Let $f:(0, \pi) \rightarrow \mathbb{R}$ be a twice differentiable function such that \[ \lim _{t \rightarrow x} \frac{f(x) \sin t-f(t) \sin x}{t-x}=\sin ^{2} x \text { for all } x \in(0, \pi) . \] If $f\left(\frac{\pi}{6}\right)=-\frac{\pi}{12}$, then which of the following statement(s) is (are) TRUE?

(B) $f(x)<\frac{x^{4}}{6}-x^{2}$ for all $x \in(0, \pi)$; (C) There exists $\alpha \in(0, \pi)$ such that $f^{\prime}(\alpha)=0$; (D) $f^{\prime \prime}\left(\frac{\pi}{2}\right)+f\left(\frac{\pi}{2}\right)=0$

(A) $f\left(\frac{\pi}{4}\right)=\frac{\pi}{4 \sqrt{2}}$

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Differential Equations

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Let

f : (0, π) \to R

be a twice differentiable function such that

t \to x lim \frac{f ( x ) sin t - f ( t ) sin x}{t - x} = sin^{2} x for all x \in (0, π) .

f (\frac{π}{6}) = - \frac{π}{12}

, then which of the following statement(s) is (are) TRUE?

(A) $f (\frac{π}{4}) = \frac{π}{4 2}$
(B) $f (x) < \frac{x ^{4}}{6} - x^{2}$ for all $x \in (0, π)$ ✓ Correct
(C)There exists $α \in (0, π)$ such that $f^{'} (α) = 0$ ✓ Correct
(D) $f^{''} (\frac{π}{2}) + f (\frac{π}{2}) = 0$ ✓ Correct

Answer: (B), (C), (D)

Solution

Using L'Hôpital's rule:

t \to x lim \frac{f ( x ) cos t - f ^{'} ( t ) sin x}{1} = sin^{2} x

\Rightarrow f (x) cos x - f^{'} (x) sin x = sin^{2} x

This is a first-order linear ODE. Dividing by

sin x

\frac{d}{d x} (\frac{f ( x )}{sin x}) = - 1

\Rightarrow \frac{f ( x )}{sin x} = - x + c

\Rightarrow f (x) = - x sin x + c sin x

Using

f (\frac{π}{6}) = - \frac{π}{12}

- \frac{π}{12} = - \frac{π}{6} \cdot \frac{1}{2} + c \cdot \frac{1}{2} \Rightarrow c = 0

Thus

f (x) = - x sin x

.

(A):

f (\frac{π}{4}) = - \frac{π}{4} \cdot \frac{1}{2} = - \frac{π}{4 2} \neq = \frac{π}{4 2}

✗

(B), (C), (D) can be verified using

f (x) = - x sin x

. ✓

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JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Differential Equations

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