JEE Advanced 2018 · Paper 2 · Mathematics Q6

← JEE Advanced 2018 · Paper 2

MathematicsMultiple correct+4 / -2Question 42 of 54

Let $f: (0, \pi) \to \mathbb{R}$ be a twice differentiable function such that

$\lim_{t \to x} \frac{f(x)\sin t - f(t)\sin x}{t - x} = \sin^2 x \quad \text{for all } x \in (0, \pi).$

If $f\left(\frac{\pi}{6}\right) = -\frac{\pi}{12}$ , then which of the following statement(s) is (are) TRUE?

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

f(x) < \frac{x^4}{6} - x^2

for all

x \in (0, \pi)

There exists

\alpha \in (0, \pi)

such that

f'(\alpha) = 0

f''\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = 0

Answer and worked solution

Official answer: B, C, D

The limit is a derivative. Differentiate the numerator in $t$ at $t=x$ :
$f(x)\cos x-f'(x)\sin x=\sin^2x\implies\left(\frac f{\sin x}\right)'=\frac{f'\sin x-f\cos x}{\sin^2x}=-1.$

Solve. $\dfrac f{\sin x}=-x+C$ . From $f(\pi/6)=-\pi/12$ : $\dfrac{-\pi/12}{1/2}=-\dfrac\pi6$ , so $C=0$ . Hence $f(x)=-x\sin x$ .

(A) $f(\pi/4)=-\dfrac\pi{4\sqrt2}$ , negative. False.
(B) $-x\sin x<\dfrac{x^4}6-x^2\iff x-\sin x<\dfrac{x^3}6$ . This holds for $x>0$ because $\sin x>x-\dfrac{x^3}6$ . True.
(C) $f'(x)=-\sin x-x\cos x=0\iff\tan x=-x$ , which has a root in $(\pi/2,\pi)$ . True.
(D) $f''=-2\cos x+x\sin x$ , so $f''(\pi/2)=\pi/2$ . With $f(\pi/2)=-\pi/2$ the sum is 0. True.

(B), (C), (D)

Topics:Linear first-order equations Limits Rolle's and mean value theorems

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