JEE Advance - Mathematics (2017 - Paper 1 Offline - No. 10)
Let f : R $$ \to $$ R be a differentiable function such that f(0) = 0, $$f\left( {{\pi \over 2}} \right) = 3$$ and f'(0) = 1.
If $$g(x) = \int\limits_x^{\pi /2} {[f'(t)\text{cosec}\,t - \cot t\,\text{cosec}\,t\,f(t)]dt} $$
for $$x \in \left( {0,\,{\pi \over 2}} \right]$$, then $$\mathop {\lim }\limits_{x \to 0} g(x)$$ =
If $$g(x) = \int\limits_x^{\pi /2} {[f'(t)\text{cosec}\,t - \cot t\,\text{cosec}\,t\,f(t)]dt} $$
for $$x \in \left( {0,\,{\pi \over 2}} \right]$$, then $$\mathop {\lim }\limits_{x \to 0} g(x)$$ =
Answer
2
Explanation
Let $$g(x) = \int\limits_x^{\pi /2} {[f'(t)\text{cosec}\,t - \cot t\,\text{cosec}\,t\,f(t)]dt} $$
$$ = \int\limits_x^{\pi /2} {{d \over {dt}}(f(t)\cos ect))} $$
So, $$g(x) = f(\pi /2)\cos ec{\pi \over 2} - f(x)\cos ecx$$
$$ = 3 - f(x)\cos ecx$$
$$\therefore$$ $$g(x) = 3 - {{f(x)} \over {\sin x}}$$
$$\mathop {\lim }\limits_{x \to 0} g(x) = 3 - \mathop {\lim }\limits_{x \to 0} {{f(x)} \over {\sin x}}$$
As the above is a 0/0 form, use L'Hospital's rule to get
$$\mathop {\lim }\limits_{x \to 0} g(x) = 3 - \mathop {\lim }\limits_{x \to 0} {{f'(x)} \over {\cos x}} = 3 - f'(0) = 3 - 1 = 2$$
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