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MATHS M.C.Q (1 Marks)

Model Paper 8 · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 18 questions.

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M.C.Q (1 Marks)

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18 questions · timed · auto-graded

Question 11 Mark
Find the area of the triangle with vertices $A(1, 1, 2) B(2, 3, 5)$ and $C(1, 5, 5)$
Answer
$(d) \frac{\sqrt{6 I}}{2}$
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
$\overrightarrow{A B}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{A C}=4 \hat{j}+3 \hat{k}$
$\therefore \overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3\end{array}\right|=-6 \hat{i}-3 \hat{j}+4 \hat{k}$
$\begin{array}{l}\Rightarrow|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{61} \end{array} $
$\Rightarrow \frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$
$=\frac{1}{2} \sqrt{61}$
Therefore, the area of triangle $\text{ABC}$ is $=\frac{1}{2} \sqrt{61}$
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Question 21 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?
Answer
$(c) \pm 2$
Explanation:  $\pm 2$
$\left|\begin{array}{cc}2 x & -3 \\ 5 & x \end{array}\right|=\left|\begin{array}{rr}10 & 1 \\ -3 & 2\end{array}\right|$
$2 x^2+15=20+3$
$2 x^2=23-15$
$2 x^2=8$
$x^2=4$
$x= \pm 2$
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Question 31 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?
Answer
(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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Question 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
Answer
$(d) y=2 x-4$
Explanation: 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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Question 51 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
Answer
(a) $p =-\frac{3}{2}, q =\frac{1}{2}$
Explanation:  $p =-\frac{3}{2}, q =\frac{1}{2}$
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Question 61 Mark
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4 Find P( A |B)
Answer
(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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Question 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 $\vec{b} ?$
Answer
$(d) \frac{19}{9}$
Explanation: 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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Question 81 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
Answer
(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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Question 91 Mark
If the direction cosines of a line are $\left(\frac{1}{a}, \frac{1}{a}, \frac{1}{a}\right)$, then:
Answer
(c) $a = \pm \sqrt{3}$
Explanation: $a = \pm \sqrt{3}$
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Question 101 Mark
If $\vec{a} \cdot \vec{b}=0$ and $\vec{a} \times \vec{b}=0$, then which one of the following is correct?
Answer
(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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Question 111 Mark
Let $A=\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|$, then
Answer
(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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Question 121 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
Answer
(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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Question 131 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
Answer
$(b) 2$
Explanation: $\because \int_0^5(1+f(x)) d x=7$
$\therefore \int_0^5 d x+\int_0^5 f(x) d x=7$
$\Rightarrow(x)_0^5+\int_0^5 f(x) d x=7$
$\Rightarrow \int_0^5 f(x) d x=7-5=2$
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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Question 141 Mark
The position of origin $(0,0)$ w.r.t. feasible region represented by $x+y \geq 1$ is
Answer
(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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Question 151 Mark
If the matrix A is both symmetric and skew symmetric, then
Answer
(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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Question 161 Mark
The value of objective function is maximum under linear constraints
Answer
(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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Question 171 Mark
The function f(x) = cot x is discontinuous on the set
Answer
(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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Question 181 Mark
If $A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$ then $A ^{-1}=$ ?
Answer
$(b) \operatorname{adj} A$
Explanation: $A=\left(\begin{array}{ll}\cos \theta & -\sin \theta
\\ \sin \theta & \cos \theta\end{array}\right)$
$|A|=\cos ^2 \theta-\left(-\sin ^2 \theta\right)$
$=\cos ^2 \theta+\left(\sin ^2 \theta\right)$
$=1 \ldots \text { (i) }$
We know that $A^{-1}=\frac{1}{|A|} \ adj \ A$
$=\operatorname{adj} A ($ From $I)$
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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip