JEE Advanced 2007 Paper 1Question 1
$$\lim_{x\to\frac{\pi}{4}}\frac{\int_{2}^{\sec^2 x} f(t)\,dt}{x^2-\frac{\pi^2}{16}}$$ equals
Limits & Continuity
Solution
Concept. At $x=\frac\pi4$, $\sec^2x=2$ (integral $=0$) and $x^2-\frac{\pi^2}{16}=0$: a $\frac00$ form. Formulas used. (1) $\frac{d}{dx}\int_2^{u(x)}f(t)\,dt=f(u)\,u'(x)$(2) $\frac{d}{dx}\sec^2x=2\sec^2x\tan x$(3) L'Hospital for $\frac00$Step 1 - Differentiate top (1), (2). $f(\sec^2x)\cdot2\sec^2x\tan x$. Step 2 - Differentiate bottom. $2x$. Step 3 - Put $x=\frac\pi4$ (3). $\sec^2\frac\pi4=2$, $\tan\frac\pi4=1$: $$\frac{f(2)\cdot2\cdot2\cdot1}{2\cdot\frac\pi4}=\frac{4f(2)}{\pi/2}=\frac8\pi f(2)$$ Answer: A