Question

Let $g$ be a differentiable function satisfying $\underset{0}{\overset{x}{\int }}(x-t+1)g(t)dt=x^4+x^2$ for all $x\ge 0$. The value of $\underset{0}{\overset{1}{\int }}\frac{12}{g^{\prime }(x)+g(x)+10}dx$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

$x\underset{0}{\overset{x}{\int }}g(t)dt+\underset{0}{\overset{x}{\int }}(1-t)g(t)dt=x^4+x^2$
differentiate w.r.t. $x$
$xg(x)+\underset{0}{\overset{x}{\int }}g(t)dt+(1-x)g(x)=4x^3+2x$
Again differentiate w.r.t $x$
$g^{\prime }(x)+g(x)=12x^2+2$
Now, ${\underset{0}{\overset{1}{\int }}\frac{12}{g^{\prime }+g+10}dx=\underset{0}{\overset{1}{\int }}\frac{dx}{x^2+1}={\tan }^{-1}x]}_0^1=\frac{\pi }{4}$.