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JEE Mains · Maths · STD 12 - 1. relation and function

વિધેય \(\mathrm{f}(\mathrm{x})=\log _{\sqrt{5}}(3+\cos \left(\frac{3 \pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}+\mathrm{x}\right)+\cos \left(\frac{\pi}{4}-\mathrm{x}\right)\) \(-\cos \left(\frac{3 \pi}{4}-\mathrm{x}\right))\) નો વિસ્તાર મેળવો.

  1. A \((0, \sqrt{5})\)
  2. B \([-2,2]\)
  3. C \(\left[\frac{1}{\sqrt{5}}, \sqrt{5}\right]\)
  4. D \([0,2]\)
Verified Solution

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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