)-(d), (iv)-(b)
(B) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d)
(C) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a)
(D) (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a)
Show Solution
Solution.
Compute each left-column matrix explicitly.
(i) Given, \(A = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right]\)
Therefore, \({A^T} = \left[ {\begin{array}{*{20}{c}} 1&3 \\ 2&4 \end{array}} \right]\)
Now, \( A + {A^T} = \left[ {\begin{array}{*{20}{c}} {1 + 1}&{2 + 3} \\ {3 + 2}&{4 + 4} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&5 \\ 5&8 \end{array}} \right]\)
(ii) \(A + B = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - 1}&1 \\ { - 2}&0 \end{array}} \right]\)
\( \Rightarrow A + B = \left[ {\begin{array}{*{20}{c}} {1 + \left( { - 1} \right)}&{2 + 1} \\ {3 + \left( { - 2} \right)}&{4 + 0} \end{array}} \right]\)
\( \Rightarrow A + B = \left[ {\begin{array}{*{20}{c}} 0&3 \\ 1&4 \end{array}} \right]\)
\(\therefore {\left( {A + B} \right)^T} = \left[ {\begin{array}{*{20}{c}} 0&1 \\ 3&4 \end{array}} \right]\)
(iii) \(AB = \left[ {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} { - 1}&1 \\ { - 2}&0 \end{array}} \right]\)
\( = \left[ {\begin{array}{*{20}{c}} {1 \times ( - 1) + 2 \times \left( { - 2} \right)}&{1 \times 1 + 2 \times 0} \\ {3 \times ( - 1) + 4 \times \left( { - 2} \right)}&{3 \times 1 + 4 \times 0} \end{array}} \right]\)
\( = \left[ {\begin{array}{*{20}{c}} { - 5}&1 \\ { - 11}&3 \end{array}} \right]\)
\(\therefore {\left( {AB} \right)^T} = \left[ {\begin{array}{*{20}{c}} { - 5}&{ - 11} \\ 1&3 \end{array}} \right]\)
So the correct mapping is:
(i) — (b), (ii) — (c), (iii) — (d), (iv) — (a).
Answer: (C) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a)
Q11. If \( X = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \), then \( X^5 \) will be:
(A) \( 36X \)
(B) \( 50X \)
(C) \( 90X \)
(D) \( 81X \)
Show Solution
Solution.
Compute each left-column matrix explicitly.
(i) Given, \(X = \left[ {\begin{array}{*{20}{c}} 3&0&0 \\ 0&3&0 \\ 0&0&3 \end{array}} \right] = 3I \) where \(I\) be the identity matrix of order \(3 \times 3\)
\( \Rightarrow {X^5} = {\left( {3I} \right)^5} = {3^5}I\;\left[ {\because {I^n} = I} \right]\)
\(\therefore {X^5} = 243I = 81\left( {3I} \right) = 81X\)
Answer: (D) \( 81X \)
Q12. Let \( S \) be a \( 2 \times m \) ordered matrix and \( T \) be a \( 3 \times n \) ordered matrix, and conformable for product \( TS \) matrix of order \( p \times 4 \). Then the values of \( m, n, p \) are:
(A) \( m=3, n=2, p=4 \)
(B) \( m=4, n=2, p=3 \)
(C) \( m=3, n=4, p=2 \)
(D) \( m=4, n=3, p=2 \)
Show Solution
Solution.
For the product \(TS\) to be defined we must have the number of columns of \(T\) equal to the number of rows of \(S\). That is \(n = 2\) because \(T\) is \(3\times n\) and \(S\) is \(2\times m\).
When \(TS\) exists its order is \((\text{rows of }T)\times(\text{columns of }S) = 3\times m\). We are told \(TS\) has order \(p\times 4\). Hence
\(3 \times m = p \times 4.\)
Comparing dimensions: the number of rows gives \(3 = p\), and the number of columns gives \(m = 4\).
So \(m=4,\; n=2,\; p=3\).
Answer: (B) \( m=4, \; n=2, \; p=3 \)
Q13. The values of \( a, b, c \) for which the matrix \(\begin{bmatrix} 1 & a+b-c & a+b+c \\ 1 & 2 & a-b+c \\ 9 & 5 & 3 \end{bmatrix}\) will be symmetric are:
(A) \( a=3, b=2, c=4 \)
(B) \( a=2, b=3, c=1 \)
(C) \( a=1, b=2, c=3 \)
(D) \( a=0, b=1, c=3 \)
Show Solution
Solution.
For the matrix to be symmetric, it must follow \( A=A^T \).
\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{a + b - c}&{a + b + c} \\ 1&2&{a - b + c} \\ 9&5&3 \end{array}} \right] = {\left[ {\begin{array}{*{20}{c}} 1&{a + b - c}&{a + b + c} \\ 1&2&{a - b + c} \\ 9&5&3 \end{array}} \right]^T}\)
\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{a + b - c}&{a + b + c} \\ 1&2&{a - b + c} \\ 9&5&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&9 \\ {a + b - c}&2&5 \\ {a + b + c}&{a - b + c}&3 \end{array}} \right]\)
From the equality of two matrices, we have the following equations:
\(a + b + c = 9 \cdots \cdots \left( i \right)\)
\(a + b - c = 1 \cdots \cdots \left( {ii} \right)\)
\(a - b + c = 5 \cdots \cdots \left( {iii} \right)\)
\( (ii) + (iii) \) we get,
\( (a+b-c) + (a-b+c) = 1 + 5\)
\(\Rightarrow 2a = 6 \)
\(\Rightarrow a = 3. \)
\(\left( i \right) - \left( {ii} \right)\) we get,
\(\left( {a + b + c} \right) - \left( {a + b - c} \right) = 9 - 1\)
\( \Rightarrow 2c = 8\)
\(\therefore c = 4\)
Use \( c=4, a=3 \) in the first equation:
\( a + b + c = 9 \Rightarrow b = 2. \)
So \( (a,b,c) = (3,2,4) \).
Answer: (A) \( a=3, b=2, c=4 \)
Q14. If \( A \) is a square matrix of order 3 and \( |A|=7 \), then the value of \( |2A^T| \) is:
(A) 32
(B) 28
(C) 16
(D) 56
Show Solution
Solution.
For any square matrix \(A\) of order \(n\) and scalar \(k\), we have \(\lvert kA \rvert = k^n \lvert A \rvert\). Also \(\lvert A^T \rvert = \lvert A \rvert\).
Here \(n=3\), \(k=2\) and \(\lvert A \rvert = 7\). Therefore
\(\lvert 2A^T \rvert = 2^3 \lvert A^T \rvert = 8 \lvert A \rvert = 8 \times 7 = 56.\)
Answer: (D) 56
Q15. If the inverse of a matrix \( A \) of order \( 3 \times 3 \) exists and \( |A| = 5 \), then the value of \( |\text{adj } A| \) is:
(A) 20
(B) 15
(C) 5
(D) 25
Show Solution
Solution.
For an invertible matrix \(A\) of order \(n\), we have \(\displaystyle |\operatorname{adj}A| = |A|^{\,n-1}.\)
Here \(n=3\) and \(|A|=5\). Thus
\(\displaystyle |\operatorname{adj}A| = 5^{3-1} = 5^2 = 25.\)
Answer: (D) 25
Q16. If \(\begin{vmatrix}-x^2 & xy & xz \\xy & -y^2 & yz \\xz & yz & -z^2\end{vmatrix}= \lambda x^2 y^2 z^2,\) then the value of \( \lambda \) is equal to:
(A) 1
(B) 2
(C) 3
(D) 4
Show Solution
Solution.
Given, \(\left| {\begin{array}{*{20}{c}} { - {x^2}}&{xy}&{xz} \\ {xy}&{ - {y^2}}&{yz} \\ {xz}&{yz}&{ - {z^2}} \end{array}} \right| = \lambda {x^2}{y^2}{z^2}\)
Taking \(x=1,y=1,z=1 \), we have
\(\left| {\begin{array}{*{20}{c}} { - 1}&{1}&{1} \\ {1}&{ - 1}&{1} \\ {1}&{1}&{ - {1}} \end{array}} \right| = \lambda {x^2}{y^2}{z^2}\)
\( \Rightarrow - 1\left| {\begin{array}{*{20}{c}} { - 1}&1 \\ 1&{ - 1} \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} 1&1 \\ 1&{ - 1} \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 1&1 \end{array}} \right| = \lambda \left( 1 \right)\left( 1 \right)\left( 1 \right)\)
\( \Rightarrow - 1\left( {1 - 1} \right) - 1\left( { - 1 - 1} \right) + 1\left( {1 + 1} \right) = \lambda \)
\( \Rightarrow - 1\left( { - 2} \right) + 1\left( 2 \right) = \lambda \)
\(\therefore \lambda = 4\)
Answer: (D) 4
Q17. The system of equations \( kx + y+ z = 1, \; x + ky + z = k, \; x + y + kz = k^2 \) will have unique solution when:
(A) \( k \neq 1 \)
(B) \( k \neq 2 \)
(C) \( k \neq 1, k \neq -2 \)
(D) \( k \neq 0 \)
Show Solution
Solution.
Write the system in matrix form \(A\mathbf{x}=\mathbf{b}\) where
\(A=\begin{bmatrix}k&1&1\\[4pt]1&k&1\\[4pt]1&1&k\end{bmatrix}.\)
The system has a unique solution iff \(\det(A)\neq0\). For a matrix with diagonal entry \(a=k\) and off-diagonal entries \(b=1\) (size \(3\times3\)), the determinant is
\( \det(A)=(k-1)^{2}(k+2). \)
Therefore \(\det(A) \neq 0\) exactly when \(k \neq 1\) or \(k \neq -2\).
Hence the system has a unique solution for
\(\boxed{k\neq 1,\;k\neq -2.}\)
Answer: (C) \( k \neq 1,\, k \neq -2 \)
Q18. Let \( f(x) = \frac{x^2 - 1}{x^3 - 1}, \; x \neq 1 \), is continuous at \( x=1 \). Then the value of \( f(1) \) is:
(A) 1
(B) \( \tfrac{1}{3} \)
(C) \( \tfrac{2}{3} \)
(D) 2
Show Solution
Solution.
We need the value of \(f(1)\) given that \(f\) is continuous at \(x=1\), so compute the limit
\( f(1)=\lim_{x\to1}\frac{x^2-1}{x^3-1}. \)
Both numerator and denominator is equal to zero (i.e. \( \frac{0}{0} \)) at \(x=1\), so apply L'Hôpital's rule:
\( \lim_{x\to1}\frac{x^2-1}{x^3-1} =\lim_{x\to1}\frac{\frac{d}{dx}(x^2-1)}{\frac{d}{dx}(x^3-1)} =\lim_{x\to1}\frac{2x}{3x^2} =\frac{2\cdot 1}{3\cdot 1}= \frac{2}{3}. \)
Answer: (C) \( \tfrac{2}{3} \)
Q19. If \( x^m y^n = (x+y)^{m+n} \), then the value of \( \tfrac{dy}{dx} \) is:
(A) 0
(B) \( \tfrac{y}{x} \)
(C) \( \tfrac{x+y}{xy} \)
(D) \( xy \)
Show Solution
Solution.
We are given: \(\; x^m y^n = (x+y)^{m+n}.\)
Taking logarithm on both sides:
\(\; m \ln x + n \ln y = (m+n) \ln(x+y).\)
Differentiating w.r.t. \(x\):
\(\Rightarrow \frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = \frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right).\)
\(\Rightarrow \frac{n}{y}\frac{dy}{dx} - \frac{m+n}{x+y}\frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x}.\)
\( \Rightarrow \frac{{dy}}{{dx}}\left( {\frac{n}{y} - \frac{{m + n}}{{x + y}}} \right) = \frac{{m + n}}{{x + y}} - \frac{m}{x}\)
\( \Rightarrow \frac{{dy}}{{dx}}\left( {\frac{{n\left( {x + y} \right) - y\left( {m + n} \right)}}{{y\left( {x + y} \right)}}} \right) = \frac{{x\left( {m + n} \right) - m\left( {x + y} \right)}}{{x\left( {x + y} \right)}}\)
\( \Rightarrow \frac{{dy}}{{dx}}\left( {\frac{{n\left( {x + y} \right) - y\left( {m + n} \right)}}{{y\left( {x + y} \right)}}} \right) = \frac{{x\left( {m + n} \right) - m\left( {x + y} \right)}}{{x\left( {x + y} \right)}}\)
\( \Rightarrow \frac{{dy}}{{dx}}\left( {\frac{{nx + ny - my - ny}}{y}} \right) = \frac{{mx + nx - mx - my}}{x}\)
\( \Rightarrow \frac{{dy}}{{dx}}\left( {\frac{{nx - my}}{y}} \right) = \frac{{nx - my}}{x}\)
\( \Rightarrow \frac{{dy}}{{dx}}\left( {\frac{1}{y}} \right) = \frac{1}{x}\)
So, \(\frac{dy}{dx} = \frac{y}{x}.\)
Answer: (B) \( \tfrac{y}{x} \)
Q20. If \( f(2) = 4, \; f'(2) = 4 \), then the value of \(\lim_{x \to 2} \frac{xf(2) - 2f(x)}{x-2}\) is:
(A) -2
(B) 2
(C) 3
(D) -4
Show Solution
Solution.
We need
\( L=\lim_{x\to2}\frac{xf(2)-2f(x)}{x-2}. \)
Since \(f(2)=4\), rewrite the numerator:
\( xf(2)-2f(x)=4x-2f(x)=2\big(2x-f(x)\big). \)
So
\( L=2\lim_{x\to2}\frac{2x-f(x)}{x-2}. \)
The expression inside the limit is of the \(0/0\) form (because \(\frac{2\cdot2-f(2)}{2-2}=\frac{4-4}{0}=\frac{0}{0}\)), so apply L'Hôpital's rule (differentiate numerator and denominator):
\( \lim_{x\to2}\frac{2x-f(x)}{x-2} =\lim_{x\to2}\frac{2-f'(x)}{1}=2-f'(2). \)
Given \(f'(2)=4\), Therefore
\( L=2 - (4)=-2. \)
Answer: (A) -2
Q21. If \( f(x) = \log_x (\log_e x) \), then the value of \( f'(e) \) is:
(A) \( e \)
(B) \( \tfrac{2}{e} \)
(C) \( \tfrac{1}{e} \)
(D) 0
Show Solution
Solution.
Write the function using natural logs:
\( f(x)=\log_x(\ln x)=\dfrac{\ln(\ln x)}{\ln x}. \)
Differentiate using the quotient rule. Let \(u=\ln(\ln x)\), \(v=\ln x\). Then \(u'=\dfrac{1}{x\ln x}\) and \(v'=\dfrac{1}{x}\). So
\[ f'(x)=\dfrac{u'v-uv'}{v^2} =\dfrac{\dfrac{1}{x\ln x}\cdot\ln x-\ln(\ln x)\cdot\dfrac{1}{x}}{(\ln x)^2} =\dfrac{\dfrac{1}{x}-\dfrac{\ln(\ln x)}{x}}{(\ln x)^2} =\dfrac{1-\ln(\ln x)}{x(\ln x)^2}. \]
Now evaluate at \(x=e\). We have \(\ln e=1\) and \(\ln(\ln e)=\ln 1=0\). Thus
\( f'(e)=\dfrac{1-0}{e\cdot 1^2}=\dfrac{1}{e}. \)
Answer: (C) \( \tfrac{1}{e} \)
Q22. If \( x = \sin^{-1}t, \; y = \sqrt{1-t^2} \), then the value of \( \tfrac{d^2y}{dx^2} \) at \( t=1 \) is:
(A) 1
(B) 0
(C) \( \tfrac{1}{2} \)
(D) -1
Show Solution
Solution.
Since \(x=\sin^{-1}t\), we have \(t=\sin x\). Then
\(y=\sqrt{1-t^2}=\sqrt{1-\sin^2 x}=\cos x\) (taking the nonnegative root for \(x\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\)).
Differentiate with respect to \(x\):
\(\dfrac{dy}{dx}=-\sin x.\)
Differentiate again:
\(\dfrac{d^2y}{dx^2}=-\cos x.\)
At \(t=1\) we have \(x=\sin^{-1}(1)=\dfrac{\pi}{2}\), so
\(\dfrac{d^2y}{dx^2}\Big|_{t=1}=-\cos\!\big(\tfrac{\pi}{2}\big)=-0=0.\)
Answer: (B) 0
Q23. If \( \tfrac{dx}{dy} = l \) and \( \tfrac{d^2x}{dy^2} = m \), then the value of \( \tfrac{d^2y}{dx^2} \) is:
(A) \( -\tfrac{m}{l^3} \)
(B) \( \tfrac{m}{l^3} \)
(C) \( \tfrac{1}{m} \)
(D) \( \tfrac{1}{l} \)
Show Solution
Solution.
Given \( \dfrac{dx}{dy}=l \). Hence
\( \dfrac{dy}{dx}=\dfrac{1}{l}. \)
Differentiate \( \dfrac{dx}{dy} \) with respect to \(y\):
\(\frac{{{d^2}x}}{{d{y^2}}} = \frac{d}{{dy}}\left( {\frac{{dx}}{{dy}}} \right) = \frac{{dl}}{{dy}}=m\)
Differentiate \( \dfrac{dy}{dx} \) with respect to \(x\):
\(\frac{d^2y}{dx^2}=\frac{d}{dx}\!\left(\frac{1}{l}\right) =-\frac{1}{l^2}\frac{dl}{dx} \)
\(\Rightarrow \frac{d^2y}{dx^2}=-\frac{1}{l^2}\frac{dl}{dy}\times \frac{dy}{dx} \)
\(\Rightarrow \frac{d^2y}{dx^2}=-\frac{1}{l^2}\times m \times \frac{1}{l} \)
Therefore, \(\frac{d^2y}{dx^2} = -\frac{m}{l^3}\)
Answer: (A) \( -\dfrac{m}{l^3} \)
Q24. Let \( f(x) = \begin{cases} x^2 + ax + b, & x < 1 \\ x, & x \geq 1 \end{cases} \). If \( f(x) \) is differentiable at \( x=1 \), then \( (a-b) \) is equal to:
(A) 0
(B) -2
(C) -6
(D) -3
Show Solution
Solution.
For \(f\) to be differentiable at \(x=1\) it must be continuous at \(x=1\) and the left and right derivatives at \(x=1\) must be equal.
Continuity:
Left-hand value at \(x=1\): \(1^2 + a\cdot1 + b = 1 + a + b\).
Right-hand value at \(x=1\): \(1 = 1\).
So \(1 + a + b = 1 \Rightarrow a + b = 0.\) \(\;(1)\)
Derivative from left: \[f'_-(1)=\frac{d}{dx}(x^2+ax+b)\big|_{x=1}=(2x+a)\big|_{x=1}=2\cdot1+a=2+a.\]
Derivative from right: \(f'_+(1)=\frac{d}{dx}(x)\big|_{x=1}=1.\)
Equate derivatives: \(2+a=1 \Rightarrow a=-1.\) \(\;(2)\)
From (1) and (2): \(b=-a=1\).
Hence \(a-b = (-1) - 1 = -2.\)
Answer: (B) -2.
Q25. The points of discontinuity of the function \( f(x) = \tfrac{x^2+4x+3}{x^3+3x^2-x-3} \) are:
(A) \( x = 1, -1, -3 \)
(B) \( x = -1, -3 \)
(C) \( x = 1, -3 \)
(D) \( x = 1, -1 \)
Show Solution
Solution.
For the given function to be discontinuous, denominator must be zero.
So, we examine the denominator:
\(x^3+3x^2-x-3\). Test \(x=1\): \(1+3-1-3=0\), so \(x-1\) is a factor.
On dividing denominator \((x^3+3x^2-x-3)\) by the factor \((x-1)\) to get
\[x^3+3x^2-x-3=(x-1)(x^2+4x+3)=(x-1)(x+1)(x+3).\]
So, Points where the original denominator is zero are \(x=1,-1,-3\).
Hence, points of discontinuity of the given function are \(x=1,-1,-3\).
Answer: (A) \( x = 1, -1, -3 \)
Q26. The volume of a spherical balloon increases at the rate of \( 10 \, \text{cm}^3/\text{sec} \). The rate of change of its surface area when its radius is \( 16 \, \text{cm} \), is:
(A) \( 1.5 \, \text{cm}^2/\text{sec} \)
(B) \( 1.8 \, \text{cm}^2/\text{sec} \)
(C) \( 2 \, \text{cm}^2/\text{sec} \)
(D) \( 1.25 \, \text{cm}^2/\text{sec} \)
Show Solution
Solution.
Volume of a sphere: \(V=\dfrac{4}{3}\pi r^3\). Given \(\dfrac{dV}{dt}=10\ \text{cm}^3/\text{s}\).
Differentiating \(V\) w.r.t. \(t\): \(\dfrac{dV}{dt}=4\pi r^2\frac{dr}{dt}\). Hence
\(\displaystyle \frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt} =\frac{10}{4\pi r^2}.\)
Surface area: \(S=4\pi r^2\).
So, on differentiating w.r.t. t, we get:
\(\displaystyle \frac{dS}{dt}=8\pi r\frac{dr}{dt}=8\pi r \times \frac{10}{4\pi r^2}=\frac{20}{r}.\)
At \(r=16\):
We have \(\dfrac{dS}{dt}=\dfrac{20}{16}=1.25.\)
Answer: (D) \(1.25\ \text{cm}^2/\text{s}\)
Q27. Which one of the following is correct for all values of \( x \) if \( x \in (0,1) \) ?
(A) \( e^x ≤ 1+x \)
(B) \( \log_e(1+x) ≤ x \)
(C) \( \sin x > x \)
(D) \( \log_e x > x \)
Show Solution
Solution (using increasing / decreasing functions).
(A) Consider \(h(x)=e^x-(1+x)\). Then \(h'(x)=e^x-1\). For \(x\in(0,1)\), \(e^x>1\) so \(h'(x)>0\); hence \(h\) is increasing on \((0,1)\). Since \(h(0)=0\) and \(h\) increases for \(x>0\),
So, we have \(h(x)>0\) for \(x\in(0,1)\). Thus \(e^x>1+x\). So option (A) is false.
(B) Let \(g(x)=\ln(1+x)-x\). Then \(g'(x)=\frac{1}{1+x}-1=\frac{-x}{1+x} ≤ 0\quad\text{for }x\in(0,1).\) So \(g\) is strictly decreasing on \((0,1)\). And \(g(0)= 0\).
Therefore for \(x\in(0,1)\) we have \(g(x)≤g(0)=0\), i.e. \(\ln(1+x) ≤ x\). Hence \(\ln(1+x)\le x\) (strict inequality for \(x>0\)), so option (B) is true.
(C) Take \(s(x)=\sin x - x\). Then \(s'(x)=\cos x -1\le 0\) on \((0,1)\) (indeed \(≤0\) for \(x\in(0,1]\)), so \(s\) is decreasing and \(s(0)=0\). Thus \(s(x)≤0\) for \(x\in(0,1)\), i.e. \(\sin x ≤ x\). Option (C) is false.
(D) Let \(t(x)=\ln x - x\). Then \(t'(x)=\frac{1}{x}-1\). For \(x\in(0,1)\) we have \(\frac{1}{x}-1>0\), so \(t\) is increasing on \((0,1)\).
But \(\lim_{x\to0^+}t(x)=-\infty\) and \(t(1)=\ln1-1=-1\), so for every \(x\in(0,1)\) we have \(t(x)≤t(1)=-1≤0\). Hence \(\ln x ≤ x\); option (D) is false.
Answer: (B) \(\log_e(1+x)\le x\)
Q28. Let \( f(x) = \tfrac{x}{1+|x|} \). Then \( f(x) \) is monotonically increasing in the interval:
(A) \( \mathbb{R} \)
(B) \( \mathbb{R} - \{-1\} \)
(C) \( (-1,1) \)
(D) \( (-\infty, 0) \)
Show Solution
Solution.
Write \(f(x)=\dfrac{x}{1+|x|}\). Consider two cases.
Case 1: \(x>0\). Here \(|x|=x\) so \(f(x)=\dfrac{x}{1+x}\).
\(\displaystyle f'(x)=\frac{(1+x)-x}{(1+x)^2}=\frac{1}{(1+x)^2}>0\) for \(x>0\).
Case 2: \(x≤0\). Here \(|x|=-x\) so \(f(x)=\dfrac{x}{1-x}.\)
\(\displaystyle f'(x)=\frac{(1-x)-x(-1)}{(1-x)^2}=\frac{1}{(1-x)^2}>0\) for \(x≤0\).
At \(x=0\) the left and right derivatives equal \(1\), so \(f\) is differentiable there and increasing through \(0\).
Thus \(f'(x)>0\) for every real \(x\), so \(f\) is strictly (monotonically) increasing on all of \(\mathbb{R}\).
Answer: (A) \(\mathbb{R}\)
Q29. If the straight line \( y = x \) and the curve \( xy = k^2 \) cut at a right angle, then:
(A) \( k = 0 \)
(B) \( k = \pm 1 \)
(C) \( -\infty < k < \infty \)
(D) \( 0 \leq k < \infty \)
Show Solution
Solution.
The solpe of the given staright line (i.e. \(y=x\)) is 1, say \(m_{1}=1\).
Given, equation of the curve \(y=\frac{k^2}{x}\), on differentiating we get,
\(\frac{dy}{dx}=-\frac{k^2}{x^2}\)
So, slope of the curve, \(m_{2}=-\frac{k^2}{x^2}\)
As, the given straight line and the curve meet at right angle. So \(m_{1} \times m_{2} = -1\).
Therefore.
\(1 \times (-\frac{k^2}{x^2})=-1 \Rightarrow k=±x\)
Since, x can have any real value except \(x=0\)(At \(x=0\), given straight line and curve does not meet at right angle). So none of the given option matches this exactly.
Note: So the correct set will be \(\{k\in\mathbb{R}:k\neq0\}\).
Q30. If the straight line \( lx - my + n = 0 \) touches the parabola \( y^2 = 4ax \), then:
(A) \( am^2 = nl \)
(B) \( an^2 = ml \)
(C) \( al^2 = mn \)
(D) \( mn = al \)
Show Solution
Solution.
Given, equation of parabola \(y^2=4ax \), differentiating w.r.t \(x \), we get
\(2y \frac{dy}{dx}=4a \)
\(\Rightarrow \frac{dy}{dx}=\frac{2a}{y} \)
As, the parametric co-ordinates of the parabola are \((at^2,\,2at)\).
Therefore, slope of the parabola at \((at^2,\,2at)\) is
\(m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = a{t^2},y = 2at}} = \frac{{2a}}{{2at}} = \frac{1}{t}\)
The equation of the tangent at the point \((at^2,2at)\) is
\(y - 2at = m\left( {x - a{t^2}} \right)\)
\( \Rightarrow y - 2at = \frac{1}{t}\left( {x - a{t^2}} \right)\)
\( \Rightarrow x - a{t^2} = ty - 2a{t^2}\)
\(\therefore x - ty + a{t^2} = 0\; \cdots \left( i \right)\)
Given, equation of the tangent is \( lx - my + n = 0 \), so
\( x- \frac{m}{l}y + \frac{n}{l}=0\; \cdots \left( ii \right) \)
Compare equation (i) and (ii), we get
\(t=\frac{m}{l}\) and \(t^2=\frac{n}{al}\)
On comparing, we get
\((\frac{m}{l})^2=\frac{n}{al}\)
Hence \( \; a m^2 = n l.\) — Answer: (A)
Q31. If \( y = \dfrac{\log_e x}{x} \), then the maximum value of \( y \) is:
(A) \( e \)
(B) \( e^2 \)
(C) \( \tfrac{1}{e} \)
(D) \( \tfrac{1}{e^2} \)
Show Solution
Solution.
Given, \( y = \dfrac{\log_e x}{x} \)
Differentiating w.r.t. \( x \), we get
\( \dfrac{dy}{dx} = \dfrac{x \cdot \dfrac{1}{x} - (\log_e x) \cdot 1}{x^2} \)
\( \Rightarrow \dfrac{dy}{dx} = \dfrac{1 - \log_e x}{x^2} \)
For maximum or minimum value,
\( \dfrac{dy}{dx} = 0 \)
\( \Rightarrow \dfrac{1 - \log_e x}{x^2} = 0 \)
Since, \( x \neq 0 \)
\( \Rightarrow 1 - \log_e x = 0 \)
\( \Rightarrow \log_e x = 1 \)
\( \Rightarrow x = e \)
Now,
\( \dfrac{d^2y}{dx^2} = \dfrac{d}{dx} \left( \dfrac{1 - \log_e x}{x^2} \right) \)
\( \Rightarrow \dfrac{x^2 \left( -\dfrac{1}{x} \right) - (1 - \log_e x)\cdot 2x}{x^4} \)
At \( x = e \),
\( \dfrac{d^2y}{dx^2}\bigg|_{x=e} = \dfrac{-e - 2e(1 - \log_e e)}{e^4} \)
\( \Rightarrow \dfrac{-e - 2e(0)}{e^4} = \dfrac{-e}{e^4} ≤ 0 \)
\( \therefore y \) is maximum at \( x = e \).
Hence, the maximum value of \( y \) is
\( y = \dfrac{\log_e e}{e} = \dfrac{1}{e} \)
Answer: (C) \( \tfrac{1}{e} \)
Q32. Statement–I: \( f(x) = 3 + |x-3| \) has a local minimum value 3.
Statement–II: \( f(x) = \sin x \) has infinite number of maximum and minimum values.
Which of the following options is correct?
(A) Statement–I is true, Statement–II is false
(B) Statement–I is false, Statement–II is true
(C) Statements–I and II both are true
(D) Statements–I and II both are false
Show Solution
Solution.
Given \( f(x) = 3 + |x-3| \).
The minimum value of \(|x-3|\) is \(0\) at \(x=3\).
\(\Rightarrow f(3) = 3 + 0 = 3\).
Thus, \( f(x) \) has a local minimum value \(3\) at \(x=3\).
\(\therefore\) Statement–I is true.
Now, consider \( f(x) = \sin x \).
We know \(\sin x\) oscillates between \(-1\) and \(1\) infinitely many times.
Hence, \(\sin x\) has infinitely many maxima and minima.
\(\therefore\) Statement–II is also true.
So, both Statement–I and Statement–II are true.
Answer: (C) Statements–I and II both are true.
Q33. If \( P(A) = \tfrac{1}{4}, \; P(B) = \tfrac{1}{3} \) and \( P(A-B) = \tfrac{1}{6} \), then the events \( A \) and \( B \) are mutually:
(A) exclusive
(B) independent
(C) dependent
(D) exhaustive
Show Solution
Solution.
We are given: \( P(A) = \tfrac{1}{4}, \; P(B) = \tfrac{1}{3}, \; P(A-B) = \tfrac{1}{6} \).
Now, \( P(A-B) = P(A) - P(A \cap B) \).
\(\Rightarrow \tfrac{1}{6} = \tfrac{1}{4} - P(A \cap B)\).
\(\Rightarrow P(A \cap B) = \tfrac{1}{4} - \tfrac{1}{6} = \tfrac{3 - 2}{12} = \tfrac{1}{12}\).
Also, \( P(A) \cdot P(B) = \tfrac{1}{4} \cdot \tfrac{1}{3} = \tfrac{1}{12} \).
Since \( P(A \cap B) = P(A) \cdot P(B) \), the events \(A\) and \(B\) are independent.
Answer: (B) independent
Q34. A biased coin is tossed \( n \) times. The probability of getting a head is \( p \) \( (0 < p < 1) \). The probability that the \( r^{th} \) \((r<n)\) head will appear in the \( n^{th} \) tossing is:
(A) \(\binom{n-1}{r-1} p^r q^{\,n-r}\)
(B) \(\binom{n}{r} p^r q^{\,n-r}\)
(C) \( p^r \)
(D) \( p^{\,n-r} \)
Show Solution
Solution.
We want the probability that the \( r^{th} \) head appears exactly on the \( n^{th} \) toss.
This means:
(i) In the first \( n-1 \) tosses, there must be exactly \( r-1 \) heads.
(ii) The \( n^{th} \) toss must be a head.
Probability of exactly \( r-1 \) heads in first \( n-1 \) tosses is
\( \binom{n-1}{r-1} p^{\,r-1} q^{\, (n-r)} \), where \( q = 1-p \).
Probability of a head on the \( n^{th} \) toss is \( p \).
Therefore, required probability =
\( \binom{n-1}{r-1} p^{\,r-1} q^{\, (n-r)} \cdot p \)
\( = \binom{n-1}{r-1} p^r q^{\, (n-r)} \).
Answer: (A) \(\binom{n-1}{r-1} p^r q^{\,n-r}\)
Q35. If three events \( X, Y, Z \) are mutually exclusive and exhaustive where \( P(X) = \tfrac{2}{3}P(Y), \; P(Z) = \tfrac{1}{3}P(Y) \), then \( P(Z) \) is equal to:
(A) \( \tfrac{1}{6} \)
(B) \( \tfrac{1}{3} \)
(C) \( \tfrac{5}{6} \)
(D) \( \tfrac{1}{2} \)
Show Solution
Solution.
Let \( P(Y)=y \).
Then \( P(X)=\tfrac{2}{3}y \) and \( P(Z)=\tfrac{1}{3}y \).
Since \( X,Y,Z \) are exhaustive, \( P(X)+P(Y)+P(Z)=1 \).
\( \Rightarrow \tfrac{2}{3}y + y + \tfrac{1}{3}y = 1 \).
\( \Rightarrow \left(\tfrac{2}{3} + 1 + \tfrac{1}{3}\right) y = 1 \Rightarrow 2y = 1 \Rightarrow y = \tfrac{1}{2} \).
Thus \( P(Z)=\tfrac{1}{3}y = \tfrac{1}{3} \cdot \tfrac{1}{2} = \tfrac{1}{6} \).
Answer: (A) \( \tfrac{1}{6} \)
Q36. \( A \) and \( B \) are two independent events. The probability of occurrence of exactly one of the two events is:
(A) \( P(A/B) \)
(B) \( P(B/A) \)
(C) \( P(A) + P(B) - P(AB) \)
(D) \( P(A) + P(B) - 2P(AB) \)
Show Solution
Solution.
We want the probability that exactly one of the two events \(A\) and \(B\) occurs.
This means either \(A\) occurs and \(B\) does not occur, or \(B\) occurs and \(A\) does not occur.
So, required probability = \( P(A \cap B') + P(B \cap A') \).
= \( P(A)P(B') + P(B)P(A') \) [since \(A\) and \(B\) are independent].
= \( P(A)(1-P(B)) + P(B)(1-P(A)) \).
= \( P(A) + P(B) - 2P(A)P(B) \).
= \( P(A) + P(B) - 2P(AB) \).
Answer: (D) \( P(A) + P(B) - 2P(AB) \)
Q37. A person regularly watches on TV either the Discovery channel or a Sports channel at night. The probability of watching Sports channel is \( \tfrac{4}{5} \). The probability of him falling asleep while watching the Discovery channel is \( \tfrac{3}{4} \) and in case of Sports channel this probability is \( \tfrac{1}{4} \). Then the probability of watching the Discovery channel if some day the person falls asleep is:
(A) \( \tfrac{3}{5} \)
(B) \( \tfrac{4}{5} \)
(C) \( \tfrac{4}{7} \)
(D) \( \tfrac{3}{7} \)
Show Solution
Solution.
Let \( D \): person watches Discovery channel, \( S \): person watches Sports channel.
We have \( P(S) = \tfrac{4}{5}, \; P(D) = 1 - \tfrac{4}{5} = \tfrac{1}{5} \).
Let \( A \): person falls asleep.
We are given: \( P(A|D) = \tfrac{3}{4}, \; P(A|S) = \tfrac{1}{4} \).
We need \( P(D|A) \).
By Bayes’ theorem,
\( P(D|A) = \dfrac{P(D)P(A|D)}{P(D)P(A|D) + P(S)P(A|S)} \).
= \( \dfrac{\tfrac{1}{5} \cdot \tfrac{3}{4}}{\tfrac{1}{5} \cdot \tfrac{3}{4} + \tfrac{4}{5} \cdot \tfrac{1}{4}} \).
= \( \dfrac{3/20}{3/20 + 4/20} = \dfrac{3/20}{7/20} \).
= \( \tfrac{3}{7} \).
Answer: (D) \( \tfrac{3}{7} \)
Q38. In a probability distribution, the mean of the random variable \( X \) is \( \tfrac{6}{5} \) and mean of \( X^2 \) is 2, then the standard deviation of \( X \) is:
(A) \( \tfrac{\sqrt{7}}{5} \)
(B) \( \tfrac{\sqrt{14}}{5} \)
(C) \( \tfrac{7}{\sqrt{5}} \)
(D) \( \tfrac{\sqrt{6}}{\sqrt{5}} \)
Show Solution
Solution.
We know, \( \text{Var}(X) = E(X^2) - [E(X)]^2 \).
Here, \( E(X) = \tfrac{6}{5}, \; E(X^2) = 2 \).
\( \text{Var}(X) = 2 - \left(\tfrac{6}{5}\right)^2 \).
= \( 2 - \tfrac{36}{25} \).
= \( \tfrac{50}{25} - \tfrac{36}{25} = \tfrac{14}{25} \).
So, standard deviation \( = \sqrt{\tfrac{14}{25}} = \tfrac{\sqrt{14}}{5} \).
Answer: (B) \( \tfrac{\sqrt{14}}{5} \)
Q39. If the probability that exactly one of the two events \( A \) and \( B \) occurs is \( x \) and the probability that both \( A \) and \( B \) occur is \( y \), then \( P(A)+P(B) \) is equal to:
(A) \( x+y \)
(B) \( x+2y \)
(C) \( 2x+2y \)
(D) \( 2x+y \)
Show Solution
Solution.
Probability that exactly one of \( A \) and \( B \) occurs = \( x \).
\(\Rightarrow P(A \cap B') + P(B \cap A') = x \).
Also, probability that both occur = \( P(A \cap B) = y \).
Now, \( P(A) + P(B) = P(A \cap B') + P(B \cap A') + 2P(A \cap B) \).
= \( x + 2y \).
Answer: (B) \( x + 2y \)
Q40. The following represents a probability distribution of a random variable \( X \):
\[ \begin{array}{c|ccccc} X=x_i & 0 & 1 & 2 & 3 & 4 \\ \hline P(X=x_i) & k & 2k & 3k & 4k & 5k \end{array} \]
Then \( P(X \geq 2) \) is:
(A) \( \tfrac{1}{5} \)
(B) \( \tfrac{2}{5} \)
(C) \( \tfrac{3}{5} \)
(D) \( \tfrac{4}{5} \)
Show Solution
Solution.
We are given the probability distribution:
\( P(X=0)=k, \; P(X=1)=2k, \; P(X=2)=3k, \; P(X=3)=4k, \; P(X=4)=5k \).
Since total probability = 1,
\( k + 2k + 3k + 4k + 5k = 1 \).
\( 15k = 1 \Rightarrow k = \tfrac{1}{15} \).
Now, \( P(X \geq 2) = P(X=2)+P(X=3)+P(X=4) \).
= \( 3k + 4k + 5k = 12k \).
= \( 12 \cdot \tfrac{1}{15} = \tfrac{12}{15} = \tfrac{4}{5} \).
Answer: (D) \( \tfrac{4}{5} \)
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