JEE Advanced 2017 · Paper 1 · Mathematics Q9

MathematicsInteger+3Question 45 of 54

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0) = 0$ , $f\left(\frac{\pi}{2}\right) = 3$ and $f'(0) = 1$ . If

$g(x) = \int_x^{\frac{\pi}{2}} \left[f'(t)\operatorname{cosec} t - \cot t\operatorname{cosec} t\, f(t)\right] dt$

for $x \in \left(0, \frac{\pi}{2}\right]$ , then $\lim_{x \to 0} g(x) =$

Your answer

Answer and worked solution

Official answer: 2

Spot the derivative.
$\frac d{dt}\left(f(t)\operatorname{cosec}t\right)=f'(t)\operatorname{cosec}t-f(t)\cot t\operatorname{cosec}t.$

Integrate.
$g(x)=\frac{f(\pi/2)}{\sin(\pi/2)}-\frac{f(x)}{\sin x}=3-\frac{f(x)}{\sin x}.$

Limit. As $x\to0$ , $\dfrac{f(x)}{\sin x}\to\dfrac{f'(0)}1=1$ (L'Hôpital, since $f(0)=0$ ). So $g\to2$ .

Answer: 2

Topics:Fundamental theorem of calculus Limits

JEE Advanced 2017 · Paper 1, Mathematics question 9