Question

Fill in the blanks.
$(i)$ Let $f$ : $R\to R$ such that $f\{x-f(y)\}=f\{(y)\}+xf(y)+f(x)-1\forall x$, $y\in R$. Then the value of $|\{f(16)\}|$ is P.
$(ii)$ If $f=\{(1$, $2)$, $(3$, $5)$, $(4$, $1)\}$ and $g=\{(2$, $3)$, $(5$, $1)$, $(1$, $3)\}$, then $gof=$ Q .
$(iii)$ If $f$ : $R\to R$ be defined by $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$, then $(fofof)(x)=$ R .

	P	Q	R
$(a)$	$126$	$(1$, $2)$, $(2$, $1)$, $(1$, $3)$	$\frac{x}{\sqrt{x^2+3}}$
$(b)$	$128$	$(1$, $4)$, $(3$, $1)$, $(4$, $3)$	$\frac{1}{\sqrt{3x^2+1}}$
$(c)$	$125$	$(-2$, $5)$, $(5$, $2)$, $(1$, $5)$	$\frac{x^2}{\sqrt{x^2+3}}$
$(d)$	$127$	$(1$, $3)$, $(3$, $1)$, $(4$, $3)$	$\frac{x}{\sqrt{3x^2+1}}$

• A. $(a)$
• B. $(b)$
• C. $(c)$
• D. $(d)$

Answer 1

Answer: (D)

$(i)$ Putting $x=f(y)=0$, then
$f(0)=f(0)+0+f(0)-1$
$\Rightarrow f(0)=1$
Putting $x=f(y)$ we get, $f(0)=f(x)+x^2+f(x)-1$
$\Rightarrow f\left(x\right)=1-\frac{x^2}{2}$
$\therefore f\left(16\right)=1-\frac{256}{2}$
$=1-128$
$=-127$
$\Rightarrow \left|f\left(16\right)\right|=127$
$(ii)$ Given, $f=\{(1$, $2)$, $(3$, $5)$, $(4$, $1)\}$ and
$g=\{(2$, $3)$, $(5$, $1)$, $(1$, $3)\}$
Now, $gof(1)=g\{f(1)\}=g(2)=3$
$gof(3)=g\{f(3)\}=g(5)=1$
$gof(4)-g\{f(4)\}=g(1)=3$
Hence, $gof=\{(1$, $3)$, $(3$, $1)$, $(4$, $3)\}$
$(iii)$ Given, $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$
Now, $\left(fofof\right)\left(x\right)=f\left[f\left\{f\left(x\right)\right\}\right]$
$=f\left[f\left(\frac{x}{\sqrt{1+x^2}}\right)\right]$
$=f\left[\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}\right]$
$=f\left[\frac{x}{\sqrt{1+2x^2}}\right]$
$=\frac{\frac{x}{\sqrt{1+2x^2}}}{\sqrt{1+\frac{x^2}{1+2x^2}}}$
$=\frac{x}{\sqrt{3x^2+1}}$