The maximum value of ${\rm{\;}}\sin \,\,\left( {x + \frac{\pi }{6}} \right) + \cos \,\,\left( {x + \frac{\pi }{6}} \right){\rm{\;}}$ in the interval $\left( {0,\,\,\frac{\pi }{2}} \right)$ is attained at

JEE MATHEMATICS MCQ: TRANSFORMATIONS

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The maximum value of

\mathrm{\;}\sin \,\,\left( {x + \frac{\pi }{6}} \right) + \cos \,\,\left( {x + \frac{\pi }{6}} \right)\mathrm{\;}

in the interval

\left( {0,\,\,\frac{\pi }{2}} \right)

is attained at

View answer and explanation

Correct option: A

x = \frac{\pi }{{12}}

Option A (

x = \frac{\pi }{{12}}

) is correct because it is the value of x where the given trigonometric function attains its maximum value in the specified interval.

Practice All Chapter MCQs →

1 + \cos \,\,56^\circ + \cos \,\,58^\circ - \cos \,\,66^\circ = \lambda \cos \,\,28^\circ \,\,\cos \,\,29^\circ \,\,\sin \,\,33^\circ

\lambda =

\cos \,\,\frac{\pi }{{11}} + \cos \,\,\frac{{3\,\pi }}{{11}} + \cos \,\,\frac{{5\,\pi }}{{11}} + \cos \,\,\frac{{7\pi }}{{11}} + \cos \,\,\frac{{9\,\pi }}{{11}}

\;\cos \,\,x + \cos \,\,y + \cos \,\,\alpha = 0\;\,\mathrm{and}\,\sin x + \sin \,\,y + \sin \,\,\alpha = 0

\cot \left( {\frac{{x + y}}{2}} \right)\;

f\left( x \right) = \sin \left( {\frac{{2x + 3}}{{6\;\pi }}} \right),

0 < x,\mathrm{ }y < \pi

\cos x + \cos y - \cos (x + y) = \frac{3}{2}

\sin x + \cos y

\mathrm{\;sin}\,\,36^\circ \,\,\sin \,\,72^\circ \,\,\sin \,\,108^\circ \,\,\sin \,\,144^\circ \;