Functions — Page 12 | JEE Mathematics

Q111

If the domain of the function

f(x)=\dfrac{1}{\sqrt{10+3 x-x^2}}+\dfrac{1}{\sqrt{x+|x|}}

is

(a, b)

, then

(1+a)^2+b^2

is equal to :

A 29

B 30

C 25

D 26

Correct Answer

Option D

Solution

x+|x|= \begin{cases}2 x, & x \geq 0 \\ 0, & x \Rightarrow \frac{1}{\sqrt{x+|x|}}, domain is x>0, as 2 x \neq 0 Similarly,

\begin{aligned} &\frac{1}{\sqrt{3 x+10-x^2}} \text { is defined when } 3 x+10-x^2>0\\ &\begin{aligned} \Rightarrow & x^2-3 x-10

Q112

\text{ If the domain of the function } f(x)=\log _e\left(\frac{2 x-3}{5+4 x}\right)+\sin ^{-1}\left(\frac{4+3 x}{2-x}\right) \text{ is }[\alpha, \beta) \text{, then } \alpha^2+4 \beta \text{ is equal to }

A 4

B 3

C 7

D 5

Correct Answer

Option A

Solution

$\dfrac{2 x-3}{4 x+5}>0$ $\begin{aligned} & \therefore \quad x \in\left(-\infty,-\dfrac{5}{4}\right) \cup\left(\dfrac{3}{2}, \infty\right) ........(i) \\ & -1 \leq \dfrac{3 x+4}{2-x} \leq 1 \\ & \dfrac{3 x+4}{2-x} \leq 1 \\ & \Rightarrow \quad \dfrac{3 x+4}{2-x}-1 \leq 0\end{aligned}$ $\begin{aligned} & \Rightarrow \quad \dfrac{3 x+4-2+x}{x-2} \geq 0 \\ & \Rightarrow \quad \dfrac{4 x+2}{x-2} \geq 0 \\ & \Rightarrow \quad x \in\left(-\infty,-\dfrac{1}{2}\right] \cup(2, \infty)........(ii)\end{aligned}$

\begin{aligned} & \frac{3 x+4}{2-x} \geq-1 \\ \Rightarrow & \frac{3 x+4}{2-x}+1 \geq 0 \\ \Rightarrow & \frac{3 x+4+2-x}{2-x} \geq 0 \\ \Rightarrow & \frac{2 x+6}{x-2} \leq 0 \\ \therefore & x \in[-3,2)........(iii) \end{aligned}

Taking intersection of (i), (ii) and (iii)

\begin{aligned} & x \in\left[-3,-\frac{5}{4}\right) \\ & \alpha=-3, \beta=-\frac{5}{4} \\ & \alpha^2+4 \beta=4 \end{aligned}

Q113

Let the domains of the functions

f(x)=\log _4 \log _3 \log _7\left(8-\log _2\left(x^2+4 x+5\right)\right)

and

\mathrm{g}(x)=\sin ^{-1}\left(\dfrac{7 x+10}{x-2}\right)

be

(\alpha, \beta)

and

[\gamma, \delta]

, respectively. Then

\alpha^2+\beta^2+\gamma^2+\delta^2

is equal to :

A 15

B 13

C 16

D 14

Correct Answer

Option A

Solution

$f(x)=\log _4\left(\log _3\left(\log _7\left(8-\log _2\left(x^2+4 x+5\right)\right)\right)\right.$

\begin{aligned} & \log _3\left(\log _1\left(8-\log _2\left(x^2+4 x+5\right)\right)\right)>0 \\ & \log _7\left(8-\log _2\left(x^2+4 x+5\right)\right)>1 \\ & 8-\log _2\left(x^2+4 x+5\right)>7 \\ & -\log _2\left(x^2+4 x+5\right)>-1 \\ & \log _2\left(x^2+4 x+5\right)<1 \\ & x^2+4 x+5<2 \\ & x^2+4 x+3<0 \\ & \Rightarrow(x+3)(x+1)<0 \quad \ldots(1) \\ & \log _7\left(8-\log _2\left(x^2+4 x+5\right)\right)>0 \\ & 8-\log _2\left(x^2+4 x+5\right)>1 \\ & \log _2\left(x^2+4 x+5\right)<9 \\ & x^2+4 x+5<2^9 \\ & x^2+4 x+5<512 \\ & \Rightarrow x^2+4 x-507<0 \\ & \Rightarrow x=-4 \pm \sqrt{16+2028} \end{aligned}

\begin{aligned} & x=\frac{-4 \pm \sqrt{2044}}{2} \quad\text{..... (2)}\\ & \Rightarrow\left(x-\left(\frac{-4+\sqrt{2044}}{2}\right)\right)\left(x-\left(\frac{-4-\sqrt{2044}}{2}\right)\right)<0 \\ & x^2+4 x+5>0 \\ & D>0 \\ & x \in R \\ & \text{ Also, } 8-\log _2\left(x^2+4 x+5\right)>0 \\ & \log _2\left(x^2+4 x+5\right)<8 \\ & x^2+4 x+5<256 \\ & \Rightarrow x^2+4 x-251<0 \\ & \Rightarrow x=-4 \pm \sqrt{16+1004} \\ & \Rightarrow x=\frac{-4 \pm \sqrt{1020}}{2} \end{aligned}

\begin{aligned} &\Rightarrow\left(x-\left(\frac{-4+\sqrt{1020}}{2}\right)\right)\left(x-\left(\frac{-4-\sqrt{1020}}{2}\right)\right)<0\\ &\therefore \text{ Intersection of (1), (2) and (3) } \end{aligned}

\begin{aligned} & \therefore x \in(-3,-1) \\ & -1 \leq \frac{7 x+10}{x-2} \leq 1 \\ & \Rightarrow x \in[-2,-1] \\ & \therefore \alpha^2+\beta^2+\gamma^2+\delta^2=(-3)^2+(-1)^2+(-2)^{-2}+(-1)^2 \\ & =9+1+4+1 \\ & =15 \end{aligned}

Q114

Let

f, g:(1, \infty) \rightarrow \mathbb{R}

be defined as

f(x)=\dfrac{2 x+3}{5 x+2}

and

g(x)=\dfrac{2-3 x}{1-x}

. If the range of the function fog:

[2,4] \rightarrow \mathbb{R}

is

[\alpha, \beta]

, then

\dfrac{1}{\beta-\alpha}

is equal to

A 56

B 2

C 29

D 68

Correct Answer

Option A

Solution

\begin{aligned} & g(2)=4, g(4)=\frac{10}{3} \\ & f \text{ is decreasing in }\left(\frac{10}{3}, 4\right) \\ & \therefore \quad \alpha=f(4)=\frac{1}{2} \\ & \beta=f\left(\frac{10}{3}\right)=\frac{29}{56} \\ & \frac{1}{\beta-\alpha}=\frac{1}{\frac{29}{56}-\frac{1}{2}}=56 \end{aligned}

Q115

Let

f

be a function such that

f(x)+3 f\left(\dfrac{24}{x}\right)=4 x, x \neq 0

. Then

f(3)+f(8)

is equal to

A 13

B 11

C 10

D 12

Correct Answer

Option B

Solution

\begin{aligned} & f(x)+3 f\left(\frac{24}{x}\right)=4 x, x \neq 0 \quad \ldots(1)\\ & \text{ replace } x \text{ by } \frac{24}{x} \\ & f\left(\frac{24}{x}\right)+3 f\left(\frac{24}{24}\right)=4\left(\frac{24}{x}\right)=\frac{96}{x} \quad \ldots(2) \\ & 3 \times(2)-(1) \\ & \Rightarrow 8 f(x)=\frac{96.3}{x}-4 x \Rightarrow f(x)=\frac{36}{x}-\frac{x}{2} \\ & f(3)+f(8)=\left(12-\frac{3}{2}\right)+\left(\frac{36}{8}-4\right) \end{aligned}

$=8+\dfrac{36}{8}-\dfrac{12}{8}=11$

Q116

If the domain of the function

f(x)=\log _7\left(1-\log _4\left(x^2-9 x+18\right)\right)

is

(\alpha, \beta) \cup(\gamma, o)

, then

\alpha+\beta+\gamma+\hat{o}

is equal to

A 17

B 15

C 16

D 18

Correct Answer

Option D

Solution

\begin{aligned} & 1-\log _4\left(x^2-9 x+18\right)>0 \\ & \log _4\left(x^2-9 x+18\right)<1 \\ & x^2-9 x+18<4 \\ & x^2-9 x+14<0 \\ & x \in(2,7) \\ & x^2-9 x+18>0 \\ & x \in(-\infty, 3) \cup(6, \infty) \end{aligned}

\begin{aligned} & x \in(2,3) \cup(6,7) \\ & \alpha+\beta+\gamma+\delta=18 \end{aligned}

Q117

For x

\in

R, x

\ne

0, Let f0(x) =

{1 \over {1 - x}}

and fn+1 (x) = f0(fn(x)), n = 0, 1, 2, . . . . Then the value of f100(3) + f1

\left( {{2 \over 3}} \right)

+ f2

\left( {{3 \over 2}} \right)

is equal to :

A

{8 \over 3}

B

{5 \over 3}

C

{4 \over 3}

D

{1 \over 3}

Correct Answer

Option B

Solution

As fn+1(x) = f0(fn(x)) $\therefore$ f1(x) = f0+1(x) = f0(f0(x)) =

{1 \over {1 - {1 \over {1 - x}}}}

=

{{x - 1} \over x}

f2(x) = f1+1(x) = f0(f1(x)) =

{1 \over {1 - {{x - 1} \over x}}}

= x f3(x) = f2+1(x) = f0(f2(x)) =

{1 \over {1 - x}}

f4(x) = f3+1(x) = f0(f3(x)) =

{1 \over {1 - {1 \over {1 - x}}}}

=

{{x - 1} \over x}

$\therefore$ f0(x) = f3(x) = f6(x) = . . . . . =

{1 \over {1 - x}}

f1(x) = f4(x) = f7(x) = . . . . . .=

{{x - 1} \over x}

f2(x) = f5(x) = f8(x) = . . . . . . = x So, f100(3) =

{{3 - 1} \over 3}

=

{2 \over 3}

f1

\left( {{2 \over 3}} \right)

=

{{{2 \over 3} - 1} \over {{2 \over 3}}}

= $-$

{1 \over 2}

f2

\left( {{3 \over 2}} \right)

=

{{3 \over 2}}

$\therefore$ f100(3) + f1

\left( {{2 \over 3}} \right)

+ f2

\left( {{3 \over 2}} \right)

=

{2 \over 3}

$-$

{1 \over 2}

+

{3 \over 2}

=

{5 \over 3}

Q118

Let f : A

\to

B be a function defined as f(x) =

{{x - 1} \over {x - 2}},

Where A = R

-

{2} and B = R

-

{1}. Then f is :

A invertible and

{f^{ - 1}}(y) =

{{3y - 1} \over {y - 1}}

B invertible and

{f^{ - 1}}\left( y \right) = {{2y - 1} \over {y - 1}}

C invertible and

{f^{ - 1}}\left( y \right) = {{2y + 1} \over {y - 1}}

D not invertible

Correct Answer

Option B

Solution

Assume, y = f(x) $\Rightarrow$ y =

{{x - 1} \over {x - 2}}

$\Rightarrow$ yx - 2y = x - 1 $\Rightarrow$ (y - 1)x = 2y - 1 $\Rightarrow$ x =

{{2y - 1} \over {y - 1}}

= f -1(y) As on the given domain the function is invertible and its inverse can be computed as shown above.

Q119

If

f(x)+2 f\left(\dfrac{1}{x}\right)=3 x, x \neq 0

, and

\mathrm{S}=\{x \in \mathbf{R}: f(x)=f(-x)\}

; then

\mathrm{S}:

A is an empty set.

B contains exactly one element.

C contains exactly two elements.

D contains more than two elements.

Correct Answer

Option C

Solution

We have, $f(x)+2 f\left(\dfrac{1}{x}\right)=3 x, \quad x \neq 0$ $\ldots$ (i) On replacing $x$ by $\dfrac{1}{x}$ in the above equation, we get

\begin{aligned} & f\left(\frac{1}{x}\right)+2 f(x) =\frac{3}{x} \\\\ \Rightarrow & \,\, 2 f(x)+f\left(\frac{1}{x}\right) =\frac{3}{x} \,\,\,\,\,...(ii) \end{aligned}

On multiplying Eq. (ii) by 2, we get

4 f(x)+2 f\left(\frac{1}{x}\right)=\frac{6}{x}\quad...(iii)

and subtracting Eq. (i) from Eq. (iii), we get

[4 f(x)+2 f\left(\frac{1}{x}\right)] - [f(x)+2 f\left(\frac{1}{x}\right)]=\frac{6}{x} - 3x

$\Rightarrow {3 f(x)=\dfrac{6}{x}-3 x}$ $\Rightarrow f(x)=\dfrac{2}{x}-x$ Now, consider $\quad f(x)=f(-x)$

\begin{aligned} &\Rightarrow \frac{2}{x}-x =-\frac{2}{x}+x \\\\ &\Rightarrow \frac{4}{x} =2 x \\\\ &\Rightarrow 2 x^2 =4 \\\\ &\Rightarrow x^2 =2 \\\\ &\Rightarrow x =\pm \sqrt{2} \end{aligned}

Hence, $S$ contains exactly two elements.

Q120

Let f(x) = 210.x + 1 and g(x)=310.x

-

1. If (fog) (x) = x, then x is equal to :

A

{{{3^{10}} - 1} \over {{3^{10}} - {2^{ - 10}}}}

B

{{{2^{10}} - 1} \over {{2^{10}} - {3^{ - 10}}}}

C

{{1 - {3^{ - 10}}} \over {{2^{10}} - {3^{ - 10}}}}

D

{{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}

Correct Answer

Option D

Solution

(fog) (x) = x $\Rightarrow$

\,\,\,

f (g(x)) = x $\Rightarrow$

\,\,\,

f (310. x $-$ 1) = x [ as g(x) = 310. x $-$ 1] $\Rightarrow$

\,\,\,

210 . (310 . x $-$ 1) + 1 = x $\Rightarrow$

\,\,\,

310 . x $-$ 1 + 2 $-$ 10 = x . 2 $-$ 10 [dividing by 210] $\Rightarrow$

\,\,\,

310 . x $-$ 2 $-$ 10 . x = 1 $-$ 2 $-$ 10 $\Rightarrow$

\,\,\,

x (310 $-$ 2 $-$ 10) = 1 $-$ 2 $-$ 10 $\Rightarrow$

\,\,\,

x =

{{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}