Skip to main contentSkip to solution

The maximum value of the function f(x)=sin⁡x(1+cos⁡x)f(x)=\sin x(1+\cos x) in the internal [0,π][0,\pi] is

Solution

✅ Correct Option: 4

f′(x)=cos⁡x+cos⁡2x=2cos⁡2x+cos⁡x−1=(2cos⁡x−1)(cos⁡x+1)f'(x)=\cos x+\cos 2x=2\cos^2x+\cos x-1=(2\cos x-1)(\cos x+1).

On [0,π][0,\pi] this vanishes at x=π3x=\frac{\pi}{3} (and x=πx=\pi).

Then f(π3)=32⋅32=334f\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2}\cdot\frac{3}{2}=\frac{3\sqrt3}{4}, which exceeds the endpoint values 00.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question