Question

Prove that f(x) = sinx + $\sqrt{3}$cosx has maximum value at x = $\frac{\mathrm{\pi }}{6}$. [NCERT EXEMPLAR]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=\sin x+\sqrt{3}\cos x \\ \Rightarrow f\text{'}\left(x\right)=\cos x+\sqrt{3}\left(-\sin x\right) \\ \Rightarrow f\text{'}\left(x\right)=\cos x\mathit{-}\sqrt{3}\sin x \\ \mathrm{For}f\left(x\right)\mathrm{to}\mathrm{have}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{value},\mathrm{we}\mathrm{must}\mathrm{have}f\text{'}\left(x\right)=0 \\ \Rightarrow \mathrm{cosx}\mathit{-}\sqrt{3}\mathrm{sinx}=0 \\ \Rightarrow \mathrm{cosx}\mathit{=}\sqrt{3}\mathrm{sinx} \\ \Rightarrow \cot x\mathit{=}\sqrt{3} \\ \Rightarrow x=\frac{\mathrm{\pi }}{6} \\ \mathrm{Also},f\text{'}\text{'}\left(x\right)=-\mathrm{sinx}\mathit{-}\sqrt{3}\cos x \\ \Rightarrow f\text{'}\text{'}\left(\frac{\mathrm{\pi }}{6}\right)=-\sin \frac{\mathrm{\pi }}{6}-\sqrt{3}\cos \frac{\mathrm{\pi }}{6}=-\frac{1}{2}-\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)=-\frac{1}{2}-\frac{3}{2}=-2<0 \\ \mathrm{So},x=\frac{\mathrm{\pi }}{6}\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{maxima}. \end{gathered}$$