Question

Let $g(x)$ be a non-constant twice differentiable function defined on $R$ (the set of all real numbers) such that $y=g(x)$ is symmetric about the line $x=2$ and $g(-2)=g^{\prime }\left(\frac{1}{2}\right)=g^{\prime }(1)=0$.
If $I_1=\underset{-\pi }{\overset{\pi }{\int }}g(2+x)\sin xdx$ and $I_2=\underset{0}{\overset{4}{\int }}\frac{1}{1+e^{g^{\prime }(x)}}dx$ then which one of the following must hold good?

• A. $I_1<I_2$
• B. $I_1>I_2$
• C. $I_1=I_2$
• D. Nothing definite can be said

Answer 1

Answer: (A)

We have $g(2-x)=g(2+x)$
Now, replacing $x$ by $(2-x)$, we get
$g(x)=g(4-x)\Rightarrow g^{\prime }(x)=-g^{\prime }(4-x)$
Now, $I_1=\underset{-\pi }{\overset{\pi }{\int }}g(2+x)\sin xdx$
Using King,
$I_1=\underset{-\pi }{\overset{\pi }{\int }}g(2-x)(-\sin x)dx$
On adding $2I_1=\underset{-\pi }{\overset{\pi }{\int }}(g(2+x)-g(2-x))(\sin x)dx=0($ As $g(2-x)=g(2+x)\forall x\in R)$ Hence $I_1=0$
Also, let $I_2=\underset{0}{\overset{4}{\int }}\frac{1}{1+e^{g^{\prime }(x)}}dx$....(1)
$=\underset{0}{\overset{4}{\int }}\frac{1}{1+e^{g^{\prime }(4-x)}}dx\text{ (Applying King property) }$
$\therefore I_2=\underset{0}{\overset{4}{\int }}\frac{1}{1+e^{-g^{\prime }(x)}}dx$....(2)
$\left({\mathrm{Asg}}^{\prime }(x)=g(4-x)\forall x\in R\right)$
So, on adding (1) and (2), we get
$2I_2=\underset{0}{\overset{4}{\int }}\left(\frac{1}{1+e^{g^{\prime }(x)}}+\frac{1}{1+e^{-g^{\prime }(x)}}\right)dx\Rightarrow 2I_2=4\Rightarrow I_2=2$
Hence, $I_2>I_1$.