M.C.Q (1 Marks)

MCQ 11 Mark

If a vector makes an angle of $\frac{\pi}{4}$ with the positive directions of both x-axis and y-axis, then the angle which it makes with positive z-axis is:

A
$0$
B
$\frac{\pi}{4}$
C
$\frac{3 \pi}{4}$
✓
$\frac{\pi}{2}$

Answer

Correct option: D.

$\frac{\pi}{2}$

(d)$\frac{\pi}{2}$
Explanation: $\frac{\pi}{2}$

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MCQ 21 Mark

The derivative of $\sin ^2 x$ w.r.t. $e^{\cos x} i$

A
$\frac{2}{e^{\cos x}}$
B
$\frac{2 \cos x}{e^{\cos x}}$
✓
$-\frac{2 \cos x}{e^{\cos x}}$
D
$\frac{e^{\cos x}}{-2}$

Answer

Correct option: C.

$-\frac{2 \cos x}{e^{\cos x}}$

(c) $-\frac{2 \cos x}{e^{\cos x}}$
Explanation: Let $u(x)=\sin ^2 x$ and $v(x)=e^{\cos x}$.
We want to find $\frac{d u}{d v}=\frac{\frac{d u}{d z}}{\frac{d v}{d x}}$.
Clearly, $\frac{d u}{d x}=2 \sin x \cos x$ and $\frac{d v}{d x}= e ^{\cos x }(-\sin x )=-(\sin x ) e ^{\cos x }$ $\frac{d u}{d v}=\frac{2 \sin x \cos x}{-\sin x e^{\cos \alpha}}=-\frac{2 \cos x}{e^{\cos \alpha}}$

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MCQ 31 Mark

Find $\lambda$ and $\mu$ if $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$

A
$5, \frac{27}{2}$
✓
$3, \frac{27}{2}$
C
$3, \frac{27}{5}$
D
$4, \frac{27}{2}$

Answer

Correct option: B.

$3, \frac{27}{2}$

(b) $3, \frac{27}{2}$
Explanation: It is given that:
$(2 \hat{i}+6 \hat{j}+27 \hat{k}) X(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$
$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu\end{array}\right|=\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=\overrightarrow{0}$, equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$ on both sides, we get
$(6 \mu-27 \lambda)=0,(2 \mu-27)=0,(2 \lambda-6)=0$.
solving, we get $\lambda=3, \mu=\frac{27}{2}$

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MCQ 41 Mark

The integrating factor of differential equation $\cos x \frac{d y}{d x}+ y \sin x =1$ is

A
$\sin x$
✓
$\sec x$
C
$\tan x$
D
$\cos x$

Answer

Correct option: B.

$\sec x$

Given that,
$\cos x \frac{d y}{d x}+y \sin x=1$
$\Rightarrow \frac{d y}{d x}+y \tan x=\sec x$
Here, $P =\tan x$ and $Q =\sec x$
$\text { IF }=e^{\int P d x}$
$=e^{\int \tan x d x}$
$=e^{\ln \sec x}$
$\therefore \text { IF }=\sec x$

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MCQ 51 Mark

The probabilities of $A, B$ and $C$ of solving a problem are $\frac{1}{6}, \frac{1}{5}$ and $\frac{1}{3}$ respectively. What is the probability that the problem is solved?

✓
$\frac{5}{9}$
B
$\frac{4}{9}$
C
$\frac{1}{3}$
D
$\frac{1}{7}$

Answer

Correct option: A.

$\frac{5}{9}$

The probability that the problem is solved
$=P(A \cup B \cup C)= P ( A )+ P ( B )+ P ( C )- P ( A \cap B )- P ( B \cap C )- P ( C$ $\cap A )+3 P ( A \cap B \cap C )$
Considering independent events, $P ( A \cap B )= P ( A ) \cdot P ( B )$,
$P(BC)=P(B) \cdot P(C), P(C \cap A)=P(C) \cdot P(A),$
$P(A \cap B \cap C)=P(A) \cdot P(B) \cdot P(C)$
Thus, $P(A \cup B \cup C)$ is,
$\Rightarrow \frac{1}{6}+\frac{1}{5}+\frac{1}{3}-\frac{1}{30}-\frac{1}{15}-\frac{1}{18}+3\left(\frac{1}{90}\right)=\frac{5}{9}$

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MCQ 61 Mark

If $\left|\begin{array}{lll}2 & 3 & 2 \\ x & x & x \\ 4 & 9 & 1\end{array}\right|+3=0$, then the value of $x$ is

A
1
B
$0$
✓
-1
D
3

Answer

Correct option: C.

-1

(c)-1
Explanation: -1

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MCQ 71 Mark

The vector with initial point P (2, -3, 5) and terminal point Q(3, -4, 7) is

A
$-\hat{i}+\hat{j}-2 \hat{k}$
✓
$\hat{i}-\hat{j}+2 \hat{k}$
C
$5 \hat{i}-7 \hat{j}+12 \hat{k}$
D
$5 \hat{i}-7 \hat{j}-12 \hat{k}$

Answer

Correct option: B.

$\hat{i}-\hat{j}+2 \hat{k}$

(b) $\hat{i}-\hat{j}+2 \hat{k}$
Explanation: To find the vector we need to find the $\overrightarrow{P Q}$
$=3 \hat{i}-4 \hat{j}+7 \hat{k}-(2 \hat{i}+3 \hat{j}-5 \hat{k})$.
Hence, the vector formed by above points is with the following $(1,-1,2)$.

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MCQ 81 Mark

The point which does not lie in the half plane $2 x+3 y-12 \leq 0$ is

A
(2,1)
B
(-3, 2)
C
(1, 2)
✓
(2, 3)

Answer

Correct option: D.

(2, 3)

(d) (2, 3)
Explanation:
Since (2, 3) does not satisfy $2 x+3 y-12 \leq 0$ as $2 \times 2+3 \times 3-12=4+9-12=1 \neq 0$

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MCQ 91 Mark

The number of all possible matrices of order 2 x 3 with each entry 1 or 2 is

A
24
✓
64
C
6
D
16

Answer

Correct option: B.

(b) 64
Explanation: The order of the matrix = 2 x 3
The number of elements = 2 x 3 = 6
Each place can have either 1 or 2. So, each place can be filled in 2 ways.
Thus, the number of possible matrices $=2^6=64$

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MCQ 101 Mark

$\int \frac{1}{x \sqrt{x^4-1}} d x=?$

A
$\operatorname{cosec}^{-1} x^2+C$
✓
$\frac{1}{2} \sec ^{-1} x^2+C$
C
$\sec ^{-1} x^2+C$
D
$2 \operatorname{cosec}^{-1} x^2+C$

Answer

Correct option: B.

$\frac{1}{2} \sec ^{-1} x^2+C$

Formula $:- \int x^n d x=\frac{x^{n+1}}{n+1}+c ; \int \frac{1}{t \sqrt{t^2-1}} d t=\sec ^{-1} t+c$
Therefore,
Put $x^2=t$
$\Rightarrow 2 xdx=dt$
$=\int \frac{1}{x \sqrt{t^2-1}} \times \frac{d t}{2 x} \Rightarrow \frac{1}{2} \int \frac{1}{t \sqrt{t^2-1}} d t$
$=\frac{1}{2} \sec ^{-1} t+c$
$=\frac{1}{2} \sec ^{-1} x^2+c$

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MCQ 111 Mark

Let $\theta=\sin ^{-1}\left(\sin \left(-600^{\circ}\right)\right)$, then value of $\theta$ is

A
$\frac{-2 \pi}{3}$
B
$\frac{2 \pi}{3}$
C
$\frac{\pi}{2}$
✓
$\frac{\pi}{3}$

Answer

Correct option: D.

$\frac{\pi}{3}$

$\sin ^{-1} \sin \left(-600 \times \frac{\pi}{180}\right)=\sin ^{-1} \sin \left(\frac{-10 \pi}{3}\right)$
$=\sin ^{-1}\left(-\sin \left(4 \pi-\frac{2 \pi}{3}\right)\right)=\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$
$=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{3}\right)\right)=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)=\frac{\pi}{3}$

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MCQ 121 Mark

The feasible region for an LPP is always a

✓
convex polygon
B
Straight line
C
concave polygon
D
type of polygon

Answer

Correct option: A.

convex polygon

(a) convex polygon
Explanation: Feasible region for an LPP is always a convex polygon.

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MCQ 131 Mark

The solution of the differential equation $\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{1}{\left(1+x^2\right)^2}$ is:

A
$\frac{y}{1+x^2}= c +\tan ^{-1} x$
B
$y\left(1+x^2\right)=c+\sin ^{-1} x$
✓
$y\left(1+x^2\right)=c+\tan ^{-1} x$
D
$y \log \left(1+x^2\right)=c+\tan ^{-1} x$

Answer

Correct option: C.

$y\left(1+x^2\right)=c+\tan ^{-1} x$

We have,$\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{1}{\left(1+x^2\right)^2}$
Which is linear differential equation.
Here, $P =\frac{2 x}{1+x^2}$ and $Q =\frac{1}{\left(1+x^2\right)^2}$
$\therefore I.F. =e^{\int \frac{2 x}{1+x^2} d x}$
$=e^{\log \left(1+x^2\right)}$
$=1+x^2$
$\therefore$ the general solution is
$y\left(1+x^2\right)=\int\left(1+x^2\right) \frac{1}{\left(1+x^2\right)^2}+C$
$\Rightarrow y\left(1+x^2\right)=\int \frac{1}{1+x^2} d x+C$
$\Rightarrow y\left(1+x^2\right)=\tan ^{-1} x+C$

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MCQ 141 Mark

The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is

✓
perpendicular to z-axis
B
parallel to z-axis
C
parallel to y-axis
D
parallel to x-axis

Answer

Correct option: A.

perpendicular to z-axis

(a) perpendicular to $z$-axis
Explanation: We have,
$\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$
Also, the given line is parallel to the vector $\vec{b}=3 \hat{i}+\hat{j}+0 \hat{k}$
Let $x \hat{i}+y \hat{j}+z \hat{k}$ be perpendicular to the given line.
Now,
$3 x+4 y+0 z=0$
It is satisfied by the coordinates of z -axis, i.e. $(0,0,1)$ Hence, the given line is perpendicular to z -axis.

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MCQ 151 Mark

Let $g ( x )=\left\{\begin{array}{ll}e^{2 x}, & x<0 \\ e^{-2 x}, & x \geq 0\end{array}\right.$ then $g ( x )$ does not satisfy the condition

A
differentiable at $x = 0$
B
continuous $\forall x \in R$
✓
continuous $\forall x \in R$ and non differentiable at $x= \pm 1$
D
not differentiable at $x = 0$

Answer

Correct option: C.

continuous $\forall x \in R$ and non differentiable at $x= \pm 1$

Given $g ( x )=\left\{\begin{array}{ll}e^{2 x}, & x<0 \\ e^{-2 x}, & x \geq 0\end{array}\right. g ^{\prime}( x )=\left\{\begin{aligned} 2 e^{2 x}, & x<0 \\ -2 e^{-2 x}, & x \geq 0\end{aligned}\right.$
$\therefore \text{LHD}$ at $x =0, g^{\prime}(0)=2 e ^{2 \times 0}=2 e ^0=2$
$\text{RHD}$ at $x=0, g^{\prime}(0)=-2 e^0=-2 \times 1=-2$
As $\text{LHD} \neq \text{RHD}$ at $x=0$
$\therefore g ( x )$ is not differentiable at $x =0$
Again $\text { RHL }=\lim _{x \rightarrow 0^{+}} g(x)$
$=\lim _{x \rightarrow 0^{+}} e^{-2 x}=e^0=1$
$\text { LHL }=\lim _{x \rightarrow 0^{-}} g(x)$
$=\lim _{x \rightarrow 0^{-}} e^{2 x}=e^0=1$
$g(0)=e^0=1$
As $\text { LHL }=\text { RHL }=f(0)$
$\therefore g ( x )$ is continuous $\forall x \in R$

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MCQ 161 Mark

The adjoint of the matrix $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ is

A
$\left[\begin{array}{cc}4 & -2 \\ -3 & -1\end{array}\right]$
✓
$\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]$
C
$\left[\begin{array}{cc}4 & -3 \\ -2 & 1\end{array}\right]$
D
$\left[\begin{array}{ll}4 & -2 \\ 1 & -3\end{array}\right]$

Answer

Correct option: B.

$\left[\begin{array}{cc}4 & -2 \\ -3 & 1\end{array}\right]$

Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ then $| A |=\left|\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right|$.
Now, cofactors of elements of $| A |$ are
$C_{11}=(-1)^{1+1} 4=4$
$C_{12}=(-1)^{1+2}(3)=-3$
$C_{21}=(-1)^{2+1}(2)=-2$
and $C_{22}=(-1)^{2+2}(1)=1$
Now, $\text{adj} (A) =\left[\begin{array}{ll}C_{11} & C_{12} \\ C_{21} & C_{22}\end{array}\right]^T$
$\begin{array}{l}=\left[\begin{array}{rr}4 & -3 \\ -2 & 1\end{array}\right]^T \\ =\left[\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right]\end{array}$

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MCQ 171 Mark

If A is skew symmetric matrix of order 3, then the value of |A| is

A
9
B
3
✓
$0$
D
27

Answer

Correct option: C.

$0$

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MCQ 181 Mark

If matrices A and B anticommute then

A
$( AB )=( BA )^{-1}$
B
AB = BA
C
$( AB )^{-1}=( BA )$
✓
AB = - B A

Answer

Correct option: D.

AB = - B A

(d) AB = -BA
Explanation: If A and B anticommute then AB = -BA

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Maths M.C.Q (1 Marks)

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