Question

Let f : R → R be a differentiable function such that
$f(\frac{π}{4})=\sqrt2,f(\frac{π}{2})=0$ and $f′(\frac{π}{2})=1$
and let
$g(x) = \int_{x}^{\frac{\pi}{4}} \left(f'(t)\sec(t) + \tan(t)\sec(t)f(t)\right) \, dt$
for$x∈(\frac{π}{4},\frac{π}{2})$ Then $\lim_{{x \to \frac{\pi}{2}^-}} g(x)$is equal to

• A. 2
• B. 3
• C. 4
• D. -3

Answer 1

Answer: (B)

3

Answer 2

The correct answer is (B) : 3
Given :
$f(\frac{π}{4})=\sqrt2,f(\frac{π}{2})=0$
and
$f′(\frac{π}{2})=1$
$g(x) = \int_{x}^{\frac{\pi}{4}} \left(f'(t) \sec(t) + \tan(t) \tan(t) \sec(t) f(t)\right) \, dt$
=$[\sec(t) + f(t)]_{x}^{\frac{\pi}{4}} = 2 - \sec(x) \cdot f(x)$
Now,
$\lim_{{x \to \frac{\pi}{2}^-}}g(x) = \lim_{{h \to 0}} g\left(\frac{\pi}{2} - h\right) = \lim_{{h \to 0}} \left[2 -cosec(h)) \cdot f\left(\frac{\pi}{2} - h\right)\right]$
$\lim_{{h \to 0}} \left[2 - \frac{f\left(\frac{\pi}{2} - h\right)} {\sin(h)}\right] = \lim_{{h \to 0}} \left[2 + \frac{f'\left(\frac{\pi}{2} - h\right)}{ \cos(h)}\right] = 3$