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 $I _1=\int\limits_{-\pi}^\pi g (2+ x ) \sin x d x$ and $I _2=\int\limits_0^4 \frac{1}{1+ e ^{ g ^{\prime}( x )}} dx$ then which one of the following must hold good?

• A. $I_1
• 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, $I_1=\int\limits_{-\pi}^\pi g(2+x) \sin x d x$
Using King,
$I_1=\int\limits_{-\pi}^\pi g(2-x)(-\sin x) d x$
On adding $2 I_1=\int\limits_{-\pi}^\pi(g(2+x)-g(2-x))(\sin x) d x=0 ($ As $g(2-x)=g(2+x) \forall x \in R)$ Hence $I_1=0$
Also, let $I _2=\int\limits_0^4 \frac{1}{1+ e ^{ g ^{\prime}( x )}} dx$....(1)
$=\int\limits_0^4 \frac{1}{1+ e ^{ g ^{\prime}(4- x )}} dx \text { (Applying King property) }$
$\therefore I _2=\int\limits_0^4 \frac{1}{1+ e ^{- g ^{\prime}( x )}} dx$....(2)
$\left(\operatorname{Asg}^{\prime}( x )= g (4- x ) \forall x \in R \right)$
So, on adding (1) and (2), we get
$2 I _2=\int\limits_0^4\left(\frac{1}{1+ e ^{ g ^{\prime}( x )}}+\frac{1}{1+ e ^{- g ^{\prime}( x )}}\right) dx \Rightarrow 2 I _2=4 \Rightarrow I _2=2$
Hence, $I_2 >I_1$.