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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 :

JEE · Math · previous-year question

  1. A.$$\left( {0,\sqrt 5 } \right)$$
  2. B.[$$-$$2, 2]
  3. C.$$\left[ {{1 \over {\sqrt 5 }},\sqrt 5 } \right]$$
  4. D.[0, 2]correct

Answer

D. [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)

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