Functions — Page 10 | JEE Mathematics

Q91

Let

f: \mathbf{R} \rightarrow \mathbf{R}

and

g: \mathbf{R} \rightarrow \mathbf{R}

be defined as

f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.

and

g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.

. Then, gof :

\mathbf{R} \rightarrow \mathbf{R}

is :

A one-one but not onto

B neither one-one nor onto

C onto but not one-one

D both one-one and onto

Correct Answer

Option B

Solution

Given, $f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.$ then $g \circ f(x)$ $=g(f(x))$ $\begin{aligned} & \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}f(x), f(x) \geq 0 \\ e^{f(x)}, f(x)<0\end{array}\right. \\\\ & \mathrm{g}(\mathrm{f}(\mathrm{x}))=\left\{\begin{array}{l}e^{-x},(-\infty, 0] \\ e^{\ln x},(0,1) \\ \ln x,[1, \infty)\end{array}\right.\end{aligned}$ $g(f(x))=\left\{\begin{array}{c}e^{-x},(-\infty, 0] \\ x,(0,1) \\ \ln x,[1, \infty)\end{array}\right.$ From the graph of $g(f(x))$ , we can say $\mathrm{g}(\mathrm{f}(\mathrm{x})) \Rightarrow$ Many one into

Q92

The function

f: \mathbf{N}-\{1\} \rightarrow \mathbf{N}

; defined by

f(\mathrm{n})=

the highest prime factor of

\mathrm{n}

, is :

A one-one only

B neither one-one nor onto

C onto only

D both one-one and onto

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{f}: \mathrm{N}-\{1\} \rightarrow \mathrm{N} \\ & \mathrm{f}(\mathrm{n})=\text{ The highest prime factor of } \mathrm{n} . \\ & \mathrm{f}(2)=2 \\ & \mathrm{f}(4)=2 \\ & \Rightarrow \text{ many one } \\ & 4 \text{ is not image of any element } \\ & \Rightarrow \text{ into } \end{aligned}

Hence many one and into Neither one-one nor onto.

Q93

Let

f: \mathbf{R}-\left\{\frac{-1}{2}\right\} \rightarrow \mathbf{R}

and

g: \mathbf{R}-\left\{\frac{-5}{2}\right\} \rightarrow \mathbf{R}

be defined as

f(x)=\dfrac{2 x+3}{2 x+1}

and

g(x)=\dfrac{|x|+1}{2 x+5}

. Then, the domain of the function fog is :

A

\mathbf{R}-\left\{-\dfrac{7}{4}\right\}

B

\mathbf{R}

C

\mathbf{R}-\left\{-\dfrac{5}{2},-\dfrac{7}{4}\right\}

D

\mathbf{R}-\left\{-\dfrac{5}{2}\right\}

Correct Answer

Option D

Solution

\begin{aligned} & f(x)=\frac{2 x+3}{2 x+1} ; x \neq-\frac{1}{2} \\ & g(x)=\frac{|x|+1}{2 x+5}, x \neq-\frac{5}{2} \end{aligned}

Domain of

f(g(x))

f(g(x))=\frac{2 g(x)+3}{2 g(x)+1}

x \neq-\frac{5}{2}

and

\frac{|x|+1}{2 x+5} \neq-\frac{1}{2}

x \in R-\left\{-\frac{5}{2}\right\}

and

x \in R

$\therefore$ Domain will be

\mathrm{R}-\left\{-\frac{5}{2}\right\}

Q94

If the function

f(x)=\left(\dfrac{1}{x}\right)^{2 x} ; x>0

attains the maximum value at

x=\dfrac{1}{\mathrm{e}}

then :

A

\mathrm{e}^\pi<\pi^{\mathrm{e}}

B

\mathrm{e}^{2 \pi}<(2 \pi)^{\mathrm{e}}

C

(2 e)^\pi>\pi^{(2 e)}

D

\mathrm{e}^\pi>\pi^{\mathrm{e}}

Correct Answer

Option D

Solution

f\left(\frac{1}{\pi}\right)

\begin{aligned} & \Rightarrow\left(\frac{1}{1}\right)^{\frac{2}{\pi}}

Q95

Let

f(x)=\dfrac{1}{7-\sin 5 x}

be a function defined on

\mathbf{R}

. Then the range of the function

f(x)

is equal to :

A

\left[\dfrac{1}{8}, \dfrac{1}{5}\right]

B

\left[\dfrac{1}{7}, \dfrac{1}{6}\right]

C

\left[\dfrac{1}{7}, \dfrac{1}{5}\right]

D

\left[\dfrac{1}{8}, \dfrac{1}{6}\right]

Correct Answer

Option D

Solution

\begin{aligned} & f(x)=\frac{1}{7-\sin 5 x} \\\\ & -1 \leq \sin 5 x \leq 1 \\\\ & -1 \leq-\sin 5 x \leq 1 \\\\ & -1+7 \leq 7-\sin 5 x \leq 1+7 \\\\ & 6 \leq 7-\sin 5 x \leq 8 \\\\ & \frac{1}{8} \leq \frac{1}{7-\sin 5 x} \leq \frac{1}{6} \\\\ & \frac{1}{8} \leq f(x) \leq \frac{1}{6} \\\\ & \text{ Range }=\left[\frac{1}{8}, \frac{1}{6}\right] \end{aligned}

Q96

If

f(x)=\dfrac{4 x+3}{6 x-4}, x \neq \dfrac{2}{3}

and

(f \circ f)(x)=g(x)

, where

g: \mathbb{R}-\left\{\frac{2}{3}\right\} \rightarrow \mathbb{R}-\left\{\frac{2}{3}\right\}

, then

(g ogog)(4)

is equal to

A

-4

B

\dfrac{19}{20}

C

-\dfrac{19}{20}

D 4

Correct Answer

Option D

Solution

To find

(g \circ g \circ g)(4),

we first need to understand the composition of

f

with itself, i.e.,

(f \circ f)(x) = f(f(x)) = g(x).

We can then repeatedly apply

g

to get the given expression. First, let's calculate

(f \circ f)(x) = g(x):

g(x) = (f \circ f)(x) = f(f(x))

= f\left(\frac{4x+3}{6x-4}\right)

To evaluate this expression, we substitute

\frac{4x+3}{6x-4}

for

x

in the function

f(x):

g(x) = f\left(\frac{4x+3}{6x-4}\right) = \frac{4\left(\frac{4x+3}{6x-4}\right) + 3}{6\left(\frac{4x+3}{6x-4}\right) - 4}

Now, we simplify the expression:

g(x) = \frac{4(4x+3) + 3(6x-4)}{6(4x+3) - 4(6x-4)}

= \frac{16x + 12 + 18x - 12}{24x + 18 - 24x + 16}

= \frac{34x}{34}

= x

So,

g(x) = x

for all

x

in the domain of

g

, which is

\mathbb{R}-\left\{\frac{2}{3}\right\}

. It's important to note that the domain restriction is preserved through the composition because

f(x)

has a vertical asymptote at

x = \frac{2}{3}

which doesn't intersect the graph. So,

g(x)

is the identity function on its domain, which means that applying

g

any number of times will result in the same input for

x

in the given domain. Hence, we have:

(g \circ g \circ g \circ g)(4) = g(g(g(g(4)))) = g(g(g(4))) = g(g(4)) = g(4) = 4

This corresponds to option D, which is

4

.

Q97

If the domain of the function

f(x)=\cos ^{-1}\left(\frac{2-|x|}{4}\right)+\left\{\log _e(3-x)\right\}^{-1}

is

[-\alpha, \beta)-\{\gamma\}

, then

\alpha+\beta+\gamma

is equal to :

A 11

B 12

C 9

D 8

Correct Answer

Option A

Solution

\begin{aligned} & -1 \leq\left|\frac{2-|x|}{4}\right| \leq 1 \\ & \Rightarrow\left|\frac{2-|x|}{4}\right| \leq 1 \\ & -4 \leq 2-|x| \leq 4 \\ & -6 \leq-|x| \leq 2 \\ & -2 \leq|x| \leq 6 \\ & |x| \leq 6 \end{aligned}

\Rightarrow x \in[-6,6]

.... (1) Now,

3-x\ne 1

And

x\ne2

.... (2) and

3-x>0

x

\begin{aligned} & \text { From (1), (2) and (3) } \\ & \Rightarrow x \in[-6,3)-\{2\} \\ & \alpha=6 \\ & \beta=3 \\ & \gamma=2 \\ & \alpha+\beta+\gamma=11 \end{aligned}$$

Q98

Let the range of the function

f(x)=\dfrac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R}

be

[a, b]

. If

\alpha

and

\beta

ar respectively the A.M. and the G.M. of

a

and

b

, then

\dfrac{\alpha}{\beta}

is equal to

A

\pi

B

\sqrt{\pi}

C

\sqrt{2}

D 2

Correct Answer

Option C

Solution

\begin{aligned} & F(x)=\frac{1}{2+\sin 3 x+\cos 3 x}, x \in \mathbb{R} \\ & \sin 3 x+\cos 3 x \in[-\sqrt{2}, \sqrt{2}] \\ & 2+\sin 3 x+\cos 3 x \in[2-\sqrt{2}, 2+\sqrt{2}] \\ & \Rightarrow \frac{1}{2+\sin 3 x+\cos 3 x} \in\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right] \\ & \Rightarrow a=\frac{1}{2+\sqrt{2}}, b=\frac{1}{2-\sqrt{2}} \\ & \alpha=\frac{a+b}{2}=\frac{\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}}}{2} \\ & =\frac{4}{2 \times 2}=1 \\ & \beta=\sqrt{a b}=\sqrt{\left(\frac{1}{2+\sqrt{2}}\right) \times\left(\frac{1}{2-\sqrt{2}}\right)} \\ & =\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \\ & \Rightarrow \frac{\alpha}{\beta}=\sqrt{2} \\ \end{aligned}

Q99

If the domain of the function

f(x)=\sin ^{-1}\left(\dfrac{x-1}{2 x+3}\right)

is

\mathbf{R}-(\alpha, \beta)

, then

12 \alpha \beta

is equal to :

A 40

B 36

C 24

D 32

Correct Answer

Option D

Solution

\begin{array}{ll} f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right) & \\ -1 \leq \frac{x-1}{2 x+3} \leq 1 & \frac{x-1}{2 x+3}+1 \geq 0 \\ \frac{x-1}{2 x+3}-1 \leq 0 & \frac{x-1+2 x+3}{2 x+3} \geq 0 \\ \frac{x-1-2 x-3}{2 x+3} \leq 0 & \frac{3 x+2}{2 x+3} \geq 0 \end{array}

\begin{aligned} & \frac{-x-4}{2 x+3} \leq 0 \\ & \frac{x+4}{2 x+3} \geq 0 \end{aligned}

\begin{aligned} & x \in(-\infty,-4] \cup\left[\frac{-2}{3}, \infty\right) \\ & \text{ Domain : } R-\left(-4, \frac{-2}{3}\right) \\ & \alpha=-4 \\ & \beta=\frac{-2}{3} \\ & 12 \times \alpha \beta=12 \times 4 \times \frac{2}{3}=32 \end{aligned}

Q100

Let

f(x)=\left\{\begin{array}{ccc}-\mathrm{a} & \text{ if } & -\mathrm{a} \leq x \leq 0 \\ x+\mathrm{a} & \text{ if } & 0 0

and

\mathrm{g}(x)=(f(|x|)-|f(x)|) / 2

. Then the function

g:[-a, a] \rightarrow[-a, a]

is

A neither one-one nor onto.

B both one-one and onto.

C one-one.

D onto

Correct Answer

Option A

Solution

\begin{aligned} & f(x)=\left\{\begin{array}{l} -a \quad \text{ if }-a \leq x \leq 0 \\ x+a \quad \text{ if } 0< x \leq a \end{array}\right. \\ & f(|x|)=\left\{\begin{array}{cc} -a & -a \leq|x| \leq 0 \\ |x|+a & \text{ if } 0 < |x| \leq a \end{array}\right. \end{aligned}

|x|<0

is not possible, so

\begin{aligned} & f(|x|)= \begin{cases}x+a & -a \leq x \leq 0 \\ -x+a & 0 < x \leq a\end{cases} \\ & |f(x)|= \begin{cases}a & -a \leq x \leq 0 \\ x+a & 0 < x \leq a\end{cases} \\ & h(x)= \begin{cases}-\frac{x}{2} & -a \leq x \leq 0 \\ 0 & 0 < x \leq a\end{cases} \end{aligned}

Neither one-one nor onto.