QJEEM21Ma3310: JEE Main 2021 Mathematics - Class 11 JEEM21Ma3310 Solution: The range of the function, is :

Question

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The range of the function,

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

is :

Accepted Answer

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

f(x) = {\log _{\sqrt 5 }}\left[ {3 + 2\cos \left( {{\pi \over 4}} \right)\cos (x) - 2\sin \left( {{{3\pi } \over 4}} \right)\sin (x)} \right]

f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]

Since

- \sqrt 2 \le \cos x - \sin x \le \sqrt 2

\Rightarrow {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 \left( { - \sqrt 2 } \right) \le f(x) \le {{\log }_{\sqrt 5 }}\left[ {3 + \sqrt 2 \left( {\sqrt 2 } \right)} \right]} \right]

\Rightarrow {\log _{\sqrt 5 }}(1) \le f(x) \le {\log _{\sqrt 5 }}(5)

So, Range of f(x) is [0, 2]

Option (d)

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