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JEE Main 202117 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
17 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2021 · 17 March · Morning Shift · Mathematics · Limits

The value of _ x → 0 ^ + ^ - 1 (x - [ x ] ^ 2 ) · ^ - 1 (x - [ x ]...

The value of \lim_{x \to {0}^{+}}\dfrac{\cos^{- 1}\left(x - [ x{]}^{2}\right) \cdot \sin^{- 1}\left(x - [ x{]}^{2}\right)}{x - {x}^{3}},where[x]where \left[x\right] denotes the greatest integer ≤x\le x is:

Options

  1. Aπ\pi
  2. B00
  3. Cπ4\dfrac{\pi}{4}
  4. Dπ2\dfrac{\pi}{2}

Correct answer

D. π2\dfrac{\pi}{2}

Step-by-step solution

Given L=lim⁡x→0+cos⁡−1(x−[x]2)⋅sin⁡−1(x−[x]2)x−x3L = \lim_{x \to {0}^{+}}\dfrac{\cos^{- 1}\left(x - [ x{]}^{2}\right) \cdot \sin^{- 1}\left(x - [ x{]}^{2}\right)}{x - {x}^{3}}

∵x∈[0,1)⇒[x]=0∵ x \in [ 0, 1 ) ⇒ \left[x\right] = 0

Then L=lim⁡x→0+cos⁡−1x(1−x2)×lim⁡x→0+sin⁡−1xxL = \lim_{x \to {0}^{+}}\dfrac{\cos^{- 1}x}{\left(1 - {x}^{2}\right)} \times \lim_{x \to {0}^{+}}\dfrac{\sin^{- 1}x}{x}

L=π2×lim⁡x→0+11−x2(Apply L’Hospital rule for00form)L = \dfrac{\pi}{2} \times \lim_{x \to {0}^{+}}\dfrac{1}{\sqrt{1 - {x}^{2}}}\left(\text{Apply L'Hospital rule for}\dfrac{0}{0}\text{form}\right)

L=π2L = \dfrac{\pi}{2}

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