M.C.Q (1 Marks)

MCQ 11 Mark

In how many ways can 5 persons occupy 3 seats?

A
30
B
15
C
60
D
20

Answer

(c) 60
Explanation: Required number of ways $={ }^5 P _3=\frac{5!}{(5-3)!}=\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}=60$

View full question & answer→

MCQ 21 Mark

$\lim _{x \rightarrow 3} \frac{x-3}{|x-3|}$ is equal to:

A
1
B
-1
C
$0$
D
None of these

Answer

(d) None of these
Explanation: $\lim _{x \rightarrow 3} \frac{x-3}{|x-3|}$
LHL at x = 3
$\lim _{x \rightarrow 3^{-}} \frac{x-3}{-(x-3)}[\because|x-3|=-(x-3) x<3]$
= -1
RHL at x = 3
$\lim _{x \rightarrow 3^{+}} \frac{x-3}{x-3}[\therefore|x-3|=x-3$, when $x>3]$
= 1
$LHL \neq RHL$

View full question & answer→

MCQ 31 Mark

The value of $\frac{2\left(\sin 2 x+2 \cos ^2 x-1\right)}{\cos x-\sin x-\cos 3 x+\sin 3 x}$ is

A
$\sin x$
B
$\cos x$
✓
$\operatorname{cosec} x$
D
$\sec x$

Answer

Correct option: C.

$\operatorname{cosec} x$

We have,
$\frac{2\left(\sin 2 x+2 \cos ^2 x-1\right)}{\cos x-\sin x-\cos 3 x+\sin 3 x}$
$=\frac{2(\sin 2 x+\cos 2 x)}{\cos x-\sin x-4 \cos ^3 x+3 \cos x+3 \sin x-4 \sin ^3 x}$
$=\frac{2(\sin 2 x+\cos 2 x)}{4 \cos x-4 \cos ^3 x+2 \sin x-4 \sin ^3 x}$
$=\frac{2(\sin 2 x+\cos 2 x)}{4 \cos x\left(1-\cos ^2 x\right)+2 \sin x\left(1-2 \sin ^2 x\right)}$
$=\frac{2(\sin 2 x+\cos 2 x)}{4 \cos x \sin ^2 x+2 \sin x \cos 2 x}$
$=\frac{2(\sin 2 x+\cos 2 x)}{2 \times 2 \sin x \cos x \sin x+2 \sin x \cos 2 x}$
$=\frac{2(\sin 2 x+\cos 2 x)}{2 \sin 2 x \sin x+2 \sin x \cos 2 x}$
$=\frac{2(\sin 2 x+\cos 2 x)}{2 \sin x(\sin 2 x+\cos 2 x)}$
$=\frac{1}{\sin x}$
$=\operatorname{cosec} X $

View full question & answer→

MCQ 41 Mark

If A = {1, 3, 5, B} and B = {2, 4}, then

A
$\{4\} \subset A$
B
None of these
C
$B \subset A$
D
$4 \in A$

Answer

(b) None of these
Explanation: $4 \notin A$
$\{4\} \not \subset A$
$B \not \subset A$
Therefore, we can say that none of these options satisfy the given relation.

View full question & answer→

MCQ 51 Mark

The solution set of $6 x-1 > 5$ 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 $\{x: x>1, x \in R\}$

View full question & answer→

MCQ 61 Mark

$(\sqrt{5}+1)^4+(\sqrt{5}-1)^4$ is

A
an irrational number
B
a negative real number
✓
a rational number
D
a negative integer

Answer

Correct option: C.

a rational number

We have $( a + b )^{ n }+( a - b )^{ n }$
$=\left[{ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+{ }^n C_3 a^{n-3} b^3+\ldots . .+{ }^n C_n b^n\right]+$
${\left[{ }^n C_0 a^n-{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2-{ }^n C_3 a^{n-3} b^3+\ldots . .+(-1)^n \cdot{ }^n C_n b^n\right]}$
$=2\left[{ }^n C_0 a^n+{ }^n C_2 a^{n-2} b^2+\ldots\right]$
Let $a=\sqrt{5}$ and $b=1$ and $n=4$
Now we get $(\sqrt{5}+1)^4+(\sqrt{5}-1)^4=2\left[{ }^4 C_0(\sqrt{5})^4+{ }^4 C_2(\sqrt{5})^2 1^2+{ }^4 C_4(\sqrt{5})^0 1^4\right]$
$=2[25+30+1]=112$

View full question & answer→

MCQ 71 Mark

$\left\{\frac{c_1}{\infty}+2 \frac{C_2}{C_1}+3 \frac{C_3}{C_2}+\ldots+n \cdot \frac{C_n}{C_{n-1}}\right\}=?$

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

Answer

(a) $\frac{1}{2} n(n+1)$
Explanation: We know that $\frac{C_r}{C_{r 1}}=\frac{n-r+1}{r}$,
Substituting $r =1,2,3, \ldots, n$, we obtain
$\frac{C_1}{C_0}+2 \cdot \frac{C_2}{C_1}+3 \cdot \frac{C_3}{C_2}+\ldots+n \cdot \frac{C_n}{C_{n-1}}= n +( n -1)+( n -2)+\ldots+1=\frac{1}{2} n(n+1)$.

View full question & answer→

MCQ 81 Mark

Let $F_1$ be the set of parallelograms, $F_2$ the set of rectangles, $F_3$ the set of rhombuses, $F_4$ the set of squares and $F_5$ the set of trapeziums in a plane. Then $F _1$ may be equal to

A
$F_2 \cap F_3$
B
$F _3 \cap F_4$
C
$F _2 \cup F_5$
D
$F _2 \cup F_3 \cup F_4 \cup F_1$

Answer

(d) $F_2 \cup F_3 \cup F_4 \cup F_1$
Explanation: We know that
Every rectangle, square and rhombus is a parallelogram
But, no trapezium is a paralleogrm
Thus, $F_1=F_2 \cup F_3 \cup F_4 \cup F_1$

View full question & answer→

MCQ 91 Mark

Which is smaller, $\sin 64^{\circ}$ or $\cos 64^{\circ}$ ?

A
$\cos 64^{\circ}$
B
$\sin 64^{\circ}$
C
cannot be compared
D
both are equal

Answer

(a) $\cos 64^{\circ}$
Explanation: In quadrant $I , \sin \theta$ is increasing.
Now, $\cos 64^{\circ}=\cos \left(90^{\circ}-26^{\circ}\right)=\sin 26^{\circ}$.
Clearly, $\sin 26^{\circ}<\sin 64^{\circ} \Rightarrow \cos 64^{\circ}<\sin 64^{\circ}$

View full question & answer→

MCQ 101 Mark

If $y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}$ then $\frac{d y}{d x}$ is equal to

A
$\frac{-4 x}{x^2-1}$
B
$\frac{1-x^2}{4 x}$
C
$\frac{-4 x}{\left(x^2-1\right)^2}$
D
$\frac{4 x}{x^2-1}$

Answer

(c) $\frac{-4 x}{\left(x^2-1\right)^2}$
Explanation: Given $y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{z^2}} \Rightarrow y=\frac{x^2+1}{x^2-1}$
$\therefore \quad \frac{d y}{d x}=\frac{\left(x^2-1\right) \cdot 2 x-\left(x^2+1\right) \cdot 2 x}{\left(x^2-1\right)^2}$
$=\frac{2 x\left(x^2-1-x^2-1\right)}{\left(x^2-1\right)^2}=\frac{2 x(-2)}{\left(x^2-1\right)^2}=\frac{-4 x}{\left(x^2-1\right)^2}$

View full question & answer→

MCQ 111 Mark

If ${ }^{20} C_{r+1}={ }^{20} C_{r-1}$, then $r$ is equal to

A
$19$
✓
$10$
C
$12$
D
$11$

Answer

Correct option: B.

$10$

$r +1+ r -1=20\left[\because{ }^n C_x={ }^n C_y \Rightarrow n = x + y\right.$ or $\left.x = y \right]$
$\Rightarrow 2 r =20$
$\Rightarrow r =10$

View full question & answer→

MCQ 121 Mark

Mark the correct answer for $(1+ i )^{-1}=$ ?

A
$\left(\frac{-1}{2}+\frac{1}{2} i\right)$
B
(2 - 3i)
C
$\left(\frac{1}{2}-\frac{1}{2} i\right)$
D
(2 - i)

Answer

(c) $\left(\frac{1}{2}-\frac{1}{2} i\right)$
Explanation: $(1+i)^{-1}=\frac{1}{(1+i)}=\frac{1}{(1+i)} \times \frac{(1-i)}{(1-i)}=\frac{(1-i)}{\left(1^2-i^2\right)}=\frac{(1-i)}{2}=\left(\frac{1}{2}-\frac{1}{2} i\right)$

View full question & answer→

MCQ 131 Mark

Distance of the point $(\alpha, \beta, \gamma)$ from $y$-axis is

A
$\sqrt{\alpha^2+\gamma^2}$
B
$|\beta|+|\gamma|$
C
$|\beta|$
D
$\beta$

Answer

(a) $\sqrt{\alpha^2+\gamma^2}$
Explanation: The foot of perpendicular from point $P(\alpha, \beta, \gamma)$ on y -axis is $Q(0, \beta, 0)$
$\therefore$ Required distance, $P Q=\sqrt{(\alpha-0)^2+(\beta-\beta)^2+(\gamma-0)^2}=\sqrt{\alpha^2+\gamma^2}$

View full question & answer→

MCQ 141 Mark

A line passes through P (1, 2) such that its intercept between the axes is bisected at P. The equation of the line is

A
x + 2y = 5
B
2x + y – 4 = 0
C
x + y – 3 = 0
D
x – y + 1 = 0

Answer

(b) 2x + y – 4 = 0
Explanation: We know that the equation of a line making intercepts a and b with x-axis and y-axis, respectively, is given by
$\frac{x}{a}+\frac{y}{b}=1$
which give a = 2 and b = 4.
Thus, now we have to find the required equation of the line is given by $\frac{x}{2}+\frac{y}{4}=1$ or $2 x+y-4=0$

View full question & answer→

MCQ 151 Mark

If $f(x)=\frac{x^n-a^n}{x-a}$ for some constant, a, then $f ^{\prime}(a)$ is equal to

A
$1 / 2$
B
does not exist
C
1
D
$0$

Answer

(b) does not exist
Explanation: Given $f(x)=\frac{x^n-a^n}{x-a}$
$f^{\prime}(x)=\frac{(x-a)\left(n \cdot x^{n-1}\right)-\left(x^n-a^n\right) \cdot 1}{(x-a)^2}$
$\therefore \quad f^{\prime}(a)=\frac{(a-a)\left(n \cdot a^{n-1}\right)-\left(a^n-a^n\right)}{(a-a)^2}$
So $f^{\prime}(a)=\frac{0}{0}=$ does not exist

View full question & answer→

MCQ 161 Mark

The mean of five numbers is $30.$ If one number is excluded, their mean becomes $28.$ The excluded number is$:$

A
$38$
B
$30$
C
$35$
✓
$28$

Answer

Correct option: D.

$28$

Let the numbers are $x _1, x _2, x _3, x$ and $x _5$.
Then$,$ we have$, \frac{x_1+x_2+x_3+x_4+x_5}{5}=30$
$\Rightarrow x _1+ x _2+ x _3+ x _4+ x _5=150 \ldots \text { (i) }$
Now$,$ suppose $x_1$ is excluded$,$ then $\frac{x_2+x_3+x_4+x_5}{4}=28\ [$given$]$
$\Rightarrow x _2+ x _3+ x _4+ x _5=112 \ldots (i)$
From $Eqs.(i)$ and $(ii), $we get $x_1 = 150 - 112 = 38$

View full question & answer→

MCQ 171 Mark

Number of relations that can be defined on the set A = {a, b, c, d} is

A
24
B
$4^4$
C
16
D
$2^{16}$

Answer

(d) $2^{16}$
Explanation: No. of elements in the set $A =4$. Therefore, the no. of elements in $A \times A=4 \times 4=16$. As, the no. of relations in $A \times A=$ no. of subsets of $A \times A=2^{16}$.

View full question & answer→

MCQ 181 Mark

Find the value of $\sec \left(\frac{-19 \pi}{3}\right)$.

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

Answer

(c) 2
Explanation: $\sec \left(\frac{-19 \pi}{3}\right)=\sec \frac{19 \pi}{3}[\because \sec (-\theta)=\sec \theta]$
$=\sec \left(6 \pi+\frac{\pi}{3}\right)=\sec \frac{\pi}{3}=2[\because \sec (2 n \pi+\theta)=\sec \theta]$

View full question & answer→

Maths M.C.Q (1 Marks)

Test yourself on this topic