Question

If $f: R \rightarrow R$ 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. $\frac{f(\sqrt{3})}{8}\left[8 \sqrt{3}-(f(\sqrt{3}))^7-4 f(\sqrt{3})\right]$ 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. $\frac{f(\sqrt{3})}{8}\left[4 \sqrt{3}-(f(\sqrt{3}))^8-4 f(\sqrt{3})\right]$ sq. units

Answer 1

Answer: (A)

$f(x)\left[\left(f(x)^6+1\right]=x\right.$
$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) $
$\Rightarrow \int\limits_0^{\sqrt{2}}\left(x^7+x\right) d x=\frac{x^8}{8}+\left.\frac{x^2}{2}\right|_0 ^{\sqrt{2}}=2+1=3$
Now, we know that $\int\limits_0^a f(x) d x+\int\limits_0^{1(2)} f^{-1}(x) d x=a f(a)$
$\text { Hence } \int\limits_0^{\sqrt{3}} f(x) d x+\int\limits_0^{1(\sqrt{3})} f^{-1}(x) d x=\sqrt{3} f(\sqrt{3})$
$\int\limits_0^{\sqrt{3}} f(x) d x=\sqrt{3} f(\sqrt{3})-\int\limits_0^{1(\sqrt{3})}\left(x^7+x\right) d x$
$\int\limits_0^{\sqrt{3}} f(x) d x=\frac{f(\sqrt{3})}{8}\left[8 \sqrt{3}-(f(\sqrt{3}))^7-4 f(\sqrt{3})\right]$