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JEE Main 20211 SeptemberEvening Shift
Exam
JEE Main
Year
2021
Date
1 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2021 · 1 September · Evening Shift · Mathematics · Functions

The range of the function f ( x ) = _ 5 (3 + cos ((3π)/(4) + x) +...

The range of the function f(x)=log⁡5f ( x ) = \log_{\sqrt{5}}\left(3 + cos\left(\dfrac{3\pi}{4} + x\right) + \cos \left(\dfrac{\pi}{4} + x\right) + \cos \left(\dfrac{\pi}{4} - x\right) - \cos \left(\dfrac{3\pi}{4} - x\right)\right) is :

Options

  1. A[15,5]\left[\dfrac{1}{\sqrt{5}}, \sqrt{5}\right]
  2. B[0,2][ 0, 2 ]
  3. C(0,5)( 0, \sqrt{5} )
  4. D[−2,2][ - 2, 2 ]

Correct answer

B. [0,2][ 0, 2 ]

Step-by-step solution

f(x)=log⁡5(3−2sin⁡x+2cos⁡x)f ( x ) = \log_{\sqrt{5}} ( 3 - \sqrt{2}\sin x + \sqrt{2}\cos x )

−2≤(−2sin⁡x+2cos⁡x)≤2- 2 \le ( - \sqrt{2}\sin x + \sqrt{2}\cos x ) \le 2

⇒1≤(3−2sin⁡x+2cos⁡x)≤5⇒ 1 \le ( 3 - \sqrt{2}\sin x + \sqrt{2}\cos x ) \le 5

⇒log⁡51≤f(x)≤log⁡55⇒ \log_{\sqrt{5}}1 \le f ( x ) \le \log_{\sqrt{5}}5

⇒0≤f(x)≤2⇒ 0 \le f ( x ) \le 2

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