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MATHS M.C.Q (1 Marks)

Model Paper 5 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 18 questions.

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M.C.Q (1 Marks)

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18 questions · timed · auto-graded

MCQ 11 Mark
The value of $\sin ^2 \frac{5 \pi}{12}-\sin ^2 \frac{\pi}{12}$ is
  • A
    $\sqrt{3} / 2$
  • B
    $\frac{1}{2}$
  • C
    $0$
  • D
    1
Answer
(a) $\sqrt{3} / 2$
Explanation: $\frac{5 \pi}{12}=75^{\circ}, \frac{\pi}{12}=15^{\circ}$
$\begin{array}{l}\sin ^2 75^{\circ}-\sin ^2 15^{\circ} \\ =\sin ^2 75^{\circ}-\cos ^2 75^{\circ}\left[\sin \left(90^{\circ}-\theta\right)=\cos \theta\right]\end{array}$
Now, $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)$
$\begin{array}{l}=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ} \\ =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2} \\ =\frac{\sqrt{3}+1}{2 \sqrt{2}}\end{array}$
$\cos 75^{\circ}=\cos \left(45^{\circ}+30^{\circ}\right)$
$\begin{array}{l}=\cos 45^{\circ} \cos 30^{\circ}-\sin 45^{\circ} \sin 30^{\circ} \\ =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2} \\ =\frac{\sqrt{3}-1}{2 \sqrt{2}}\end{array}$
Hence, 
$\begin{array}{l}\sin ^2 75^{\circ}-\cos ^2 75^{\circ}=\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)^2-\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^2 \\ =\frac{3+1+2 \sqrt{3}-3-1+2 \sqrt{3}}{8} \\ =\frac{4 \sqrt{3}}{8} \\ =\frac{\sqrt{3}}{2}\end{array}$
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MCQ 21 Mark
Let R be set of points inside a rectangle of sides a and b (a, b > 1) with two sides along the positive direction of x-axis and y-axis. Then
Answer
(c) $R =\{( x , y ): 0< x < a , 0< y < b \}$
Explanation: We have, R be set of points inside a rectangle of sides a and b
Since, $a , b >1$
$a$ and $b$ cannot be equal to 0
Thus, $R=\{(x, y): 0<x<a,0<y<b\}$
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MCQ 31 Mark
If P(n, r) = C(n, r) then
  • A
    r = 0 or 2
  • B
    r = 1 or n
  • C
    r = 0 or 1
  • D
    n = r
Answer
(c) $r =0$ or 1
Explanation: Given $P ( n , r )= C ( n , r )$
$\begin{array}{l}\Rightarrow \frac{n!}{(n-r)!}=\frac{n!}{r!(n-r)!} \\ \Rightarrow 1=\frac{1}{r!} \\ \Rightarrow r!=1 \\ \Rightarrow r=0 \text { or } r=1[\because 0!=1,1!=1]\end{array}$
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MCQ 41 Mark
If $f(x)=\sqrt{1-x^2}, x \in(0,1)$, then $f^{\prime}(x)$, is equal to
  • A
    $\sqrt{1-x^2}$
  • B
    $\sqrt{x^2-1}$
  • C
    $\frac{1}{\sqrt{1-x^2}}$
  • D
    $\frac{-x}{\sqrt{1-x^2}}$
Answer
(d) $\frac{-x}{\sqrt{1-x^2}}$
Explanation: $f ( x )=\sqrt{1-x^2}$
$f^{\prime}(x)=\frac{1}{2 \sqrt{1-x^2}}-2 x=\frac{-x}{\sqrt{1-x^2}}$
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MCQ 61 Mark
The value of $\sin 78^{\circ}-\sin 66^{\circ}-\sin 42^{\circ}+\sin 60^{\circ}$ is
  • A
    None of these
  • B
    $\frac{1}{2}$
  • C
    -1
  • D
    $\frac{-1}{2}$
Answer
(a) None of these 
Explanation: $\sin 78^{\circ}-\sin 66^{\circ}-\sin 42^{\circ}+\sin 60^{\circ}$
$\begin{array}{l}=\sin 78^{\circ}-\sin 42^{\circ}-\sin 66^{\circ}+\sin 60^{\circ} \\ =2 \sin \left(\frac{78^{\circ}-42^{\circ}}{2}\right) \cos \left(\frac{78^{\circ}+42}{2}\right)-\sin 66^{\circ}+\sin 60^{\circ}\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right] \\ =2 \sin 18^{\circ} \cos 60^{\circ}-\sin 66^{\circ}+\sin 60^{\circ} \\ =2 \times \frac{1}{2} \sin 18^{\circ}-\sin 66^{\circ}+\frac{\sqrt{3}}{2} \\ =\sin 18^{\circ}-\sin 66^{\circ}+\frac{\sqrt{3}}{2} \\ =0.309-0.914+0.866 \\ =0.261\end{array}$
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MCQ 71 Mark
$\left\{C_1+2 C_2+3 C_3+\ldots+n C_n\right\}=?$
  • A
    $( n -1) \cdot 2^{ n }$
  • B
    $n \cdot 2^n$
  • C
    $( n +1) \cdot 2^n$
  • D
    $n \cdot 2^{n-1}$
Answer
(d) $n \cdot 2^{ n -1}$
Explanation: $C _1+2 C _2+3 C _3+\ldots+n C _{ n }= n +2 \cdot \frac{ n ( n -1)}{2}+3 \cdot \frac{n(n-1)(n-2)}{3!}+\ldots+n$
$\begin{array}{l}= n \cdot\left[1+( n -1) \frac{( n -1)( n -2)}{2!}+\ldots+1\right] \\ = n \cdot\left[{ }^{( n -1)} C _0+{ }^{( n -1)} C _1+{ }^{( n -1)} C _2+\ldots+{ }^{( n -1)} C _{ n -1}\right] \\ = n \cdot(1+1)^{ n -1}= n \cdot 2 ^{ n -1}\end{array}$
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MCQ 81 Mark
$\sum_{r=0}^n 4^r \cdot{ }^n C_r$ is equal to
  • A
    $6^{ n }$
  • B
    $5^{-n}$
  • C
    $4^{ n }$
  • D
    $5^{ n }$
Answer
(d) $5^{ n }$
Explanation: $\sum_{r=0}^n 4^{r \cdot n} C_r=4^0 \cdot{ }^n C_0+4^1 \cdot{ }^n C_1+4^2 \cdot{ }^n C_2+\ldots+4^n \cdot{ }^n C_n$
$\begin{array}{l}=1+4 \cdot^n C_1+4^2 \cdot{ }^n C_2+\ldots .+4^n \cdot{ }^n C_n \\ =(1+4)^n=5^n\end{array}$
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MCQ 91 Mark
Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of m and n are
  • A
    7,4
  • B
    6,4
  • C
    3,3
  • D
    6,3
Answer
(d) 6,3
Explanation: Let A and B be two sets having m and n elements respectively. Then,
Number of subsets of $A =2^{ m }$, Number of subsets of $B =2^{ n }$
It is given that $2^{ m }-2^{ n }=56$
So, $2^{ n }\left(2^{ m - n }-1\right)=2^3\left(2^3-1\right)$
$n =3$ and $m - n =3 \Rightarrow n =3$ and $m =6$.
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MCQ 101 Mark
A pendulum swings through an angle of $42^{\circ}$ in describing an arc of length 55 cm . The length of the pendulum is
  • A
    88cm
  • B
    60cm
  • C
    56cm
  • D
    75cm
Answer
(d) 75 cm
Explanation: $\theta=42^{\circ}=\left(42 \times \frac{\pi}{180}\right)^c=\left(\frac{7 \pi}{30}\right)^c$ and $1=55 cm$.
$\therefore r=\frac{l}{\theta}=\left(55 \times \frac{30}{7 \pi}\right) cm =\left(55 \times \frac{30}{7} \times \frac{7}{22}\right) cm =75 cm$.
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MCQ 111 Mark
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\cos x}$ is equal to
  • A
    1
  • B
    $0$
  • C
    -1
  • D
    does not exit
Answer
(b) $0$
Explanation: $\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\cos x}=\lim _{y \rightarrow 0} \frac{1-\sin \left(\frac{\pi}{2}-y\right)}{\cos \left(\frac{\pi}{2}-y\right)} \quad$ taking $\frac{\pi}{2}-x=y$
$\begin{array}{l}=\lim _{y \rightarrow 0} \frac{1-\cos y}{\sin y}=\lim _{y \rightarrow 0} \frac{2 \sin ^2 \frac{y}{2}}{2 \sin \frac{y}{2} \cos \frac{y}{2}} \\ =\lim _{y \rightarrow 0} \tan \frac{y}{2}=0\end{array}$
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MCQ 121 Mark
A polygon has 44 diagonals. The number of its sides is
  • A
    8
  • B
    11
  • C
    12
  • D
    10
Answer
(b) 11 
Explanation: We have an n sided polygon has n vertices. If you join every distinct pair of vertices you will get lines. These lines account for the in sides of the polygon as well as for the diagonals.
So the number of diagonals is given by ${ }^n C_2-n=\frac{n(n-1)}{2}-n=\frac{n(n-3)}{2}$
But number of diagonals = 44 
$\begin{array}{l}\Rightarrow 44=\frac{n(n-3)}{2} \\ \Rightarrow 88=n(n-3) \\ \Rightarrow n^2-3 n-88=0 \\ \Rightarrow(n-11)(n+8)=0 \\ \Rightarrow n=11,-8\end{array}$
Since n cannot be negative, we get n = 11   
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MCQ 131 Mark
If $x+i y=\frac{3+5 i}{7-6 i}$, then $y=$
  • A
    $\frac{43}{85}$
  • B
    $\frac{-9}{85}$
  • C
    $\frac{9}{85}$
  • D
    $\frac{53}{85}$
Answer
(d) $\frac{53}{85}$
Explanation: $\frac{53}{85}$ 
$x + iy =\frac{3+5 i}{7-6 i}$
$\begin{array}{l}\Rightarrow x + iy =\frac{3+5 i}{7-6 i} \times \frac{7+6 i}{7+6 i} \\ \Rightarrow x + iy =\frac{21+53 i+30 i^2}{49-36 i^2} \\ \Rightarrow x + iy =\frac{21-30+53 i}{49-36 i} \\ \Rightarrow x + iy =\frac{-9}{85}+i \frac{53}{85}\end{array}$
On comparing both the sides: 
$y=\frac{53}{85}$
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MCQ 141 Mark
Perpendicular distance of the point (3, 4, 5) from the y-axis is,
  • A
    $\sqrt{34}$
  • B
    4
  • C
    $\sqrt{41}$
  • D
    5
Answer
(a) $\sqrt{34}$
Explanation: Distance of $(\alpha, \beta, \gamma)$ from y -axis is given by $d =\sqrt{\alpha^2+\gamma^2}$
$\therefore$ Distance $( d )$ of $(3,4,5)$ 
from $y$-axis is $d=\sqrt{3^2+5^2}=\sqrt{9+25}=\sqrt{34}$
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MCQ 151 Mark
The equation of the line passing through (1, 2) and perpendicular to x + y + 7 = 0 is
  • A
    y – x – 1 = 0
  • B
    y – x + 1 = 0
  • C
    y – x + 2 = 0
  • D
    y – x – 2 = 0.
Answer
(a) y – x – 1 = 0 
Explanation: Suppose the slope of the line be m. Then, its equation passing through (1, 2) is given by
y – 2 = m (x – 1) ... (1)
Again, this line is perpendicular to the given line x + y + 7 = 0 whose slope is – 1
Thus, we have m ( – 1) = – 1 or m = 1 
Thus, the required equation of the line is obtained by substituting thethe value of m in (1), i.e., 
y – 2 = x – 1 or y – x – 1 = 0
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MCQ 161 Mark
If $f(x)=x \sin x$, then $f^{\prime}\left(\frac{\pi}{2}\right)$ is equal to
  • A
    1
  • B
    $\frac{1}{2}$
  • C
    -1
  • D
    $0$
Answer
(a) 1
Explanation: $f^{\prime}(x)=x \cos x+\sin x$
So, $f^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi}{2} \cos \frac{\pi}{2}+\sin \frac{\pi}{2}=1$
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MCQ 171 Mark
The domain of the function $f(x)=\sqrt{x-1}+\sqrt{6-x}$
  • A
    $(-\infty, 6)$
  • B
    $[2,6]$
  • C
    $[1,6]$
  • D
    $[-2,6]$
Answer
(c) $[1,6]$
Explanation: For $f(x)$ to be real, we must have,
$\begin{array}{l}x-1 \geqslant 0 \text { and } 6-x \geqslant 0 \\ \Rightarrow x \geqslant \varphi \text { and } x-6 \leqslant 6 \\ \therefore \text { Domain }=[1,6]\end{array}$
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MCQ 181 Mark
If the mean of the squares of first n natural numbers be 11, then n is equal to
  • A
    5
  • B
    $\frac{-13}{2}$
  • C
    11
  • D
    13
Answer
(a) 5 
Explanation: Mean $=\frac{\frac{n(n+1)(2 n+1)}{6}}{n}=\frac{(n+1)(2 n+1)}{6}$
$\begin{array}{l}\Rightarrow 11=\frac{(n+1)(2 n+1)}{6} \\ \Rightarrow 66=(n+1)(2 n+1) \\ \Rightarrow 2 n^2+3 n-65=0 \\ \Rightarrow 2 n^2+13 n-10 n-65=0 \\ \Rightarrow(2 n+13)(n-5)=0 \\ \Rightarrow n=5, \frac{-13}{2}\end{array}$
So, n = 5
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M.C.Q (1 Marks) - MATHS STD 11 Science Questions - Vidyadip