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Question: The function \(f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}\) is not defined at \(x = \pi....

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

A

−12- \frac{1}{2}

B

12\frac{1}{2}

C

– 1

D

1

Answer

– 1

Explanation

Solution

lim⁡x→πf(x)=lim⁡x→π2cos⁡2x2−2sin⁡x2cos⁡x22cos⁡2x2+2sin⁡x2cos⁡x2\lim_{x \rightarrow \pi}f(x) = \lim_{x \rightarrow \pi}\frac{2\cos^{2}\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^{2}\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}}

=lim⁡x→πcos⁡x2−sin⁡x2cos⁡x2+sin⁡x2=lim⁡x→πtan⁡(π4−x2)= \lim_{x \rightarrow \pi}\frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}} = \lim_{x \rightarrow \pi}\tan\left( \frac{\pi}{4} - \frac{x}{2} \right) x2−1>0x ^ { 2 } - 1 > 0 At

x=π,f(π)=−tan⁡π4=−1x = \pi,f(\pi) = - \tan\frac{\pi}{4} = - 1.