Question
The maximum value of $$1 + \sin \left( {\frac{\pi }{4} + \theta } \right) + 2\cos \left( {\frac{\pi }{4} - \theta } \right)$$ for real values of $$\theta $$ is
A.
3
B.
5
C.
4
D.
None of these
Answer :
4
Solution :
Value $$ = 1 + \frac{1}{{\sqrt 2 }}\left( {\cos \theta + \sin \theta } \right) + \sqrt 2 \left( {\cos \theta + \sin \theta } \right)$$
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \left( {\frac{1}{{\sqrt 2 }} + \sqrt 2 } \right) \cdot \left( {\cos \theta + \sin \theta } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \left( {\frac{1}{{\sqrt 2 }} + \sqrt 2 } \right) \cdot \sqrt 2 \cos \left( {\theta - \frac{\pi }{4}} \right). \cr} $$
∴ the maximum value $$ = 1 + \left( {\frac{1}{{\sqrt 2 }} + \sqrt 2 } \right)\sqrt 2 = 4.$$