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 $h\left(x\right)=\sin \left(e^{-x}\ln \right(f\left(x\right)\left)\right)$.

If length of the tangent drawn to the curve $y=h\left(x\right)$ at its point $P(x_0,y_0)$ is $\frac{1}{2\sqrt{2}},$ then $(|x_0|+|y_0|)$ equals

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $\frac{5}{4}$

Answer 1

The correct option is C $\frac{3}{4}$

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

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


$h\left(x\right)=x^2$
$h^{\prime }\left(x_0\right)=2x_0$
$L_T=\left|\frac{y_0}{m}\sqrt{1+m^2}\right|$
$\therefore \frac{x_0^2}{2x_0}\sqrt{1+4x_0^2}=\frac{1}{2\sqrt{2}}$
$\Rightarrow x_0^2(1+4x_0^2)=\frac{1}{2}$
$\Rightarrow 8x_0^4+2x_0^2-1=0\Rightarrow (4x_0^2-1)(2x_0^2+1)=0$
$\Rightarrow x_0=\pm \frac{1}{2},y_0=\frac{1}{4}$
$\therefore |x_0|+|y_0|=\frac{3}{4}$