MCQ 11 Mark
If the line $\frac{x-2}{2 k}=\frac{y-3}{3}=\frac{z+2}{-1}$ and $\frac{x-2}{8}=\frac{y-3}{6}=\frac{z+2}{-2}$ are parallel, value of k is
Answer(b) 2
Explanation: Given lines are $\frac{x-2}{2 k}=\frac{y-3}{3}=\frac{2+2}{-1}$ and $\frac{x-2}{8}=\frac{y-3}{6}=\frac{z+2}{-2}$
The direction ratio of the first line is (2k, 3, -1) and the direction ratio of second line is (8, 6, -2). Lines are parallel:
So,
$\frac{2 k}{8}=\frac{3}{6}=\frac{-1}{-2}$
$\Rightarrow \frac{k}{4}=\frac{1}{2}=\frac{1}{2}$
∴ k = 2
View full question & answer→MCQ 21 Mark
If $y=\left(1+\frac{1}{x}\right)^x$, then $\frac{d y}{d x}=$
- A
$\left(1+\frac{1}{x}\right)^x \log \left(1+\frac{1}{x}\right)$
- B
$\left(x+\frac{1}{x}\right)^x\left\{\log \left(1+\frac{1}{x}\right)+\frac{1}{x+1}\right\}$
- ✓
$\left(1+\frac{1}{x}\right)^x\left\{\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right\}$
- D
$\left(x+\frac{1}{x}\right)^x\left\{\log (x+1)-\frac{x}{x+1}\right\}$
AnswerCorrect option: C. $\left(1+\frac{1}{x}\right)^x\left\{\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right\}$
$y=\left(1+\frac{1}{x}\right)^x$
Taking log on both sides,
$\log y=\log \left(1+\frac{1}{x}\right)^z$
$\log y=x \log \left(1+\frac{1}{x}\right)$
Differentiate with respect to $x,$
$\frac{1}{y} \frac{d y}{d x}=x \times \frac{1}{1+\frac{1}{x}} \times \frac{-1}{x^2}+\log \left(1+\frac{1}{x}\right)$
$\frac{1}{y} \frac{d y}{d x}=\frac{x^2}{x+1} \times \frac{-1}{x^2}+\log \left(1+\frac{1}{x}\right)$
$\frac{d y}{d x}=y\left(\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right)\right)$
$\frac{d y}{d x}=\left(1+\frac{1}{x}\right)^x\left(\frac{-1}{x-1}+\log \left(1+\frac{1}{x}\right)\right)$
$\frac{d y}{d x}=\left(1+\frac{1}{x}\right)^x\left(\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right)$
View full question & answer→MCQ 31 Mark
If $\vec{a}, \vec{b}$ and $(\vec{a}+\vec{b})$ are all unit vectors and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then the value of $\theta$ is:
- ✓
$\frac{2 \pi}{3}$
- B
$\frac{\pi}{3}$
- C
$\frac{5 \pi}{6}$
- D
$\frac{\pi}{6}$
AnswerCorrect option: A. $\frac{2 \pi}{3}$
(a) $\frac{2 \pi}{3}$
Explanation: $\frac{2 \pi}{3}$
View full question & answer→MCQ 41 Mark
Find the particular solution for $2 x y+y^2-2 x^2 \frac{d y}{d x}=0 ; y=2$ when $x =1$
- ✓
$y=\frac{2 x}{1-\log |x|}(x \neq 0, x \neq e)$
- B
$y=\frac{3 x}{1-\log |x|}(x \neq 0, x \neq e)$
- C
$y=\frac{3 x}{1-\log |x|}(x \neq 0, x \neq e)$
- D
$y=\frac{5 x}{1+\log |x|}(x \neq 0, x \neq e)$
AnswerCorrect option: A. $y=\frac{2 x}{1-\log |x|}(x \neq 0, x \neq e)$
Let $y = vx$
$\frac{d y}{d x}=v+x \frac{d}{d x}$
Question becomes $v+x \frac{d y}{d z}=\frac{2 x+x^2}{2}$
$x \frac{d x}{d x}=\frac{2 v+v^2}{2}-t$
$x \frac{d x}{d x}=\frac{2 v+v^2-2}{2}$
$2 \int \frac{d}{x^2}=\int \frac{d x}{z}$
$\frac{-2}{w}=\log x+c$
When $x=1 y =2$ we get
$\frac{-2 x}{y}=\log x+c$
$\frac{-2}{2}=\log 1+c$
$\Rightarrow c=-1$
$\frac{-2 x}{y}=\log x-1$
$y=\frac{2 x}{1-\log x}$
View full question & answer→MCQ 51 Mark
If for two events $A$ and $B, P(A-B)=\frac{1}{5}$ and $P(A)=\frac{3}{5}$, then $P\left(\frac{B}{A}\right)$ is equal to
- A
$\frac{1}{2}$
- B
$\frac{2}{5}$
- ✓
$\frac{2}{3}$
- D
$\frac{3}{5}$
AnswerCorrect option: C. $\frac{2}{3}$
(c) $\frac{2}{3}$
Explanation: $\frac{2}{3}$
View full question & answer→MCQ 61 Mark
Let $X =\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right], A =\left[\begin{array}{rrr}1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1\end{array}\right]$ and $B =\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$. If $A = B,$ then $X$ is equal to
- A
$\left[\begin{array}{l}0 \\ 2 \\ 1\end{array}\right]$
- ✓
$\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right]$
- C
$\left[\begin{array}{l}-1 \\ -2 \\ -3\end{array}\right]$
- D
$\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right]$
Given that
$ X=\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right], A=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 2 & 0 & 1 \\
3 & 2 & 1 \end{array}\right] $ and $ B=\left[\begin{array}{l} 3 \\ 1 \\ 4 \end{array}\right]$
Also $A X=B$ and we have to find the value of $X,$
Pre $-$ multiplying $A ^{-1}$ both sides we gel,
$A^{-1} AX=A^{-1} B$
$IX=A^{-1} B\ \left(\because A^{-1} A=I\right)$
$X=A^{-1} B\ (\because IX=X) ..... (i)$
Now,
$ |A|=\left|\begin{array}{ccc} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{array}\right|$
$=1(0-2)+1(2-3)+2(4-0)$
$=-2-1+8=5$
And $\operatorname{adj} A=\left[\begin{array}{ccc}-2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2\end{array}\right]$
$ \therefore\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right] =\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right] $
On comparing both sides we get, $x _1=-1, x _2=2$ and $x _3=3$.
View full question & answer→MCQ 71 Mark
A unit vector â makes equal but acute angles on the co-ordinate axes. The projection of the vector a on the vector $\vec{b}=5 \hat{ i }+7 \hat{ j }-\hat{ k }$ is
- A
$\frac{3}{5 \sqrt{3}}$
- ✓
$\frac{11}{15}$
- C
$\frac{4}{5}$
- D
$\frac{11}{5 \sqrt{3}}$
AnswerCorrect option: B. $\frac{11}{15}$
(b) $\frac{11}{15}$
Explanation: $\frac{11}{15}$
View full question & answer→MCQ 81 Mark
The corner points of the feasible region determined by the following system of linear inequalities:
$2 x+y \leq 10, x+3 y \leq 15, x, y \geq 0$ are $(0,0),(5,0),(3,4)$ and $(0,5)$. Let $Z=p x+q y$, where $p, q \geq 0$.
Condition on $p$ and $q$ so that the maximum of $Z$ occurs at both $(3, 4)$ and $(0, 5)$ is
- A
$p = 3q$
- ✓
$q = 3p$
- C
$p = q$
- D
$p = 2q$
AnswerCorrect option: B. $q = 3p$
The maximum value of $Z$ is unique.
It is given that the maximum value of $Z$ occurs at two points $(3,4)$ and $(0,5)$
$\therefore$ Value of $Z \text { at }(3,4)=$ Value of $Z \text { at }(0,5)$
$\Rightarrow p(3)+q(4)=p(0)+q(5)$
$\Rightarrow 3 p+4 q=5 q$
$\Rightarrow q=3 p$
View full question & answer→MCQ 91 Mark
Let $A$ be a skew-symmetric matrix of order 3. If $|A|=x$, then $(2023)^x$ is equal to:
- A
- B
$(2023)^2$
- C
$\frac{1}{2023}$
- ✓
View full question & answer→MCQ 101 Mark
$\int \frac{1+\tan x}{1-\tan x} d x$ is equal to:
- A
$\sec ^2\left(\frac{\pi}{4}-x\right)+C$
- B
$\sec ^2\left(\frac{\pi}{4}+x\right)+C$
- C
$\log \left|\sec \left(\frac{\pi}{4}-x\right)\right|+C$
- ✓
$\log \left|\sec \left(\frac{\pi}{4}+x\right)\right|+C$
AnswerCorrect option: D. $\log \left|\sec \left(\frac{\pi}{4}+x\right)\right|+C$
(d) $\log \left|\sec \left(\frac{x}{4}+x\right)\right|+C$
Explanation: $\log \left|\sec \left(\frac{x}{4}+x\right)\right|+C$
View full question & answer→MCQ 111 Mark
If the angle between the vectors $\overrightarrow{ a }$ and $\overrightarrow{ b }$ is $\frac{\pi}{4}$ and $|\vec{a} \times \vec{b}|=1$, then $\vec{a} \cdot \vec{b}$ is equal to
- A
- B
$\frac{1}{\sqrt{2}}$
- C
$\sqrt{2}$
- ✓
View full question & answer→MCQ 121 Mark
A Linear Programming Problem is as follows:
Maximize/Minimize objective function $Z = 2x - y +5$
Subject to the constraints
$3 x+4 y \leq 60$
$x+3 y \leq 30$
$x \leq 0, y \geq 0$
In the corner points of the feasible region are $A(0, 10), B(12, 6), C(20, 0)$ and $O(0,0),$ then which of the following is true?
- ✓
Minimum value of $Z$ is $-5$
- B
At two corner points, value of $Z$ are equal
- C
Maximum value of $Z$ is $40$
- D
Difference of maximum and minimum values of $Z$ is $35$
AnswerCorrect option: A. Minimum value of $Z$ is $-5$
| Corner Points |
Value of $Z = 2x - y + 5$ |
| $A(0, 10)$ |
$Z=2(0)-10+5 = -5 ($ Minimum$)$ |
| $B(12, 6)$ |
$Z2(12)-6+ 5 = 23$ |
| $C(20, 0)$ |
$Z=2(20)-0+5 = 45 ($Maximum$)$ |
| $O(0,0)$ |
$Z=0(0)-0+5=5$ |
So the minimum value of $Z$ is $-5.$ View full question & answer→MCQ 131 Mark
The integrating factor of the differential equation $\frac{d y}{d x}+y=\frac{1+y}{x}$ is:
- A
$e ^{ X }$
- B
$\frac{x}{e^x}$
- ✓
$\frac{e^z}{x}$
- D
$xe ^{ x }$
AnswerCorrect option: C. $\frac{e^z}{x}$
We have, $\frac{d y}{d x}+y=\frac{1+y}{x}$
$\Rightarrow \frac{d y}{d x}=\frac{1}{x}+\frac{y(1-x)}{x}$
$\Rightarrow \frac{d y}{d x}-\left(\frac{1-x}{x}\right) y=\frac{1}{x}$
This is a linear differential equation.
On comparing it with $\frac{d y}{d x}+P y=Q,$ we get
$P=\frac{-(1-x)}{x}, Q=\frac{1}{x}$
$\text { I.F. }=e^{\int P d x}=e^{-\int \frac{1-x}{x} d x}$
$=e^{x-\log x}$ or $ \frac{e^x}{x}$
View full question & answer→MCQ 141 Mark
The Cartesian equations of a line are $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-5}{-1}$. Its vector equation is
- ✓
$\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$
- B
$\vec{r}=(2 \hat{i}+3 \hat{j}-\hat{k})+\lambda(\hat{i}-2 \hat{j}+5 \hat{k})$
- C
$\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-4 \hat{k})$
- D
$\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}-3 \hat{j}-4 \hat{k})$
AnswerCorrect option: A. $\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$
(a) $\vec{r}=(\hat{i}-2 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$
Explanation: Fixed point $(1,-2,5)$ and the parallel vector is $2 \hat{\imath}+3 \hat{\jmath}-\hat{ k }$
View full question & answer→MCQ 151 Mark
If $A=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|,$ then $A^2$ is:
Answer$ A^2 = A \times A$
$=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right| \cdot\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|=\left|\begin{array}{lll}1+1+1 & 1+1+1 & 1+1+1 \\ 1+1+1 & 1+1+1 & 1+1+1 \\ 1+1+1 & 1+1+1 & 1+1+1\end{array}\right| $
$ =\left|\begin{array}{lll}3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3\end{array}\right|=3\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|$
$=3A$
View full question & answer→MCQ 161 Mark
Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj} A|=64$. Then $|A|$ is equal to:
- A
$-8$ only
- B
$64$
- C
$8$ only
- ✓
$8$ or $-8$
AnswerCorrect option: D. $8$ or $-8$
We know that $|\operatorname{Adj} A|=|A|^{ n -1}, n$ is the order of the matrix.
$\because 64=|A|^{3-1}$
$|A|^2=64$
$|A|= \pm 8$
View full question & answer→MCQ 171 Mark
The value of the determinant $\left|\begin{array}{ccc}6 & 0 & -1 \\ 2 & 1 & 4 \\ 1 & 1 & 3\end{array}\right|$ is
Answer$\left|\begin{array}{ccc}6 & 0 & -1 \\ 2 & 1 & 4 \\ 1 & 1 & 3\end{array}\right|$
$=6(3-4)-0(6-4)+(-1)(2-1)$
$=6(-1)+0+(-1)$
$=-6-1$
$=-7$
View full question & answer→MCQ 181 Mark
If $|A|=|k A|$, where $A$ is a square matrix of order 2 , then sum of all possible values of $k$ is
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