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MATHS M.C.Q (1 Marks)

Model Paper 2 · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 18 questions.

Questions

M.C.Q (1 Marks)

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

Question 11 Mark
If $A=\left[\begin{array}{rr}2 & -1 \\ 1 & 3\end{array}\right]$, then $A ^{-1}= ?$
Answer
$A^{-1}=\frac{1}{|A|}$ adj $A \ldots (i)$
$ |A|=3 \times 2-(1) \times(-1) =7$
$C_{11}=3, C_{12}=-1$
$C_{21}=1, C_{22}=2$
Co$-$factor matrix $A=\left(\begin{array}{ll}2 & 1 \\ 4 & 3\end{array}\right)$
Adj $A=\left(\begin{array}{ll}3 & -1 \\ 1 & 2\end{array}\right)^{\prime}$
$=\left(\begin{array}{cc}3 & 1 \\-1 & 2\end{array}\right)$
Putting in $1$
$\begin{array}{l} A^{-1}=\frac{1}{|7|}\left(\begin{array}{cc}3 & 1 \\-1 & 2\end{array}\right) \end{array} $
$ =\left(\begin{array}{cc} \frac{3}{7} & \frac{1}{7} \\ \frac{-1}{7} & \frac{2}{7} \end{array}\right) $
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Question 21 Mark
A matrix $A=\left[a_{i j}\right]_{3 \times 3}$ is defined by $a_{i j}=\left\{\begin{array}{cc}2 i+3 j & , \quad i<j \\ 5 & , \quad i=j \\ 3 i-2 j & , \quad i>j\end{array}\right.$
The number of elements in A which are more than 5 , is 4 :
Answer
(c) 4
Explanation: Here, A $=\left[\begin{array}{ccc}5 & 8 & 11 \\ 4 & 5 & 13 \\ 7 & 5 & 5\end{array}\right]$
Thus, number of elements more than 5 , is 4 .
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Question 31 Mark
The straight line $\frac{x-2}{3}=\frac{y-3}{1}=\frac{z+1}{0}$ is
Answer
(b) perpendicular to the $z$-axis
Explanation: It is perpendicular to z-axis.
Given, direction ratios of the line : $a_1=3, a_2=1, a_3=0$ & direction ratios of $z$-axis is $b_1=0, b_2=0, b_3=1$. Now, $a _1 a _2+ b _1 b_2+ c _1 c _2=3.0+1.0+0.1=0$ which implies that line is perpendicular to z -axis.
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Question 41 Mark
If A and B are independent events, then $P(\bar{A} / \bar{B})= ?$
Answer
$(b)  1- P ( A )$
$ \text { Explanation: } P(\overline{A} / \overline{ B })$
$=\frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}$
$=\frac{P(\bar{A}) P(\bar{B})}{1-P(B)}$
$=1- P ( A )$
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Question 51 Mark
The area of the triangle, whose vertices are $(3, 8), (-4, 2)$ and $(5, 1),$ is
Answer
$(b)\ \frac{61}{2}\ \text{sq. units} $
The area of triangle is given by
$\Delta=\frac{1}{2}\left|\begin{array}{rrr}3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1\end{array}\right| $
$ =\frac{1}{2}[3(2-1)-8(-4-5)+1(-4-10)]  $
$=\frac{1}{2}(3+72-14)$
$=\frac{61}{2}\ \text{sq. units}$
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Question 61 Mark
The general solution of the $D E x^2 \frac{d y}{d x}=x^2+x y+y^2$ is
Answer
(a) $\tan ^{-1} \frac{y}{x}=\log x+C$
Explanation: We have,
$x^2 \frac{d y}{d z}=x^2+x y+y^2$
$\frac{d y}{d x}=1+\frac{y}{x}+\frac{y^2}{x^2}$Let $y = vx$
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$1+v+v^2=v+x \frac{d v}{d x}$
$1+v^2=x \frac{d v}{d x}$
$\frac{d x}{x}=\frac{d v}{v^2+1}$
On integrating on both sides, we obtain
$\log x=\tan ^{-1} v+C$
$\tan ^{-1} \frac{y}{x}=\log x+c$
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Question 71 Mark
If the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$ and $|\vec{a} \times \vec{b}|=3 \sqrt{3}$, then the value of $\vec{a} \cdot \vec{b}$ is
Answer
(d) 3
Explanation: 3
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Question 81 Mark
If $\overrightarrow{ a }+\overrightarrow{ b }=\hat{ i }$ and $\overrightarrow{ a }=2 \hat{ i }-2 \hat{ j }+2 \hat{ k }$, then $|\overrightarrow{ b }|$ equals:
Answer
(d) 3
Explanation: 3
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Question 101 Mark
Which of the following is a convex set?
Answer
(c) $\{(x, y): x \geq 2, y \leq 4\}$
Explanation: $\{(x, y): x \geq 2, y \leq 4\}$is the region between two parallel lines, so any line segment joining any two points in it lies in it. Hence, it is a convex set.
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Question 111 Mark
For any $2-$ rowed square matrix $A,$ if $A \cdot(\operatorname{adj} A)=\left[\begin{array}{ll}8 & 0 \\ 0 & 8\end{array}\right]$ then the value of $| A |$ is
Answer
$(a) \ 8$
$(\operatorname{adj} A)=\left(\begin{array}{ll} 8 & 0 \\ 0 & 8 \end{array}\right) $
$ =8\left(\begin{array}{ll} 1 & 0 \\0 & 1\end{array}\right) $
$=|A| I $
$|A|=8$
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Question 121 Mark
If the vectors $3 \hat{i}+\lambda \hat{j}+\hat{k}$ and $2 \hat{i}-\hat{j}+8 \hat{k}$ are perpendicular, then $\lambda$ is equal to
Answer
$(d) 14$
Explanation: given vectors $3 \hat{i}+\lambda \hat{j}+\hat{k}$ and $2 \hat{i}-\hat{j}+8 \hat{k}$ are perpendicular to each other
$ \Longrightarrow(3 \hat{i}+\lambda \hat{j}+\hat{k}) \cdot(2 \hat{i}-\hat{j}+8 \hat{k})=0$
$\Longrightarrow 6-\lambda+8=0 \Longrightarrow \lambda=14$
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Question 131 Mark
The equation of a line passing through point (2, -1, 0) and parallel to the line $\frac{ x }{1}=\frac{ y -1}{2}=\frac{2- z }{2}$ is:
Answer
(c) $\frac{ x -2}{1}=\frac{ y +1}{2}=\frac{ z }{-2}$
Explanation: $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z}{-2}$
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Question 141 Mark
What are the order and degree respectively of the differential equation whose solution is $y=c x+c^2-3 c^{3 / 2}+2$ where c is a parameter?
Answer
Given, $y=c x+c^2-3 c^{3 / 2}+2....(i)$
On differentiating both sides w.r.t. x, we get
$\frac{d y}{d x}= C ...(ii)$ From Eqs. $(i)$ and $(ii)$, we have
$ y=\frac{d y}{d x} \times x+\left(\frac{d y}{d x}\right)^2-3\left(\frac{d y}{d x}\right)^{3 / 2}+2$
$\Rightarrow y-x \frac{d y}{d x}-\left(\frac{d y}{d x}\right)^2-2=-3\left(\frac{d y}{d x}\right)^{3 / 2}$
$\Rightarrow\left[y-x\left(\frac{d y}{d x}\right)-\left(\frac{d y}{d x}\right)^2-2\right]^2=9\left(\frac{d y}{d x}\right)^3 $
Hence, order is $1$ and degree is $4 $.
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Question 161 Mark
A point out of following points lie in plane represented by
Answer
$(0, 3)$ satisfy the equation $2 x+3 y \leq 12$
$2 \times 0+3 \times 3 \leq 12$
$9 \leq 12$
$\text { But }(3,3),(4,3),(0,5) \text { does not satisfy } 2 x+3 y \leq 12$
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Question 171 Mark
If $x = a \sec \theta, y = b \tan \theta$ then $\frac{d y}{d x}= ?$
Answer
 $x = a \sec \theta$, we get
$\therefore \frac{d x}{d \theta}=\operatorname{asec} \theta \cdot \tan \theta$
$\therefore \frac{d \theta}{d x}=\frac{1}{asec \theta \cdot \tan \theta}$
$y=b \tan \theta \cdot we \text { get }$
$\therefore \frac{d y}{d \theta}=b \cdot \sec ^2 \theta$
$\Rightarrow \frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}$
$\Rightarrow \frac{d y}{d x}=b \cdot \sec ^2 \theta \times \frac{1}{asec \theta \cdot \tan \theta}$
$\Rightarrow \frac{d y}{d x}=\frac{b \sec \theta}{atan \theta}$
$\Rightarrow \frac{dy}{dx}=\frac{b \cdot \frac{1}{\cos \theta}}{a \cdot \frac{\sin \theta}{\cos \theta}}$
$\Rightarrow \frac{d y}{d x}=\frac{b}{a} \operatorname{cosec} \theta $
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Question 181 Mark
$\int_0^\pi \frac{1}{1+\sin x} dx$ equals
Answer
$ \text { Explanation: } \int_0^\pi \frac{1}{1+\sin x} d x$
$=\int_0^\pi \frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} d x$
$=\int_0^\pi \frac{1-\sin x}{1-\sin ^2 x} d x$
$=\int_0^\pi \frac{1-\sin x}{\cos ^2 x} d x$
$=\int_0^\pi\left(\sec ^2 x-\sec x \tan x\right) d x$
$=[\tan x-\sec x]_0^\pi$
$=0+1-0+1$
$=2 $
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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip