M.C.Q (1 Marks)

MCQ 11 Mark

${ }^{36} C_{34}=?$

A
610
B
630
C
1224
D
612

Answer

(b) 630
Explanation: ${ }^{36} C _{34}={ }^{36} C _{(36-34)}={ }^{36} C _2=\frac{36 \times 35}{2}=630$

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MCQ 21 Mark

Mark the correct answer for $(2-3 i)(-3+4 i)=$ ?

A
(6 + 17i)
B
(-6 + 17i)
C
(6 - 15i)
D
(6 - 17i)

Answer

(a) (6 + 17i)
Explanation: $(2-3 i)(-3+4 i)=\left(-6+8 i+9 i-12 i^2\right)=(-6+17 i+12)=(6+17 i)$

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MCQ 31 Mark

$\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}=?$

A
cosec x
B
sec x
C
cot x
D
tan x

Answer

(d) tan x
Explanation: $\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}=\frac{2 \sin ^2 x+\sin x}{2 \sin x \cos x+\cos x}=\frac{\sin x(2 \sin x+1)}{\cos x(2 \sin x+1)}=\tan x$

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MCQ 41 Mark

For any two sets $A$ and $B , A \cap(A \cup B)=$_________

A
$\neq \phi$
B
B
C
$\phi$
D
A

Answer

(d) A
Explanation: Common between set A and $(A \cup B)$ is set A itself

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MCQ 51 Mark

If x < 5 , then

A
− x > − 5
B
none of these
C
− x < 5
D
x > − 5

Answer

(a) − x > − 5
Explanation: Given x < 5
Multiplying both sides of the above inequality by -1,we get
− x > − 5 (The sign of the inequality is to be reversed if both sides of an inequality are multiplied by the same negative real number)

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MCQ 61 Mark

$\left\{C_0+2 C_1+3 C_2+\ldots+(n+1) C_n\right\}=?$

A
$( n +1). 2^{ n }$
B
$( n +2) . 2^{ n +1}$
✓
$( n +2). 2^{ n -1}$
D
$n . 2^{ n -1}$

Answer

Correct option: C.

$( n +2). 2^{ n -1}$

Here, $C _0+2 C _1+3 C _2+\ldots+( n +1) C _{ n }$
$=\left(C_0+C_1+C_2+\ldots+C_n\right)+\left(C_1+2 C_2+3 C_3+\ldots+n C_n\right)$
$=2^n+n \cdot 2^{n-1}$
$=(n+2) \cdot 2^{n-1}$

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MCQ 71 Mark

$(3+6+12+\ldots+1536)=?$

A
1023
B
3069
C
2046
D
4092

Answer

(b) 3069
Explanation: This is a GP in which $a =3, r =\frac{6}{3}=2$ and $l =1536$
$\therefore$ required sum $=\frac{(l r-a)}{(r-1)}=\frac{(1536 \times 2-3)}{(2-1)}=(3072-3)=3069$.

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MCQ 81 Mark

If a set A has n elements then the total number of subsets of A is

A
2n
B
n
C
$2^n$
D
$n^2$

Answer

(c) $2^{ n }$
Explanation: The total no of subsets $=2^{ n }$

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MCQ 91 Mark

The radian measure of $50^{\circ} 37^{\prime} 30^{\prime \prime}$ is

A
$\left(\frac{5 \pi}{16}\right)^c$
B
$\left(\frac{9 \pi}{32}\right)^c$
C
$\left(\frac{7 \pi}{18}\right)^c$
D
$\left(\frac{11 \pi}{36}\right)^c$

Answer

(b)$\left(\frac{9 \pi}{32}\right)^c$
Explanation: $50^{\circ} 37^{\prime} 30^{\prime \prime}=50^{\circ}+\left(37 \frac{30}{60}\right)^{\prime}=50^{\circ}+\left(\frac{75}{2}\right)^{\prime}=50^{\circ}+\left(\frac{75}{2 \times 60}\right)^{\circ}=\left(50 \frac{5}{8}\right)^{\circ}=\left(\frac{405}{8}\right)^{\circ}$
$180^{\circ}=\pi^c \Rightarrow 1^{\circ}=\left(\frac{\pi}{180}\right)^c \Rightarrow\left(\frac{405}{8}\right)^{\circ}=\left(\frac{\pi}{180} \times \frac{405}{8}\right)^c=\left(\frac{9 \pi}{32}\right)^c$

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MCQ 101 Mark

The solution set for $(x+3)+4>-2 x+5$ :

A
$(-\infty, 2)$
✓
$\left(\frac{-2}{3}, \infty\right)$
C
$(-\infty,-2)$
D
$(2, \infty)$

Answer

Correct option: B.

$\left(\frac{-2}{3}, \infty\right)$

$(x+3)+4>-2 x+5$
$\Rightarrow x+7>-2 x+5$
$\Rightarrow x+7+2 x>-2 x+5+2 x$
$\Rightarrow 3 x+7>5$
$\Rightarrow 3 x+7-7>5-7$
$\Rightarrow 3 x>-2$
$\Rightarrow x>\frac{-2}{3}$
$\Rightarrow x \in\left(\frac{-2}{3}, \infty\right)$

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MCQ 111 Mark

Let R be a relation on N defined by x + 2y = 8. The domain of R is

A
(1, 2, 3, 4)
B
(2, 4, 8)
C
(2, 4, 6, 8)
D
(2, 4, 6)

Answer

(d) (2, 4, 6)
Explanation: We have , x + 2y = 8
$y=\frac{8-x}{2}$
since, x and y are Natural numbers, So x must be an even number.
if x = 2, y = 3;
if x = 4, y = 2;
if x = 6, y = 1.
|So, relation R = {(2, 3), (4, 2), (6, 1)}
Hence, the domain of R is (2, 4, 6).

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MCQ 121 Mark

If $z=\left(\frac{1+i}{1-i}\right)$, then $z^4$ equals.

A
$0$
B
$-1$
C
$2$
✓
$1$

Answer

Correct option: D.

$1$

$\text { Let } z =\frac{1+i}{1-i}$
$z=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$
$\Rightarrow z=\frac{1+ i ^2+2 i}{1-i^2}$
$\Rightarrow z=\frac{2 i}{2}$
$\Rightarrow z=i$
$\Rightarrow z^4=i^4$
Since $i^2 = -1,$ we have:
$\Rightarrow z^4=i^2 \times i^2$
$\Rightarrow z^4=1$

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MCQ 131 Mark

Let A = (a, b, c), B = (a, b), C = (a, b, d), D = (c, d) and E = (d). Then which of the following statement is not correct?

A
$D \supseteq E$
B
$C - B = E$
C
$B \cup E=C$
D
$C - D = E$

Answer

(d) C - D = E
Explanation: C - D = (a, b, c) - (c, d) = (a, b)
But $E=\{d\}$
Hence $C-D \neq E$

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MCQ 141 Mark

The centroid of a triangle is (2, 7) and two of its vertices are (4, 8) and (-2, 6). The third vertex is

A
$(0,0)$
B
(4, 7)
C
(7, 7)
D
(7, 4)

Answer

(b) (4, 7)
Explanation: Let A (4, 78) and B (-2, 6) be the given vertex. Let C(h, k) be the third vertex.
The centroid of $\triangle ABC$ is $\left(\frac{4-2+h}{3}, \frac{8+6+k}{3}\right)$
It is given that the centroid of triangle ABC is (2, 7) as obtained from above formula,
$\therefore \frac{4-2+h}{3}=2, \frac{8+6+k}{3}=7$
$\Rightarrow h=4, k=7$
Thus, the third vertex is $(4,7)$

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MCQ 151 Mark

$\underset{{X \rightarrow 0}}{\lim} \frac{\tan 2 x-x}{3 x-\sin x}$ is equal to

A
$\frac{1}{2}$
B
2
C
$\frac{1}{4}$
D
$-\frac{1}{2}$

Answer

(a) $\frac{1}{2}$
Explanation: Given, $\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0} \frac{x\left[\frac{\tan 2 x}{x}-1\right]}{x\left[3-\frac{\sin x}{x}\right]}$
$\lim _{x \rightarrow 0} \frac{\frac{\tan 2 x}{2 x} \times 2-1}{3-\frac{\sin x}{x}}=\frac{1.2-1}{3-1}=\frac{2-1}{2}=\frac{1}{2}$

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MCQ 161 Mark

Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is

A
$\frac{1}{221}$
B
$\frac{1}{26}$
C
$\frac{1}{13}$
D
$\frac{4}{223}$

Answer

(a) $\frac{1}{221}$
Explanation: Total number of ways drawing 2 cards successively without replacement
$={ }^{52} C _1 \times{ }^{51} C _1$ and number of ways 2 aces without replacement $={ }^4 C _1 \times{ }^3 C _1$
$\therefore$ Required probability $=\frac{{ }^4 C_1 \times{ }^3 C_1}{{ }^{32} C_1 \times{ }^{51} C_1}=\frac{4 \times 3}{52 \times 51}$
$=\frac{1}{13 \times 17}=\frac{1}{22}$

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MCQ 171 Mark

Domain of definition of the function $f ( x )=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)$ is

✓
$(-1,0) \cup(1,2) \cup(2, \infty)$
B
$(1,2) \cup(2, \infty)$
C
$(-1,0) \cup(1,2)$
D
$(1,2)$

Answer

Correct option: A.

$(-1,0) \cup(1,2) \cup(2, \infty)$

For $f(x)$ to be real, we must have
$4-x^2 \neq 0$ and $x^3-x>0$
$\Rightarrow x^2 \neq 4$ and $x\left(x^2-1\right)>0$
$\Rightarrow x \neq 2,-2$ and $x(x-1)(x+1)>0$
$\Rightarrow x \neq 2,-2$ and $-1$

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MCQ 181 Mark

If $\sin \theta+\operatorname{cosec} \theta=2$, then $\sin ^2 \theta+\operatorname{cosec}^2 \theta$ is equal to

✓
$2$
B
$1$
C
$3$
D
$4$

Answer

Correct option: A.

$2$

Given $\sin \theta+\operatorname{cosec} \theta=2$
Squaring on both sides, we get
$\sin ^2 \theta+\csc ^2 \theta+2 \sin \theta \csc \theta=4[\because \sin \theta \operatorname{cosec} \theta=1]$
$\Rightarrow \sin ^2 \theta+\operatorname{cosec}^2 \theta=4-2=2$

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Maths M.C.Q (1 Marks)

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