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JEE Advanced 2017Paper 1
Exam
JEE Advanced
Year
2017
Paper
Paper 1
Subject
Mathematics
Chapter
Definite Integration

JEE Advanced 2017 · Paper 1 · Mathematics · Definite Integration

Let f: ℝ → ℝ be a differentiable function such that f(0) = 0,...

Let f:R→Rbeadifferentiablefunctionsuchthatf: \mathbb{R} \to \mathbb{R} be a differentiable function such thatf\left(0\right) = 0, f\left(\dfrac{\pi}{2}\right) = 3andand{f}^{′}\left(0\right) = 1 .IfIfg\left(x\right) = {∫}_{x}^{\dfrac{\pi}{2}}\left[{f}^{′}\left(t\right)cosect - \cot t \cos ectf\left(t\right)\right]dt,for, forx \in (0,π2]\left(0, \dfrac{\pi}{2}\right] then lim⁡x→0g(x)=\lim_{x \to 0} g\left(x\right) =

Correct answer

2

Step-by-step solution

g(x)=∫xπ2[f′(t)cosect−f(t)cosectcott]dtg( x ) = {∫}_{x}^{\dfrac{\pi}{2}}[ {f}^{'}( t )cosect - f( t )cosect\text{cot}t ]dt

=[f(t)cosect]xπ2= {\left[f ( t ) cosect\right]}_{x}^{\dfrac{\pi}{2}}

=f(π2)cosec(π2)−f(x)sinx=3−f(x)sinx= f\left(\dfrac{\pi}{2}\right)cosec\left(\dfrac{\pi}{2}\right) - \dfrac{f\left(x\right)}{sin x} = 3 - \dfrac{f\left(x\right)}{sin x}

∴lim⁡x→0g(x)=3−lim⁡x→0f(x)sinx=lim⁡x→0f′(x)cosx;asf′(0)=1∴ \lim_{x \to 0} g\left(x\right) = 3 - \lim_{x \to 0} \dfrac{f\left(x\right)}{sin x} = \lim_{x \to 0} \dfrac{{f}^{'}\left(x\right)}{cos x}; as{f}^{'}\left(0\right) = 1

⇒lim⁡x→0g(x)=3−1=2⇒ \lim_{x \to 0} g\left(x\right) = 3 - 1 = 2

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