Functions

$ \begin{aligned} & \text { Given, } f(x)=\sqrt{\frac{1-x^2}{1+x^2}} \\ & \qquad \begin{array}{ll} & \frac{1-x^2}{1+x^2} \geq 0 \\ \Rightarrow & 1-x^2 \geq 0 \quad\left[\because 1+x^2>0\right] \\ \Rightarrow & x^2 \leq 1 \\ \Rightarrow & x \in[-1,1] \\ \therefore & D=[-1,1] \\ \text { Let } y=\frac{1-x^2}{1+x^2} \\ & y+x^2 y=1-x^2 \end{array} \end{aligned} $

          $ \begin{aligned} &\begin{array}{ll} \Rightarrow & x^2(y+1)=1-y \\ \Rightarrow & x^2=\frac{1-y}{1+y} \\ \Rightarrow & x= \pm \sqrt{\frac{1-y}{1+y}} \\ \Rightarrow & \frac{1-y}{1+y} \geq 0 \end{array}\\ &\text { As we know that } y \geq 0,1+y>0\\ &\begin{aligned} \Rightarrow \quad 1-y & \geq 0 \\ y & \leq 1 \\ y & \in[0,1] \\ G & =[0,1] \\ D \cap G & =[0,1] \end{aligned} \end{aligned} $

Given,

$ \begin{array}{rc} & f(x)=\log _2\left(2^x-2\right)+\sqrt{1-x} \\ & 1-x \geq 0 \\ \Rightarrow &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x \leq 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Also, $\quad 2^x-2>0$

$ \begin{aligned}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ 2^x & >2 \\ x & >1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i) and (ii), it can be concluded that $x$ contains no real values.

$ \begin{aligned} & \text { } \begin{aligned} Given,\,\,\,\,\, f(x) & =1-x, g(x)=\frac{1}{1-x} \\ h(x) & =1 / x \\ f(F(h(x))) & =g(x) \\ \Rightarrow \quad f\left(F\left(\frac{1}{x}\right)\right) & =\frac{1}{1-x} \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow & & 1-F\left(\frac{1}{x}\right) & =\frac{1}{1-x} \\ \Rightarrow & & F\left(\frac{1}{x}\right) & =1-\frac{1}{1-x} \\ \Rightarrow & & F\left(\frac{1}{x}\right) & =\frac{-x}{1-x} \end{aligned} $

Substitute $x=\frac{1}{2022}$ to determine $F(2022)$

$ \begin{aligned} F(2022) & =\frac{-\frac{1}{2022}}{1-\frac{1}{2022}} \\ & =-\frac{1}{2021}=\frac{1}{1-2022} \\ & =g(2022) \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} & \cos [\sinh (\log x)+\cosh (\log x)] \\ & =\cos \left[\frac{e^{\log x}-e^{-\log x}}{2}+\frac{e^{\log x}+e^{-\log x}}{2}\right] \\ & {\left[\because \sinh x=\frac{e^x-e^{-x}}{2}\right. \text { and }} \left.\cosh x=\frac{e^x+e^{-x}}{2}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} &=\cos \left[\frac{2 x}{2}\right]=\cos x\\ &\text { The minimum value of } \cos x=-1\\ &\begin{aligned} \therefore &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, k=-1 \\ \text { And, } \cosh (k+1) & =\frac{e^{k+1}+e^{-(k+1)}}{2} \\ & =\frac{e^0+e^0}{2}=1 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} & f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R \\ & f(x)=\sec x+\tan x \\ &=\frac{1+\tan ^2 x / 2}{1-\tan ^2 x / 2}+\frac{2 \tan x / 2}{1-\tan ^2 x / 2} \\ &=\frac{(1+\tan x / 2)^2}{1-\tan ^2 x / 2}=\frac{1+\tan x / 2}{1-\tan x / 2} \\ &=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right) \\ & f^{\prime}(x)=\sec ^2\left(\frac{\pi}{4}+\frac{x}{2}\right) \cdot \frac{1}{2}>0 \forall \\ & x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{aligned} $

$\therefore f(x)$ is an increasing function.

Hence, $f(x)$ is one-one function.

$ \begin{aligned} & f:[0, \infty) \rightarrow R, f(x)=x^2 \\ & f^{\prime}(x)=2 x \geq 0 \forall x \in[0, \infty) \end{aligned} $

$\therefore f(x)$ is one-one function.

Both Statements I and II are true.

$f(x)=\left\{\begin{array}{cc}2 x-5, & x<-3 \\ x+2, & -3 \leq x<5 \\ 3 x+1, & x \geq 5\end{array}\right.$

$ \begin{aligned}(A) f(-5)+f(0)+ & f(-1)=(2(-5)-5) +(0+2)+(-1+2) \\ = & -15+2+1=-12 \end{aligned} $

$ \begin{aligned}(B) & f((f(5))+10 f(-3)) \\ & \quad=f((3 \times 5+1)+10 \cdot(-3+2)) \\ & \quad=f(6)=3 \times 6+1=19 \end{aligned} $

(C) $f(|f(-4)|)=f(|-13|)=3 \times 13+1=40$

$ \begin{aligned}(D) f(f(f(1))) & =f(f(3))=f(5) \\ & =3 \times 5+1=16 \end{aligned} $

$ \begin{aligned} & \text { } f(x)=\frac{\sqrt{6 x^2+5 x-6}}{\sqrt{4-x}-\sqrt{x+4}} \\ & 6 x^2+5 x-6 \geq 0 \\ & \Rightarrow \quad 6 x^2+9 x-4 x-6 \geq 0 \\ & \Rightarrow \quad 3 x(2 x+3)-2(2 x+3) \geq 0 \\ & \Rightarrow \quad(3 x-2)(2 x+3) \geq 0 \\ & \end{aligned} $

$ \begin{array}{lc} & x \in\left(-\infty, \frac{-3}{2}\right] \cup\left[\frac{2}{3}, \infty\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & 4-x \geq 0 \Rightarrow x \leq 4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\ \Rightarrow & x \in(-\infty, 4] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow & x+4 \geq 0 \Rightarrow x \geq-4 \end{array} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x \in[-4, \infty) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ \therefore \quad x \in\left[-4, \frac{-3}{2}\right] \cup\left[\frac{2}{3}, 4\right] $

$ \begin{aligned} & \text { Given, } f(x)=\frac{1}{\sqrt{[x]^2+[x]-2}} \\ & =\frac{1}{\sqrt{\left([x]+\frac{1}{2}\right)^2-\frac{9}{4}}} \end{aligned} $

Maximum value of $\left([x]+\frac{1}{2}\right)^2$ is $\infty$.

So, $f(x)=\frac{1}{\sqrt{\infty-\frac{9}{4}}} \rightarrow 0$

$\left([x]+\frac{1}{2}\right)_{\min }^2-\frac{9}{4}$ to be positive, $[x]$ must be 2 .

$ \begin{aligned} & \Rightarrow \quad\left(2+\frac{1}{2}\right)^2-\frac{9}{4}=\frac{25}{4}-\frac{9}{4}=\frac{16}{4}=4 \\ & \Rightarrow \quad f(x)=\frac{1}{\sqrt{4}}=\frac{1}{2} \\ & \text { Range : }\left(0, \frac{1}{2}\right] \end{aligned} $

Let $x$ and $y$ be any two elements in the domain (Z), such that

$ \begin{equation*} f(x+y)=f(x)+f(y) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Differentiating above expression w.r.t. $y$, keeping $x$ constant, we get

$ f^{\prime}(x+y)=f^{\prime}(y) $

Let $y=0 \Rightarrow f^{\prime}(x+0)=f^{\prime}(0)$

and Assume $f^{\prime}(0)=K$

$ \therefore \quad f^{\prime}(x)=K $

Integrating on both sides, we get

$ \begin{aligned} & f(x)=K x+C \\ & \quad\{C \rightarrow \text { integration constant }\}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Now putting $x=y=0$ in Eq. (i), we get $f(0+0)=f(0)+f(0) \Rightarrow f(0)=2 f(0)$ or $f(0)=0$

Let $x=0$, from Eq. (ii)

$ \begin{aligned} f(0) & =K \times 0+C \text { or } 0=0+C \text { or } C=0 \\ \therefore f(x) & =k x \end{aligned} $

Case (i) If $k>0$, then $f(x)$ is strictly increasing.

Case (ii) If $k<0$, then $f(x)$ is strictly decreasing

So, function is injective

Also $f(x)$ where every element in the codomain is a valid output of the function i.e., range is equal to codomain.

So, function is surjective also. Therefore, there are two bijective functions.

Given that, $A_n=\{(n+1) k \mid k \in \mathbf{N}\}$

If $n=1, A_1=\{2 k \mid k \in \mathbf{N}\}$

for, $k=1,2,3 \ldots A_1=\{2,4,6,8, \ldots\}$

If $n=2, A_2=\{3 k \mid k \in \mathbf{N}\}$

for $k=1,2,3, \ldots \quad A_2=\{3,6,9, \ldots\}$

If $n=3, A_3=\{4 k \mid k \in \mathbf{N}\}$

for $k=1,2,3, \ldots A_3=\{4,8,12, \ldots\}$

Now, $X=\bigcup_{n \in \mathbf{N}} A_n$

or $X=A_1 \cup A_2 \cup A_3 \ldots$

or $\quad X=\{2,3,4,5,6, \ldots\}$

Here, $f: X \rightarrow N$ and $f(x)=x \forall x \in \mathbf{X}$

$\because f(x)$ is linear function i.e., it is one-one function but there is no value of $x$ in domain where $f(x)=1$

$\therefore f(x)$ is not onto function.

We have,

$ f(n)= \begin{cases}2 n, & \text { if } n>0 \\ 1, & \text { if } n=0 \\ -2 n-1, & \text { if } n<0\end{cases} $

When, $n>0$

$ f(n)=2,4,6,8 \ldots \ldots $

When, $n<0$

$ f(n)=1,3,5,7 \ldots . . $

∴ Range of $f(n)$ is $\{1,2,3,4, \ldots \ldots\}$.

$\therefore \quad$ Codomain $=$ Range

Hence, $f(n)$ is onto.

Here, $f(0)-f(-1)=1$

$\therefore f$ is not one one.

We have,

$ \begin{aligned} \frac{x^5-5}{x^5+x^2} & =f(x)+\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \\ \frac{x^5-5}{x^3+x^2} & =x^2-x+1+\frac{-x^2-5}{x^3+x^2} \\ \therefore \frac{-x^2-5}{x^3+x^2} & =\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \\ -x^2-5 & =A x(x+1)+B(x+1)+C\left(x^2\right) \\ -x^2-5 & =(A+C) x^2+(A+B) x+B \\ A+C & =-1, A+B=0, B=-5 \end{aligned} $

Solving we get, $A=5, B=-5, C=-6$

$ \begin{array}{ll} \therefore & f(x)=x^2-x+1 \\ \therefore & f(K)=K^2-K+1 \end{array} $

Given, $f(K)+A+B+C=1$

$ \begin{aligned} & & K^2-K+1+5-5-6 & =1 \\ \Rightarrow & & K^2-K-6 & =0 \\ \Rightarrow & & (K+2)(K-3) & =0 \\ \Rightarrow & & K & =3,-2 \end{aligned} $

Largest value of $K=3$

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } f(x)=x-\frac{1}{x} \mathrm{gl} \\ & \therefore(f(x))^3=x^3-\frac{1}{x^3}-3(x)\left(\frac{1}{x}\right)\left(x-\frac{1}{x}\right) \\ & =f\left(x^3\right)-3\left(x-\frac{1}{x}\right) \\ & \Rightarrow \quad(f(x))^3=f\left(x^3\right)-3 f(x) \\ & \Rightarrow \quad 3 f(x)=f\left(x^3\right)-[f(x)]^3 \end{aligned} \end{aligned} $

We have,

$ f(x)=[x] \text { and } g(x)=3\left[\frac{x}{3}\right] $

Given, $f(x)=g(x)$

$ \therefore \quad[x]=3\left[\frac{x}{3}\right] $

Here, $[x]$ and $\left[\frac{x}{3}\right]$ is an integers

Let $\quad\left[\frac{x}{3}\right]=k$

$ \begin{array}{lc} \because {[x]=3 k} \\ \because x \in[3 k, 3 k+1) \\ \{x \in \mathbf{R} / 3 k \leq x<3 k+1, k \in \mathbf{Z}\} \end{array} $

$ \begin{aligned} & \text { Given, } f(x+y)=f(x) \cdot f(y) \\ & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\quad f(x)=a^x \\ & \,\,\,\,\,\,\,\,\,\,\,\,f(1)=a^1=2 \Rightarrow a=2 \end{aligned} $

$ \begin{aligned} &\therefore \quad f(x)=2^x\\ &\text { Area enclosed by the lines }\\ &\begin{aligned} & 2|x|+5|y| \leq 4 \\ & 4\left(\frac{1}{2} \times 2 \times \frac{4}{5}\right)=\frac{16}{5} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{16}{1+(2)^2} \\ =\frac{(2)^4}{1+(2)^2}= & \frac{f(4)}{1+f(2)} \end{aligned} $

We have,

$ f(x)= \begin{cases}\frac{60-5 x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3 x, & 7 \leq x \leq 10\end{cases} $

Now, $f(x)=10 \forall x \in[6,7]$

So, $f(x)$ is not one-one.

Again,

$ \begin{array}{ll} & 0 \leq x \leq 6 \\ \Rightarrow & -30 \leq-5 x \leq 0 \\ \Rightarrow & 60-30 \leq 60-5 x \leq 60 \\ \Rightarrow & \frac{30}{3} \leq \frac{60-5 x}{3} \leq \frac{60}{3} \\ \Rightarrow & 10 \leq \frac{60-5 x}{3} \leq 20 \text { and } 7 \leq x \leq 10 \\ \Rightarrow & -30 \leq-3 x \leq-21 \\ \Rightarrow & 1 \leq 31-3 x \leq 10 \end{array} $

∴ Range of $f(x)=[1,20]$

It is given that co-domain of $f(x)=[1,20]$

∴ Range of $f(x)=$ co-domain of $f(x)$

So, $f(x)$ is onto.

We have,

$ f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)} $

For $f(x)$ to be defined

$ \begin{aligned} & \log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0 \text { and } \frac{5 x-x^2}{4}>0 \\ \Rightarrow & \quad \frac{5 x-x^2}{4} \geq 1 \text { and } \frac{5 x-x^2}{4}>0 \\ \Rightarrow & \quad \frac{5 x-x^2}{4} \geq 1 \\ \Rightarrow & \quad 5 x-x^2 \geq 4 \\ \Rightarrow & x^2-5 x+4 \leq 0 \\ \Rightarrow & (x-4)(x-1) \leq 0 \end{aligned} $

$ \begin{aligned} &\therefore \quad x \in[1,4]\\ &\text { So, domain of } f(x) \text { is }[1,4] \text {. } \end{aligned} $