JEE Mains · Maths · STD 12 - 1. relation and function
The range of the function, \(\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}-\mathrm{x}\right)\) \(-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))\) is :
- A \((0, \sqrt{5})\)
- B \([-2,2]\)
- C \(\left[\frac{1}{\sqrt{5}}, \sqrt{5}\right]\)
- D \([0,2]\)
Answer & Solution
Correct Answer
(D) \([0,2]\)
Step-by-step Solution
Detailed explanation
\(f(x)=\log _{\sqrt{5}}\) \((3+\cos \left(\frac{3 \pi}{4}+x\right)+\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)\) \(-\cos \left(\frac{3 \pi}{4}-x\right))\)…
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