Functions

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

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 $R-\{0\}$.