Question

If $f\left(x\right)={\int }_a^x\left[f\right(x){]}^{-1}dx$ and ${\int }_a^1\left[f\right(x){]}^{-1}dx=\sqrt{2},$ then

• A. $f\left(2\right)=2$
• B. $f^{\prime }\left(2\right)=1/2$
• C. $f^{-1}\left(2\right)=2$
• D. ${\int }_0^{1}f\left(x\right)dx=\sqrt{2}$

Answer 1

Answer: (A), (B), (C)

$f\left(2\right)=2$
$f^{-1}\left(2\right)=2$
$f^{\prime }\left(2\right)=1/2$

$f\left(x\right)={\int }_a^x\frac{1}{f\left(x\right)}dx\Rightarrow f^{\prime }\left(x\right)=\frac{1}{f\left(x\right)}\cdot 1-0\Rightarrow f\left(x\right)f^{\prime }\left(x\right)=1$
$\Rightarrow \int f\left(x\right)f^{\prime }\left(x\right)dx=\int 1dx$
$\Rightarrow \frac{1}{2}\left[f\right(x){]}^2=x+c$
Now given that ${\int }_a^1\left[f\right(x){]}^{-1}dx=\sqrt{2}\Rightarrow f(1)=\sqrt{2}$
$\Rightarrow$ From $\left(1\right),\frac{1}{2}\left[f\right(1){]}^2=1+c\Rightarrow c=0$
$\Rightarrow f\left(x\right)=\pm \sqrt{2x}$
But $f\left(1\right)=\sqrt{2}\Rightarrow f\left(x\right)=\sqrt{2x}\Rightarrow f\left(2\right)=2$
Also, $f^{\prime }\left(x\right)=\frac{1}{\sqrt{2x}}\Rightarrow f^{\prime }\left(2\right)=1/2$
${\int }_0^{1}f\left(x\right)dx={\int }_0^1\sqrt{2x}dx={\left[\frac{(2x{)}^{3/2}}{3}\right]}_0^1=\frac{(2{)}^{3/2}}{3}$
Also, $f^{-1}\left(x\right)=\frac{x^2}{2}\Rightarrow f^{-1}\left(2\right)=2$