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 $x=0$ and $x=\sqrt{3}$ is

• A. $\frac{f(\sqrt{3})}{8}\left[8\sqrt{3}-(f(\sqrt{3}){)}^7-4f(\sqrt{3})\right]$ sq. units
• B. $\frac{f(\sqrt{3})}{4}\left[8\sqrt{3}-(f(\sqrt{3}){)}^7\right]$ sq. units
• C. $\left(\sqrt{3}f(\sqrt{3})-\frac{93}{8}\right)$ sq. units
• D. $\frac{f(\sqrt{3})}{8}\left[4\sqrt{3}-(f(\sqrt{3}){)}^8-4f(\sqrt{3})\right]$ sq. units

Answer 1

Answer: (A)

$f(x)[\left(f(x{)}^6+1\right]=x$
$f(0)\left[(f(0){)}^6+1\right]=0\Rightarrow f(0)=0$
$\text{ and }7(f(x){)}^6\cdot f^{\prime }(x)+f^{\prime }(x)=1$
$\Rightarrow f^{\prime }(x)\left[7(f(x){)}^6+1\right]=1$
$\Rightarrow f^{\prime }(x)>0\forall x\in R$
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 $\underset{0}{\overset{a}{\int }}f(x)dx+\underset{0}{\overset{1(2)}{\int }}{f}^{-1}(x)dx=af(a)$
$\text{ Hence }\underset{0}{\overset{\sqrt{3}}{\int }}f(x)dx+\underset{0}{\overset{1(\sqrt{3})}{\int }}{f}^{-1}(x)dx=\sqrt{3}f(\sqrt{3})$
$\underset{0}{\overset{\sqrt{3}}{\int }}f(x)dx=\sqrt{3}f(\sqrt{3})-\underset{0}{\overset{1(\sqrt{3})}{\int }}\left(x^7+x\right)dx$
$\underset{0}{\overset{\sqrt{3}}{\int }}f(x)dx=\frac{f(\sqrt{3})}{8}\left[8\sqrt{3}-(f(\sqrt{3}){)}^7-4f(\sqrt{3})\right]$