Question

Consider $f\left(x\right)=1+\underset{0}{\overset{2x}{\int }}\frac{e^{\frac{t}{2}}\cdot f\left(\frac{t}{2}\right)\left(2t\right)}{\sqrt{16-t^4}}dt-\underset{x}{\overset{0}{\int }}f\left(t\right)\cdot e^t{\sin }^{-1}\left(t^2\right)dt$ and $h\left(x\right)=\sin \left(e^{-x}\ln \right(f\left(x\right)\left)\right)$.
Then the range of $y=h\left(x\right)+4x+5$ is

• A. $[1,\infty )$
• B. $(1,\infty )$
• C. $[2,10]$
• D. $(2,10)$

Answer 1

The correct option is D $(2,10)$

$f\left(x\right)=1+\underset{0}{\overset{2x}{\int }}\frac{e^{\frac{t}{2}}\cdot f\left(\frac{t}{2}\right)\left(2t\right)}{\sqrt{16-t^4}}dt-\underset{x}{\overset{0}{\int }}f\left(t\right)\cdot e^t{\sin }^{-1}\left(t^2\right)dt$
By using Newton-Leibnitz theorem, we have
$f^{\prime }\left(x\right)=\frac{e^x\cdot f\left(x\right)\cdot 2\left(2x\right)}{\sqrt{16-16x^4}}\times 2-(0-f(x\left)e^x{\sin }^{-1}{x}^2\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2x{e}^{x}f\left(x\right)}{\sqrt{1-x^4}}+f\left(x\right)e^x{\sin }^{-1}{x}^2$
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{2x{e}^x}{\sqrt{1-x^4}}+e^x{\sin }^{-1}{x}^2\cdots \left(1\right)$

Integrating both sides,
$\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int e^x[{\sin }^{-1}{x}^2+\frac{2x}{\sqrt{1-x^4}}]dx$
$\Rightarrow \ln \left(f\right(x\left)\right)=e^x{\sin }^{-1}\left(x^2\right)+c[\because \int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+c]$

Now, $f\left(0\right)=1\Rightarrow c=0$
$\therefore \sin \left(e^{-x}\ln \right(f\left(x\right)\left)\right)=x^2=h\left(x\right)$
From $\left(1\right),$ $x\in (-1,1)$
$\therefore h\left(x\right)=x^2$ where $x\in (-1,1)$

$h\left(x\right)+4x+5$
$=x^2+4x+5$
$=(x+2{)}^2+1=g(x)$ (say)
$g^{\prime }\left(x\right)=2(x+2)>0$ for all $x\in (-1,1)$
$\Rightarrow g\left(x\right)$ is increasing in $(-1,1)$
$\Rightarrow g\left(x\right)\in \left(g\right(-1),g(1\left)\right)$
$\therefore$ Range of $g\left(x\right)$ is $(2,10)$