MHT CET · Maths · Application of Derivatives
Values of \(\mathrm{c}\) as per Rolle's theorem for \(f(x)=\sin x+\cos x+6\) on \([0,2 \pi]\) are
- A \(\frac{\pi}{3}, \frac{5 \pi}{3}\)
- B \(\frac{\pi}{6}, \frac{5 \pi}{6}\)
- C \(\frac{\pi}{4}, \frac{5 \pi}{4}\)
- D \(\frac{\pi}{4}, \frac{7 \pi}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4}, \frac{5 \pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(x)=\sin x+\cos x+6 \)
\( \therefore \mathrm{f}^{\prime}(x)=\cos x-\sin x \)
\( \text { Now, } \mathrm{f}^{\prime}(\mathrm{c})=0 \)
\( \Rightarrow \cos \mathrm{c}-\sin \mathrm{c}=0 \)
\( \Rightarrow \cos \mathrm{c}=\sin \mathrm{c} \)
\( \Rightarrow \tan \mathrm{c}=1 \)
\( \Rightarrow \mathrm{c}=\frac{\pi}{4}, \frac{5 \pi}{4} \quad \ldots[\because x \in[0,2 \pi]]\)
\( \therefore \mathrm{f}^{\prime}(x)=\cos x-\sin x \)
\( \text { Now, } \mathrm{f}^{\prime}(\mathrm{c})=0 \)
\( \Rightarrow \cos \mathrm{c}-\sin \mathrm{c}=0 \)
\( \Rightarrow \cos \mathrm{c}=\sin \mathrm{c} \)
\( \Rightarrow \tan \mathrm{c}=1 \)
\( \Rightarrow \mathrm{c}=\frac{\pi}{4}, \frac{5 \pi}{4} \quad \ldots[\because x \in[0,2 \pi]]\)
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