MCQ

MCQ 512 Marks

Range of the function $f(x)=\frac{1}{3 x+2}$ is

A
R
✓
$R -\{0\}$
C
$(0, \infty)$
D
$R-\left\{-\frac{2}{3}\right\}$

Answer

Correct option: B.

$R -\{0\}$

(B)
$\operatorname{Dom}(f)=R-\left\{-\frac{2}{3}\right\}$
For Range(f), let $y= f (x)=\frac{1}{3 x+2}$
$\therefore \quad 3 x+2=\frac{1}{y} \Rightarrow x=\frac{1}{3}\left(\frac{1}{y}-2\right)$
$x$ is real if $y \neq 0$.
Hence, $R _{ f }= R -\{0\}$

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MCQ 522 Marks

The domain of the function $\cos ^{-1}\left(\log _2\left(x^2+5 x+8\right)\right)$ is

A
[2,3]
B
[-2,2]
C
[3,1]
✓
[-3,-2]

Answer

Correct option: D.

[-3,-2]

(D)
$-1 \leq \log _2\left(x^2+5 x+8\right)<1$
$\Rightarrow \frac{1}{2} \leq\left(x^2+5 x+8\right) \leq 2$
$\Rightarrow x^2+5 x+\frac{15}{2} \geq 0$
$\Rightarrow x^2+2\left(\frac{5}{2}\right) x+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2+\frac{15}{2} \geq 0$
$\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{5}{4} \geq 0$ and $x^2+5 x+6 \leq 0$
$\Rightarrow(x+3)(x+2) \leq 0$
$\Rightarrow x \in[-3,-2]$

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MCQ 532 Marks

The domain of the function $f (x)=\sqrt{\cos ^{-1}\left(\frac{1-|x|}{2}\right)}$ is

A
(-3,3)
✓
[-3,3]
C
$(-\infty,-3) \cup(3, \infty)$
D
$(-\infty,-3] \cup[3, \infty)$

Answer

Correct option: B.

[-3,3]

(B)
$f (x)=\sqrt{\cos ^{-1}\left(\frac{1-|x|}{2}\right)}$
$\therefore -1 \leq \frac{1-|x|}{2} \leq 1$
$\Rightarrow-2-1 \leq-|x| \leq 2-1$
$\Rightarrow-3 \leq-|x| \leq 1$
$\Rightarrow-1 \leq|x| \leq 3$
$\Rightarrow x \in[-3,3]$

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MCQ 542 Marks

Domain of the function $f(x)=\sin ^{-1}\left(1+3 x+2 x^2\right)$ is

A
$(-\infty, \infty)$
B
$(-1,1)$
✓
$\left[-\frac{3}{2}, 0\right]$
D
$\left(-\infty, \frac{-1}{2}\right) \cup(2, \infty)$

Answer

Correct option: C.

$\left[-\frac{3}{2}, 0\right]$

(C)
$-1 \leq 1 + 3 x + 2 x^2 \leq 1$
Case I : $2 x^2+3 x+1 \geq-1 ; 2 x^2+3 x+2 \geq 0$
$x=\frac{-3 \pm \sqrt{9-16}}{6}=\frac{-3 \pm i \sqrt{7}}{6}$ (imaginary).
Case II :$2 x^2+3 x+1 \leq 1$
$\Rightarrow 2 x^2+3 x \leq 0 \Rightarrow 2 x\left(x+\frac{3}{2}\right)<0$
$\Rightarrow \frac{-3}{2} \leq x \leq 0 \Rightarrow x \in\left[-\frac{3}{2}, 0\right]$
In case I, we get imaginary value hence, rejected
∴ Domain of function $=\left[\frac{-3}{2}, 0\right]$.

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MCQ 552 Marks

The domain of the function $f (x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}$ is

A
[1,2)
✓
( 2, 3)
C
[1,2]
D
[2, 3]

Answer

Correct option: B.

( 2, 3)

(B)
To define $f (x), 9-x^2>0 \Rightarrow|x|<3$
$\Rightarrow-3< x< 3,$ ...(i)
and $-1 \leq(x-3) \leq 1$
$\Rightarrow 2 \leq x \leq 4$ ...(ii)
From (i) and (ii), $2 \leq x<3$ i.e., $[2,3)$.

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MCQ 562 Marks

The domain of the function $f (x)=\exp \left(\sqrt{5 x-3-2 x^2}\right)$ is

A
$\left[1, \frac{-3}{2}\right]$
B
$\left[\frac{3}{2}, \infty\right]$
C
$(-\infty, 1]$
✓
$\left[1, \frac{3}{2}\right]$

Answer

Correct option: D.

$\left[1, \frac{3}{2}\right]$

(D)
$f (x)= e ^{\sqrt{5 x-3-2 x^2}}$
$\Rightarrow 5 x-3-2 x^2 \geq 0$
$\Rightarrow(x-1)\left(x-\frac{3}{2}\right) \leq 0$
$\therefore \quad D _{ f }=\left[1, \frac{3}{2}\right]$

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MCQ 572 Marks

The domain of the function $f (x)=\log _{3+x}\left(x^2-1\right)$ is

A
$(-3,-1) \cup(1, \infty)$
B
$[-3,-1) \cup[1, \infty)$
✓
$(-3,-2) \cup(-2,-1) \cup(1, \infty)$
D
$[-3,-2) \cup(-2,-1) \cup[1, \infty)$

Answer

Correct option: C.

$(-3,-2) \cup(-2,-1) \cup(1, \infty)$

(C)
$f (x)$ is to be defined when $x^2-1>0$
$\Rightarrow x^2>1, \Rightarrow x<-1$ or $x>1$ and $3+x>0$
$\therefore \quad x>-3$ and $x \neq-2$
$\therefore \quad D_f=(-3,-2) \cup(-2,-1) \cup(1, \infty)$

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MCQ 582 Marks

The domain of the function $f(x)=\sqrt{\log \frac{1}{|\sin x|}}$ is

A
$R-\{2 n \pi, n \in I\}$
✓
$R -\{ n \pi, n \in I )$
C
$R -(-\pi, \pi)$
D
$(-\infty, \infty)$

Answer

Correct option: B.

$R -\{ n \pi, n \in I )$

(B)
$f (x)=\sqrt{\log \frac{1}{|\sin x|}}$
$\Rightarrow \sin x \neq 0 \Rightarrow x \neq n \pi+(-1)^{ n } 0 \Rightarrow x \neq n \pi$
Domain of $f (x)= R -\{ n \pi, n \in I \}$.

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MCQ 592 Marks

The domain of definition of $f(x)=\sqrt{\frac{1-|x|}{2-|x|}}$ is

A
$(-\infty,-1) \cup(2, \infty)$
✓
$[-1,1] \cup(2, \infty) \cup(-\infty,-2)$
C
$(-\infty, 1) \cup(2, \infty)$
D
$[-1,1] \cup(2, \infty)$

Answer

Correct option: B.

$[-1,1] \cup(2, \infty) \cup(-\infty,-2)$

(B)
$\frac{1-|x|}{2-|x|} \geq 0$
$\Rightarrow \frac{|x|-1}{|x|-2} \geq 0$
$\Rightarrow|x| \leq 1$ as $|x|>2$
$\Rightarrow x \in(-\infty,-2) \cup(2, \infty) \cup[-1,1]$

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MCQ 602 Marks

Domain of $f(x)=\log |\log x|$ is

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

Answer

Correct option: C.

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

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MCQ 612 Marks

The domain of the function $y= f (x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}$ is

✓
$[-2,1)$, excluding 0
B
$[-3,-2]$, excluding -2.5
C
$[0,1]$, excluding 0
D
none of these

Answer

Correct option: A.

$[-2,1)$, excluding 0

(A)
$D_f=D_g \cap D_h$
where $g(x)=\frac{1}{\log _{10}(1-x)}$ and $h(x)=\sqrt{2+x}$
Now, $D _{ g }=\left\{x \in R : 1-x>0, \log _{10}(1-x) \neq 0\right\}$
$-\{x \in R : x<1,1-x \neq 1\}$
$=\{x \in R : x<1, x \neq 0\}$
and $D _{ h }=\{x \in R : x+2 \geq 0\}$
$=\{x \in R : x \geq-2\}$
$\therefore \quad D_f=[(-\infty, 1)-\{0\}] \cap[-2, \infty)$
$=[-2,1)-\{0\}$

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MCQ 622 Marks

The domain of $f(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is

A
$R-\{-2\}$
B
$(-2,+\infty)$
C
$R -\{-1,-2,-3\}$
✓
$(-3, \infty)-\{-1,-2\}$

Answer

Correct option: D.

$(-3, \infty)-\{-1,-2\}$

(D)
Here, $x+3>0$ and $x^2+3 x+2 \neq 0$
$\therefore \quad x>-3$ and $(x+1)(x+2) \neq 0$, i.e., $x \neq-1,-2$.
$\therefore \quad$ Domain $=(-3, \infty)-\{-1,-2\}$.

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MCQ 632 Marks

The domain of the fuction $\sqrt{\log \left(x^2-6 x+6\right)}$ is

A
$(-\infty, \infty)$
B
$(-\infty, 3-\sqrt{3}) \cup(3+\sqrt{3}, \infty)$
✓
$(-\infty, 1] \cup[5, \infty)$
D
$[0, \infty)$

Answer

Correct option: C.

$(-\infty, 1] \cup[5, \infty)$

(C)
The function $f (x)=\sqrt{\log \left(x^2-6 x+6\right)}$ is defined, when $\log \left(x^2-6 x+6\right) \geq 0$
$\Rightarrow x^2-6 x+6 \geq 1 \Rightarrow(x-5)(x-1) \geq 0$
This inequality holds, if $x \leq 1$ or $x \geq 5$.
Hence, the domain of the function will be $(-\infty, 1] \cup[5, \infty)$.

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MCQ 642 Marks

The domain of the function
$f(x)=\sin ^{-1}\left(\frac{8.3^{x-2}}{1-3^{2 (x-1)}}\right)$ is

A
$(-\infty, 0]$
B
$[2, \infty)$
✓
$(-\infty, 0) \cup[2, \infty)$
D
$(-\infty,-1) \cup[1, \infty)$

Answer

Correct option: C.

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

(C)
$f (x)$ is defined for
$-1 \leq \frac{8.3^{x-2}}{1-3^{2(x-1)}} \leq 1$
$\Rightarrow-1 \leq \frac{\left(3^2-1\right)\left(3^{x-2}\right)}{1-3^{2 x-2}} \leq 1$
$\Rightarrow-1 \leq \frac{3^x-3^{x-2}}{1-3^{2 x-2}} \leq 1$
$\Rightarrow \frac{3^x-3^{x-2}}{1-3^{2 x-2}}+1 \geq 0$ and $\frac{3^x-3^{x-2}}{1-3^{2 x-2}}-1 \leq 0$
$\Rightarrow \frac{1+3^x-3^{x-2}-3^{2 x-2}}{1-3^{2 x-2}} \geq 0$ and
$\frac{3^x-3^{x-2}-1+3^{2 x-2}}{1-3^{2 x-2}} \leq 0$
$\Rightarrow \frac{\left(3^x+1\right)\left(3^{x-2}-1\right)}{\left(3^x \cdot 3^{x-2}-1\right)} \geq 0$ and $\frac{\left(3^x-1\right)\left(3^{x-2}+1\right)}{\left(3^{2 x-2}-1\right)} \geq 0$
$\Rightarrow \frac{\left(3^{x-2}-1\right)}{\left(3^x \cdot 3^{x-2}-1\right)} \geq 0$ and $\frac{\left(3^x-1\right)}{\left(3^{2 x-2}-1\right)} \geq 0$
$\Rightarrow \frac{\left(3^x-3^2\right)}{\left(3^{2 x}-3^2\right)} \geq 0$ and $\frac{\left(3^x-1\right)}{\left(3^{2 x}-3^2\right)} \geq 0$
$\Rightarrow \frac{\left(3^x-3^2\right)}{\left(3^x-3\right)} \geq 0$ and $\frac{\left(3^x-1\right)}{\left(3^x-3\right)} \geq 0$
$\Rightarrow x \in(-\infty, 1] \cup[2, \infty)$ and $x \in(-\infty, 0] \cup(1, \infty)$
$\Rightarrow x \in(-\infty, 0] \cup[2, \infty)$

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MCQ 652 Marks

The domain of the function
$f(x)=\sqrt{x^2-5 x+6}+\sqrt{2 x+8-x^2}$, is

A
[2,3]
B
[-2,4]
✓
$[-2,2] \cup[3,4]$
D
$[-2,1] \cup[2,4]$

Answer

Correct option: C.

$[-2,2] \cup[3,4]$

(C)
$f (x)$ is defined, if
$x^2-5 x+6 \geq 0$ and $2 x+8-x^2 \geq 0$
$\Rightarrow(x-2)(x-3) \geq 0$ and $(x-4)(x+2) \leq 0$
$\therefore \quad x \in(-\infty, 2] \cup[3, \infty)$ and $x \in[-2,4]$
$\therefore \quad x \in[-2,2] \cup[3,4]$

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MCQ 662 Marks

The domain of $\sin ^{-1}\left[\log \left(\frac{x}{3}\right)\right]$ is

✓
[1,9]
B
(-1,9)
C
[-9,1]
D
[-9,-1]

Answer

Correct option: A.

[1,9]

(A)
$y=\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]$
$\therefore \quad-1 \leq \log _3\left(\frac{x}{3}\right) \leq 1$
$\therefore \quad \frac{1}{3} \leq \frac{x}{3} \leq 3$
$\therefore \quad 1 \leq x \leq 9$
$\therefore \quad x \in[1,9]$

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MCQ 672 Marks

Domain of the function $\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$ is

A
(-1,1)
B
(- 1, 1) - {0}
C
[-1,1]
✓
[- 1, 1] - {0}

Answer

Correct option: D.

[- 1, 1] - {0}

(D)
$1+x \geq 0$
$\Rightarrow x \geq-1 ; 1-x \geq 0$
$\Rightarrow x \leq 1, x \neq 0$
Hence, domain is $[-1,1]-\{0\}$.

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MCQ 682 Marks

Domain of the function $f(x)=\sqrt{\frac{x}{1+x}}$ is

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

Answer

Correct option: A.

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

(A)
For domain, take $\frac{x}{1+x} \geq 0$
$\therefore D _{ f }=(-\infty,-1) \cup[0, \infty)$

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MCQ 692 Marks

If If f: R → S defined by $f(x)=\sin x-\sqrt{3} \cos \ x+1$ is onto, them the interval of S is

✓
[-1, 3]
B
[1, 1]
C
[0, 1]
D
[0, -1]

Answer

Correct option: A.

[-1, 3]

(A)
$-\sqrt{1+(-\sqrt{3})^2} \leq(\sin x-\sqrt{3} \cos x) \leq \sqrt{1+(-\sqrt{3})^2}$
$\therefore -2 \leq(\sin x-\sqrt{3} \cos x) \leq 2$
$\therefore-2+1 \leq(\sin x-\sqrt{3} \cos x+1) \leq 2+1$
$\therefore-1 \leq(\sin x-\sqrt{3} \cos x+1) \leq 3$
i.e., range $=[-1,3]$
$\therefore $ For $f$ to be onto $S=[-1,3]$.

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MCQ 702 Marks

If the number of elements in the sets G and A are 3 and 4 respectively, then match the items of List I with those of List II.

	List I		List II
(a)	The number of non bijective functions from GG to G	I.	24
(b)	The number of bijective functions from A to A	II.	0
(c)	The number of functions from G to GA	III.	1728
(d)	The number of surjective functions from A to AA	IV.	12
		V.	19683

✓
a b c d
V I III II
B
a b c d
V III IV II
C
a b c d
III IV V II
D
a b c d
I II III IV

Answer

Correct option: A.

a	b	c	d
V	I	III	II

(A)
(a) $n( G \times G )=9$
The number of functions from a finite set A into a finite set $B =[ n ( B )]^{ n ( A )}$
we get, The number of bijections from a finite set A onto a finite set B is
$n ( A )!; \quad$ if $n ( A )= n ( B )$
0 ; $\quad$otherwise
The number of non-biective functions from $G \times G$ to G is $3^9-0=19,683$
(b) The number of bijections from a finite set A onto a finite set B is
$n ( A )!; \quad$ if $n ( A )= n ( B )$
0 ; $\quad$otherwise
we get, The Number of bijective functions from A to $A =4!=24$
(c) $n ( G \times A )=12$
The number of functions from a finite set A into a finite set $B =[ n ( B )]^{ n ( A )}$
we get, The number of functions G to $G \times A$
$=12^3=1728$
(d) $n ( A \times A )=16$
∴ The number of surjective functions from
A to $A \times A$ is 0 .

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MCQ 712 Marks

The function f : R $\rightarrow$ R defined by $f (x)= e ^x$ is

A
Onto
B
Many-one
✓
One-one and into
D
Many one and onto

Answer

Correct option: C.

One-one and into

(C)
Let $f \left(x_1\right)= f \left(x_2\right)$ for $x_1, x_2 \in R$
$\Rightarrow e ^{x_1}= e ^{x_2}$
$\Rightarrow x_1=x_2$
⇒ f is one one
Let $y= e ^x$
$\Rightarrow \log y=x$
Note that the negative numbers and zero has no pre-image in the domain of the function.
⇒ function f is into.

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MCQ 722 Marks

A function f from the set of natural numbers to integers defined by $f(n)=\left\{\begin{array}{l}\frac{n-1}{2}, \text { when } n \text { is odd } \\ -\frac{n}{2}, \text { when } n \text { is even }\end{array}\right.$, is

A
One-one but not onto
B
Onto but not one-one
C
One-one and onto both
D
Neither one-one nor onto

Answer

(C)
$f : N \rightarrow I$
$\begin{array}{l}f(1)-0, f(2)=-1, f(3)=1, f(4)=-2, f(5)=2 \\ \text {and } f(6)=-3 \text { so on. }\end{array}$

In this type of function every element of set A has unique image in set B and there is no element left in set B. Hence f is one-one and onto function.

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MCQ 732 Marks

Let f : N ⟶ N defined by
$f(n)=\left\{\begin{array}{l}\frac{n+1}{2} \text { if } n \text { is odd } \\ \frac{n}{2} \text { if } n \text { is even }\end{array}\right.$ then $f$ is even

✓
onto but not one-one
B
one-one and onto
C
neither one-one nor onto
D
Neither one-one nor onto

Answer

Correct option: A.

onto but not one-one

(A)
$f : N \rightarrow N$
$f(n)=\left\{\begin{array}{ll}\frac{n+1}{2} & \text { if } n \text { isodd } \\ \frac{n}{2} & \text { if niseven }\end{array}\right.$
Now for $n=1, f(1)=\frac{1+1}{2}=1$
and if $n=2, f(2)=\frac{2}{2}=1$
$\therefore f(1)=f(2)$, But $1+2$.
$\therefore f (x)$ is not one-one.
$f (x)=\frac{ n +1}{2}$ if n is odd
if $y=\frac{ n +1}{2}$ then $n =2 y-1, \forall y$
Also, $f (x)=\frac{ n }{2}$ if n is even i.e., $y=\frac{ n }{2}$
or $n =2 y \forall y$
$\therefore f (x)$ is onto.

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MCQ 742 Marks

$f: R \rightarrow R, f(x)=x^2+3 x+4$ is __________

A
one-one and onto
B
one-one and not onto
✓
many-one and not onto
D
many-one and not onto

Answer

Correct option: C.

many-one and not onto

(C)
$f (x)=x^2+3 x+4$
$=x^2+3 x+\frac{9}{4}-\frac{9}{4}+4$
$=\left(x^2+3 x+\frac{9}{4}\right)+\frac{7}{4}$
$=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}$
Hence, $f$ is many-one and not onto

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MCQ 752 Marks

Let f : R $\rightarrow$ R be defined by $f (x)=x^4$, then

A
f is one-one and onto
B
f may be one-one and onto
C
f is one-one but not onto
✓
f is neither one-one nor onto

Answer

Correct option: D.

f is neither one-one nor onto

(D)
Let $x_1, x_2 \in R$ such that $f \left(x_1\right)= f \left(x_2\right)$
$\Rightarrow x_1= \pm x_2$
$\therefore f \left(x_1\right)= f \left(x_2\right)$ does not impty that $x_1=x_2$
$\therefore f$ is not one-one.
Consider an element 2 in the co-domain R .
There doesnot exist any $x$ is domain R such that $f (x)=2$.
$\therefore f$ is not onto.

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MCQ 762 Marks

Which one of the following is a bijective function on the set of real numbers?

✓
$2 x-5$
B
$|x|$
C
$x^2$
D
$x^2+1$

Answer

Correct option: A.

$2 x-5$

(A)
$|x|$ is not one-one; $x^2$ is not one-one;
$x^2+1$ is not one-one.
But $2 x-5$ is one-one because
$f (x)= f (y) \Rightarrow 2 x-5=2 y-5 \Rightarrow x=y$
Now, $f (x)=2 x-5$ is onto.
$\therefore f (x)=2 x-5$ is bijective.

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MCQ 772 Marks

Set A has 3 elements and set B has 4 elements. The number of injection that can be defined from A to B is

A
144
B
12
✓
24
D
64

Answer

Correct option: C.

(C)
The number of one-one functions that can be defined from a set A into a finite set B is
${ }^{ n ( B )} P _{ n ( A )} \quad ;$ if $n ( B ) \geq n ( A )$
0 ; otherwise
the total number of injective functions from a set A containing 3 elements to a set B containing 4 elements $={ }^4 P_3=24$.

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MCQ 782 Marks

Number of bijective function from a set of 10 elements to itself is

A
5!
✓
10!
C
15!
D
8!

Answer

Correct option: B.

10!

(B)
The number of bijections from a finite set A onto a finite set B is
$n ( A )!; \quad$ if $n ( A )= n ( B )$
0 ;$\quad$ otherwise
the number of bijective function from a set of 10 elements to itself $=10$ !

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MCQ 792 Marks

If $A =\{x \mid x \in N, x \leq 5\}$,$B=\left\{x \mid x \in Z, x^2-5 x+6=0\right\},$then the number of onto functions from $A$ to $B$ is

A
23
B
32
C
2
✓
30

Answer

Correct option: D.

(D)
$A =\{1,2,3,4,5\}$
$B=\{2,3\}$
The number of onto functions, that can be defined from a finite set A , containing n elements onto a finite set B , containing 2 elements $=2^n-2$
the number of onto functions from A to B
$=2^5-2=32-2=30$

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MCQ 802 Marks

A is a set having 6 distinct elements. The number of distinct functions from A to A which are not bijection is

A
$6!-6$
B
$6^6-6$
✓
$6^6-6!$
D
$6!$

Answer

Correct option: C.

$6^6-6!$

(C)
The number of functions from a finite set A into a finite set $B =[ n ( B )]^{ n ( A )}$
The number of bijections from a finite set A onto a finite set B is
$n ( A )!; \quad$ if $n ( A )= n ( B )$
0 ; $\quad$otherwise
total number of distinct functions from
$A \rightarrow A = n ^{ n }=6^6$, and
number of bijections $=n!=6!$
∴ Number of functions which are not bijections
$=6^6-6!$

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MCQ 812 Marks

Mapping f : R $\rightarrow$ R which is defined as f(x) = cos x, x $\in$ R will be

✓
Neither one-one nor onto
B
One-one
C
Onto
D
One-one onto

Answer

Correct option: A.

Neither one-one nor onto

(A)
Let $x_1, x_2 \in R$, then $f\left(x_1\right)-\cos x_1$, &
$f \left(x_2\right)=\cos x_2$, Now $f \left(x_1\right)= f \left(x_2\right)$
$\Rightarrow \cos x_1=\cos x_2 \Rightarrow x_1=2 n \pi \pm x_2$
$\Rightarrow x_1 \neq x_2$,
$\therefore \quad$ it is not one-one.
Again the value of f-image of $x$ lies in between - 1 to 1
$\Rightarrow f [ R ]=\{ f (x):-1 \leq f (x) \leq 1\}$
So other numbers of co-domain (besides -1 and 1) is not f-image. $f[R] \in R$, so it is also not onto. So this mapping is neither one-one nor onto.

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MCQ 822 Marks

If R denotes the set of all real numbers, then the function f : R $\rightarrow$ R defined by f(x) = [x] is

A
One-one only
B
Onto only
C
Both one-one and onto
✓
Neither one-one nor onto

Answer

Correct option: D.

Neither one-one nor onto

(D)
Let $f \left(x_1\right)= f \left(x_2\right) \Rightarrow\left[x_1\right]=\left[x_2\right] \nRightarrow x_1=x_2$
{For example, if $x_1=1.4, x_2=1.5$, then
$[1.4]=[1.5]=1\}$
$\therefore f$ is not one-one.
Also, $f$ is not onto as its range $I$ (set of integers) is a proper subset of its co-domain R.

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MCQ 832 Marks

If $f : R -\{3\} \rightarrow R -\{1\}$ be defined by $f(x)=\frac{x-2}{x-3}$, then $f$ is

A
one-one into
✓
one-one onto
C
many-one into
D
many-one onto

Answer

Correct option: B.

one-one onto

(B)
$f(x)-\frac{x-2}{x-3}, x\neq3$
Let $y= f (x) \Rightarrow y=\frac{x-2}{x-3} \Rightarrow x=\frac{2-3 y}{1-y}$
$\Rightarrow y \neq 1 \Rightarrow$ Range of $f (x)$ is $R -\{1\}$
So, f is onto
For one-one, let $f \left(x_1\right)= f \left(x_2\right)$
$\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \Rightarrow x_1=x_2$
Hence, $f$ is one-one.

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MCQ 842 Marks

If $f:[0, \infty) \rightarrow[0, \infty)$ and $f(x)=\frac{x}{1+x}$, then $f$ is

A
One-one and onto
✓
One-one but not onto
C
Onto but not one-one
D
Neither one-one nor onto

Answer

Correct option: B.

One-one but not onto

(B)
For $0 \in[0, \infty)$ in co-domain we cannot find
any $x \in[0, \infty)$ in domain such that $f (x)=0$
⇒ function is one-one but not onto.

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MCQ 852 Marks

The function $f(x)=\sin \left(\log \left(x+\sqrt{\left.x^2+1\right)}\right)\right.$ is

A
Even function
✓
Odd function
C
Neither even nor odd
D
peridic function

Answer

Correct option: B.

Odd function

(B)
$f (x)=\sin \left(\log \left(x+\sqrt{1+x^2}\right)\right)$
$\rightarrow f (-x)=\sin \left[\log \left(-x+\sqrt{1+x^2}\right)\right]$
$\Rightarrow f (-x)=\sin \log \left(\left(\sqrt{1+x^2}-x\right) \frac{\left(\sqrt{1+x^2}+x\right)}{\left(\sqrt{1+x^2}+x\right)}\right)$
$\Rightarrow f (-x)=\sin \log \left[\frac{1}{x+\sqrt{1+x^2}}\right]$
$\Rightarrow f (-x)=\sin \left[-\log \left(x+\sqrt{1+x^2}\right)\right]$
$\Rightarrow f (-x)=-\sin \left[\log \left(x+\sqrt{1+x^2}\right)\right]$
$\Rightarrow f (-x)=- f (x)$
$\therefore f (x)$ is odd function.

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MCQ 862 Marks

The function $f (x)=\sec \left[\log \left(x+\sqrt{1+x^2}\right)\right]$ is

A
Odd
✓
Even
C
Neither odd not even
D
Constant

Answer

Correct option: B.

Even

(B)
$f (-x)=\sec \left[\log \left(-x+\sqrt{1+(-x)^2}\right)\right]$
$=\sec \left[\log \left(-x+\sqrt{1+x^2}\right)\right]$
$=\sec \left[\log \left(\sqrt{1+x^2}-x\right)\right]$
$=\sec \left[\log \left(\frac{1+x^2-x^2}{\sqrt{1+x^2}+x}\right)\right]$
$=\sec \left[\log \left(\frac{1}{\sqrt{1+x^2}+x}\right)\right]$
$=\sec \left[-\log \left(\sqrt{1+x^2}+x\right)\right]$
$=\sec \left[\log \left(\sqrt{1+x^2}+x\right)\right]$
$\therefore f (x)$ is an even function.

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MCQ 872 Marks

If $f (x)=\log \frac{1+x}{1-x}$, then $f (x)$ is

A
Even function
B
$f \left(x_1\right) f \left(x_2\right)= f \left(x_1+x_2\right)$
C
$\frac{ f \left(x_1\right)}{ f \left(x_2\right)}= f \left(x_1-x_2\right)$
✓
Odd function

Answer

Correct option: D.

Odd function

(D)
Here, $f (x)=\log \frac{1+x}{1-x}$
and $f(-x)=\log \left(\frac{1-x}{1+x}\right)=\log \left(\frac{1+x}{1-x}\right)^{-1}$
$=-\log \left(\frac{1+x}{1-x}\right)=-f(x)=f(-x)$
$\Rightarrow f (x)$ is an odd function.

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MCQ 882 Marks

If the real valued function $f(x)=\frac{a^x-1}{x^x\left(a^x+1\right)}$ is even, then n equals

A
2
B
$\frac{-2}{3}$
C
$\frac{1}{4}$
✓
$-\frac{1}{3}$

Answer

Correct option: D.

$-\frac{1}{3}$

(D)
Since $f (x)$ is even, $f (-x)= f (x)$
$\therefore \quad \frac{ a ^{-x}-1}{(-x)^{ n }\left( a ^{-x}+1\right)}=\frac{ a ^x-1}{x^{ n }\left( a ^x+1\right)}$
$\Rightarrow \frac{1- a ^x}{(-1)^{ n } x^{ n }\left(1+ a ^x\right)}=\frac{ a ^x-1}{x^{ n }\left( a ^x+1\right)}$
$\Rightarrow \frac{-1}{(-1)^{ n }}=1 \Rightarrow-1=(-1)^{ n }$
$\therefore \quad n =-\frac{1}{3}$ can satisfy the equation.

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MCQ 892 Marks

Which of the following functions is an odd function?

✓
$f (x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$
B
$f (x)=x\left(\frac{ a ^x+1}{ a ^x-1}\right)$
C
$f (x)=\log _{10}\left(\frac{1-x^2}{1+x^2}\right)$
D
$f (x)= k ($ constant $)$

Answer

Correct option: A.

$f (x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$

(A)
Let $f (x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$, then
$f (-x)=\sqrt{1-x+x^2}-\sqrt{1+x+x^2}$
Here, $f (-x)=- f (x)$
$\therefore f (x)$ is an odd function.

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MCQ 902 Marks

Which of the following is an even function?

A
$\sqrt{x}$
✓
$x^2+\sin ^2 x$
C
$\sin ^3 x$
D
all of above

Answer

Correct option: B.

$x^2+\sin ^2 x$

(B)
Let $f(x)=x^2+\sin ^2 x$
Here, $f (-x)= f (x)$
$\therefore f (x)$ is an even function.

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MCQ 912 Marks

Consider the function $f (x)=\cos x^2$. Then

A
f is of period $2 \pi$
B
f is of period $\sqrt{2 \pi}$
✓
f is not periodic
D
f is of periodic $\pi$

Answer

Correct option: C.

f is not periodic

(C)
Let $f (x)$ is periodic with period T.
Then, $\cos (x+ T )^2=\cos x^2$ for all $x \in R$
$\Rightarrow \cos (x+ T )^2-\cos x^2=0$
$\Rightarrow-2 \sin \left(\frac{(x+ T )^2+x^2}{2}\right) \sin \left(\frac{(x+ T )^2-x^2}{2}\right)=0$
$\forall x \in R$
$\Rightarrow(x+ T )^2-x^2= n \pi$ or $(x+ T )^2+x^2= n \pi$
$\forall x \in R$
Here $T$ is denendent on the value of $r$
$\Rightarrow f (x)$ is not periodic.

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MCQ 922 Marks

Let f be a real valued function, satisfying $f (x+y)= f (x) f (y)$ for all $x, y \in R$ Such that, $f (1)= a$. Then, $f (x)=$

✓
$a^x$
B
$a x$
C
$x^{ a }$
D
$\log x$

Answer

Correct option: A.

$a^x$

(A)
The general expression for the function
satisfying $f (x+y)= f (x) f (y)$ for all $x, y \in R$ is
$f (x)=[ f (1)]^x= a ^x$ for all $x, y \in R$.
$\ldots[\because f(1)=a]$

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MCQ 932 Marks

If $[x]$ denotes the greatest integer $\leq x$, then $\left[\frac{2}{3}\right]+\left[\frac{2}{3}+\frac{1}{99}\right]+\left[\frac{2}{3}+\frac{2}{99}\right]+\ldots .+\left[\frac{2}{3}+\frac{98}{99}\right]=$

A
99
B
98
✓
66
D
65

Answer

Correct option: C.

(C)
Given expression
$=\sum_{i=0}^{98}\left[\frac{2}{3}+\frac{i}{99}\right]$
$=\sum_{i=0}^{32}\left[\frac{2}{3}+\frac{i}{99}\right]+\sum_{i=33}^{98}\left[\frac{2}{3}+\frac{i}{99}\right]$
$=0+\sum_{i=33}^{98}\left[\frac{2}{3}+\frac{i}{99}\right]$
$\ldots\left[\because \frac{2}{3} \leq \frac{2}{3}+\frac{ i }{99}<1\right.$ for $\left.i =0,1,2, \ldots, 32\right]$
$=66$
$\ldots\left[\begin{array}{c}\because \text { each term in the summation is one or more } \\ \text { but less than } 2 \text { when } i=33,34,35, \ldots, 98\end{array}\right]$

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MCQ 942 Marks

If $f (x)+2 f \left(\frac{1}{x}\right)=3 x, x \neq 0$, and $S =\{x \in R : f (x)= f (-x)\} ;$ then S

A
Contains exactly one element
✓
Contains exactly two elements
C
Contains more than two elements
D
Is an empty set

Answer

Correct option: B.

Contains exactly two elements

(B)
$f(x)+2 f\left(\frac{1}{x}\right)-3 x$....(i)
$\therefore f\left(\frac{1}{x}\right)+2 f(x)=\frac{3}{x}$....(ii)
From (i) and (ii), we get
$3 f (x)=\frac{6}{x}-3 x$
$\Rightarrow f (x)=\frac{2}{x}-x \Rightarrow f (-x)=-\frac{2}{x}+x$
Since, $f (x)= f (-x)$
$\therefore \quad \frac{2}{x}-x=-\frac{2}{x}+x$
$\Rightarrow \frac{4}{x}=2 x \Rightarrow x^2=2 \Rightarrow x= \pm \sqrt{2}$
$\therefore$ option (B) is the correct answer.

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MCQ 952 Marks

The graph of the function y = f(x) is symmetrical about the line x = 2, then

A
$f(x)=-f(-x)$
✓
$f (2+x)= f (2-x)$
C
$f(x)=f(-x)$
D
$f (x+2)= f (x-2)$

Answer

Correct option: B.

$f (2+x)= f (2-x)$

(B)
$f (x)= f (-x) \Rightarrow f (0+x)= f (0-x)$ is
symmetrical about $x=0$.
$\therefore f (2+x)= f (2-x)$ is symmetrical about $x=2$.

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MCQ 962 Marks

If $f(x)=\cos \left[\pi^2\right] x+\cos \left[-\pi^2\right] x$, then

A
$f\left(\frac{\pi}{4}\right)=2$
B
$f(-\pi)=2$
C
$f(\pi)=1$
✓
$f\left(\frac{\pi}{2}\right)=-1$

Answer

Correct option: D.

$f\left(\frac{\pi}{2}\right)=-1$

(D)
$f (x)=\cos \left[\pi^2\right] x+\cos \left[-\pi^2\right] x$
$f(x)=\cos (9 x)+\cos (-10 x)$
$\cdots\left[\begin{array}{c}\because \pi=3.14 \Rightarrow[9.85]=9 \\ \text { and }[-9.85]=-10\end{array}\right]$
$=\cos (9 x)+\cos (10 x)$
$=2 \cos \left(\frac{19 x}{2}\right) \cos \left(\frac{x}{2}\right)$
$\therefore \quad f\left(\frac{\pi}{2}\right)=2 \cos \left(\frac{19 \pi}{4}\right) \cos \left(\frac{\pi}{4}\right) ;$
$\therefore \quad f\left(\frac{\pi}{2}\right)=2 \times \frac{-1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=-1$

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MCQ 972 Marks

If $e ^{ f (x)}=\frac{10+x}{10-x}, x \in(-10,10)$ and $f ( x )= kf \left(\frac{200 x}{100+x^2}\right)$, then $k =$

✓
0.5
B
0.6
C
0.7
D
0.8

Answer

Correct option: A.

0.5

(A)
$e ^{ f (x)}=\frac{10+x}{10-x}, x \in(-10,10)$
$\Rightarrow f (x)=\log \left(\frac{10+x}{10-x}\right)$
$\Rightarrow f \left(\frac{200 x}{100+x^2}\right)=\log \left[\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right]$
$=\log \left[\frac{10(10+x)}{10(10-x)}\right]^2$
$=2 \log \left(\frac{10+x}{10 x}\right)$
$=2 f (x)$
$\therefore f (x)=\frac{1}{2} f \left(\frac{200 x}{100+x^2}\right) \Rightarrow k =\frac{1}{2}=0.5$

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MCQ 982 Marks

If $f (x)=\log \left[\frac{1+x}{1-x}\right]$, then $f \left[\frac{2 x}{1+x^2}\right]$ is equal to

A
$[ f (x)]^2$
B
$[f(x)]^3$
✓
$2 f (x)$
D
$3 f (x)$

Answer

Correct option: C.

$2 f (x)$

(C)
$f (x)=\log \left[\frac{1+x}{1-x}\right]$
$\therefore f\left(\frac{2 x}{1+x^2}\right)=\log \left[\frac{1+\frac{2 x}{1+x^2}}{1-\frac{2 x}{1+x^2}}\right]$
$=\log \left[\frac{x^2+1+2 x}{x^2+1-2 x}\right]$
$=\log \left[\frac{1+x}{1-x}\right]^2$
$=2 \log \left[\frac{1+x}{1-x}\right]$
$=2 f (x)$

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MCQ 992 Marks

Given the function $f (x)=\frac{ a + a }{2}, a > 2$, then

✓
$2 f (x) \cdot f (y)$
B
$f (x) \cdot f (y)$
C
$\frac{ f (x)}{ f (y)}$
D
$\frac{1}{2} f (x) f (y)$

Answer

Correct option: A.

$2 f (x) \cdot f (y)$

(A)
$f (x+y)+ f (x-y)$
$=\frac{1}{2}\left[ a ^{x+y}+ a ^{-x-y}+ a ^{x-y}+ a ^{-x+y}\right]$
$=\frac{1}{2}\left[ a ^x\left( a ^y+ a ^{-y}\right)+ a ^{-x}\left( a ^y+ a ^{-y}\right)\right]$
$=\frac{1}{2}\left( a ^x+ a ^{-x}\right)\left( a ^y+ a ^{-y}\right)$
$=2 f (x) f (y)$

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MCQ 1002 Marks

If $f(x)=\cos (\log x)$, then$f(x) f(y)-\frac{1}{2}[f(x / y)+f(x y)]=$

A
-1
B
$\frac{1}{2}$
C
-2
✓
$0$

Answer

Correct option: D.

$0$

(D)
Given, $f (x)=\cos (\log x) \Rightarrow f (y)=\cos (\log y)$
Then, $f (x) . f (y)-\frac{1}{2}\left[ f \left(\frac{x}{y}\right)+ f (x y)\right]$
$=\cos (\log x) \cos (\log y)$
$-\frac{1}{2}\left[\cos \left(\log \frac{x}{y}\right)+\cos (\log x y)\right]$
$=\cos (\log x) \cos (\log y)$
$-\frac{1}{2}[2 \cos (\log x) \cos (\log y)]=0$

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Maths MCQ