Values of $\mathrm{c}$ as per Rolle's theorem for $f(x)=\sin x+\cos x+6$ on $[0,2 \pi]$ are
Values of $\mathrm{c}$ as per Rolle's theorem for $f(x)=\sin x+\cos x+6$ on $[0,2 \pi]$ are
- $\frac{\pi}{3}, \frac{5 \pi}{3}$
- $\frac{\pi}{6}, \frac{5 \pi}{6}$
- $\frac{\pi}{4}, \frac{5 \pi}{4}$
- $\frac{\pi}{4}, \frac{7 \pi}{4}$
Solution
$\begin{array}{ll} & \mathrm{f}(x)=\sin x+\cos x+6 \\ \therefore \quad & \mathrm{f}^{\prime}(x)=\cos x-\sin x \\ & \text { Now, } \mathrm{f}^{\prime}(\mathrm{c})=0 \\ \Rightarrow \cos \mathrm{c}-\sin \mathrm{c}=0 \\ \Rightarrow \cos \mathrm{c}=\sin \mathrm{c} \\ \Rightarrow \tan \mathrm{c}=1 \\ \Rightarrow \mathrm{c}=\frac{\pi}{4}, \frac{5 \pi}{4} \quad \ldots[\because x \in[0,2 \pi]]\end{array}$
Asked in: MHT CET 2023 (13 May Shift 1)
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