Question

Let $g$ be a differentiable function satisfying $\int\limits_0^x(x-t+1) g(t) d t=x^4+x^2$ for all $x \geq 0$. The value of $\int\limits_0^1 \frac{12}{g^{\prime}(x)+g(x)+10} d x$ is equal to

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

Answer 1

Answer: (C)

$ x \int\limits_0^x g(t) d t+\int\limits_0^x(1-t) g(t) d t=x^4+x^2$
differentiate w.r.t. $x$
$x g(x)+\int\limits_0^x g(t) d t+(1-x) g(x)=4 x^3+2 x$
Again differentiate w.r.t $x$
$g^{\prime}(x)+g(x)=12 x^2+2$
Now, $\left. \int\limits_0^1 \frac{12}{g^{\prime}+g+10} d x=\int\limits_0^1 \frac{d x}{x^2+1}=\tan ^{-1} x\right]_0^1=\frac{\pi}{4}$.