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 direction cosines of a line are $\left(\frac{1}{a}, \frac{1}{a}, \frac{1}{a}\right)$, then:

A
$0 < a < 1$
B
$a > 2$
✓
$a = \pm \sqrt{3}$
D
$a > 0$

Answer

Correct option: C.

$a = \pm \sqrt{3}$

(c) $a = \pm \sqrt{3}$
Explanation: $a = \pm \sqrt{3}$

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MCQ 21 Mark

The value of p and q for which the function $f ( x )=\left\{\begin{array}{cl}\frac{\sin (p+1) x+\sin x}{x} & , x<0 \\ \frac{q}{x^2} & , x=0 \\ \frac{\sqrt{x+b x^2}-\sqrt{x}}{x^{\frac{3}{2}}} & , x>0\end{array}\right.$ is continuous for all $x \in R$, are

✓
$p =-\frac{3}{2}, q =\frac{1}{2}$
B
$p=-\frac{3}{2}, q=-\frac{1}{2}$
C
$p =\frac{5}{2}, q =\frac{7}{2}$
D
$p=\frac{1}{2}, q=\frac{3}{2}$

Answer

Correct option: A.

$p =-\frac{3}{2}, q =\frac{1}{2}$

(a) $p =-\frac{3}{2}, q =\frac{1}{2}$
Explanation: $p =-\frac{3}{2}, q =\frac{1}{2}$

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MCQ 31 Mark

If $\vec{a} \cdot \vec{b}=0$ and $\vec{a} \times \vec{b}=0$, then which one of the following is correct?

A
$\vec{a}$ is parallel to $\vec{b}$
✓
$\vec{a}=0$ or $\vec{b}=0$
C
$\vec{a}$ is perpendicular to $\vec{b}$
D
$\vec{a}$ and $\vec{b} \neq 0$

Answer

Correct option: B.

$\vec{a}=0$ or $\vec{b}=0$

(b) $\vec{a}=0$ or $\vec{b}=0$
Explanation: Given that, $\vec{a} \cdot \vec{b}=0$,
i.e. $\vec{a}$ and $\vec{b}$ are perpendicular to each other and $\vec{a} \times \vec{b}=0$
i.e. $\vec{a}$ and $\vec{b}$ are parallel to each other. So, both conditions are possible iff $\vec{a}=0$ and $\vec{b}=0$

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MCQ 41 Mark

A solution of the differential equation $\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}+y=0$ is

A
$y=2 x^2-4$
B
$y =2 x$
C
$y=2$
✓
$y=2 x-4$

Answer

Correct option: D.

$y=2 x-4$

Let, $\frac{d y}{d x}=p$
$\therefore p^2-x p+y=0$
$y=x p-p^2 \ldots \text { (i) }$
$\Rightarrow \frac{d y}{d x}=(x-2 p) \frac{d y}{d x}+p$
$\Rightarrow p=(x-2 p) \frac{d p}{d x}+p$
$\therefore \frac{d p}{d x}=0$
$\Rightarrow P$ is constant
from Eqn. $(i), y=x \cdot c-c^2$
$\therefore y=2 x-4$ is the correct option

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MCQ 51 Mark

Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4 Find P( A |B)

A
0.27
✓
0.3
C
0.2
D
0.33

Answer

Correct option: B.

0.3

(b) 0.3
Explanation:
Let A and B be independent events with $P ( A )=0.3$ and $P(B)=0.4 P ( A / B )= P ( A )=0.3$.

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MCQ 61 Mark

If the value of a third-order determinant is 12, then the value of the determinant formed by replacing each element by its cofactor will be

A
-12
B
12
C
13
✓
144

Answer

Correct option: D.

144

(d) 144
Explanation: Let A is the determinant.
$\therefore|A|=12$
Also, we know that, if $A$ is a square matrix of order $n$, then $|\operatorname{adj} A|=|A|^{ n -1}$.For $n=3$, $|\operatorname{adj} A|=|A|^{3-1}=|A|^2$.
$\therefore|\operatorname{adj} A|=(12)^2=144 .$

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MCQ 71 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 scalar projection of $f \vec{a}$ on $\vec{b} ?$

A
$\frac{23}{9}$
B
$\frac{17}{9}$
C
1
✓
$\frac{19}{9}$

Answer

Correct option: D.

$\frac{19}{9}$

Scalar projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
$=\frac{(\hat{i}-2 \hat{j}+\hat{k}) \cdot(4 \hat{i}-4 \hat{j}+7 \hat{k})}{|4 i-4 \hat{j}+7 \hat{k}|}$
$=\frac{(4+8+7)}{\sqrt{(4)^2+(-4)^2+(7)^2}}$
$=\frac{19}{9}$
which is the required scalar projection of $\vec{a}$ and $\vec{b}$.

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MCQ 81 Mark

The value of objective function is maximum under linear constraints

A
at (0, 0)
✓
at any vertex of feasible region
C
the vertex which is maximum distance from (0,0)
D
at the centre of feasible region

Answer

Correct option: B.

at any vertex of feasible region

(b) at any vertex of feasible region
Explanation: In linear programming problem we substitute the coordinates of vertices of feasible region in the objective function and then we obtain the maximum or minimum value. Therefore, the value of objective function is maximum under linear constraints at any vertex of feasible region.

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MCQ 91 Mark

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

A
$A^2=0$
✓
$A^2=A$
C
$A^2=I$
D
$A^2=4$

Answer

Correct option: B.

$A^2=A$

(b) $A ^2= A$
Explanation: $A =\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|$, then $A ^2=\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|=\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|= A$

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MCQ 101 Mark

If $\int_{-2}^5 f(x) d x=4, \int_0^5(1+f(x)) d x=7$, then the value of the integral $\int_{-2}^0 f(x) dx$ is equal to

A
$-3$
✓
$2$
C
$3$
D
$5$

Answer

Correct option: B.

$2$

$\because \int_0^5(1+f(x)) d x=7$
$\therefore \int_0^5 d x+\int_0^5 f(x) d x=7$
$\Rightarrow(x)_0^5+\int_0^5 f(x) d x=7$
$\Rightarrow \int_0^5 f(x) d x=7-5=2$
Also, $\int_{-2}^5 f(x) d x=4$
$\Rightarrow \int_{-2}^0 f(x) d x+\int_0^5 f(x) d x=4$
$\Rightarrow \int_{-2}^0 f(x) d x=2$

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MCQ 111 Mark

Find the area of the triangle with vertices $A(1, 1, 2) B(2, 3, 5)$ and $C(1, 5, 5)$

A
$\frac{\sqrt{65}}{3}$
B
$\frac{\sqrt{65}}{2}$
C
$\frac{\sqrt{61}}{3}$
✓
$\frac{\sqrt{61}}{2}$

Answer

Correct option: D.

$\frac{\sqrt{61}}{2}$

Given position vector of $A , \overrightarrow{O A}=\hat{i}+\hat{j}+2 \hat{k}$ position vector of $B , \overrightarrow{O B}=2 \hat{i}+3 \hat{j}+5 \hat{k}$ and that of $C , \overrightarrow{O C}=\hat{i}+5 \hat{j}+5 \hat{k}$
therefore, $\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=(2 \hat{i}+3 \hat{j}+5 \hat{k})-(\hat{i}+\hat{j}+2 \hat{k})=\hat{i}+2 \hat{j}+3 \hat{k} $
$($by triangle law of vector addition$)$ thus we may write
$\overrightarrow{A B}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{A C}=4 \hat{j}+3 \hat{k}$
$\therefore \overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3\end{array}\right|=-6 \hat{i}-3 \hat{j}+4 \hat{k}$
$\Rightarrow|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{61}$
$\Rightarrow \frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|=\frac{1}{2} \sqrt{61}$
Therefore, the area of triangle $\text{ABC}$ is $=\frac{1}{2} \sqrt{61}$

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MCQ 121 Mark

The position of origin $(0,0)$ w.r.t. feasible region represented by $x+y \geq 1$ is

A
on the line x + y = 0
B
on the line x - y = 0
✓
not in the region
D
in the region

Answer

Correct option: C.

not in the region

(c) not in the region
Explanation: Since $(0,0)$ does not satisfy $x+y \geq 1$
i.e., $0+0 \neq 1$
$\Rightarrow(0,0)$ not lie in feasible region represented by $x+y \geq 1$.

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MCQ 131 Mark

The degree of the differential equation $\left(1+\frac{d y}{d x}\right)^3=\left(\frac{d^2 y}{d x^2}\right)^2$ is

✓
2
B
1
C
3
D
4

Answer

Correct option: A.

2

(a) 2
Explanation: We have $\left[1+\left(\frac{d y}{d x}\right)^2\right]^{1 / 2}=\frac{d^2 y}{d x^2}$
$\Rightarrow\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=\left(\frac{d^2 y}{dx^2}\right)^2$
So, the degree of differential equation is 2.

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MCQ 141 Mark

The Cartesian equations of a line are $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z-3}{-2}$. What is its vector equation?

A
$\vec{r}=(2 \hat{i}-3 \hat{j}-2 \hat{k})$
B
$\vec{r}=(2 \hat{i}+3 \hat{j}-2 \hat{k})+\lambda(2 \hat{i}-\hat{j}+3 \hat{k})$
✓
$\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(2 i+3 j-2 k)$
D
$\vec{r}=(2 \hat{i}+3 \hat{j}-2 \hat{k})$

Answer

Correct option: C.

$\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(2 i+3 j-2 k)$

(c) $\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(2 i+3 j-2 k)$
Explanation: Fixed point is $2 \hat{\imath}-\hat{\jmath}+3 \hat{ k }$ and the vector is $2 \hat{\imath}+3 \hat{\jmath}-2 \hat{k}$
Equation $(2 \hat{\imath}-\hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+3 \hat{\jmath}-2 \hat{k})$

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MCQ 151 Mark

The function f(x) = cot x is discontinuous on the set

A
$\{x=2 n \pi: n \in Z \}$
B
$\left\{x=(2 n+1) \frac{\pi}{2} ; n \in Z \right\}$
C
$\left\{x=\frac{n \pi}{2} ; n \in Z \right\}$
✓
$\{x=n \pi: n \in Z \}$

Answer

Correct option: D.

$\{x=n \pi: n \in Z \}$

(d) $\{x=n \pi: n \in Z \}$
Explanation: We have $f ( x )=\cot x$ is continuous in $R-\{n \pi: n \in Z\}$
Since, $f ( x )=\cot x=\frac{\cos x}{\sin x}$ (since, $\sin x =0$ at $n \pi, n \in Z$ )
Hence, $f ( x )=\cot x$ is discontinuous on the set $\{x=n \pi: n \in Z\}$

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MCQ 161 Mark

If $A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$ then $A ^{-1}= ?$

A
$-adj \ A$
✓
$adj \ A$
C
$-A$
D
$A$

Answer

Correct option: B.

$adj \ A$

$A=\left(\cos \theta -\sin \theta \sin \theta \cos \theta\right)$
$|A|=\cos ^2 \theta-\left(-\sin ^2 \theta\right)$
$=\cos ^2 \theta+\left(\sin ^2 \theta\right)$
$=1 \ldots \text { (i) }$
We know that $A^{-1}=\frac{1}{|A|}$ adj A
$=\operatorname{adj} A ($ From $I)$

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MCQ 171 Mark

For which value of $x$, are the determinants $\left|\begin{array}{cc}2 x & -3 \\ 5 & x \end{array}\right|$ and $\left|\begin{array}{rr}10 & 1 \\ -3 & 2\end{array}\right|$ equal?

A
$\pm 3$
B
2
✓
$\pm 2$
D
-3

Answer

Correct option: C.

$\pm 2$

$\pm 2$
$\left|\begin{array}{cc}2 x & -3 \\ 5 & x \end{array}\right|=\left|\begin{array}{rr}10 & 1 \\ -3 & 2\end{array}\right|$
$2 x^2+15=20+3$
$2 x^2=23-15$
$2 x^2=8$
$x^2=4$
$x= \pm 2$

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MCQ 181 Mark

If the matrix A is both symmetric and skew symmetric, then

A
A is a null matrix
✓
A is a zero matrix
C
A is a square matrix
D
A is a diagonal matrix

Answer

Correct option: B.

A is a zero matrix

(b) A is a zero matrix
Explanation: Only a null matrix can be symmetric as well as skew symmetric.
In Symmetric Matrix $A^{ T }= A$,
Skew Symmetric Matrix $A^{ T }=- A$,
Given that the matrix is satisfying both the properties.
Therefore, Equating the RHS we get $A =- A$ i.e, $2 A=0$.
Therefore $A =0$, which is a null matrix.

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