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JEE Main 20179 AprilOnline
Exam
JEE Main
Year
2017
Date
9 April
Mode
Online
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2017 · 9 April · Online · Mathematics · Continuity and Differentiability

The value of k which the function f(x) = \ cases ((4)/(5))( 4x)/(...

The value of kk which the function f\left(x\right) = \left\{\begin{cases}\left(\dfrac{4}{5}\right)\dfrac{\tan 4x}{\tan 5x}, & 0 < x < \dfrac{\pi}{2} \\ k + \dfrac{2}{5}, & x = \dfrac{\pi}{2}\end{cases}\right) is continuous atx = \dfrac{\pi}{2},$ is

Options

  1. A25\dfrac{2}{5}
  2. B−25- \dfrac{2}{5}
  3. C1720\dfrac{17}{20}
  4. D35\dfrac{3}{5}

Correct answer

D. 35\dfrac{3}{5}

Step-by-step solution

Given function is continuous at x=π2x = \dfrac{\pi}{2}

⇒f(π2)=f(π+2)=f(π−2)⇒ f\left(\dfrac{\pi}{2}\right) = f\left(\dfrac{{\pi}^{+}}{2}\right) = f\left(\dfrac{{\pi}^{-}}{2}\right)

⇒k+25=(45)tan⁡4xtan⁡5x⇒ k + \dfrac{2}{5} = \left(\dfrac{4}{5}\right)\dfrac{\tan 4x}{\tan 5x}

⇒k+25=(45)tan⁡(2π)cot⁡(5π2)⇒ k + \dfrac{2}{5} = {\left(\dfrac{4}{5}\right)}^{\tan \left(2\pi\right)\cot \left(\dfrac{5\pi}{2}\right)}

⇒k+25=(45)0⇒ k + \dfrac{2}{5} = {\left(\dfrac{4}{5}\right)}^{0}

⇒k+25=1⇒ k + \dfrac{2}{5} = 1

⇒k=35⇒ k = \dfrac{3}{5}

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