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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Mathematics
Chapter
Limits

BITSAT 2019 · Mathematics · Limits

The value of _ x → (3π)/(4) 4 ^ 2 xcosx - x + x x + x is

The value of lim⁡x→3π44sin⁡2xcosx−cos⁡x+sin⁡xsin⁡x+cos⁡x\lim_{x \to \dfrac{3\pi}{4}}\dfrac{4\sin^{2}xcosx - \cos x + \sin x}{\sin x + \cos x} is

Options

  1. A−1- 1
  2. B00
  3. C11
  4. DNone of these

Correct answer

A. −1- 1

Step-by-step solution

L=lim⁡x→3π44sin⁡2xcosx−cos⁡x+sin⁡xsin⁡x+cos⁡xL = \lim_{x \to \dfrac{3\pi}{4}}\dfrac{4\sin^{2}xcosx - \cos x + \sin x}{\sin x + \cos x}

L=lim⁡x→3π42sin⁡x(sin⁡2x+cos⁡2x+2sin⁡xcos⁡x−1)−cos⁡x+sin⁡xsin⁡x+cos⁡xL = \lim_{x \to \dfrac{3\pi}{4}}\dfrac{2\sin x\left(\sin^{2}x + \cos^{2}x + 2\sin x \cos x - 1\right) - \cos x + \sin x}{\sin x + \cos x}

L=lim⁡x→3π42sin⁡x[(sin⁡x+cos⁡x)2−1]−cos⁡x+sin⁡xsin⁡x+cos⁡xL = \lim_{x \to \dfrac{3\pi}{4}}\dfrac{2\sin x\left[( \sin x + \cos x{)}^{2} - 1\right] - \cos x + \sin x}{\sin x + \cos x}

L=lim⁡x→3π42sin⁡x(sin⁡x+cos⁡x)2−2sin⁡x−cos⁡x+sin⁡xsin⁡x+cos⁡xL = \lim_{x \to \dfrac{3\pi}{4}}\dfrac{2\sin x ( \sin x + \cos x{)}^{2} - 2\sin x - \cos x + \sin x}{\sin x + \cos x}

L=lim⁡x→3π4(sin⁡x+cos⁡x){2sin⁡x(sin⁡x+cos⁡x)−1}sin⁡x+cos⁡xL = \lim_{x \to \dfrac{3\pi}{4}}\dfrac{\left(\sin x + \cos x\right)\left\{2\sin x\left(\sin x + \cos x\right) - 1\right\}}{\sin x + \cos x}

L=lim⁡x→3π42sin⁡x(sin⁡x+cos⁡x)−1=−1L = \lim_{x \to \dfrac{3\pi}{4}}2\sin x\left(\sin x + \cos x\right) - 1 = - 1

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