Question

If $f:R\to R$ be a differentiable function such that $(f(x){)}^7=x-f(x)$ then
The area bounded by curve $y=f(x)$ between the ordinates $x=0,x=\sqrt{3}$ and $x$-axis is

• A. $\frac{f(\sqrt{3})}{8}\left[8\sqrt{3}-(f\sqrt{3}{)}^7-4f(\sqrt{3})\right]$ sq. units
• B. $\frac{f(\sqrt{3})}{8}\left[8\sqrt{3}-(f(\sqrt{3}){)}^p\right]$ sq. units
• C. $\left(\sqrt{3}f(\sqrt{3})-\frac{93}{8}\right)$ sq. units
• D. none of these

Answer 1

Answer: (A)

$f(x)\left[f(x{)}^6+1\right]=x$
$f(0)\left[f(0{)}^6+1\right]=0\Rightarrow f(0)=0$
and $(f(x){)}^6\cdot f^{\prime }(x)=1-f^1(x)$
$\Rightarrow f^{\prime }(x)\left[7f(x{)}^6+1\right]=1\Rightarrow f^{\prime }(x)>0\forall x\in R$
Hence, $f(x)$ is increasing function $\forall x\in R$
So there exists an inverse of $f(x)$ such that
$f^{-1}(0)=0$
$\Rightarrow x^7+x=f^{-1}(x)\Rightarrow \underset{0}{\overset{\sqrt{2}}{\int }}\left(x^7+x\right)dx=\frac{x^8}{8}+{\frac{x^2}{2}|}_0^{\sqrt{2}}=2+1=3$
Now, we know that
$\underset{0}{\overset{a}{\int }}f(x)dx+\underset{0}{\overset{f(a)}{\int }}{f}^{-1}(x)dx=af(a)$
$\text{ Hence, }{\int }_0^{\sqrt{3}}f(x)dx+\underset{0}{\overset{f(\sqrt{3})}{\int }}{f}^{-1}(x)dx=\sqrt{3}f(\sqrt{3})$
$\underset{0}{\overset{\sqrt{3}}{\int }}f(x)dx=\sqrt{3}f(\sqrt{3})-\underset{0}{\overset{f(\sqrt{3})}{\int }}\left(x^7+x\right)dx$
$\underset{0}{\overset{\sqrt{3}}{\int }}f(x)dx=\frac{f(\sqrt{3})}{8}\left[8\sqrt{3}-(f(\sqrt{3}){)}^7-4f(\sqrt{3})\right]$