$\lim_{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos \left(\frac{1}{t^{3}}\right) dt\right)$ is equal to

Limits — JEE Main · 2024 · 29 Jan 2024 · Shift 1 Question with Solution

lim_{x \to \frac{π}{2}} (\frac{1}{( x - \frac{π}{2} ) ^{2}} \int_{x^{3}}^{(\frac{π}{2})^{3}} cos (\frac{1}{t ^{3}}) d t)

is equal to

Answer: (C)

L = x \to π /2 lim \frac{1}{( x - \frac{π}{2} ) ^{2}} \int_{x^{3}}^{(π /2)^{3}} cos (t^{1/3}) d t

Let

F (x) = \int_{x^{3}}^{(π /2)^{3}} cos (t^{1/3}) d t

. By the Leibniz rule, differentiating only the lower limit:

F^{'} (x) = - cos ((x^{3})^{1/3}) \cdot \frac{d}{d x} (x^{3}) = - 3 x^{2} cos x

x = π /2

F (π /2) = 0

(equal limits) and

F^{'} (π /2) = - 3 (\frac{π}{2})^{2} cos \frac{π}{2} = 0

, so the ratio

F (x) / (x - π /2)^{2}

is a genuine

0/0

form at second order. Applying L'Hopital's rule twice:

L = x \to π /2 lim \frac{F ^{'} ( x )}{2 ( x - π /2 )} = x \to π /2 lim \frac{F ^{''} ( x )}{2}

F^{''} (x) = \frac{d}{d x} (- 3 x^{2} cos x) = - 6 x cos x + 3 x^{2} sin x

x = π /2

cos (π /2) = 0

sin (π /2) = 1

, so

F^{''} (π /2) = 3 (\frac{π}{2})^{2} (1) = \frac{3 π ^{2}}{4}

L = \frac{F ^{''} ( π /2 )}{2} = \frac{3 π ^{2} /4}{2} = \frac{3 π ^{2}}{8}

Key confirmed: C.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option