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JEE Main 202226 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
26 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2022 · 26 June · Morning Shift · Mathematics · Limits

_ x → 1 2 ( ^ - 1 x) - x 1 - tan ( ^ - 1 x) is equal to

\lim_{x \to \dfrac{1}{\sqrt{2}}}\dfrac{\sin \left(\cos^{- 1}x\right) - x}{1 - tan\left(\cos^{- 1}x\right)} is equal to

Options

  1. A12\dfrac{1}{\sqrt{2}}
  2. B−12\dfrac{- 1}{\sqrt{2}}
  3. C2\sqrt{2}
  4. D−1- 1

Correct answer

B. −12\dfrac{- 1}{\sqrt{2}}

Step-by-step solution

Let cos⁡−1x=t\cos^{- 1}x = t i.e. x=cos⁡tx = \cos t

So lim⁡t→π4sin⁡t−cos⁡t1−tan⁡t\lim_{t \to \dfrac{\pi}{4}}\dfrac{\sin t - \cos t}{1 - \tan t} =lim⁡t→π4(−cos⁡t)=−12= \lim_{t \to \dfrac{\pi}{4}}\left(- \cos t\right) = \dfrac{- 1}{\sqrt{2}}

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