M.C.Q (1 Marks)

MCQ 11 Mark

In how many ways can a committee of 5 members be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies?

A
50
B
200
C
25
D
100

Answer

(b) 200
Explanation:Number of ways of selecting 3 men out of 6 and 2 ladies out of $5=\left({ }^6 C_3 \times{ }^5 C_2\right)$
$=\left(\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1}\right)=200$

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MCQ 21 Mark

For any positive integer $n,(-\sqrt{-1})^{4 n+3}=$ ?

A
1
B
i
C
-i
D
-1

Answer

(b) i
Explanation: $(-\sqrt{-1})^{4 n+3}=(- i )^{4 n +3}=\left\{(- i )^4\right)^{ n }(-1)^3=1 \times(- i ) \times(- i ) \times(- i )= i ^2 \times(- i )=-1 \times(- i )= i$

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MCQ 31 Mark

Mark the Correct alternative in the following: $8 \sin \frac{x}{8} \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}$ is equal to

✓
$\sin x$
B
$8 \cos x$
C
$\cos x$
D
$8 \sin x$

Answer

Correct option: A.

$\sin x$

$8 \sin \frac{x}{8} \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}$
$4\left(2 \sin \frac{x}{8} \cos \frac{x}{8}\right) \cos \frac{x}{2} \cos \frac{x}{4} \text { [by rearranging terms] }$
$4\left(2 \sin \frac{x}{8} \cos \frac{x}{8}\right) \cos \frac{x}{2} \cos \frac{x}{4} \text { [using the formula } \sin 2 \theta=2 \sin \theta \cos \theta \text { ] }$
$=4\left(\sin \frac{x}{4}\right) \cos \frac{x}{2} \cos \frac{x}{4}$
$=2\left(2 \sin \frac{x}{4} \cos \frac{x}{4}\right) \cos \frac{x}{2}$
$=2\left(\sin \frac{2 x}{4}\right) \cos \frac{x}{2}$
$=\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)$
$=\sin x$
Hence $\sin \frac{x}{8} \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8}=\sin x$

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MCQ 41 Mark

If $A$ and $B$ are two given sets, then $A \cap(A \cap B)^C$ is equal to

A
B
B
A
C
$A \cap B ^{ C }$
D
$\phi$

Answer

(c) $A \cap B ^{ C }$
Explanation: $A \cap B ^{ C }$
A and B are two sets.
$A \cap B$ is the common region in both the sets.
$\left(A \cap B^c\right)$ is all the region in the universal set except $A \cap B$
Now, $A \cap(A \cap B)^c=A \cap B^C$

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MCQ 51 Mark

The solution set of $6x - 1 > 5$ is :

A
$\{x: x>1, x \in N\}$
✓
$\{x: x>1, x \in R\}$
C
$\{x: x<1, x \in N\}$
D
$\{x: x<1, x \in W\}$

Answer

Correct option: B.

$\{x: x>1, x \in R\}$

$6 x-1>5$
$\Rightarrow 6 x-1+1>5+1$
$\Rightarrow 6 x>6$
$\Rightarrow x>1$
Hence the solution set is $\{x: x>1, x \in R\}$

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MCQ 61 Mark

If $C_r$ denotes ${ }^n C_r$ in the expansion of $(1+x)^n$, then $C_0+C_1+C_2+\ldots+C_n=$ ?

A
2n
B
$2^{ n }$
C
$\frac{1}{3} n(2 n+1)$
D
$2^{ n }$

Answer

(d) $2^n$
Explanation: Here, we know that $C _0+ C _1+ C _2+\ldots+ C _{ n }=2^{ n }$

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MCQ 71 Mark

The sum of first $10$ terms of a $G.P.$ is equal to $244$ times the sum of its first five terms. Then the common ratio is

A
$7$
B
$5$
C
$4$
✓
$3$

Answer

Correct option: D.

$3$

Let $r$ be the common ratio of the $G.P.$
Given $S _{10}=244 S_5$
$\Rightarrow \frac{S_{10}}{S_5}=244$
We have $S_n=\frac{a\left(r^n-1\right)}{r-1}$
$\Rightarrow \frac{\frac{a\left(r^{10}-1\right)}{r-1}}{\frac{a\left(r^5-1\right)}{r-1}}=244, r-1 \neq 0$
$\Rightarrow \frac{r^{10}-1}{r^5-1}=244$
$\Rightarrow r^{10}-1-244 r^5+244=0$
$\Rightarrow\left(r^5\right)^2-244 r^5+243=0$
$\Rightarrow(r 5)^2-243 r^5-1 r^5+243=0$
$\Rightarrow r^5\left(r^5-1\right)-243\left(r^5-1\right)=0$
$\Rightarrow r^5=243$ or $r^5=1$
Since $r - 1 \neq 0, r$ cannot be $1$
$\Rightarrow r=\sqrt[5]{243}=3$

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MCQ 81 Mark

Two finite sets have $m$ and $n$ elements respectively. The total number of subsets of first set is $56$ more than the total number of subsets of the second set. The values of $m$ and $n$ respectively are.

A
$5, 1$
B
$7, 6$
C
$8, 7$
✓
$6, 3$

Answer

Correct option: D.

$6, 3$

Since, let $A$ and $B$ be such sets,
i.e., $n (A) = m,$ and $n(B) = n$
Thus, $n ( P ( A ))=2^{ m }, n ( P ( B ))=2^{ n }$
Therefore, $n ( P ( A ))- n ( P ( B ))=56$,
i.e., $2^{ m }-2^{ n }=56$
$\Rightarrow 2^n\left(2^{m-n}-1\right)=2^3 7$
$\Rightarrow n=3,2^{m-n}-1=7$
$\Rightarrow m=6$

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MCQ 91 Mark

$\cos 405^{\circ}=?$

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

Answer

(d)$\frac{1}{\sqrt{2}}$
Explanation: $\cos 405^{\circ}=\cos \left(360^{\circ}+45^{\circ}\right)=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$

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MCQ 101 Mark

The solution set of the inequation $3x < 5$, when $x$ is a natural number is

A
$(1, 2)$
✓
$(1)$
C
$(4)$
D
$(0, 1)$

Answer

Correct option: B.

$(1)$

$3x < 5$
$\Rightarrow x<\frac{5}{3}$
$\Rightarrow x<1 \frac{2}{3}$
Hence solution set $=\left\{x: x<1 \frac{2}{3}, x \in N\right\}=\{1\}$

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MCQ 111 Mark

If $R=\left\{(x, y) ; x, y \in Z, x^2+y^2 \leq 4\right\}$ is a relation on $Z$, then domain of $R$ is

A
(-2, -1, 0, 1, 2)
B
(0, -1, -2)
C
(-1, 0, 1)
D
(0, 1, 2)

Answer

(a) $\{-2,-1,0,1,2\}$
Explanation: Domain of R is a set constituting all values of x .
Here, possible values for $x$ by equation $x^2+y^2 \leq 4$ will be $0,1,-1,2,-2$.
So, Domain of R is : $\{-2,-1,0,1,2\}$.

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MCQ 121 Mark

If $\left(\frac{1+i}{1-i}\right)^x=1$, then

✓
$x = 4n$
B
$x = 2n$
C
$x = 2n+1$
D
$x =4 n +1$, where $n \in N$

Answer

Correct option: A.

$x = 4n$

$\Rightarrow\left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^x=1 $
$\Rightarrow\left[\frac{1+2 i+i^2}{1-i^2}\right]^x=1 $
$\Rightarrow\left[\frac{2 i}{1+1}\right]^x=1$
$\Rightarrow I ^{ x }=1$
$\Rightarrow x =4 n , n \in N $

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MCQ 131 Mark

If $A=\left\{(x, y): x^2+y^2=25\right\}$ and $B=\left\{(x, y): x^2+9 y^2=144\right\}$ then $A \cap B$ contains

A
three points
B
two points
C
one point
✓
four points

Answer

Correct option: D.

four points

From $A, x^2+y^2=25$ and from $B, x^2+9 y^2=144$
$\therefore$ From $B, \left(x^2+y^2\right)+8 y^2=144$
$\Rightarrow 25+8 y^2=144$
$\Rightarrow 8 y^2=119$
$\Rightarrow y = \pm \sqrt{\frac{119}{8}}$
$\therefore x^2+y^2=25$
$\Rightarrow x^2$
$=25-y^2$
$=25-\frac{119}{8}$
$=\frac{81}{8}$
$\Rightarrow x = \pm \sqrt{\frac{81}{8}}$
Since we solved equations simultaneously, therefore $A \cap B$ has four points $A$ has $2$ elements $B$ has $2$ elements.

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MCQ 141 Mark

The acute angle between the lines $y = 2x$ and $y = - 2x$ is

A
greater than $60^{\circ}$
B
$90^{\circ}$
✓
less than $60^{\circ}$
D
$60^{\circ}$

Answer

Correct option: C.

less than $60^{\circ}$

The angle between two straight lines is given by $\frac{\left|m_1-m_2\right|}{1+m_1 m_2}$
Here $m _1=2$ and $m _2=-2$
Sustituting the values we get,
$\tan \theta=\frac{2-(-2)}{1+2 \cdot(-2)}$
$=\frac{4}{5}<60^{\circ}$

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MCQ 151 Mark

$\lim _{x \rightarrow 0} \frac{|\sin x|}{x}$ is equal to

A
$0$
✓
Does not exist
C
$1$
D
$-1$

Answer

Correct option: B.

Does not exist

Given, $\lim _{x \rightarrow 0} \frac{|\sin x|}{x}$
$ LHL =\lim _{x \rightarrow 0^{-}} \frac{-\sin x}{x}=-1 \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$RHL =\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x}=1$
$\text{LHL} \neq {RHL,}$ So the limit does not exist.

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MCQ 161 Mark

If P(E) denotes the probability of an event E, then E is called certain event, if

A
P(E) is either 0 or 1
B
P(E) = 0
C
P(E) = 1
D
$P ( E )=\frac{1}{2}$Z

Answer

(c) P(E) = 1
Explanation: P(E) = 1

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MCQ 171 Mark

If R is a relation on the set $A=\{1,2,3,4,5,6,7,8,9\}$ given by $x R y \Leftrightarrow y=3 x$, then $R=$

A
((3, 1), (2, 6), (3, 9))
B
((3, 1), (6, 2), (9, 3))
C
((3, 1), (6, 2), (8, 2), (9, 3))
D
none of these

Answer

(d) none of these
Explanation: For A = {1, 2, 3, 4, 5, 6, 7, 8, 9} the satisfying complete relation is:
R = {(1, 3), (2, 6), (3, 9)}

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MCQ 181 Mark

If $\cos A = m \cos B$, then $\cot \frac{A+B}{2} \cot \frac{B-A}{2}=$

A
$\frac{m-1}{m+1}$
B
$\frac{m-2}{m+2}$
C
$\frac{m+2}{m-2}$
✓
$\frac{m+1}{m-1}$

Answer

Correct option: D.

$\frac{m+1}{m-1}$

Given:
$\cos A = m \cos B $
$\Rightarrow \frac{\cos A}{\cos B}=\frac{m}{1}$
$\Rightarrow \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1}$
$\Rightarrow \frac{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}=\frac{m+1}{m-1}$
$\left[\because \cos A+\cos B =2 \cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right) \text { and } \cos A -\cos B =-2 \sin \left(\frac{A+B}{2}\right) \cos \right.$
$\left.\left(\frac{A-B}{2}\right)\right]$
$\Rightarrow \frac{\cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}{-\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}=\frac{m+1}{m-1}$
$\Rightarrow-\cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)=\frac{m+1}{m-1}$
$\Rightarrow \cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)=\frac{1+m}{1-m}$

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Maths M.C.Q (1 Marks)

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