NDA 2017
Exam NDA
Year 2017
Subject Mathematics
Chapter Trigonometric Ratios & Identities NDA 2017 · Mathematics · Trigonometric Ratios & Identities
The maximum value of left(x+ (π)/(6))+ lleft(x+(π)/(6))in the...
The maximum value of
sin \l \sin \l sin \l left(x+
π 6 ) + cos \l l l e f t ( x + π 6 ) i n t h e i n t e r v a l ( 0 , π 2 ) \frac{\pi}{6})+\cos\l lleft(x+\frac{\pi}{6})in the interval \left(0, \frac{\pi}{2}\right) 6 π ) + cos \l ll e f t ( x + 6 π ) in t h e in t er v a l ( 0 , 2 π ) is attained at
Options A π 12 \frac{\pi}{12} 12 π B π 6 \frac{\pi}{6} 6 π C π 3 \frac{\pi}{3} 3 π D π 2 \frac{\pi}{2} 2 π Correct answer A. π 12 \frac{\pi}{12} 12 π
Step-by-step solution
sin ( x + π 6 ) + cos ( x + π 6 ) = 2 [ 1 2 sin ( x + π 6 ) + 1 2 cos ( x + π 6 ) ] = 2 [ sin ( x + π 6 ) cos π 4 + cos ( x + π 6 ) sin π 4 ] = 2 [ sin ( x + π 6 + π 4 ) ] ( ∵ sin ( A + B ) = sin A cos B + cos A sin B = 2 [ sin ( x + 5 π 12 ) ] \begin{aligned}
& \sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right) \\
& =\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin \left(x+\frac{\pi}{6}\right)+\frac{1}{\sqrt{2}} \cos \left(x+\frac{\pi}{6}\right)\right] \\
& =\sqrt{2}\left[\sin \left(x+\frac{\pi}{6}\right) \cos \frac{\pi}{4}+\cos \left(x+\frac{\pi}{6}\right) \sin \frac{\pi}{4}\right] \\
& =\sqrt{2}\left[\sin \left(x+\frac{\pi}{6}+\frac{\pi}{4}\right)\right] \\
& (\because \sin (A+B)=\sin A \cos B+\cos A \sin B \\
& =\sqrt{2}\left[\sin \left(x+\frac{5 \pi}{12}\right)\right]
\end{aligned} sin ( x + 6 π ) + cos ( x + 6 π ) = 2 [ 2 1 sin ( x + 6 π ) + 2 1 cos ( x + 6 π ) ] = 2 [ sin ( x + 6 π ) cos 4 π + cos ( x + 6 π ) sin 4 π ] = 2 [ sin ( x + 6 π + 4 π ) ] ( ∵ sin ( A + B ) = sin A cos B + cos A sin B = 2 [ sin ( x + 12 5 π ) ]
Given interval is
( 0 , π 2 ) \left(0, \frac{\pi}{2}\right) ( 0 , 2 π )
For, maximum value
x + 5 π 12 = π 2 ⇒ x = π 2 − 5 π 12 = 6 π − 5 π 12 = π 12 \mathrm{x}+\frac{5 \pi}{12}=\frac{\pi}{2} \Rightarrow \mathrm{x}=\frac{\pi}{2}-\frac{5 \pi}{12}=\frac{6 \pi-5 \pi}{12}=\frac{\pi}{12} x + 12 5 π = 2 π ⇒ x = 2 π − 12 5 π = 12 6 π − 5 π = 12 π
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