Question:
The range of the function,
\(f(x)=\log_{\sqrt5}\left(3+\cos\left(\frac{3\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}-x\right)-\cos\left(\frac{3\pi}{4}-x\right)\right)\)
is :
The range of the function,
\(f(x)=\log_{\sqrt5}\left(3+\cos\left(\frac{3\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}-x\right)-\cos\left(\frac{3\pi}{4}-x\right)\right)\)
is :
\(f(x)=\log_{\sqrt5}\left(3+\cos\left(\frac{3\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}+x\right)+\cos\left(\frac{\pi}{4}-x\right)-\cos\left(\frac{3\pi}{4}-x\right)\right)\)
is :
Show Hint
Expand the four cosines; they combine to \(2\cos\left(x+\frac{\pi}{4}\right)\), so the log argument lies in \([1,5]\).
Updated On: Nov 4, 2026
- \(\left(0,\sqrt5\right)\)
- \([-2,2]\)
- \(\left[\frac{1}{\sqrt5},\sqrt5\right]\)
- \([0,2]\)
Show Solution
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The Correct Option is D
Solution and Explanation
\(\cos\left(\frac{3\pi}{4}+x\right)=\frac{-\cos x-\sin x}{\sqrt2}\), \(\cos\left(\frac{\pi}{4}+x\right)=\frac{\cos x-\sin x}{\sqrt2}\), \(\cos\left(\frac{\pi}{4}-x\right)=\frac{\cos x+\sin x}{\sqrt2}\), \(-\cos\left(\frac{3\pi}{4}-x\right)=\frac{\cos x-\sin x}{\sqrt2}\).
Sum \(=\frac{2\cos x-2\sin x}{\sqrt2}=\sqrt2(\cos x-\sin x)=2\cos\left(x+\frac{\pi}{4}\right)\in[-2,2]\).
Argument of the log \(=3+2\cos\left(x+\frac{\pi}{4}\right)\in[1,5]\), which is positive, so \(f\) is defined for all real \(x\).
\(\log_{\sqrt5}\) is increasing: \(\log_{\sqrt5}1=0\), \(\log_{\sqrt5}5=2\). Range \(=[0,2]\).
Answer: (4) [0, 2]
Sum \(=\frac{2\cos x-2\sin x}{\sqrt2}=\sqrt2(\cos x-\sin x)=2\cos\left(x+\frac{\pi}{4}\right)\in[-2,2]\).
Argument of the log \(=3+2\cos\left(x+\frac{\pi}{4}\right)\in[1,5]\), which is positive, so \(f\) is defined for all real \(x\).
\(\log_{\sqrt5}\) is increasing: \(\log_{\sqrt5}1=0\), \(\log_{\sqrt5}5=2\). Range \(=[0,2]\).
Answer: (4) [0, 2]
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