Question

The value of ${\int }_1^2{\left[f\left\{g\left(x\right)\right\}\right]}^{-1}.f^{\prime }\left\{g\left(x\right)\right\}.g^{\prime }\left(x\right)dx,$ where $g\left(1\right)=g\left(2\right),$ is equal to?

• A. $1$
• B. $2$
• C. $0$
• D. none of these

Answer 1

Answer: (C)

$0$

Let $I={\int }_1^2{\left[f\left\{g\left(x\right)\right\}\right]}^{-1}.f^{\prime }\left\{g\left(x\right)\right\}.g^{\prime }\left(x\right)dx$

Substitute $\left[f\left\{g\left(x\right)\right\}\right]=t\Rightarrow f^{\prime }\left\{g\left(x\right)\right\}.g^{\prime }\left(x\right)dx=dt$

$\therefore I={\int }_{f\left(g\left(1\right)\right)}^{f\left(g\left(2\right)\right)}tdt={\left[\frac{t^2}{2}\right]}_{f\left(g\left(1\right)\right)}^{f\left(g\left(2\right)\right)}=0\because g\left(1\right)=g\left(2\right)$