Question

(i) The maximum value of sin$(x+\frac{\pi }{6})$+ cos $(x+\frac{\pi }{6})$ in the interval (0,$\pi$/2) is attained when x =

• A. $\pi /12$
• B. $\pi /6$
• C. $\pi /3$
• D. $\pi /2$

Answer 1

Answer: (A)

$\pi /12$

sin$(x+\frac{\pi }{6})$+ cos $(x+\frac{\pi }{6})$ =$\sqrt{2}cos(x+\frac{\pi }{6}-\frac{\pi }{4})$ = $\sqrt{2}cos(x-\frac{\pi }{12})$etc.
It is max., when x- $\pi /12$ =0 i.e, x= $\pi /12$