MHT CET · Maths · Continuity and Differentiability
If \(f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}\), for \(x \neq \pi\) is continuous at \(x=\pi\), then \(f(\pi)=\)
- A \(-1\)
- B 2
- C 0
- D 1
Answer & Solution
Correct Answer
(A) \(-1\)
Step-by-step Solution
Detailed explanation
(A)
\(f\) is continuous at \(x=\pi\)
\(\begin{aligned} f(\pi) &=\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{1-\sin x+\cos x}{1+\sin x+\cos x} \\ &=\lim _{x \rightarrow \pi} \frac{-\cos x-\sin x}{\cos x-\sin x} \\ &=\frac{-\cos \pi-\sin \pi}{\cos \pi-\sin \pi}=\frac{-(-1)-0}{-1-0}=\frac{1}{-1}=-1 \end{aligned}\)
\(f\) is continuous at \(x=\pi\)
\(\begin{aligned} f(\pi) &=\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{1-\sin x+\cos x}{1+\sin x+\cos x} \\ &=\lim _{x \rightarrow \pi} \frac{-\cos x-\sin x}{\cos x-\sin x} \\ &=\frac{-\cos \pi-\sin \pi}{\cos \pi-\sin \pi}=\frac{-(-1)-0}{-1-0}=\frac{1}{-1}=-1 \end{aligned}\)
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