Question 11 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
Answer$(b) q=3 p$
Explanation: The maximum value of $Z$ is unique.
It is given that the maximum value of $Z$ occurs at two points $(3,4)$ and $(0,5)$
$\therefore \text { Value of } Z \text { at }(3,4)=\text { Value of } Z \text { at }(0,5)$
$\Rightarrow p(3)+q(4)=p(0)+q(5)$
$\Rightarrow 3 p+4 q=5 q$
$\Rightarrow q=3 p$
View full question & answer→Question 21 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?
Answer$(a)$ 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→Question 31 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
Answer(b) $\frac{11}{15}$
Explanation: $\frac{11}{15}$
View full question & answer→Question 41 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→Question 51 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:
Answer(a) $\frac{2 \pi}{3}$
Explanation: $\frac{2 \pi}{3}$
View full question & answer→Question 61 Mark
If $y=\left(1+\frac{1}{x}\right)^x$, then $\frac{d y}{d x}=$
Answer$(c) \ \left(1+\frac{1}{x}\right)^x\left\{\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right\}$
$y=\left(1+\frac{1}{x}\right)^x$
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→Question 71 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
Answer(c) $\frac{2}{3}$
Explanation: $\frac{2}{3}$
View full question & answer→Question 81 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
AnswerGiven 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] \text { 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→Question 91 Mark
If $|A|=|k A|$, where $A$ is a square matrix of order 2 , then sum of all possible values of $k$ is
View full question & answer→Question 101 Mark
The integrating factor of the differential equation $\frac{d y}{d x}+y=\frac{1+y}{x}$ is :
Answer$(c)\ \frac{e^x}{x}$
Explanation: 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
$P=\frac{-(1-x)}{x}, Q=\frac{1}{x}$
$\text { I.F. }=e^{\int P d x}=e^{-\int \frac{1-x}{x} d x}$
$=e^{x-\log x} \text { or } \frac{e^x}{x}$
View full question & answer→Question 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
View full question & answer→Question 121 Mark
The value of the determinant $\left|\begin{array}{ccc}6 & 0 & -1 \\ 2 & 1 & 4 \\ 1 & 1 & 3\end{array}\right|$ is
Answer$(b) -7$
Explanation: $-7$
$\left|\begin{array}{ccc}6 & 0 & -1 \\ 2 & 1 & 4 \\ 1 & 1 & 3\end{array}\right|$
$=6(3-4)-0(6-4)+(-1)(2-1)$
$=6(-1)+0+(-1)$
$=-6-1$
$=-7$
View full question & answer→Question 131 Mark
If $A=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|$, then $A^2$ is$:$
Answer${l}=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right| \cdot\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right|$
$=\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→Question 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
Answer(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→Question 151 Mark
Let $A$ be a $3 \times 3$ matrix such that $|\operatorname{adj} A|=64$. Then $|A|$ is equal to:
Answer$(d) 8$ or $-8$
Explanation: $8$ 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→Question 161 Mark
$\int \frac{1+\tan x}{1-\tan x} d x$ is equal to:
Answer(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→Question 171 Mark
Let $A$ be a skew-symmetric matrix of order 3. If $|A|=x$, then $(2023)^x$ is equal to:
View full question & answer→Question 181 Mark
Find the particular solution for $2 x y+y^2-2 x^2 \frac{d y}{d x}=0 ; y=2$ when $x =1$
Answer$(a) y=\frac{2}{1-\log x=1}(x \neq 0, x \neq e)$
Explanation: 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→