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 $x =0, x =\sqrt{3}$ and $x$-axis is

• A. $\frac{ f (\sqrt{3})}{8}\left[8 \sqrt{3}-( f \sqrt{3})^7-4 f (\sqrt{3})\right]$ sq. units
• B. $\frac{ f (\sqrt{3})}{8}\left[8 \sqrt{3}-( f (\sqrt{3}))^p\right]$ 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 $(f(x))^6 \cdot f^{\prime}(x)=1-f^1(x)$
$\Rightarrow f ^{\prime}( x )\left[7 f ( x )^6+1\right]=1 \Rightarrow f ^{\prime}( x )>0 \forall x \in R$
Hence, $f ( x )$ is increasing function $\forall x \in R$
So there exists an inverse of $f(x)$ such that
$f ^{-1}(0)=0$
$\Rightarrow x ^7+ x = f ^{-1}( x ) \Rightarrow \int\limits_0^{\sqrt{2}}\left( x ^7+ x \right) dx =\frac{ x ^8}{8}+\left.\frac{ x ^2}{2}\right|_0 ^{\sqrt{2}}=2+1=3$
Now, we know that
$\int\limits_0^a f(x) d x+\int\limits_0^{f(a)} f^{-1}(x) d x=a f(a) $
$\text { Hence, } \int_0^{\sqrt{3}} f(x) d x+\int\limits_0^{f(\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^{f(\sqrt{3})}\left(x^7+x\right) d x $
$\int\limits_0^{\sqrt{3}} f(x) d x=\frac{f(\sqrt{3})}{8}\left[8 \sqrt{3}-(f(\sqrt{3}))^7-4 f(\sqrt{3})\right]$