Functions

$f(x)=\sqrt{\frac{[x]-x}{x-[x]}}$

$x \notin I$ : $\quad[\because$ for $x \in I$, function is indeterminate form]

For $x \in$ non integer values, we have

$ f(x)=\sqrt{\frac{-(x-[x])}{x-[x]}}=\sqrt{-1} $

For $x \in$ non integer values, $f(x)$ is not real.

Therefore, $x \in \phi$.

$f(x)=\frac{1}{\sqrt{x-[x]}}$

We know, $x-[x]=\{x\}$ and $0 \leq\{x\}<1$

$ \begin{aligned} f(x) & =\frac{1}{\sqrt{\{x\}}} \\ \because \quad 0 \leq\{x\}<1 & \Rightarrow 0 \leq \sqrt{\{x\}}<1 \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{1}{0}>\frac{1}{\sqrt{\{x\}}}>\frac{1}{1} \\ \Rightarrow & \infty>\frac{1}{\sqrt{\{x\}}}>1 \\ \Rightarrow & \infty>f(x)>1 \\ \Rightarrow & f(x) \in(1, \infty) \end{array} $

Assertion

$ \begin{aligned} \operatorname{coth} x & =\frac{1-k}{1+k}(0 < k<2) \\ \because \quad \operatorname{coth} x & =\frac{e^x+e^{-x}}{e^x-e^{-x}} \\ & =\frac{1+e^{-2 x}}{1-e^{-2 x}}=\frac{1-\left(-e^{-2 x}\right)}{1+\left(-e^{-2 x}\right)} \end{aligned} $

Here, $k=-e^{-2 x}$

$ -e^{-2 x} \in(-\infty, 0) \Rightarrow k \in(-\infty, 0) $

Therefore, Assertion (A) is false.

Reason $y=\tanh x$

Here, $y$ lies between -1 and 1 , Reason is true.

$ \begin{aligned} & \text { Given, } f(x)=\sqrt{\frac{2 x^2-7 x+5}{3 x^2-5 x-2}} \\ & \quad \frac{2 x^2-7 x+5}{3 x^2-5 x-2} \geq 0 \\ & \Rightarrow \quad \frac{2 x^2-2 x-5 x+5}{3 x^2-6 x+x-2} \geq 0 \\ & \Rightarrow \frac{2 x(x-1)-5(x-1)}{3 x(x-2)+1(x-2)} \geq 0 \\ & \Rightarrow \quad \frac{(2 x-5)(x-1)}{(3 x+1)(x-2)} \geq 0 \\ & \Rightarrow \quad x \neq-\frac{1}{3}, x \neq 2 \end{aligned} $

$ \begin{aligned} &\text { Domain : }\\ &x \in\left(-\infty,-\frac{1}{3}\right) \cup[1,2) \cup\left[\frac{5}{2}, \infty\right) \end{aligned} $

Given, $f(x)=|x-2|+|x-3|$

We know that for $f(x)=|x-a|+|x-b|$, the minimum value of $f(x)$,

we get either at $x=a$

or at $x=b$

$ \begin{aligned}So\,\, for\,\,\,\, & f(x)=|x-2|+|x-3| \\ & f(2)=0+1=1 \\ & f(3)=1+0=1 \end{aligned} $

So, $f(x)_{\text {min }}=1$ and maximum value will be $\infty$ at $x=\infty$.

$ \because \quad f_{\max }=\infty+\infty=\infty $

So, range $=[1, \infty)$

$ \begin{aligned} &\text { Given, } f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right)\\ &\begin{aligned} & \text { and } g(x)=\sqrt{3+4 x-4 x^2} \\ & \Rightarrow \quad 3+4 x-4 x^2 \geq 0 \\ & \Rightarrow \quad 3-4\left(x^2-x\right) \geq 0 \\ & \Rightarrow 3-4\left(x^2-x+\frac{1}{4}\right)+1 \geq 0 \\ & \Rightarrow \quad 4-4\left(x-\frac{1}{2}\right)^2 \geq 0 \\ & \Rightarrow \quad\left(x-\frac{1}{2}\right)^2 \leq 1 \\ & \Rightarrow \quad-1 \leq x-\frac{1}{2} \leq 1 \\ & \Rightarrow \quad-1+\frac{1}{2} \leq x \leq 1+\frac{1}{2} \\ & \Rightarrow \quad-\frac{1}{2} \leq x \leq \frac{3}{2} \Rightarrow x \in\left[-\frac{1}{2}, \frac{3}{2}\right] \end{aligned} \end{aligned} $

Now, $f(x)=\frac{1}{2}-\tan \left(\frac{\pi x}{2}\right)$

As $f: A \rightarrow B$ and $g: B \rightarrow C$, so domain of $g(x)=$ range of $f(x)$

∴ Range of $f(x)=\left[-\frac{1}{2}, \frac{3}{2}\right]$.

$ \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} $

$f(x)=\log \left(\left(\frac{2 x^2-3}{x}\right)+\sqrt{\frac{4 x^4-11 x^2+9}{|x|}}\right)$

$\begin{aligned} & f(-x)=\log \left(\frac{2(-x)^2-3}{(-x)}+\sqrt{\frac{4(-x)^4-11(-x)^2+9}{|-x|}}\right) \\ & f(-x)=\log \left[\frac{2 x^2-3}{-x}+\sqrt{\frac{4 x^4-11 x^2+9}{|x|}}\right] \end{aligned}$

We know that, if function $f(x)$ is an odd function. Then,

$\begin{aligned} & f(x)+f(-x)=0 \\ & \log \left[\left(\frac{2 x^2-3}{x}\right)+\sqrt{\frac{4 x^4-11 x^2+9}{|x|}}\right] \\ & +\log \left[\sqrt{\frac{4 x^4-11 x^2+9}{|x|}}-\frac{\left(2 x^2-3\right)}{x}\right] \\ & =\log \left[\frac{4 x^4-11 x^2+9}{x^2}-\frac{\left(2 x^2-3\right)^2}{x^2}\right] \\ & =\log \left[\frac{4 x^4-11 x^2+9-4 x^4-9+12 x^2}{x^2}\right] \\ & =\log 1=0 \end{aligned}$

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

$f(f(x))=-x$ [Given]

$\begin{aligned} & \Rightarrow \quad \frac{f(x)-2}{2(f(x))+1}=-x \\ & \Rightarrow \quad \frac{\frac{x-2}{2 x+1}-2}{\frac{2 x-4}{2 x+1}+1}=-x \\ & \Rightarrow \frac{x-2-4 x-2}{2 x-4+2 x+1}=-x \\ & \Rightarrow \quad \frac{-3 x-4}{4 x-3}=-x \Rightarrow \frac{3 x+4}{4 x-3}=x \\ & \Rightarrow \quad 3 x+4=4 x^2-3 x \\ & \Rightarrow \quad 4 x^2-6 x-4=0 \\ & \Rightarrow \quad 2 x^2-3 x-2=0 \\ \end{aligned}$

Now, $\alpha+\beta=\frac{3}{2}, \alpha \beta=-1$

Now, $(\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta$

$\Rightarrow \quad \frac{9}{4}=\alpha^2+\beta^2-2 \quad[\because \alpha \beta=-1]$

$\begin{aligned} & \Rightarrow \alpha^2+\beta^2=\frac{17}{4} \\ & \Rightarrow 4\left(\alpha^2+\beta^2\right)=17 \end{aligned}$

Given, function $f(x)=\sin \left(\log \left(\frac{\sqrt{4-x^2}}{1-x}\right)\right)$ is defined for $x \neq 1$ and $x>-2$ and also $x<1$. Thus, the domain of above function is $D(f)=\{x \in R ; x \in(-2,1)\}$

$f(x)=\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}$

Let $y=\frac{x^2+2 x+8}{x^2+2 x+4}\quad [\because f(x)=\sqrt{y}]$

$\begin{aligned} & \Rightarrow y x^2+2 x y+4 y=x^2+2 x+8 \\ & \Rightarrow(y-1) x^2+2 x(y-1)+4 y-8=0 \end{aligned}$

For $x$ to be real, discriminant of equation must be greater than equals to zero.

$\begin{array}{rlrl} & {[2(y-1)]^2-4(y-1)(4 y-8)} & \geq 0 \\ \Rightarrow & 4(y-1)^2-4(y-1) 4(y-2) & \geq 0 \\ \Rightarrow & (y-1)^2-(y-1) 4(y-2) & \geq 0 \\ \Rightarrow & (y-1)[y-1-4 y+8] & \geq 0 \\ \Rightarrow & (y-1)(-3 y+7) & \geq 0 \\ & y =1, \frac{7}{3} \end{array}$

$\begin{aligned} & \therefore \quad(y-1)(-3 y+7) \geq 0 \Rightarrow y \in\left(1, \frac{7}{3}\right] \\ & {\left[\because \text { At } y=1, x^2+2 x+4 \neq x^2+2 x+8\right]} \\ & \because \quad y \in\left(1, \frac{7}{3}\right] \\ & \therefore \quad f(x) \in\left(1, \sqrt{\frac{7}{3}}\right] \\ & \end{aligned}$

$\begin{aligned} & f(x)=\sqrt{2-x^2} \text { and } g(x)=\log (1-x) \\ & \begin{aligned} (f+g)(x)= & f(x)+g(x) \\ = & \sqrt{2-x^2}+\log (1-x) \end{aligned} \end{aligned}$

$\begin{aligned} 2-x^2 \geq 0 \\ \Rightarrow x^2 \leq 2 \\ \Rightarrow |x| \leq \sqrt{2} \\ \Rightarrow -\sqrt{2} \leq x \leq \sqrt{2} \end{aligned}$

Also, $1-x>0 \Rightarrow x<1$

Required domain : $-\sqrt{2} \leq x \leq \sqrt{2} \cap x<1$

$\Rightarrow \quad-\sqrt{2} \leq x<1$

We have, ${1 \over 2}gof(x) = 2{x^2} - 5x + 2$

$ \Rightarrow g(f(x)) = 4{x^2} - 10x + 4$

$ \Rightarrow {(f(x))^2} + f(x) - 2 = 4{x^2} - 10x + 4$

$ \Rightarrow {(f(x))^2} + f(x) - (4{x^2} - 10x + 6) = 0$

$ \Rightarrow f(x) = {{ - 1\, \pm \,\sqrt {1 + 4(4{x^2} - 10x + 6)} } \over 2}$

$ \Rightarrow f(x) = {{ - 1\, \pm \,\sqrt {16{x^2} - 40x + 25} } \over 2}$

$ \Rightarrow f(x) = {{ - 1\, \pm \,(4x - 5)} \over 2} = 2x - 3, - 2x + 2$

Hence, $f(x) = 2x - 3$

Let $f(x)=y$, then $x=f^{-1}(y)$, here

$y=(x+2)^2-2$

or

$\begin{aligned} & y+2=(x+2)^2 \\ & x+2=\sqrt{y+2} \end{aligned}$

$\begin{aligned} & \Rightarrow \quad x=\sqrt{y+2}-2 \\ & \therefore \quad f^{-1}(y)=\sqrt{y+2}-2 \\ & \therefore \quad f^{-1}(x)=\sqrt{x+2}-2 \\ \end{aligned}$

$f(x)=[x], g(x)=|x|$

Now, $f\left(-\frac{5}{3}\right)=\left[-\frac{5}{3}\right]$

$=[-1.667]=-2$

Then, $\quad g\left(f\left(-\frac{5}{3}\right)\right)=g(-2)$

$(g \circ f)\left(-\frac{5}{3}\right)=|-2|=2$

$\therefore \quad(g \circ f)\left(-\frac{5}{3}\right)=2$

Given, $f(x)=a x+b$ and $g(x)=c x^3+d$

$\begin{aligned} f(g(x)) & =f\left(c x^3+d\right) \\ & =a\left(c x^3+d\right)+b \\ (f \circ g)(x) & =a c x^3+a d+b \quad .....(\text{i}) \end{aligned}$

Let $(f \circ g)(x)=y$, then

$x=(f \circ g)^{-1} y$

From Eq. (i), we get

$y=a c x^3+a d+b$

i.e.

$\begin{aligned} a c x^3 & =y-a d-b \\ x^3 & =(y-a d-b) / a c \\ x & =\left[\frac{(y-a d-b)}{a c}\right]^{1 / 3} \end{aligned}$

$\begin{array}{ll} \therefore & (f \circ g)^{-1}(y)=\left[\frac{(y-a d-b)}{a c}\right]^{1 / 3} \\ \text { or } & (f \circ g)^{-1}(x)=\left[\frac{(x-a d-b)}{a c}\right]^{1 / 3}\end{array}$

$f(10-x)=3 x^2+4 x-5$ and $f(x)=p x^2+q x+r$

Let $10-x=y$, then $x=10-y$

$\begin{aligned} f(y) & =3(10-y)^2+4(10-y)-5 \\ & =3\left(100+y^2-20 y\right)+40-4 y-5 \\ & =340+3 y^2-60 y-4 y-5 \\ & =3 y^2-64 y+335 \end{aligned}$

Replace $y$ by $x \Rightarrow f(x)=3 x^2-64 x+335$

compare with $f(x)=p x^2+q x+r$, we obtain

$\begin{aligned} & p=3, q=-64, r=335 \\ \Rightarrow \quad & p+q+r=274 \end{aligned}$

$\begin{aligned} f(x) & =\sin x+\cos x \\ g(x) & =x^2-1 \\ g[f(x)] & =(\sin x+\cos x)^2-1 \\ & =1+\sin 2 x-1=\sin 2 x \end{aligned}$

Among the given options, $\sin 2 x$ is monotonous (here strictly increasing) in $-\frac{\pi}{4} \leq 2 \leq \frac{\pi}{4}$.

$\begin{aligned} f(x) & =x^9-11 x^8-2 x^7+22 x^6+x^4 \\ & -12 x^3+11 x^2+x-3, \quad \forall x \in Z \\ & =x^8(x-11)-2 x^6(x-11)+x^3(x-11) \\ & -x^2(x-11)+(x-11)+8 \\ & =(x-11)\left[x^8-2 x^6+x^3-x^2+1\right]+8 \end{aligned}$

So, $f(11)=0+8=8$

$f(x)=x^3$ and $g(x)=3^x$

$\begin{aligned} & \Rightarrow \quad f[g(x)]=g[f(x)] \\ & \Rightarrow \quad\left(3^x\right)^3=3^{x^3} \Rightarrow 3^{3 x}=3^{x^3} \\ & \Rightarrow \quad 3 x=x^3 \Rightarrow x\left(x^2-3\right)=0 \\ \end{aligned}$

As, $\quad x \neq 0$, So $x^2-3=0$

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

Now, $f(-x)=\frac{-x}{\frac{1}{e^x}-1}-\frac{x}{2}+1=\frac{-x e^x}{1-e^x}-\frac{x}{2}+1$

$\begin{aligned} & =\frac{-2 x e^x-x\left(1-e^x\right)}{2\left(1-e^x\right)}+1 \\ & =\frac{-2 x e^x-x+x e^x}{2\left(1-e^x\right)}+1 \\ & =\frac{-x e^x-2 x+x}{2\left(1-e^x\right)}+1 \\ & =\frac{-2 x+x\left(1-e^x\right)}{2\left(1-e^x\right)}+1 \\ & =\frac{-x}{1-e^x}+\frac{x}{2}+1 \\ & =\frac{x}{e^x-1}+\frac{x}{2}+1=f(x) \end{aligned}$

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

$f(x)=\frac{1}{[x]-1}$

$f(x)$ is not defined when

$\begin{array}{ll} & {[x]-1=0} \\ \Rightarrow & {[x]=1 \Rightarrow 1 \leq x<2} \\ \Rightarrow & x \in[1,2) \\ \text { Domain } f=R-[1,2) \end{array}$