Question

f(x) = $\sin +\sqrt{3}\cos x$ is maximum when x =

(a) $\frac{\mathrm{\pi }}{3}$

(b) $\frac{\mathrm{\pi }}{4}$

(c) $\frac{\mathrm{\pi }}{6}$

(d) 0

Answer 1

$$\begin{gathered} \left(\mathrm{c}\right)\frac{\pi }{6} \\ \mathrm{Given}:f\left(x\right)=\sin x+\sqrt{3}\cos x \\ \Rightarrow f\text{'}\left(x\right)=\cos x-\sqrt{3}\sin x \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow \cos x-\sqrt{3}\sin x=0 \\ \Rightarrow \cos x=\sqrt{3}\sin x \\ \Rightarrow \tan x=\frac{1}{\sqrt{3}} \\ \Rightarrow x=\frac{\pi }{6} \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=-\sin x-\sqrt{3}\cos x \\ \Rightarrow \Rightarrow f\text{'}\text{'}\left(\frac{\pi }{2}\right)=-\sin \frac{\pi }{2}-\sqrt{3}\cos \frac{\pi }{2}\frac{-1}{2}-\frac{3}{2}=-2<0 \\ \mathrm{So},x=\frac{\pi }{2}\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{maxima}. \end{gathered}$$