IIT JEE Advanced 2008 Paper 1

We have f(x) = f(1 – x) (1) Substituting x = $\frac{1}{2} + x$ , we get f $\left(\frac{1}{2} + x\right)$ = f $\left(\frac{1}{2} - x\right)$ That is,f $\left(\frac{1}{2} + x\right)$ is an even function and sin x . f $\left(\frac{1}{2} + x\right)$ is an odd function. Therefore, $\int\limits_{-\frac{1}{2}}^{\frac{1}{2}} f\left(x+\frac{1}{2}\right)$ sin x = 0 Differentiating Eq. (1), we get f ′(x) = - f ′(1- x) (2) Substituting x = 1 / 2, we get $f'\left(\frac{1}{2}\right)$ = - $f'\left(\frac{1}{2}\right)$ ⇒ $f'\left(\frac{1}{2}\right)$ = 0 Now, substituting x = 1/4in Eq. (2), we get $f'\left(\frac{1}{4}\right)$ = $-f'\left(\frac{3}{4}\right)$ = 0 ⇒ $f'\left(\frac{1}{4}\right)$ = $f'\left(\frac{1}{2}\right)$ = $f'\left(\frac{3}{4}\right)$ Using Rolle’s theorem, f ″(x) = 0 has at least one solution in $\left(\frac{1}{4}, \frac{1}{2}\right)$ and one solution in $\left(\frac{1}{2}.\frac{3}{4}\right)$. Therefore, f ′′(x) = 0 vanishes at least twice on [0, 1]. Now, $\int\limits_{0}^{\frac{1}{2}} f(t) e^{\sin \pi t} \, dt$ Substituting t = 1 – x, we get $\int\limits_{1}^{\frac{1}{2}} f(1-x) e^{\sin \pi (1-x)}$ (- dx) = $\int\limits_{\frac{1}{2}}^{1} f(1-x) e^{\sin \pi x}$ . dx