Functions

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

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

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

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

$f(x) = {{{4^x}} \over {{4^x} + 2}}$

$f(1 - x) = {{{4^{1 - x}}} \over {{4^{1 - x}} + 2}} = {{{4 \over {{4^x}}}} \over {{4 \over {{4^x}}} + 2}}$

$ = {4 \over {4 + 2\,.\,{4^x}}}$

$f(1 - x) = {2 \over {{4^x} + 2}}$ ...... (i)

and $1 - f(x) = 1 - {{{4^x}} \over {{4^x} + 2}}$

$ = {{{4^x} + 2 - {4^x}} \over {{4^x} + 2}} = {2 \over {{4^x} + 2}} = f(1 - x)$

$ \Rightarrow f(1 - x) + f(x) = 1$ ..... (ii)

Put $x = {1 \over 4}$ in Eq. (ii), we get

$f\left( {{3 \over 4}} \right) + f\left( {{1 \over 4}} \right) = 1$ ...... (iii)

Now, put $x = {1 \over 2}$ in Eq. (ii), we get

$f\left( {{1 \over 2}} \right) + f\left( {{1 \over 2}} \right) = 1$

or $2f\left( {{1 \over 2}} \right) = 1$ ..... (iv)

Adding Eqs. (iii) and (iv), we have

$f\left( {{1 \over 4}} \right) + 2f\left( {{1 \over 2}} \right) + f\left( {{3 \over 4}} \right) = 2$

Given,

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

Then,

$\begin{aligned} g \circ f(x) & =g[f(x)] \\ & =g(2 x+1)=(2 x+1)^2-2 \\ & =4 x^2+4 x+1-2=4 x^2+4 x-1 \end{aligned}$

Given, $f(x)=\frac{a^x+a^{-x}}{2}$

Then,

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

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

$\because\{x\}=x-[x]$

$f:(0,1) \rightarrow(0,1)$

Defined by $f(x)=\min \{x-[x],-x-[x]\}$ which is an identity function, as $x \in(0,1)$ and $\{x\} \in(0,1)$ Where, $\{\cdot\}$ is fractional part function.

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

$\because$ ${({x^2} + 5x + 5)^{x + 5}} = 1$

Case I $x + 5 = 0$

$ \Rightarrow x = - 5$

Case II ${x^2} + 5x + 5 = 1$

or, ${x^2} + 5x + 4 = 0$

or, $(x + 1)(x + 4) = 0$

$ \Rightarrow x = - 1, - 4$

$x = - 1, - 4$ and $ - 5$ are the three integers satisfying given equation.

$\frac{x^4}{(x-1)(x-2)}=\frac{x^4}{x^2-3 x+2}$ is an improper fraction.

$\therefore \frac{x^4}{(x-1)(x-2)}=x^3+3 x+7+\frac{15 x-14}{(x-1)(x-2)}$

$=x^2+3 x+7+\frac{A}{x-1}+\frac{B}{x-2}$

On comparing with given equation,

$\frac{x^4}{(x-1)(x-2)}=f(x)+\frac{A}{x-1}+\frac{B}{x-2}$, we have $f(x)=x^2+3 x+7$

$\begin{aligned} \text { (i) } f(x) & =x|x|= \begin{cases}x(-x), & x<0 \\ x(x), & x \geq 0\end{cases} \\ f(x) & =\left\{\begin{array}{cl} -x^2, & x<0 \\ x^2, & x \geq 0 \end{array}\right. \\ f^{\prime}(x) & =\left\{\begin{aligned} -2 x, & x<0 \\ 2 x, & x \geq 0 \end{aligned}\right. \end{aligned}$

$f^{\prime}(x)>0, \forall x \in R-\{0\}$

$\Rightarrow f(x)$ is strictly increasing on $R-\{0\}$.