MCQ 11 Mark
Direction cosines of a line perpendicular to both x-axis and z-axis are
- ✓0, 1, 0
- B0, 0, 1
- C1, 1, 1
- D1, 0, 1
Answer
View full question & answer→Correct option: A.
0, 1, 0
(a) 0, 1, 0
Explanation: 0, 1, 0
Explanation: 0, 1, 0
Questions
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18 questions · timed · auto-graded
MCQ 21 Mark
If $f(x)=\left\{\begin{array}{cl}\frac{1}{1+e^{1 x}} & , x \neq 0 \\ 0 & , x=0\end{array}\right.$ then $f ( x )$ is
- ✓none of these
- Bdifferentiable but not continuous at $x = 0$
- Ccontinuous but not differentiable at $x = 0$
- Dcontinuous as well as differentiable at $x = 0$
Answer
View full question & answer→Correct option: A.
none of these
Given that $f(x)=\left\{\frac{1}{1+e^{\frac{1}{x}}}, x \neq 0 0, x=0\right\}$
Checking continuity at $x =0$,
$\text{LHL}: \lim _{x \rightarrow 0^{-}} \frac{1}{1+e^{\frac{1}{x}}}=1$
But $f(x = 0) = 0$
Hence, function is neither continuous nor differentiable at $x = 0$
Checking continuity at $x =0$,
$\text{LHL}: \lim _{x \rightarrow 0^{-}} \frac{1}{1+e^{\frac{1}{x}}}=1$
But $f(x = 0) = 0$
Hence, function is neither continuous nor differentiable at $x = 0$
MCQ 31 Mark
If the position vectors of P and Q are $\hat{i}+3 \hat{j}-7 \hat{k}$ and $5 \hat{i}-2 \hat{j}+4 \hat{k}$ respectively, then the cosine of the angle between $\overrightarrow{P Q}$ and $y$-axis is
- A$\frac{4}{\sqrt{162}}$
- B$\frac{11}{\sqrt{162}}$
- C$\frac{5}{\sqrt{162}}$
- ✓$-\frac{5}{\sqrt{162}}$
Answer
View full question & answer→Correct option: D.
$-\frac{5}{\sqrt{162}}$
(d) $-\frac{5}{\sqrt{162}}$
Explanation: Given position vectors $\overrightarrow{O P}=\hat{i}+3 \hat{j}-7 \hat{k}$ and $\overrightarrow{O Q}=5 \hat{i}-2 \hat{j}+4 \hat{k}$
$\Longrightarrow d r s$ of $\overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P}=4 \hat{i}-5 \hat{j}+11 \hat{k}$ and drs along with y-axis are $(0,1,0)$ or $\hat{j}$ direction cosines between $\overrightarrow{P Q}$ and y-axis is $\frac{(4 \hat{i}-5 \dot{j}+11 k) \cdot \hat{j}}{\sqrt{16+25+121}}=\frac{-5}{\sqrt{162}}$
Explanation: Given position vectors $\overrightarrow{O P}=\hat{i}+3 \hat{j}-7 \hat{k}$ and $\overrightarrow{O Q}=5 \hat{i}-2 \hat{j}+4 \hat{k}$
$\Longrightarrow d r s$ of $\overrightarrow{P Q}=\overrightarrow{O Q}-\overrightarrow{O P}=4 \hat{i}-5 \hat{j}+11 \hat{k}$ and drs along with y-axis are $(0,1,0)$ or $\hat{j}$ direction cosines between $\overrightarrow{P Q}$ and y-axis is $\frac{(4 \hat{i}-5 \dot{j}+11 k) \cdot \hat{j}}{\sqrt{16+25+121}}=\frac{-5}{\sqrt{162}}$
MCQ 41 Mark
Which of the following is the general solution of $\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+y=0 ?$
- A$y=A \cos x+B \sin x$
- B$y=(A x+B) e^{-x}$
- C$y=A e^x+B e^{-x}$
- ✓$y=(A x+B) e^x$
Answer
View full question & answer→Correct option: D.
$y=(A x+B) e^x$
For $y=(A x+B) e^x$
$\frac{d y}{d x}=( Ax + B ) e ^{ x }+ Ae ^{ x }$
$=( Ax + A + B ) e ^{ x }$
$\Rightarrow \frac{d^2 y}{d x^2}=( Ax + A + B ) e ^{ x }+ Ae ^{ x }$
$=( Ax +2 A+ B ) e ^{ x }$
$\therefore \frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}\right)+y$
$=( Ax +2 A+ B ) e ^{ x }-2( Ax + A + B ) e ^{ x }+( Ax + B ) e ^{ x }$
$=0$
$\frac{d y}{d x}=( Ax + B ) e ^{ x }+ Ae ^{ x }$
$=( Ax + A + B ) e ^{ x }$
$\Rightarrow \frac{d^2 y}{d x^2}=( Ax + A + B ) e ^{ x }+ Ae ^{ x }$
$=( Ax +2 A+ B ) e ^{ x }$
$\therefore \frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}\right)+y$
$=( Ax +2 A+ B ) e ^{ x }-2( Ax + A + B ) e ^{ x }+( Ax + B ) e ^{ x }$
$=0$
MCQ 51 Mark
You are given that $A$ and $B$ are two events such that $P ( B )=\frac{3}{5}, P ( A \mid B )=\frac{1}{2}$ and $P ( A \cup B )=\frac{4}{5}$, then $P ( A )$ equals
- ✓$\frac{1}{2}$
- B$\frac{1}{5}$
- C$\frac{3}{5}$
- D$\frac{3}{10}$
Answer
View full question & answer→Correct option: A.
$\frac{1}{2}$
We have,
$P(B)=\frac{3}{5}, P(A \mid B)$ and $P(A \cup B)=\frac{4}{5}$
Now, We know that
$P(A \mid B) \times P(B)=P(A \cap B)$
$[$Property of conditional Probability$]$
$\Rightarrow \frac{1}{2} \times \frac{3}{5}=P(A \cap B)$
$\Rightarrow P(A \cap B)=\frac{3}{10}$
Now,
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$($Additive Law of Probability$)$
$\therefore \frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}$
$\Rightarrow P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}$
$\Rightarrow P(A)=\frac{1}{5}+\frac{3}{10}$
$\Rightarrow P(A)=\frac{2+3}{10}$
$\Rightarrow P(A)=\frac{5}{10}=\frac{1}{2}$
$P(B)=\frac{3}{5}, P(A \mid B)$ and $P(A \cup B)=\frac{4}{5}$
Now, We know that
$P(A \mid B) \times P(B)=P(A \cap B)$
$[$Property of conditional Probability$]$
$\Rightarrow \frac{1}{2} \times \frac{3}{5}=P(A \cap B)$
$\Rightarrow P(A \cap B)=\frac{3}{10}$
Now,
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$($Additive Law of Probability$)$
$\therefore \frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}$
$\Rightarrow P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}$
$\Rightarrow P(A)=\frac{1}{5}+\frac{3}{10}$
$\Rightarrow P(A)=\frac{2+3}{10}$
$\Rightarrow P(A)=\frac{5}{10}=\frac{1}{2}$
MCQ 61 Mark
If, $A$ is square matrix of order $3$ such that $|A|=3$, then $|\operatorname{adj} A|$ is equal to$........$
- A$3$
- ✓$9$
- C$27$
- D$10$
Answer
View full question & answer→Correct option: B.
$9$
$\because|\operatorname{adj} A|=|A|^{n-1}$
Here, $n =3$
$\therefore|\operatorname{adj} A|=|A|^{3-1}$
$=|A|^2=(3)^2=9$
Here, $n =3$
$\therefore|\operatorname{adj} A|=|A|^{3-1}$
$=|A|^2=(3)^2=9$
MCQ 71 Mark
What is the projection of $\vec{a}=(2 \hat{i}-\hat{j}+\hat{k})$ on $\vec{b}=(\hat{i}-2 \hat{j}+\hat{k}) ?$
- A$\frac{3}{\sqrt{5}}$
- B$\frac{4}{\sqrt{5}}$
- ✓$\frac{5}{\sqrt{6}}$
- D$\frac{2}{\sqrt{3}}$
Answer
View full question & answer→Correct option: C.
$\frac{5}{\sqrt{6}}$
(c) $\frac{5}{\sqrt{6}}$
Explanation: Given vectors $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$
Now, projection $\vec{a}$ on $\vec{b}$ is $\vec{a} . \hat{b}=\frac{2+2+1}{\sqrt{4+1+1}}=\frac{5}{\sqrt{6}}$
Explanation: Given vectors $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$
Now, projection $\vec{a}$ on $\vec{b}$ is $\vec{a} . \hat{b}=\frac{2+2+1}{\sqrt{4+1+1}}=\frac{5}{\sqrt{6}}$
MCQ 81 Mark
A Linear Programming Problem is as follows:
Minimize Z = 2x + y
Subject to the constraints $x \geq 3, x \leq 9, y \geq 0$
$x-y \geq 0, x+y \leq 14$
The feasible region has
Minimize Z = 2x + y
Subject to the constraints $x \geq 3, x \leq 9, y \geq 0$
$x-y \geq 0, x+y \leq 14$
The feasible region has
- A5 corner points including (0, 0) and (9, 5)
- B5 corner points including (7, 7) and (3, 3)
- C5 corner points including (3, 6) and (9, 5)
- D5 corner points including (14, 0) and (9, 0)
MCQ 91 Mark
Let $A=\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|, $ then
- A$A^2=I$
- B$A ^2=4$
- ✓$A ^2= A$
- D$A ^2=0$
Answer
View full question & answer→Correct option: C.
$A ^2= A$
$A^2=\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|=\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|=A$
MCQ 101 Mark
$\int \sec ^2(7-4 x) d x=?$
- A$-4 \tan (7-4 x)+C$
- ✓$\frac{-1}{4} \tan (7-4 x)+C$
- C$4 \tan (7-4 x)+C$
- D$\frac{1}{4} \tan (7-4 x)+C$
Answer
View full question & answer→Correct option: B.
$\frac{-1}{4} \tan (7-4 x)+C$
Given integral is $\int \sec ^2(7-4 x) d x=$ ?
Let, $7-4 x = z$
$\Rightarrow-4 dx=dz$
So, $\int \sec ^2(7-4 x) dx=?$
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$\int \sec ^2(7-4 x) d x$ where $c$ is the integrating constant.
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$=-\frac{1}{4} \tan z+c$
$=-\frac{1}{4} \tan (7-4 x)+c$
Let, $7-4 x = z$
$\Rightarrow-4 dx=dz$
So, $\int \sec ^2(7-4 x) dx=?$
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$\int \sec ^2(7-4 x) d x$ where $c$ is the integrating constant.
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$=-\frac{1}{4} \tan z+c$
$=-\frac{1}{4} \tan (7-4 x)+c$
MCQ 111 Mark
Consider the vectors $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$ and $b =4 \hat{i}-4 \hat{j}+7 \hat{k}$.
What is the vector perpendicular to both the vectors?
What is the vector perpendicular to both the vectors?
- A$-10 \hat{i}+3 \hat{j}+4 \hat{k}$
- B$10 \hat{i}-3 \hat{j}+4 \hat{k}$
- ✓$-10 \hat{i}-3 \hat{j}+4 \hat{k}$
- D$10 \hat{i}-3 \hat{j}-4 \hat{k}$
Answer
View full question & answer→Correct option: C.
$-10 \hat{i}-3 \hat{j}+4 \hat{k}$
The vector perpendicular to both the vectors $\vec{a}$ and $\vec{b}=\vec{a} \times \vec{b}$
$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & -2 & 1 \\4 & -4 & 7\end{array}\right| $
$=i(-14+4)-\hat{j}(7-4)+\hat{k}(-4+8) \\=-10 \hat{i}-3\hat{j}+4\hat{k}$
$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & -2 & 1 \\4 & -4 & 7\end{array}\right| $
$=i(-14+4)-\hat{j}(7-4)+\hat{k}(-4+8) \\=-10 \hat{i}-3\hat{j}+4\hat{k}$
MCQ 121 Mark
Minimize $Z=5 x+10$ y subject to $x+2 y \leq 120, x+y \geq 60, x-2 y \geq 0, x, y \geq 0$
- AMinimum Z = 310 at (60, 0)
- BMinimum Z = 320 at (60, 0)
- CMinimum Z = 330 at (60, 0)
- ✓Minimum Z = 300 at (60, 0)
Answer
View full question & answer→Correct option: D.
Minimum Z = 300 at (60, 0)
(d) Minimum Z = 300 at (60, 0)
Explanation: Objective function is Z = 5x + 10y ................. (1)
The given constraints are$: x+2 y \leq 120, x+y \geq 60, x-2 y \geq 0, x, y \geq 0$.
The corner points are obtained by drawing the lines x + 2y = 120 x + y = 60 and x - 2y = 0.
The points so obtained are (60,30), (120,0), (60,0) and (40,20)
Here, Z = 300 is minimum at (60, 0).
Explanation: Objective function is Z = 5x + 10y ................. (1)
The given constraints are$: x+2 y \leq 120, x+y \geq 60, x-2 y \geq 0, x, y \geq 0$.
The corner points are obtained by drawing the lines x + 2y = 120 x + y = 60 and x - 2y = 0.
The points so obtained are (60,30), (120,0), (60,0) and (40,20)
| Corner points | Z = 5x + 10y |
| D(60.3) | 600 |
| A(120, 0) | 600 |
| B(60, 0) | 300………………(min.) |
| C(40,20) | 400 |
MCQ 131 Mark
What is the degree of the differential equation $y =x \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{-1} ?$
- A$-1$
- B$1$
- Cdoes not exist
- ✓$2$
Answer
View full question & answer→Correct option: D.
$2$
Given differential equation is
$ y = x \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{-1}$
$\Rightarrow y = x \frac{d y}{d x}+\frac{1}{(d y / d x)}$
$\Rightarrow y \left(\frac{d p}{d x}\right)= x \left(\frac{d y}{d x}\right)^2+1$
$\therefore$ Degree $=$ Power of highest derivative $=2$
$ y = x \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{-1}$
$\Rightarrow y = x \frac{d y}{d x}+\frac{1}{(d y / d x)}$
$\Rightarrow y \left(\frac{d p}{d x}\right)= x \left(\frac{d y}{d x}\right)^2+1$
$\therefore$ Degree $=$ Power of highest derivative $=2$
MCQ 141 Mark
The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is
- A$5 \sqrt{30}$
- B$\sqrt{30}$
- C$2 \sqrt{30}$
- ✓$3 \sqrt{30}$
Answer
View full question & answer→Correct option: D.
$3 \sqrt{30}$
(d) $3 \sqrt{30}$
Explanation: Use formula for shortest distance between two skew lines.
Explanation: Use formula for shortest distance between two skew lines.
MCQ 151 Mark
If $f(x)=x^2 \sin \frac{1}{x}$ where $x \neq 0$ then the value of the function $f$ at $x =0$, so that the function is continuous at $x = 0$ , is
- A$-1$
- B$1$
- ✓$0$
- D$2$
Answer
View full question & answer→Correct option: C.
$0$
We have, $f(x)=x^2 \sin \left(\frac{1}{x}\right)$, where $x \neq 0$
Since the function is continuous at $x=0$,
we have $f(0)=\lim _{x \rightarrow 0} x^2 \sin \left(\frac{1}{x}\right)....(i)$
Now, $-1 \leq \sin \frac{1}{x} \leq 1$
$\Rightarrow-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$
$\Rightarrow \lim _{x \rightarrow 0}\left(-x^2\right) \leq \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0}\left(x^2\right)$
$\Rightarrow 0 \leq \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right) \leq 0$
Therefore by squeeze principle, we have$
f(0)=\lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right)=0$
Hence, value of the function $f$ at $x=0$ so that it is continuous at $x=0$ is $0 .$
Since the function is continuous at $x=0$,
we have $f(0)=\lim _{x \rightarrow 0} x^2 \sin \left(\frac{1}{x}\right)....(i)$
Now, $-1 \leq \sin \frac{1}{x} \leq 1$
$\Rightarrow-x^2 \leq x^2 \sin \frac{1}{x} \leq x^2$
$\Rightarrow \lim _{x \rightarrow 0}\left(-x^2\right) \leq \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0}\left(x^2\right)$
$\Rightarrow 0 \leq \lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right) \leq 0$
Therefore by squeeze principle, we have$
f(0)=\lim _{x \rightarrow 0}\left(x^2 \sin \frac{1}{x}\right)=0$
Hence, value of the function $f$ at $x=0$ so that it is continuous at $x=0$ is $0 .$
MCQ 161 Mark
If $A=\left|\begin{array}{lll}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right|$ then what is the value of $A^{-1}$.
- ✓$A^2$
- B$A ^3$
- C$A$
- D$I$
Answer
View full question & answer→Correct option: A.
$A^2$
$A=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right|,|A|=0+1=1 $
$A^2=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\ \end{array}\right|\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right|$
$=\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right| $
$\operatorname{adj}(A)=\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right|$
$\therefore A^{-1}=\frac{\operatorname{aj} / A}{|A|}=\frac{1}{1}\left[\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\
1 & -1 & 1 \end{array}\right]$
$=\left[\begin{array}{ccc}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right]=A^2$
$A^2=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\ \end{array}\right|\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right|$
$=\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right| $
$\operatorname{adj}(A)=\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right|$
$\therefore A^{-1}=\frac{\operatorname{aj} / A}{|A|}=\frac{1}{1}\left[\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\
1 & -1 & 1 \end{array}\right]$
$=\left[\begin{array}{ccc}0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right]=A^2$
MCQ 171 Mark
If A is a 2-rowed square matrix and IAI = 6 then A adj A = ?
- A$\left[\begin{array}{ll}\frac{1}{6} & 0 \\ 0 & \frac{1}{6}\end{array}\right]$
- B$\left[\begin{array}{ll}3 & 0 \\ 3 & 0\end{array}\right]$
- C$\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$
- ✓$\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]$
Answer
View full question & answer→Correct option: D.
$\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]$
(d) $\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]$
Explanation: A.(adj A) = |A|I
$=6\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
$=\left(\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right)$
Explanation: A.(adj A) = |A|I
$=6\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
$=\left(\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right)$
MCQ 181 Mark
If $A, B$ are square matrices of order $3, A$ is non$-$singular and $AB = 0$ then $B$ is a
- Anon$-$singular matrix
- ✓null matrix
- Csingular matrix
- Dunit matrix
Answer
View full question & answer→Correct option: B.
null matrix
Since $AB = 0$ Given A is non$-$singular.
Therefore
$A^{-1}$ exist
Now, $A B=0$
$\Rightarrow A \times(A B)=0$
$\Rightarrow(A \times A) B=0$
$\Rightarrow I . B=0$
$\Rightarrow B=0$
Thus, $B$ must be null matrix.
Therefore
$A^{-1}$ exist
Now, $A B=0$
$\Rightarrow A \times(A B)=0$
$\Rightarrow(A \times A) B=0$
$\Rightarrow I . B=0$
$\Rightarrow B=0$
Thus, $B$ must be null matrix.