M.C.Q (1 Marks)

MCQ 11 Mark

The lines $l_1$ and $l_2$ intersect. The shortest distance between them is

A
infinity
B
negative
C
positive
✓
zero

Answer

Correct option: D.

zero

Since the lines intersect. Hence they have a common point in them. Hence the distance will be zero.

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MCQ 21 Mark

If $f(x)=\left\{\begin{array}{l}k x+5 \text {, when } x \leq 2 \\ x+1, \text { when } x>2\end{array}\right.$ is continuous at $x=2$ then $k= ?$

A
$-2$
✓
$-1$
C
$2$
D
$-3$

Answer

Correct option: B.

$-1$

For continuity left hand limit must be equal to right hand limit and value at the point.
Continuous at $x =2..$
$\text { L.H.L }=\lim _{x \rightarrow 2^{-}}(k x+5)$
$\Rightarrow \lim _{h \rightarrow 0}(k(2-h)+5)$
$\Rightarrow k(2-0)+5=2 k+5$
$\text { R.H.L }=\lim _{x \rightarrow 2^{+}}(x+1)$
$\Rightarrow \lim _{h \rightarrow 0}(2+h+1)$
$\Rightarrow 2+0+1$
$=3$
As $f(x)$ is continuous, we get
$\because 2 k+5=3$
$k=-1$

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MCQ 31 Mark

The area of a triangle with vertices A, B, C is given by

A
$|\overrightarrow{A B} \times \overrightarrow{A C}|$
B
$\frac{1}{8}|\overrightarrow{A C} \times \overrightarrow{A B}|$
C
$\frac{1}{4}|\overrightarrow{A C} \times \overrightarrow{A B}|$
✓
$\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$

Answer

Correct option: D.

$\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$

(d) $\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$
Explanation: The area of the parallelogram with adjacent sides $A B$ and $A C=|\overrightarrow{A B} \times \overrightarrow{A C}|$. Hence, the area of the triangle with vertices $A , B , C =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$.

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MCQ 41 Mark

The solution of the differential equation $\frac{d y}{d x}=e^{x-y}+x^2 e^{-y}$ is

A
$e ^{ x }+ e ^{ y }=\frac{x^3}{3}+ c$
B
$e ^{ x }- e ^{ y }=\frac{x^3}{3}+ c$
C
$y=e^{x-y}-x^2 e^{-y}+c$
✓
$e ^{ y }- e ^{ x }=\frac{x^3}{3}+C$

Answer

Correct option: D.

$e ^{ y }- e ^{ x }=\frac{x^3}{3}+C$

We have, $\frac{d y}{d x}=e^{x-y}+x^2 e^{-y}$
$\Rightarrow e^{y} dy=\left(e^{x}+x^2\right) dx$
$\Rightarrow \int e^{y} d y=\int\left(e^x+x^2\right) d x$
$\Rightarrow e^{y}=e^{x}+\frac{x^3}{3}+c$
$\Rightarrow e^{y}-e^{x}=\frac{x^3}{3}+c$

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MCQ 51 Mark

If $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{3}{5}$, then $P(B / A)+P(A / B)$ equals

A
$\frac{1}{3}$
B
$\frac{1}{4}$
C
$\frac{5}{12}$
✓
$\frac{7}{12}$

Answer

Correct option: D.

$\frac{7}{12}$

Here, $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{3}{5}$
$P(B / A)+P(A / B)=\frac{P(B \cap A)}{P(A)}+\frac{P(A \cap B)}{P(B)}$
$=\frac{P(A)+P(B)-P(A \cup B)}{P(A)}+\frac{P(A)+P(B)-P(A) B)}{P(B)}$
$=\frac{\frac{3}{10}+\frac{2}{5}-\frac{3}{5}}{\frac{3}{10}}+\frac{\frac{3}{10}+\frac{2}{5}-\frac{3}{5}}{\frac{2}{5}}$
$=\frac{\frac{1}{10}}{\frac{3}{10}}+\frac{\frac{1}{10}}{\frac{2}{5}}$
$=\frac{1}{3}+\frac{1}{4}$
$=\frac{7}{12}$

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MCQ 61 Mark

$\left|\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right|=?$

A
$\left(a^2-b^2-c^2-d^2\right)$
B
$\left(a^2+b^2+c^2-d^2\right)$
C
$\left(a^2-b^2+c^2-d^2\right)$
✓
$\left(a^2+b^2+c^2+d^2\right)$

Answer

Correct option: D.

$\left(a^2+b^2+c^2+d^2\right)$

(d) $\left(a^2+b^2+c^2+d^2\right)$
Explanation: $\Delta=( a + ib )( a - ib )+( c - id )( c + id )=\left( a ^2+ b ^2+ c ^2+ d ^2\right)$

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MCQ 71 Mark

If the vectors $4 \hat{i}+11 \hat{j}+m \hat{k}, 7 \hat{i}+2 \hat{j}+6 \hat{k}$ and $\hat{i}+5 \hat{j}+4 \hat{k}$ are coplanar, then $m =$

A
$38$
B
$-10$
C
$0$
✓
none of these

Answer

Correct option: D.

none of these

Given vectors $4 \vec{i}+11 \vec{j}+m \vec{k} \cdot 7 \vec{i}+2 \vec{j}+6 \vec{k}$ and $\vec{i}+5 \vec{j}+4 \vec{k}$ are coplanar then
$ \left|\begin{array}{ccc} 4 & 11 & m \\ 7 & 2 & 6 \\ 1 & 5 & 4 \end{array}\right|=0$
$\Longrightarrow 4(8-30)-11(28-6)+m(35-2)=0$
$-88-242+33 m=0$
$-330+33 m=0$
$m=10$

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MCQ 81 Mark

The solution set of the inequality $3 x+5 y<4$ is

A
an open half-plane not containing the origin.
✓
an open half-plane containing the origin.
C
a closed half plane containing the origin.
D
the whole XY-plane not containing the line
$
3 x+5 y=4
$

Answer

Correct option: B.

an open half-plane containing the origin.

(b) an open half-plane containing the origin.
Explanation: The strict inequality represents an open half plane and it contains the origin as $(0,0)$ satisfies it.

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MCQ 91 Mark

the order of the single matrix obtained from
$\left[\begin{array}{cc}1 & -1 \\ 0 & 2 \\ 2 & 3\end{array}\right]_{3 \times 2}\left\{\left[\begin{array}{ccc}-1 & 0 & 2 \\ 2 & 0 & 1\end{array}\right]_{2 \times 3}-\left[\begin{array}{lll}0 & 1 & 23 \\ 1 & 0 & 21\end{array}\right]_{2 \times 3}\right\}$ is

A
$2 \times 3$
B
$3 \times 3$
C
$3 \times 2$
D
$2 \times 2$

Answer

(b) $3 \times 3$$
\begin{array}{l}
\text { Explanation: }\left[\begin{array}{cc}
1 & -1 \\
0 & 2 \\
2 & 3
\end{array}\right]_{3 \times 2}\left\{\left[\begin{array}{ccc}
-1 & 0 & 2 \\
2 & 0 & 1
\end{array}\right]_{2 \times 3}-\left[\begin{array}{lll}
0 & 1 & 23 \\
1 & 0 & 21
\end{array}\right]_{2 \times 3}\right\} \\
=\left[\begin{array}{cc}
1 & -1 \\
0 & 2 \\
2 & 3
\end{array}\right]_{3 \times 2}\left[\begin{array}{ccc}
-1 & -1 & -21 \\
1 & 0 & -20
\end{array}\right]_{2 \times 3} \\
=\left[\begin{array}{ccc}
-2 & -1 & -1 \\
2 & 0 & -40 \\
1 & -2 & -102
\end{array}\right]_{3 \times 3}
\end{array}
$

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MCQ 101 Mark

$\int_0^{\frac{\pi}{4}} \frac{e^{\tan x}}{\cos ^2 x} d x=?$

✓
$(e-1)$
B
$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
C
$(e+1)$
D
$\left(\frac{1}{e}-1\right)$

Answer

Correct option: A.

$(e-1)$

Explanation: $I=\int_0^{\frac{\pi}{4}} e^{\tan x} \sec ^2 x d x$
Let, $\tan x = t$,
Differentiating both side with respect to $t$
$\sec ^2 x \frac{d x}{d t}=1$
$\Rightarrow \sec ^2 x d x=d t$
At $x=0, t=0$
At $x=\frac{\pi}{4}, t=1$
$I=\int_0^1 e^t d t$
$=e^{t_0^1}$
$= e ^1- e ^0$
$= e -1$

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MCQ 111 Mark

Which of the following is the principal value branch of $\operatorname{cosec}^{-1} x$ ?

A
$\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
✓
$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
C
$[0, \pi]-\left\{\frac{\pi}{2}\right\}$
D
$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

Answer

Correct option: B.

$\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$

(b) $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
Explanation: We know that the principal value branch of $\operatorname{cosec}^{-1} x$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$

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MCQ 121 Mark

The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20) is

✓
q=3p
B
q=2p
C
p=q
D
p=2q

Answer

Correct option: A.

q=3p

(a) $q=3 p$
Explanation: Since $Z$ occurs maximum at $(15,15)$ and $(0,20)$, therefore, $15 p+15 q=0 p+20 q \Rightarrow q=3 p$.

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MCQ 131 Mark

What is integrating factor of $\frac{d y}{d z}+y \sec x=\tan x$ ?

A
sec x
B
$e^{\sec x}$
✓
$\sec x+\tan x$
D
$\log (\sec x+\tan x)$

Answer

Correct option: C.

$\sec x+\tan x$

(c) $\sec x+\tan x$
Explanation: We have,
$
\frac{d y}{d x}+y \sec x=\tan x
$
Comparing with $\frac{d y}{d x}+ Py = Q$
$
=\sec x, Q=\tan x
$
I. $F \cdot=e^{\int \sec x d x}=e^{\log (\sec x+\tan x)}=\sec x+\tan x$

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MCQ 141 Mark

The direction ratios of a line parallel to z -axis are:

A
<0,0,0>
B
<1,1,0>
C
<1,1,1>
✓
<0,0,1>

Answer

Correct option: D.

<0,0,1>

(d) $\langle 0,0,1\rangle$
Explanation: $\langle 0,0,1\rangle$

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MCQ 151 Mark

If $y=\log \left(\sin e^x\right)$, then $\frac{d y}{d x}$ is:

A
$\cot e^{x}$
✓
$e ^{ x } \cot e ^{ x }$
C
$\operatorname{cosec} e ^{ x }$
D
$e ^{ x } \operatorname{cosec} e ^{ x }$

Answer

Correct option: B.

$e ^{ x } \cot e ^{ x }$

$e ^{ x } \cot e ^{ x }$
$y=\log \left(\sin e^{x}\right)$
$\frac{d y}{d x}=\frac{d}{d x} \log \left(\sin e^{x}\right)$
$=\frac{1}{\sin e^2} \frac{d}{d x} \sin e^{x}$
$=\frac{1}{\sin e^2} \cos e^x \frac{d}{d x} e^{x}$
$=\cot e^x\left(e^x\right)$
$=e^{x} \cot e^{x}$

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MCQ 161 Mark

If $A$ is a $3 \times 3$ matrix and $|A|=-2$, then value of $|A(\operatorname{adj} A)|$ is

A
-2
B
8
C
2
✓
-8

Answer

Correct option: D.

-8

(d)-8
Explanation:-8

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MCQ 171 Mark

Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\operatorname{det}\left(A B A^{\prime}\right)=8$ and $\operatorname{det}\left(A B^{-1}\right)=8$, then $\operatorname{det}\left(B A^{-1} B^{\prime}\right)$ is equal to

A
16
B
$\frac{1}{4}$
C
1
✓
$\frac{1}{16}$

Answer

Correct option: D.

$\frac{1}{16}$

(d) $\frac{1}{16}$
Explanation: $\frac{1}{16}$

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MCQ 181 Mark

If $A=\left[a_{i j}\right]$ is a scalar matrix of order $n \times n$ such that $a_{i j}=k$ for all $i$, then trace of $A$ is equal to

A
$\frac{n}{k}$
B
$n-k$
✓
$nk$
D
$n + k$

Answer

Correct option: C.

$nk$

$\because A =\left[ a _{ ij }\right]_{ n \times n }$
Trace of A, i.e., $\operatorname{tr}( A )=\sum a_{i j}^n i=1= a _{11}+ a _{22}+\ldots \ldots \ldots+ a _{ nm }$
$ =k+k+k+k+k+\ldots(n \text { times })$
$=k(n)$
$=nk $

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Mathematics M.C.Q (1 Marks)

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