Question

$f\left(x\right)=\sin x\left(1+\cos x\right),x\in \left(0,\frac{\pi }{2}\right)$Then, f(x) has a maxima at .

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is D $\frac{\pi }{3}$

$f\left(x\right)=\sin x\left(1+\cos x\right),x\in \left(0,\frac{\pi }{2}\right)\therefore f\text{'}\left(x\right)=\cos x\left(1+\cos x\right)-{\sin }^{2}x\Rightarrow f\text{'}\left(x\right)=\cos x+{\cos }^{2}x-{\sin }^{2}x\Rightarrow f\text{'}\left(x\right)=\cos x+\cos 2x\mathrm{and}f\text{'}\text{'}\left(x\right)=-\sin x-2\sin 2x\mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{value}\mathrm{of}f\left(x\right),f\text{'}\left(x\right)=0\Rightarrow \cos x+\cos 2x=0\Rightarrow \cos x=-\cos 2x\Rightarrow \cos x=\cos \left(\pi \pm 2x\right)\Rightarrow x=\left(\pi \pm 2x\right)\Rightarrow x=\frac{\pi }{3}\mathrm{Now},f\text{'}\text{'}\left(\frac{\pi }{3}\right)=-2\sin \frac{2\pi }{3}-\sin \frac{\pi }{3}=-2\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=-\mathrm{ve}.\mathrm{Hence},f\left(x\right)\mathrm{has}a\mathrm{maxima}\mathrm{at}x=\frac{\pi }{3}.$