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JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Limits XII

lim_{x \to \frac{π}{4}} \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}}

equals

(A) $\frac{8}{π} f (2)$ ✓ Correct
(B) $\frac{2}{π} f (2)$
(C) $\frac{2}{π} f (\frac{1}{2})$
(D) $4 f (2)$

Answer: (A)

Solution

This is a

\frac{0}{0}

form, so we apply L'Hôpital's rule:

x \to \frac{π}{4} lim \frac{\int _{2}^{s e c^{2} x} f ( t ) d t}{x ^{2} - \frac{π ^{2}}{16}} = x \to \frac{π}{4} lim \frac{f ( sec ^{2} x ) \cdot 2 sec ^{2} x tan x}{2 x}

At

x = \frac{π}{4}

:

sec^{2} (\frac{π}{4}) = 2

,

tan (\frac{π}{4}) = 1

= \frac{f ( 2 ) \cdot 2 \cdot 2 \cdot 1}{2 \cdot \frac{π}{4}} = \frac{4 f ( 2 )}{\frac{π}{2}} = \frac{8}{π} f (2)

JEE Advanced · 2007 · 08 Apr 2007 · Paper 1 · Single correct option

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