Question

Let $f\left(x\right)$ be a differentiable function satisfying $f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}\forall x,y\in R$, where $f\left(0\right)=0$ and $I=\underset{0}{\overset{2\pi }{\int }}{\left(f\right(x)-\sin x)}^{2}dx.$
Which of the following is/are correct?

• A. When $I$ is minimum, then $f(–4{\pi }^2)=3$
• B. $f\left(x\right)$ is a odd function
• C. $I$ is constant.
• D. When $I$ is minimum, then $f(–4{\pi }^2)=-3$

Answer 1

Answer: (A), (B)

$f\left(x\right)$ is a odd function

Given : $f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}\cdots \left(1\right)$
We know that
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(\frac{2x+2h}{2}\right)-f\left(x\right)}{h}\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{f\left(2x\right)+f\left(2h\right)}{2}-f\left(x\right)}{h}$
Putting $x\to 2x,y=0$ in equation $\left(1\right)$, we get
$f\left(x\right)=\frac{f\left(2x\right)}{2}$,
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\frac{f\left(2x\right)+f\left(2h\right)}{2}-\frac{f\left(2x\right)}{2}}{h}\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(2h\right)}{2h}\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(2h\right)-f\left(0\right)}{2h}\Rightarrow f^{\prime }\left(x\right)=f^{\prime }\left(0\right)=k\Rightarrow f\left(x\right)=kx+c$
As $f\left(0\right)=0$, so
$f\left(x\right)=kx$
Which is odd function.

Now,
$I=\underset{0}{\overset{2\pi }{\int }}{(kx-\sin x)}^{2}dx\Rightarrow I=\underset{0}{\overset{2\pi }{\int }}(k^2{x}^2-2kx\sin x+{\sin }^{2}x)dx\Rightarrow I=k^2{\left[\frac{x^3}{3}\right]}_0^{2\pi }-2k\underset{0}{\overset{2\pi }{\int }}x\sin xdx+\underset{0}{\overset{2\pi }{\int }}{\sin }^{2}xdx\Rightarrow I=\frac{8{\pi }^3{k}^2}{3}-2k\underset{0}{\overset{2\pi }{\int }}x\sin xdx+\pi$
Now,
$I_1=\underset{0}{\overset{2\pi }{\int }}x\sin xdx\Rightarrow I_1=[-x\cos x+\sin x{]}_0^{2\pi }\Rightarrow I_1=-2\pi$

Therefore,
$I=\left(\frac{8{\pi }^3}{3}\right)k^2+\left(4\pi \right)k+\pi$
This is a quadratic in $k$ and its minimum value occurs when
$k=\frac{-4\pi \cdot 3}{16{\pi }^3}=-\frac{3}{4{\pi }^2}$
So,
$f\left(x\right)=-\frac{3x}{4{\pi }^2}\therefore f(–4{\pi }^2)=3$