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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Continuity and Differentiability

NDA 2015 · Mathematics · Continuity and Differentiability

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

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

Options

  1. A-1/2
  2. B1/2
  3. C-1
  4. D1

Correct answer

C. -1

Step-by-step solution

lim⁡x→π1−sin⁡x+cos⁡x1+sin⁡x+cos⁡x\lim _{x \rightarrow \pi} \frac{1-\sin x+\cos x}{1+\sin x+\cos x} Using L'hospital's rule ⇒lim⁡x→π−cos⁡x−sin⁡xcos⁡x−sin⁡x⇒−cos⁡π−sin⁡πcos⁡π−sin⁡π⇒−(−1)−0−1−0\begin{aligned} & \Rightarrow \lim _{x \rightarrow \pi} \frac{-\cos x-\sin x}{\cos x-\sin x} \\ & \Rightarrow \frac{-\cos \pi-\sin \pi}{\cos \pi-\sin \pi} \\ & \Rightarrow \frac{-(-1)-0}{-1-0} \end{aligned}\Rightarrow-1$
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