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MATHS M.C.Q (1 Marks)

Model Paper 10 · 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 · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark
If $f(x)=\left\{\begin{array}{l}k x+5 \text {, when } x \leq 2 \\ x+1, \text { when } x>2\end{array}\right.$ is continuous at $x=2$ then $k= ?$
Answer
For continuity left hand limit must be equal to right hand limit and value at the point.
Continuous at $x =2..$
$\text { L.H.L }=\lim _{x \rightarrow 2^{-}}(k x+5)$
$\Rightarrow \lim _{h \rightarrow 0}(k(2-h)+5)$
$\Rightarrow k(2-0)+5=2 k+5$
$\text { R.H.L }=\lim _{x \rightarrow 2^{+}}(x+1)$
$\Rightarrow \lim _{h \rightarrow 0}(2+h+1)$
$\Rightarrow 2+0+1$
$=3$
As $f(x)$ is continuous, we get
$\because 2 k+5=3$
$k=-1$
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Question 21 Mark
If $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{3}{5}$, then $P(B / A)+P(A / B)$ equals
Answer
(d) $\frac{7}{12}$
Explanation: Here, $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{3}{5}$
$P(B / A)+P(A / B)=\frac{P(B \cap A)}{P(A)}+\frac{P(A \cap B)}{P(B)}$
$=\frac{P(A)+P(B)-P(A \cup B)}{P(A)}+\frac{P(A)+P(B)-P(A) B)}{P(B)}$
$=\frac{\frac{3}{10}+\frac{2}{5}-\frac{3}{5}}{\frac{3}{10}}+\frac{\frac{3}{10}+\frac{2}{5}-\frac{3}{5}}{\frac{2}{5}}$
$=\frac{\frac{1}{10}}{\frac{3}{10}}+\frac{\frac{1}{10}}{\frac{2}{5}}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$
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Question 31 Mark
the order of the single matrix obtained from
$\left[\begin{array}{cc}1 & -1 \\ 0 & 2 \\ 2 & 3\end{array}\right]_{3 \times 2}\left\{\left[\begin{array}{ccc}-1 & 0 & 2 \\ 2 & 0 & 1\end{array}\right]_{2 \times 3}-\left[\begin{array}{lll}0 & 1 & 23 \\ 1 & 0 & 21\end{array}\right]_{2 \times 3}\right\}$ is
Answer
$\begin{array}{l} \text { Explanation: }\left[\begin{array}{cc}1 & -1 \\0 & 2 \\2 & 3\end{array}\right]_{3 \times 2}\left\{\left[\begin{array}{ccc}-1 & 0 & 2 \\2 & 0 & 1 \end{array}\right]_{2 \times 3}-\left[\begin{array}{lll}0 & 1 & 23 \\1 & 0 & 21\end{array}\right]_{2 \times 3}\right\} \end{array}$
$=\left[\begin{array}{cc}1 & -1 \\0 & 2 \\2 & 3\end{array}\right]_{3 \times 2}\left[\begin{array}{ccc}-1 & -1 & -21 \\1 & 0 & -20\end{array}\right]_{2 \times 3}$
$=\left[\begin{array}{ccc}-2 & -1 & -1 \\2 & 0 & -40 \\1 & -2 & -102\end{array}\right]_{3 \times 3}$
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Question 41 Mark
If $y=\log \left(\sin e^x\right)$, then $\frac{d y}{d x}$ is:
Answer
(b) $e ^x \cot e ^x$
Explanation: $e ^{ x } \cot e ^{ x }$
$y=\log \left(\sin e^{x}\right)$
$\frac{d y}{d x}=\frac{d}{d x} \log \left(\sin e^{x}\right)$
$=\frac{1}{\sin e^2} \frac{d}{d x} \sin e^{x}$
$=\frac{1}{\sin e^2} \cos e^x \frac{d}{d x} e^{x}$
$=\cot e^x\left(e^x\right)$
$=e^{x} \cot e^{x}$
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Question 51 Mark
If the vectors $4 \hat{i}+11 \hat{j}+m \hat{k}, 7 \hat{i}+2 \hat{j}+6 \hat{k}$ and $\hat{i}+5 \hat{j}+4 \hat{k}$ are coplanar, then $m =$
Answer
$(d)$ None of these
Given vectors $4 \vec{i}+11 \vec{j}+m \vec{k} \cdot 7 \vec{i}+2 \vec{j}+6 \vec{k}$ and $\vec{i}+5 \vec{j}+4 \vec{k}$ are coplanar then
$\left|\begin{array}{ccc}
4 & 11 & m \\
7 & 2 & 6 \\
1 & 5 & 4
\end{array}\right|=0 $
$\Longrightarrow 4(8-30)-11(28-6)+m(35-2)=0$
$-88-242+33 m=0$
$-330+33 m=0$
$m=10$
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Question 61 Mark
The lines $l_1$ and $l_2$ intersect. The shortest distance between them is
Answer
Since the lines intersect. Hence they have a common point in them.
Hence the distance will be zero.
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Question 71 Mark
The area of a triangle with vertices A, B, C is given by
Answer
(d) $\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$
Explanation: The area of the parallelogram with adjacent sides $A B$ and $A C=|\overrightarrow{A B} \times \overrightarrow{A C}|$. Hence, the area of the triangle with vertices $A , B , C =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|$.
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Question 81 Mark
The solution of the differential equation $\frac{d y}{d x}=e^{x-y}+x^2 e^{-y}$ is
Answer
$(d) e ^{ y }- e ^{ x }=\frac{x^3}{3}+ C$
Explanation: We have, $\frac{d y}{d x}=e^{x-y}+x^2 e^{-y}$
$\Rightarrow e^{y} dy=\left(e^{x}+x^2\right) dx$
$\Rightarrow \int e^{y} d y=\int\left(e^x+x^2\right) d x$
$\Rightarrow e^{y}=e^{x}+\frac{x^3}{3}+c$
$\Rightarrow e^{y}-e^{x}=\frac{x^3}{3}+c$
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Question 91 Mark
$\left|\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right|=?$
Answer
(d) $\left(a^2+b^2+c^2+d^2\right)$
Explanation: $\Delta=( a + ib )( a - ib )+( c - id )( c + id )=\left( a ^2+ b ^2+ c ^2+ d ^2\right)$
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Question 101 Mark
The solution set of the inequality $3 x+5 y<4$ is
Answer
(b) an open half-plane containing the origin.
Explanation: The strict inequality represents an open half plane and it contains the origin as $(0,0)$ satisfies it.
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Question 111 Mark
The direction ratios of a line parallel to z -axis are:
Answer
(d) $\langle 0,0,1\rangle$
Explanation: $\langle 0,0,1\rangle$
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Question 121 Mark
Which of the following is the principal value branch of $\operatorname{cosec}^{-1} x$ ?
Answer
(b) $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
Explanation: We know that the principal value branch of $\operatorname{cosec}^{-1} x$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$
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Question 131 Mark
What is integrating factor of $\frac{d y}{d z}+y \sec x=\tan x$ ?
Answer
(c) $\sec x+\tan x$
Explanation: We have,
$
\frac{d y}{d x}+y \sec x=\tan x
$
Comparing with $\frac{d y}{d x}+ Py = Q$
$
=\sec x, Q=\tan x
$
I. $F \cdot=e^{\int \sec x d x}=e^{\log (\sec x+\tan x)}=\sec x+\tan x$
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Question 141 Mark
The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20) is
Answer
(a) $q=3 p$
Explanation: Since $Z$ occurs maximum at $(15,15)$ and $(0,20)$, therefore, $15 p+15 q=0 p+20 q \Rightarrow q=3 p$.
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Question 151 Mark
If $A$ is a $3 \times 3$ matrix and $|A|=-2$, then value of $|A(\operatorname{adj} A)|$ is
Answer
(d)-8
Explanation:-8
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Question 161 Mark
$\int_0^{\frac{\pi}{4}} \frac{e^{\tan x}}{\cos ^2 x} d x=?$
Answer
$(a) ( e -1)$
Explanation: $I=\int_0^{\frac{\pi}{4}} e^{\tan x} \sec ^2 x d x$
Let, $\tan x = t$,
Differentiating both side with respect to $t$
$\sec ^2 x \frac{d x}{d t}=1$
$\Rightarrow \sec ^2 x d x=d t$
$\text { At } x=0, t=0$
$\text { At } x=\frac{\pi}{4}, t=1$
$I=\int_0^1 e^t d t$
$=e^{t_0^1}$
$= e ^1- e ^0$
$= e -1$
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Question 171 Mark
Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\operatorname{det}\left(A B A^{\prime}\right)=8$ and $\operatorname{det}\left(A B^{-1}\right)=8$, then $\operatorname{det}\left(B A^{-1} B^{\prime}\right)$ is equal to
Answer
(d) $\frac{1}{16}$
Explanation: $\frac{1}{16}$
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Question 181 Mark
If $A=\left[a_{i j}\right]$ is a scalar matrix of order $n \times n$ such that $a_{i j}=k$ for all $i$, then trace of $A$ is equal to
Answer
$(c) nk$
Explanation: $\because A =\left[ a _{ ij }\right]_{ n \times n }$
Trace of A, i.e., $\operatorname{tr}( A )=\sum a_{i j}^n i=1= a _{11}+ a _{22}+\ldots \ldots \ldots+ a _{ nm }$
$=k+k+k+k+k+\ldots(n \text { times })$
$=k(n)$
$=nk$
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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip