M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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18 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 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 $A^{-1}$.

Answer

$(a)A ^2$
Explanation : $ A=\left|\begin{array}{ccc} 1 & -1 & 1 \\2 & -1 & 0 \\ 1 & 0 & \end{array}\right|,$
$|A|=0+1=1 \\ A^2=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right| \left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right| =\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right|$
$\operatorname{adj}(A)=\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right|$
$\therefore A^{-1}=\frac{\operatorname{aj} / A}{|A|}=\frac{1}{1}\left[\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\1 & -1 & 1 \end{array}\right]$
$=\left[\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right]$
$=A^2$

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Question 21 Mark

Minimize $Z=5 x+10$ y subject to $x+2 y \leq 120, x+y \geq 60, x-2 y \geq 0, x, y \geq 0$

Answer

(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)

Corner points	Z = 5x + 10y
D(60.3)	600
A(120, 0)	600
B(60, 0)	300………………(min.)
C(40,20)	400

Here, Z = 300 is minimum at (60, 0).

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Question 31 Mark

Direction cosines of a line perpendicular to both x-axis and z-axis are

Answer

(a) 0, 1, 0
Explanation: 0, 1, 0

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Question 41 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

Answer

Given that $f(x)=\left\{\begin{array}{l}\frac{1}{1+e^{\frac{1}{x}}}, x \neq 0 \\ 0, x=0\end{array}\right\}$
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 $x = 0$

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Question 51 Mark

Which of the following is the general solution of $\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+y=0 ?$

Answer

$(d) y=(A x+B) e^x$
Explanation: For $y=(A x+B) e^x$
$\frac{d y}{d x}=( Ax + B ) e ^{ x }+ Ae ^{ x }=( Ax + A + B ) e ^{ x }$
$\Rightarrow \frac{d^2 y}{d x^2}=( Ax + A + B ) e ^{ x }+ Ae ^{ x }=( Ax +2 A+ B ) e ^{ x }$
$\therefore \frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}\right)+y$
$=( Ax +2 A+ B ) e ^{ x }-2( Ax + A + B ) e ^{ x }+( Ax + B ) e ^{ x }$
$=0$

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Question 61 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

Answer

(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 $\frac{(4 \hat{i}-5 \dot{j}+11 k) \cdot \hat{j}}{\sqrt{16+25+121}}=\frac{-5}{\sqrt{162}}$

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Question 71 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

Answer

$(a) \frac{1}{2}$
Explanation: We have,
$P(B)=\frac{3}{5}, P(A \mid B) \text { and } P(A \cup B)=\frac{4}{5}$
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)
$\therefore \frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}$
$\Rightarrow P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}$
$\Rightarrow P(A)=\frac{1}{5}+\frac{3}{10}$
$\Rightarrow P(A)=\frac{2+3}{10}$
$\Rightarrow P(A)=\frac{5}{10}=\frac{1}{2}$

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Question 81 Mark

What is the projection of $\vec{a}=(2 \hat{i}-\hat{j}+\hat{k})$ on $\vec{b}=(\hat{i}-2 \hat{j}+\hat{k}) ?$

Answer

(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 $\vec{a} . \hat{b}=\frac{2+2+1}{\sqrt{4+1+1}}=\frac{5}{\sqrt{6}}$

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Question 91 Mark

If, $A$ is square matrix of order 3 such that $|A|=3$, then $|\operatorname{adj} A|$ is equal to$ ......$

Answer

$(b) 9$
Explanation:$\because|\operatorname{adj} A|=|A|^{n-1}$
Here, $n =3$
$\therefore|\operatorname{adj} A|=|A|^{3-1}$
$=|A|^2=(3)^2=9$

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Question 101 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

Answer

(b) 5 corner points including (7, 7) and (3, 3)
Explanation:

On plotting the constraints x = 3 x = 9 x = y and x + y = 14 we get the following graph. From the graph given below it, clear that feasible region is ABCDEA, including corner points A(9, 0) B(3, 0) C(3,3) D(7, 7) and E(9, 5) Thus feasible region has 5 corner points including (7, 7) and (3, 3).

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Question 111 Mark

Let $A=\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|,$ then

Answer

$A^2=\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|$
$=\left|\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right|$
$=A$

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Question 121 Mark

$\int \sec ^2(7-4 x) d x=?$

Answer

$(b) \frac{-1}{4} \tan (7-4 x)+C$
Explanation: Given integral is $\int \sec ^2(7-4 x) d x=$ ?
Let, $7-4 x = z$
$\Rightarrow-4 dx=dz$
So,
$\int \sec ^2(7-4 x) dx=?$
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$\int \sec ^2(7-4 x) d x \text { where } c \text { is the integrating constant. }$
$=\int \sec ^2 z \frac{d z}{-4}$
$=-\frac{1}{4} \int \sec ^2 z d z$
$=-\frac{1}{4} \tan z+c$
$=-\frac{1}{4} \tan (7-4 x)+c$

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Question 131 Mark

What is the degree of the differential equation $y =x \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{-1}$ ?

Answer

$(d) 2$
Explanation: 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 $=2$

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Question 141 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?

Answer

$(c) -10 \hat{i}-3 \hat{j}+4 \hat{k}$
The vector perpendicular to both the vectors $\vec{a}$ and $\vec{b}=\vec{a} \times \vec{b}$
${l}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & -2 & 1 \\4 & -4 & 7\end{array}\right|$
$=i(-14+4)-\hat{j}(7-4)+\hat{k}(-4+8) $
$=-10 \hat{i}-3\hat{j}+4\hat{k}$

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Question 151 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

Answer

(d) $3 \sqrt{30}$
Explanation: Use formula for shortest distance between two skew lines.

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Question 161 Mark

If $A, B$ are square matrices of order $3, A$ is non-singular and $AB = 0$ then $B$ is a

Answer

$(b)$ null matrix
Explanation: Since $AB = 0$ Given $A$ is non$-$singular. Therefore
$A^{-1} \text { exist }$
$\text { Now, } A B=0$
$\Rightarrow A \times(A B)=0$
$\Rightarrow(A \times A) B=0$
$\Rightarrow I . B=0$
$\Rightarrow B=0$
Thus, $B$ must be null matrix.

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Question 171 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

Answer

$(c) 0$
Explanation: 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 $0 .$

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Question 181 Mark

If A is a 2-rowed square matrix and IAI = 6 then A adj A = ?

Answer

(d) $\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]$
Explanation: A.(adj A) = |A|I
$=6\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
$=\left(\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right)$

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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip