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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

The function f(x) = (1 - x + x)/(1 + x + x) is not defined at x = π...

The function f(x)=1−sin⁡x+cos⁡x1+sin⁡x+cos⁡xf\left(x\right) = \dfrac{1 - \sin x + \cos x}{1 + \sin x + \cos x} is not defined at x=πx = \pi. The value of f(π)f ( \pi ), so that f(x)f ( x ) is continuous at x=πx = \pi, is

Options

  1. A−12- \dfrac{1}{2}
  2. B12\dfrac{1}{2}
  3. C−1- 1
  4. D11

Correct answer

C. −1- 1

Step-by-step solution

lim⁡x→π(1+cos⁡x)−sin⁡x(1+cos⁡x)+sin⁡x\lim_{x \to \pi}\dfrac{( 1 + \cos x ) - \sin x}{( 1 + \cos x ) + \sin x}

=lim⁡x→π2cos⁡2x2−2sinx2cosx22cos⁡2x2+2sinx2cosx2= \lim_{x \to \pi}\dfrac{2\cos^{2}\dfrac{x}{2} - 2sin\dfrac{x}{2}cos\dfrac{x}{2}}{2\cos^{2}\dfrac{x}{2} + 2sin\dfrac{x}{2}cos\dfrac{x}{2}}

=lim⁡x→π1−tanx21+tanx2= \lim_{x \to \pi}\dfrac{1 - tan\dfrac{x}{2}}{1 + tan\dfrac{x}{2}}

=lim⁡x→πtan⁡(π4−x2)=−1= \lim_{x \to \pi}\tan \left(\dfrac{\pi}{4} - \dfrac{x}{2}\right) = - 1

Since, f(x)f ( x ) is continuous at x=πx = \pi

∴f(π)=lim⁡x→πf(x)=−1∴ f\left(\pi\right) = \lim_{x \to \pi}f\left(x\right) = - 1
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