$\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^{3}}^{(\pi / 2)^{3}}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^{2}}\right)$ is equal to :

Limits — JEE Main · 2024 · 09 Apr 2024 · Shift 2 Question with Solution

lim_{x \to \frac{π}{2}} (\frac{\int _{x^{3}}^{(π /2)^{3}} ( s i n ( 2 t ^{1/3} ) + c o s ( t ^{1/3} ) ) d t}{( x - \frac{π}{2} ) ^{2}})

is equal to :

Answer: (A)

Let

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

. As

x \to π /2

F (x) \to 0

and

(x - \frac{π}{2})^{2} \to 0

: a

\frac{0}{0}

form.

F^{'} (x) = - 3 x^{2} [sin (2 x) + cos x]

(differentiating the lower limit, with

g (x^{3}) = sin (2 x) + cos x

after substituting

t^{1/3} = x

).

At

x = π /2

sin (2 x) = sin π = 0

cos x = 0

, so

F^{'} (π /2) = 0

— still

\frac{0}{0}

against the denominator's derivative

2 (x - \frac{π}{2})

, which is also

0

x = π /2

. Differentiate again:

F^{''} (x) = - 6 x [sin 2 x + cos x] - 3 x^{2} [2 cos 2 x - sin x]

x = π /2

: first term

= - 6 \cdot \frac{π}{2} \cdot (0 + 0) = 0

; second term

= - 3 (\frac{π}{2})^{2} [2 (- 1) - 1] = - 3 \cdot \frac{π ^{2}}{4} \cdot (- 3) = \frac{9 π ^{2}}{4}

.

So

F^{''} (π /2) = \frac{9 π ^{2}}{4}

, and the denominator's second derivative is

2

(constant). By L'Hôpital's rule applied twice:

x \to π /2 lim \frac{F ( x )}{( x - π /2 ) ^{2}} = \frac{F ^{''} ( π /2 )}{2} = \frac{9 π ^{2} /4}{2} = \frac{9 π ^{2}}{8}

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option