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, $\left(fofof\right)\left(x\right)=f \left[f \left\{f \left(x\right)\right\}\right]$
$=f \left[f \left(\frac{x}{\sqrt{1+x^{2}}}\right)\right]$
$=f \left[\frac{\frac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+x^{2}}}}\right]$
$=f \left[\frac{x}{\sqrt{1+2x^{2}}}\right]$
$=\frac{\frac{x}{\sqrt{1+2x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+2x^{2}}}}$
$=\frac{x}{\sqrt{3x^{2}+1}}$