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 · auto-graded multiple-choice test.

MCQ 11 Mark

If the lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular to each other then $k =$ ?

✓
$\frac{-10}{7}$
B
$\frac{5}{7}$
C
$\frac{-5}{7}$
D
$\frac{10}{7}$

Answer

Correct option: A.

$\frac{-10}{7}$

If the lines are perpendicular to each other then the angle between these lines will be
$\frac{\pi}{2}$, then the cosine will be $0$
$\vec{a}=-3 \hat{i}+2 k \hat{j}+2 \hat{k}$
$\Rightarrow|\vec{a}|=\sqrt{3^2+(2 k)^2+2^2}$
$=\sqrt{13+4 k^2}$
$\overrightarrow{b}=3 k \hat{i}+\hat{j}-5 \hat{k}$
$\Rightarrow|\overrightarrow{b}|=\sqrt{(3 k)^2+1+5^2}$
$=\sqrt{9 k^2+26}$
$\cos \left(\frac{\pi}{2}\right)=\frac{(3 k \hat{i}+\hat{j}-5 \hat{k}) \cdot(-3 \hat{i}+2 k \hat{j}+2 \hat{k})}{\sqrt{13+4 k^2} \times \sqrt{9 k^2+26}}$
$0=\frac{-9 k+2 k-10}{\sqrt{13+4 k^2} \times \sqrt{9 k^2+26}}$
$\Rightarrow k=-\frac{10}{7}$

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MCQ 21 Mark

If $x = a \sec \theta, y = b \tan \theta$ then $\frac{d y}{d x}=$ ?

A
$\frac{b}{a} \sec \theta$
B
$\frac{b}{a} \tan \theta$
✓
$\frac{b}{a} \operatorname{cosec} \theta$
D
$\frac{b}{a} \cot \theta$

Answer

Correct option: C.

$\frac{b}{a} \operatorname{cosec} \theta$

$x = a \sec \theta$, we get
$\therefore \frac{d x}{d \theta}=a \sec \theta \cdot \tan \theta$
$\therefore \frac{d \theta}{d x}=\frac{1}{asec \theta \cdot \tan \theta}$
$y=b \tan \theta,$ we get
$\therefore \frac{d y}{d \theta}=b \cdot \sec ^2 \theta$
$\Rightarrow \frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}$
$\Rightarrow \frac{d y}{d x}=b \cdot \sec ^2 \theta \times \frac{1}{asec \theta \cdot \tan \theta}$
$\Rightarrow \frac{d y}{d x}=\frac{b \sec \theta}{a \tan \theta}$
$\Rightarrow \frac{dy}{dx}=\frac{b \cdot \frac{1}{\cos \theta}}{a \cdot \frac{\sin \theta}{\cos \theta}}$
$\Rightarrow \frac{d y}{d x}=\frac{b}{a} \operatorname{cosec} \theta$

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MCQ 31 Mark

If $|\vec{a}|=4$ and $-3 \leq \lambda \leq 2$, then the range of $|\lambda \vec{a}|$ is

✓
$[0,12]$
B
$[0,8]$
C
$[8,12]$
D
$[-12,8]$

Answer

Correct option: A.

$[0,12]$

Given that, $|\vec{a}|=4$ and $-3 \leq \lambda \leq 2$
We know that, $|\lambda \vec{a}|=|\lambda||\vec{a}|$
$\Rightarrow|\lambda \vec{a}|=|-3||\vec{a}|=3.4=12 \text { at } \lambda=-3$
$\Rightarrow|\lambda \vec{a}|=|0||\vec{a}|=0.4=0 \text { at } \lambda=0$
$\Rightarrow|\lambda \vec{a}|=|2||\vec{a}|=2.4=8 \text { at } \lambda=2$
Hence, the range of $|\lambda \vec{a}|$ is $(0,12)$.

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MCQ 41 Mark

The general solution of a differential equation of the type $\frac{d x}{d y}+ P _1 x= Q _1$ is

✓
$x e^{\int P _1 d y}=\int\left( Q _1 e^{\int P _1 d y}\right) d y+ C$
B
$y e^{\int P _1 d y}=\int\left( Q _1 e^{\int P _1 d y}\right) d y+ C$
C
$y \cdot e^{\int_P d x}=\int\left(Q_1 e^{\int P_1 d x}\right) d x+C$
D
$x e^{\int P^1 d x}=\int\left(Q_1 e^{\int P_1 d x}\right) d x+C$

Answer

Correct option: A.

$x e^{\int P _1 d y}=\int\left( Q _1 e^{\int P _1 d y}\right) d y+ C$

The integrating factor of the given differential equation
$\frac{d x}{d y}+P_1 x=Q_1$ is $e^{\int P_1 d y}$
Thus, the general solution of the differential equation is given by,
$x(\text { I. F. })=\int\left(Q_1 \times \text { I.F. }\right) d y+C$
$\Rightarrow x \cdot e^{\int P_1 d y}=\int\left(Q_1 e^{\int P_1 d y}\right) d y+C$

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MCQ 51 Mark

$X$ and $Y$ are independent events such that $P(X \cap \bar{Y})=\frac{2}{5}$ and $P(X)=\frac{3}{5}$. Then $P(Y)$ is equal to:

A
$\frac{2}{3}$
✓
$\frac{1}{3}$
C
$\frac{1}{5}$
D
$\frac{2}{5}$

Answer

Correct option: B.

$\frac{1}{3}$

(b) $\frac{1}{3}$
Explanation:$\frac{1}{3}$

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MCQ 61 Mark

Adj.(KA) =____________

✓
$K ^{ n -1} Adj . A$
B
$K ^{ n +1}$ Adj. A
C
K Adj. A
D
$K ^{ n }$ Adj.A

Answer

Correct option: A.

$K ^{ n -1} Adj . A$

(a) $K ^{ n -1}$ Adj. A
Explanation: Adj. $( KA )= K ^{ n -1}$ Adj. A , where K is a scalar and A is a $n \times n$ matrix.

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MCQ 71 Mark

If the projection of $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ on $\overrightarrow{ b }=2 \hat{ i }+\lambda \hat{ k }$ is zero, then the value of $\lambda$ is:

A
$0$
B
$1$
C
$\frac{-3}{2}$
✓
$\frac{-2}{3}$

Answer

Correct option: D.

$\frac{-2}{3}$

$\frac{\vec{a} \cdot \vec{b}}{|b|}=0$
$\vec{a} \cdot \vec{b}=0$
$(\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(2 \hat{ i }+\lambda \hat{ k })=0$
$2 \times 1+3 \lambda=0$
$2+3 \lambda=0$
$3 \lambda=-2$
$\lambda=\frac{-2}{3}$

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MCQ 81 Mark

Which of the following statements is correct?
a. Every LPP admits an optimal selection.
b. A LPP admits unique optimal solution.
c. If a LPP admits two optimal solutions it has an infinite solution.
d. The set of all feasible solutions of a LPP is not a convex set.

A
Option (d)
B
Option (a)
C
Option (b)
✓
Option C

Answer

Correct option: D.

Option C

(d) Option (c)
Explanation: If a LPP admits two optimal solutions it has an infinite solution.

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MCQ 91 Mark

If $A=\left[\begin{array}{ll}5 & x \\ y & 0\end{array}\right]$ and $A=A^T$, where $A^T$ is the transpose of the matrix $A$, then

A
x = 0 y = 5
B
x = 5 y = 0
✓
x = y
D
x + y = 5

Answer

Correct option: C.

x = y

(c) $x = y$
Explanation: $A = A ^{ T }$
$\left[\begin{array}{ll}5 & x \\ y & 0\end{array}\right]=\left[\begin{array}{ll}5 & y \\ x & 0\end{array}\right]$
x=y

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MCQ 101 Mark

$\int e ^{5 \log x} dx$ is equal to:

A
$\frac{x^5}{5}+C$
B
$6 x^5+C$
✓
$\frac{x^6}{6}+C$
D
$5 x^4+C$

Answer

Correct option: C.

$\frac{x^6}{6}+C$

(c) $\frac{x^6}{6}+C$
Explanation: $\frac{x^6}{6}+C$

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MCQ 111 Mark

In $\triangle A B C, \overrightarrow{A B}=\hat{i}+\hat{j}+2 \hat{k}$ and $\overrightarrow{A C}=3 \hat{i}-\hat{j}+4 \hat{k}$. If $D$ is mid-point of $B C$, then vector $\overrightarrow{A D}$ is equal to:

A
$\hat{i}-\hat{j}+\hat{k}$
B
$2 \hat{i}-2 \hat{j}+2 \hat{k}$
C
$4 \hat{ i }+6 \hat{ k }$
✓
$2 \hat{ i }+3 \hat{ k }$

Answer

Correct option: D.

$2 \hat{ i }+3 \hat{ k }$

(d) $2 \hat{ i }+3 \hat{ k }$
Explanation: $2 \hat{ i }+3 \hat{ k }$

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MCQ 121 Mark

The corner points of the feasible region for a Linear Programming problem are P(0,5) Q(1, 5) R(4, 2) and S(12, 0) The minimum value of the objective function Z = 2x + 5y is at the point.

A
Q
B
S
✓
R
D
P

Answer

Correct option: C.

R

(c) R
Explanation:

Corner points	Value of Z = 2x + 5y
P(0, 5)	Z=2(0)+5(5) = 25
Q(1, 5)	Z=2(1)+5(5) = 27
R(4, 2)	Z=2(4)+5(2) = 18 → Minimum
S(12, 0	Z2(12)+5(0) = 24

Thus, minimum value of Z occurs ar R(4, 2)

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MCQ 131 Mark

The general solution of the differential equation $\frac{d y}{d x}=\frac{y}{x}$ is

✓
$y = kx$
B
$\log(y) - kx$
C
$\cos x$
D
$\tan x$

Answer

Correct option: A.

$y = kx$

We have,
$\frac{d y}{d x}=\frac{y}{x}$
$\Rightarrow \frac{d y}{y}=\frac{d x}{x}$
Integrating on both sides,
$\int \frac{d y}{y}=\int \frac{d x}{x}$
$\log |y|=\log |x|+\log k$
$\Rightarrow \log \left(\frac{y}{x}\right)=\log k$
$\Rightarrow y = kx$

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MCQ 141 Mark

If $\left(a_1, b_1, c_1\right)$ and $\left(a_2, b_2, c_2\right)$ be the direction ratios of two parallel lines then

A
$a_1^2+b_1^2+c_1^2=a_2^2+b_2^2+c_2^2$
✓
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
C
$a_1, a_2, b_1=b_2, c_1=c_2$
D
$a_1 a_2+b_1 b_2+c_1 c_2=0$

Answer

Correct option: B.

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

(b) $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Explanation: We know that if there are two parallel lines then their direction ratios must have a relation
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

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MCQ 151 Mark

If A is a non singular matrix and A' denotes the transpose of A, then

A
$\left| AA ^{\prime}\right| \neq\left| A ^2\right|$
B
$| A |-\left| A ^{\prime}\right| \neq 0$
✓
$| A |+\left| A ^{\prime}\right| \neq 0$
D
$| A | \neq\left| A ^{\prime}\right|$

Answer

Correct option: C.

$| A |+\left| A ^{\prime}\right| \neq 0$

(c) $| A |+\left| A ^{\prime}\right| \neq 0$
Explanation: Because, the determinant of a matrix and its transpose are always equal that is $|A|=\left|A^{\prime}\right|$

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MCQ 161 Mark

If $A \cdot(\operatorname{adj} A)=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$, then the value of $|A|+|\operatorname{adj} A|$ is equal to:

✓
$12$
B
$3$
C
$27$
D
$9$

Answer

Correct option: A.

$12$

$A \cdot(\operatorname{adj} A)=\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]$
we know that $A \cdot(\operatorname{Adj} A)=I .|A|$
$\begin{array}{l}\Longrightarrow\left[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right]=| A | I \\ \Longrightarrow 3\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=| A | I \end{array}$
$\Longrightarrow 3 I =| A | I$
$\Longrightarrow| A |=3 \cdots(1)$
$|\operatorname{Adj} A |=| A |^{3-1}[\text { Since order } n =3]$
$|\operatorname{Adj} A |=(3)^2=9$
$|\operatorname{adj}(A)|=9 ......(2)$
Now,
$|A|+|\operatorname{adj} A|=3+9=12$

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MCQ 171 Mark

Three points $P(2 x, x+3), Q(0, x)$ and $R(x+3, x+6)$ are collinear, then $x$ is equal to:

A
$2$
B
$0$
C
$3$
✓
$1$

Answer

Correct option: D.

$1$

As points are collinear
$\Rightarrow$ Area of triangle formed by $3$ points is zero.
$ \Rightarrow \frac{1}{2}\left|\begin{array}{cc} \left(x_1-x_2\right) & \left(x_2-x_3\right) \\ \left(y_1-y_2\right) & \left(y_2-y_3\right)\end{array}\right|=0 $
$\Rightarrow \frac{1}{2}\left|\begin{array}{cc} (2 x-0) & \{0-(x+3)\} \\ (x+3-x) & \{x-(x+6)\} \end{array}\right|=0$
$\Rightarrow\left|\begin{array}{cc} 2 x & -(x+3) \\ 3 & -6\end{array}\right|=0 $
$\Rightarrow-12 x+3(x+3)=0$
$\Rightarrow-12 x+3 x+9=0$
$\Rightarrow-9 x=-9$
$\Rightarrow x=1$

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MCQ 181 Mark

If $A=\left[a_{i j}\right]$ is a square matrix of order 2 such that $a_{i j}=\left\{\begin{array}{ll}1, & \text { when } i \neq j \\ 0, & \text { when } i=j\end{array}\right.$ then $A^2$ is

A
$\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right]$
✓
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
C
$\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$
D
$\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$

Answer

Correct option: B.

$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

(b) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Explanation:$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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