JEE MATHEMATICS MCQ: TRANSFORMATIONS

1 Easy

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Analyzing +4 | -1

The range of the function,

f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right)

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Correct option: D

[0,\,\,2]

Option D ([0, 2]) is correct because it represents the range of the function after simplifying the trigonometric expression and applying logarithmic properties.

More Questions from TRANSFORMATIONS

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

, then

\lambda =

The value of

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

, is If

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

then

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

is equal to The period of the function

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

is If

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

and

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

, then

\sin x + \cos y

is equal to The value of

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

is equal to