The maximum value of the function fx=sinx+π6+cosx+π6 in the interval 0,π2 occurs at

f(x)=22sin⁡x+π6+cos⁡x+π6=2sin⁡x+π6cos⁡π4+cos⁡x+π6sin⁡π4=2sin⁡x+π6+π4 =2sin⁡x+5π12 x+5π12=π2⇒x=π2−5π12⇒x=6π−5π12⇒x=π12Maximum when x=π12

The maximum value of the function fx=sinx+π6+cosx+π6 in the interval 0,π2 occurs at

f(x)=22sin⁡x+π6+cos⁡x+π6=2sin⁡x+π6cos⁡π4+cos⁡x+π6sin⁡π4=2sin⁡x+π6+π4 =2sin⁡x+5π12 x+5π12=π2⇒x=π2−5π12⇒x=6π−5π12⇒x=π12Maximum when x=π12

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The maximum value of the function $f (x) = \sin (x + \frac{π}{6}) + \cos (x + \frac{π}{6})$ in the interval $(0, \frac{π}{2})$ occurs at

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$\frac{π}{3}$

$\frac{π}{12}$

$\frac{π}{6}$

answer is A.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{\sqrt{2}}{\sqrt{2}} [\sin (x + \frac{π}{6}) + \cos (x + \frac{π}{6})] \\ = \sqrt{2} [\sin (x + \frac{π}{6}) \cos \frac{π}{4} + \cos (x + \frac{π}{6}) \sin \frac{π}{4}] \\ = \sqrt{2} \sin (x + \frac{π}{6} + \frac{π}{4}) = \sqrt{2} \sin (x + \frac{5 π}{12}) \\ x + \frac{5 π}{12} = \frac{π}{2} \Rightarrow x = \frac{π}{2} - \frac{5 π}{12} \Rightarrow x = \frac{6 π - 5 π}{12} \Rightarrow x = \frac{π}{12} \\ M a x i m u m when x = \frac{π}{12} \end{array}$