JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 12)
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 :
$$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$$ is :
$$\left( {0,\sqrt 5 } \right)$$
[$$-$$2, 2]
$$\left[ {{1 \over {\sqrt 5 }},\sqrt 5 } \right]$$
[0, 2]
Explanation
$$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$$
$$f(x) = {\log _{\sqrt 5 }}\left[ {3 + 2\cos \left( {{\pi \over 4}} \right)\cos (x) - 2\sin \left( {{{3\pi } \over 4}} \right)\sin (x)} \right]$$
$$f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$$
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)
$$f(x) = {\log _{\sqrt 5 }}\left[ {3 + 2\cos \left( {{\pi \over 4}} \right)\cos (x) - 2\sin \left( {{{3\pi } \over 4}} \right)\sin (x)} \right]$$
$$f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$$
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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