Q33 / 69 · JEE Advanced 2008 (Paper 1) · Mathematics
Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty, \infty)$, such that $f(x)=f(1-x)$ and $f^{\prime}\left(\frac{1}{4}\right)=0$. Then,
- Correct answerA$f^{\prime}(x)$ vanishes atleast twice on $[0,1]$
- Correct answerB$f^{\prime}\left(\frac{1}{2}\right)=0$
- Correct answerC$\int_{\frac{1}{2}}^{\frac{1}{2}} f\left(x+\frac{1}{2}\right) \sin x d x=0$
- Correct answerD$\int_0^{\frac{1}{2}} f(t) e^{\sin \pi t} d t=\int_{\frac{1}{2}}^1 f(1-t) e^{\sin \pi t} d t$
Solution
Given that, $f(x)=f(1-x)$
On differentiating w.r.t. $x$, we get
$$
f^{\prime}(x)=-f^{\prime}(1-x)
$$
Let us put $\quad x=\frac{1}{2}$
$$
\Rightarrow \quad 2 f^{\prime}\left(\frac{1}{2}\right)=0 \Rightarrow f^{\prime}\left(\frac{1}{2}\right)=0
$$
Since, $f^{\prime}\left(\frac{1}{2}\right)=0$ and $f^{\prime}\left(\frac{1}{4}\right)=0$
$\Rightarrow f^{\prime \prime}(x)=0$ at two points in $[0,1]$.
Now, $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$
As, $f\left(x+\frac{1}{2}\right) \sin x$ is an odd function which is clear from the following explanation.
Let $g(x)=f\left(x+\frac{1}{2}\right) \sin x$,
$$
g(-x)=f\left(\frac{1}{2}-x\right) \sin (-x)=-\sin x f\left(1-\left(\frac{1}{2}-x\right)\right)=-\sin x f\left(\frac{1}{2}+x\right)=-g(x)
$$
Moreover, $\int_{1 / 2}^1 f(1-t) e^{\sin (\pi t)} d t=\int_0^{1 / 2} f(u) \cdot e^{\sin \pi u} d u$ where, $1-t=u$.
Hence, the correct answer is options: 'A', 'B', 'C', 'D'.