IIT-JEE 2008 · Paper 1 · Mathematics Q10

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MathematicsMultiple correct+4 / -1Question 10 of 69

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,

f''(x)

vanishes at least twice on

[0,1]

f'\left(\frac12\right)=0

\displaystyle\int_{-1/2}^{1/2}f\left(x+\frac12\right)\sin x\,dx=0

\displaystyle\int_0^{1/2}f(t)\,e^{\sin\pi t}\,dt=\int_{1/2}^1f(1-t)\,e^{\sin\pi t}\,dt

Answer and worked solution

Official answer: A, B, C, D

Symmetry. $f(x)=f(1-x)$ gives $f'(x)=-f'(1-x)$ .

(B) At $x=\frac12$ : $f'(\frac12)=-f'(\frac12)$ , so $f'(\frac12)=0$ . True.
(A) Also $f'(\frac34)=-f'(\frac14)=0$ . So $f'$ vanishes at $\frac14$ , $\frac12$ , $\frac34$ , and by Rolle's theorem $f''$ vanishes at least once in each of $(\frac14,\frac12)$ and $(\frac12,\frac34)$ . True.
(C) $f\left(x+\frac12\right)=f\left(\frac12-x\right)$ is even in $x$ , and $\sin x$ is odd, so the integrand is odd and the integral is 0. True.
(D) In the right-hand integral put $t=1-u$ : it becomes $\int_0^{1/2}f(u)e^{\sin\pi(1-u)}du=\int_0^{1/2}f(u)e^{\sin\pi u}du$ , the left-hand side. True.

(A), (B), (C), (D)

Topics:Rolle's and mean value theorems Properties of definite integrals

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IIT-JEE 2008 · Paper 1, Mathematics question 10