M.C.Q (1 Marks)

MCQ 11 Mark

If $P(n, r) = C(n, r)$ then

A
$r = 0$ or $2$
B
$r = 1$ or $ n$
✓
$r = 0$ or $1$
D
$n = r$

Answer

Correct option: C.

$r = 0$ or $1$

Given $P ( n , r )= C ( n , r )$
$\Rightarrow \frac{n!}{(n-r)!}=\frac{n!}{r!(n-r)!}$
$\Rightarrow 1=\frac{1}{r!}$
$\Rightarrow r!=1$
$\Rightarrow r=0 \text { or } r=1[\because 0!=1,1!=1]$

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MCQ 21 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 31 Mark

The value of $\sin 78^{\circ}-\sin 66^{\circ}-\sin 42^{\circ}+\sin 60^{\circ}$ is

✓
None of these
B
$\frac{1}{2}$
C
$-1$
D
$\frac{-1}{2}$

Answer

Correct option: A.

None of these

$\sin 78^{\circ}-\sin 66^{\circ}-\sin 42^{\circ}+\sin 60^{\circ}$
$=\sin 78^{\circ}-\sin 42^{\circ}-\sin 66^{\circ}+\sin 60^{\circ}$
$=2 \sin \left(\frac{78^{\circ}-42^{\circ}}{2}\right) \cos \left(\frac{78^{\circ}+42}{2}\right)-\sin 66^{\circ}+\sin 60^{\circ}\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right]$
$=2 \sin 18^{\circ} \cos 60^{\circ}-\sin 66^{\circ}+\sin 60^{\circ}$
$=2 \times \frac{1}{2} \sin 18^{\circ}-\sin 66^{\circ}+\frac{\sqrt{3}}{2}$
$=\sin 18^{\circ}-\sin 66^{\circ}+\frac{\sqrt{3}}{2}$
$=0.309-0.914+0.866$
$=0.261$

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MCQ 41 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

A
$R=\{(x, y): 0 \leq x \leq a, 0 \leq y \leq b\}$
B
$R=\{(x, y): 0 \leq x<a, 0 \leq y \leq b\}$
C
$R=\{(x, y): 0<x<a, 0<y<b\}$
D
$R=\{(x, y): 0 \leq x \leq a, 0<y<b\}$

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 51 Mark

Solve: $3 x+5< x-7$, when $x$ is a real number

A
$x>-12$
B
$x<-12$
✓
$x<-6$
D
$x>-6$

Answer

Correct option: C.

$x<-6$

$3 x+5$
$\Rightarrow 3 x+5-x$
$\Rightarrow 2 x+5<-7$
$\Rightarrow 2 x+5-5<-7-5$
$\Rightarrow 2 x<-12$
$\Rightarrow \frac{2 x}{2}<-\frac{12}{2}$
$\Rightarrow x<-6$

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MCQ 61 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
$( n +1) \cdot 2^n$
✓
$n \cdot 2^{n-1}$

Answer

Correct option: D.

$n \cdot 2^{n-1}$

$C _1+2 C _2+3 C _3+\ldots+n C _{ n }$
$= n +2 \cdot \frac{ n ( n -1)}{2}+3 \cdot \frac{n(n-1)(n-2)}{3!}+\ldots+n$
$= n \cdot\left[1+( n -1) \frac{( n -1)( n -2)}{2!}+\ldots+1\right]$
$= n \cdot\left[{ }^{( n -1)} C _0+{ }^{( n -1)} C _1+{ }^{( n -1)} C _2+\ldots+{ }^{( n -1)} C _{ n -1}\right]$
$= n \cdot(1+1)^{ n -1}= n \cdot 2 ^{ n -1}$

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MCQ 71 Mark

$\sum_{r=0}^n 4^r \cdot{ }^n C_r$ is equal to

A
$6^{ n }$
B
$5^{-n}$
C
$4^{ n }$
✓
$5^{ n }$

Answer

Correct option: D.

$5^{ n }$

$\sum_{r=0}^n 4^{r \cdot n} C_r=4^0 \cdot{ }^n C_0+4^1 \cdot{ }^n C_1+4^2 \cdot{ }^n C_2+\ldots+4^n \cdot{ }^n C_n$
$=1+4 \cdot^n C_1+4^2 \cdot{ }^n C_2+\ldots .+4^n \cdot{ }^n C_n$
$=(1+4)^n$
$=5^n$

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MCQ 81 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 91 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 101 Mark

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\cos x}$ is equal to

A
$1$
✓
$0$
C
$-1$
D
does not exit

Answer

Correct option: B.

$0$

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\cos x}$
$=\lim _{y \rightarrow 0} \frac{1-\sin \left(\frac{\pi}{2}-y\right)}{\cos \left(\frac{\pi}{2}-y\right)}$ taking $\frac{\pi}{2}-x=y$
$=\lim _{y \rightarrow 0} \frac{1-\cos y}{\sin y}$
$=\lim _{y \rightarrow 0} \frac{2 \sin ^2 \frac{y}{2}}{2 \sin \frac{y}{2} \cos \frac{y}{2}}$
$=\lim _{y \rightarrow 0} \tan \frac{y}{2}=0$

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MCQ 111 Mark

A polygon has $44$ diagonals. The number of its sides is

A
$8$
✓
$11$
C
$12$
D
$10$

Answer

Correct option: B.

$11$

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 $= 44$
$\Rightarrow 44=\frac{n(n-3)}{2}$
$\Rightarrow 88=n(n-3)$
$\Rightarrow n^2-3 n-88=0$
$\Rightarrow(n-11)(n+8)=0$
$\Rightarrow n=11,-8$
Since $n$ cannot be negative,
we get $n = 11$

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MCQ 121 Mark

If $x+i y=\frac{3+5 i}{7-6 i}$, then $y=$

A
$\frac{43}{85}$
B
$\frac{-9}{85}$
C
$\frac{9}{85}$
✓
$\frac{53}{85}$

Answer

Correct option: D.

$\frac{53}{85}$

$x + iy =\frac{3+5 i}{7-6 i}$
$\Rightarrow x + iy =\frac{3+5 i}{7-6 i} \times \frac{7+6 i}{7+6 i}$
$\Rightarrow x + iy =\frac{21+53 i+30 i^2}{49-36 i^2}$
$\Rightarrow x + iy =\frac{21-30+53 i}{49-36 i}$
$\Rightarrow x + iy =\frac{-9}{85}+i \frac{53}{85}$
On comparing both the sides:
$y=\frac{53}{85}$

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MCQ 131 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 141 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 151 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 161 Mark

If the mean of the squares of first $n$ natural numbers be $11,$ then n is equal to

✓
$5$
B
$\frac{-13}{2}$
C
$11$
D
$13$

Answer

Correct option: A.

$5$

Mean $=\frac{\frac{n(n+1)(2 n+1)}{6}}{n}=\frac{(n+1)(2 n+1)}{6}$
$\Rightarrow 11=\frac{(n+1)(2 n+1)}{6}$
$\Rightarrow 66=(n+1)(2 n+1)$
$\Rightarrow 2 n^2+3 n-65=0$
$\Rightarrow 2 n^2+13 n-10 n-65=0$
$\Rightarrow(2 n+13)(n-5)=0$
$\Rightarrow n=5, \frac{-13}{2}$
So, $n = 5$

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MCQ 171 Mark

The domain of the function $f(x)=\sqrt{x-1}+\sqrt{6-x}$

A
$(-\infty, 6)$
B
$[2,6]$
✓
$[1,6]$
D
$[-2,6]$

Answer

Correct option: C.

$[1,6]$

For $f(x)$ to be real, we must have,
$x-1 \geqslant 0 \text { and } 6-x \geqslant 0$
$\Rightarrow x \geqslant \varphi \text { and } x-6 \leqslant 6$
$\therefore \text { Domain }=[1,6]$

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MCQ 181 Mark

The value of $\sin ^2 \frac{5 \pi}{12}-\sin ^2 \frac{\pi}{12}$ is

✓
$\sqrt{3} / 2$
B
$\frac{1}{2}$
C
$0$
D
$1$

Answer

Correct option: A.

$\sqrt{3} / 2$

$\sin ^2 75^{\circ}-\sin ^2 15^{\circ}$
$=\sin ^2 75^{\circ}-\cos ^2 75^{\circ}\left[\sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
Now, $\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)$
$=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}$
$=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
$\cos 75^{\circ}=\cos \left(45^{\circ}+30^{\circ}\right)$
$=\cos 45^{\circ} \cos 30^{\circ}-\sin 45^{\circ} \sin 30^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}$
$=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
Hence, $\sin ^2 75^{\circ}-\cos ^2 75^{\circ}$
$=\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)^2-\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^2$
$=\frac{3+1+2 \sqrt{3}-3-1+2 \sqrt{3}}{8}$
$=\frac{4 \sqrt{3}}{8}$
$=\frac{\sqrt{3}}{2}$

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Maths M.C.Q (1 Marks)

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