Let $f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$ be a continuous function such that $$ f(0)=1 \text { and } \int_{0}^{\frac{\pi}{3}} f(\mathrm{t}) \mathrm{dt}=0 $$ Then which of the following statements is (are) TRUE?

(A) The equation $f(x)-3 \cos 3 x=0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$; (B) The equation $f(x)-3 \sin 3 x=-\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$; (C) $\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{x} \int_{0}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}}{1-\mathrm{e}^{\mathrm{x}^{2}}}=-1$

(D) $\lim _{\mathrm{x} \rightarrow 0} \frac{\sin \mathrm{x} \int_{0}^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}}{\mathrm{x}^{2}}=-1$

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Continuity and Differentiability

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Let

f : [- \frac{π}{2}, \frac{π}{2}] \to R

be a continuous function such that

f (0) = 1 and \int_{0}^{\frac{π}{3}} f (t) dt = 0

Then which of the following statements is (are) TRUE?

(A)The equation $f (x) - 3 cos 3 x = 0$ has at least one solution in $(0, \frac{π}{3})$ ✓ Correct
(B)The equation $f (x) - 3 sin 3 x = - \frac{6}{π}$ has at least one solution in $(0, \frac{π}{3})$ ✓ Correct
(C) $lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) dt}{1 - e ^{x^{2}}} = - 1$ ✓ Correct
(D) $lim_{x \to 0} \frac{s i n x \int _{0}^{x} f ( t ) dt}{x ^{2}} = - 1$

Answer: (A), (B), (C)

Solution

(A) Let

g (x) = f (x) - 3 cos 3 x

. Then

\int_{0}^{π /3} g (x) dx = \int_{0}^{π /3} f (x) dx - 3 \int_{0}^{π /3} cos 3 x dx = 0 - 3 [sin 3 x /3]_{0}^{π /3} = 0

. By the mean value theorem for integrals, g(x) = 0 has at least one root in

(0, π /3)

. (A) is correct.

(B) Let

h (x) = f (x) - 3 sin 3 x + \frac{6}{π}

. Then

\int_{0}^{π /3} h (x) dx = 0 - 3 [- cos 3 x /3]_{0}^{π /3} + \frac{6}{π} \cdot \frac{π}{3} = 0 + 2 - 2 + 2 = 0

. So h(x) = 0 has a root in

(0, π /3)

. (B) is correct.

(C)

lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) dt}{1 - e ^{x^{2}}} = lim_{x \to 0} \frac{x ^{2}}{1 - e ^{x^{2}}} \cdot \frac{\int _{0}^{x} f ( t ) dt}{x}

. The first factor approaches

- 1

and by L'Hôpital, the second approaches

f (0) = 1

. Product is

- 1

. (C) is correct.

(D) Similarly,

lim_{x \to 0} \frac{s i n x}{x} \cdot \frac{\int _{0}^{x} f ( t ) dt}{x} = 1 \cdot 1 = 1

, not

- 1

. (D) is incorrect.

Correct answers: (A), (B), (C).

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Continuity and Differentiability

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