JEE Advanced 2008 Paper 1Question 1
Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty,\infty)$ such that $f(x)=f(1-x)$ and $f'\left(\frac14\right)=0$. Then,
Differentiation and its Application
Solution
Concept. The graph is symmetric about $x=\frac12$. Differentiate the identity and use Rolle; use the symmetry in the integrals. Formulas used. (1) $f(x)=f(1-x)\Rightarrow f'(x)=-f'(1-x)$(2) Rolle's theoremStep 1 - (B). By (1) at $x=\frac12$: $f'\left(\frac12\right)=-f'\left(\frac12\right)$, so $f'\left(\frac12\right)=0$. True. Step 2 - (A). By (1), $f'\left(\frac34\right)=-f'\left(\frac14\right)=0$. So $f'$ vanishes at $\frac14,\frac12,\frac34$. By (2), $f''$ vanishes in $\left(\frac14,\frac12\right)$ and in $\left(\frac12,\frac34\right)$. True. Step 3 - (C). $f\left(\frac12+x\right)=f\left(\frac12-x\right)$, so $f\left(x+\frac12\right)$ is even and $f\left(x+\frac12\right)\sin x$ is odd. Its integral over $\left[-\frac12,\frac12\right]$ is $0$. True. Step 4 - (D). Put $t=1-u$ on the right: $$\int_{1/2}^{1}f(1-t)e^{\sin\pi t}\,dt=\int_0^{1/2}f(u)e^{\sin\pi(1-u)}\,du=\int_0^{1/2}f(u)e^{\sin\pi u}\,du.$$ True. Answer: (A), (B), (C), (D)