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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2019 · 12 January · Morning Shift · Mathematics · Limits

_ x → (π)/(4) co t ^ 3 x - x cos (x + (π)/(4)) is

\lim_{x \to \dfrac{\pi}{4}}\dfrac{co{t}^{3}x - \tan x}{cos\left(x + \dfrac{\pi}{4}\right)} is

Options

  1. A424\sqrt{2}
  2. B828\sqrt{2}
  3. C44
  4. D88

Correct answer

D. 88

Step-by-step solution

lim⁡x→π4cot⁡3x−tan⁡xcos⁡(x+π4)\lim_{x \to \dfrac{\pi}{4}}\dfrac{\cot^{3}x - \tan x}{\cos \left(x + \dfrac{\pi}{4}\right)}

=lim⁡x→π4(1tan⁡3x−tan⁡x)(12cos⁡x−12sin⁡x)[∵cotθ=1tanθ]= \lim_{x \to \dfrac{\pi}{4}}\dfrac{\left(\dfrac{1}{\tan^{3}x} - \tan x\right)}{\left(\dfrac{1}{\sqrt{2}}\cos x - \dfrac{1}{\sqrt{2}}\sin x\right)}\left[∵ cot\theta = \dfrac{1}{tan\theta}\right]

=lim⁡x→π4(1−tan4x)tan3x.(12cos⁡x−12sin⁡x)= \lim_{x \to \dfrac{\pi}{4}}\dfrac{\left(1 - ta{n}^{4}x\right)}{ta{n}^{3}x . \left(\dfrac{1}{\sqrt{2}}\cos x - \dfrac{1}{\sqrt{2}}\sin x\right)}

=lim⁡x→π4(1−tan⁡2x)(1+tan⁡2x)tan⁡3x.(12cos⁡x−12sin⁡x)= \lim_{x \to \dfrac{\pi}{4}}\dfrac{\left(1 - \tan^{2}x\right)\left(1 + \tan^{2}x\right)}{\tan^{3}x . \left(\dfrac{1}{\sqrt{2}}\cos x - \dfrac{1}{\sqrt{2}}\sin x\right)}

=lim⁡x→π42[(1−tan⁡x)(1+tan⁡x)sec⁡2x]tan3x.(cos⁡x−sin⁡x)[∵1+tan⁡2x=sec⁡2x and a2−b2=(a−b)(a+b)]= \lim_{x \to \dfrac{\pi}{4}}\dfrac{\sqrt{2}\left[\left(1 - \tan x\right)\left(1 + \tan x\right)\sec^{2}x\right]}{ta{n}^{3}x . \left(\cos x - \sin x\right)}\left[∵ 1 + \tan^{2}x = \sec^{2}x \text{ and } {a}^{2} - {b}^{2} = \left(a - b\right)\left(a + b\right)\right]

=lim⁡x→π42[(1−sin⁡xcos⁡x)(1+tan⁡x)sec⁡2x]tan⁡3x.(cos⁡x−sin⁡x)= \lim_{x \to \dfrac{\pi}{4}}\dfrac{\sqrt{2}\left[\left(1 - \dfrac{\sin x}{\cos x}\right)\left(1 + \tan x\right)\sec^{2}x\right]}{\tan^{3}x . \left(\cos x - \sin x\right)}

=lim⁡x→π42(cos⁡x−sin⁡x).(1+tan⁡x).sec2xcos⁡x.tan3x.(cos⁡x−sin⁡x)= \lim_{x \to \dfrac{\pi}{4}}\dfrac{\sqrt{2}\left(\cos x - \sin x\right) . \left(1 + \tan x\right) . se{c}^{2}x}{\cos x . ta{n}^{3}x . \left(\cos x - \sin x\right)}

=lim⁡x→π42(1+tan⁡x)cos⁡x.sin⁡3xcos⁡3x.cos⁡2x= \lim_{x \to \dfrac{\pi}{4}}\dfrac{\sqrt{2}\left(1 + \tan x\right)}{\cos x . \dfrac{\sin^{3}x}{\cos^{3}x} . \cos^{2}x}

=lim⁡x→π42(1+tan⁡x)sin⁡3x= \lim_{x \to \dfrac{\pi}{4}}\dfrac{\sqrt{2}\left(1 + \tan x\right)}{\sin^{3}x}

=2(1+tanπ4)sin⁡3π4= \dfrac{\sqrt{2}\left(1 + tan\dfrac{\pi}{4}\right)}{\sin^{3}\dfrac{\pi}{4}}

=8= 8.

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