Question
The maximum value of $$f\left( x \right) = 3{\cos ^2}x + 4{\sin ^2}x + \cos \frac{x}{2} + \sin \frac{x}{2}$$ is :
A.
4
B.
$$3 + \sqrt 2 $$
C.
$$4 + \sqrt 2 $$
D.
none of these
Answer :
$$4 + \sqrt 2 $$
Solution :
$$\eqalign{
& f\left( x \right) = 4 - {\cos ^2}x + \cos \frac{x}{2} + \sin \frac{x}{2} \cr
& \therefore f'\left( x \right) = \sin \,2x - \frac{1}{2}\left( {\sin \frac{x}{2} - \cos \frac{x}{2}} \right) \cr
& = 2\sin \,x.\left( {{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}} \right) + \frac{1}{2}\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right) \cr
& = \left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)\left\{ {2\sin \,x\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right) + \frac{1}{2}} \right\} \cr
& = \left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)\left\{ {2\sqrt 2 \sin \,x.\sin \left( {\frac{x}{2} + \frac{\pi }{4}} \right) + \frac{1}{2}} \right\} \cr
& \therefore \,\,f'\left( x \right) = 0\,\,\,\, \Rightarrow \cos \frac{x}{2} - \sin \frac{x}{2} = 0 \cr
& \Rightarrow x = \frac{\pi }{2} \cr
& f''\left( x \right) = 2\cos \,2x - \frac{1}{2}.\frac{1}{2}\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right) \cr
& \therefore f''\left( {\frac{\pi }{2}} \right) = - 2 - \frac{1}{4}\left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right) < 0 \cr
& \therefore \max \,f\left( x \right) = f\left( {\frac{\pi }{2}} \right) = 4 - 0 + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = 4 + \sqrt 2 \cr} $$