Question:
\(f(x)=\left|x \log_{e} x\right|\), \(x>0\), is monotonically decreasing in :
\(f(x)=\left|x \log_{e} x\right|\), \(x>0\), is monotonically decreasing in :
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Remove the modulus by checking the sign of x ln x on (0,1) and on (1, infinity), then study the sign of the derivative.
Updated On: Nov 12, 2026
- \(\left(0, \frac{1}{e}\right)\)
- \(\left[\frac{1}{e}, 1\right]\)
- \((1, e)\)
- \((e, \infty)\)
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The Correct Option is B
Solution and Explanation
For \(0<x<1\), \(\ln x<0\), so \(f(x)=-x\ln x\) and \(f'(x)=-(\ln x+1)\).
\(f'(x)>0\) when \(\ln x<-1\), i.e. \(0<x<\frac1e\) (increasing); \(f'(x)<0\) when \(\frac1e<x<1\) (decreasing).
For \(x>1\), \(f(x)=x\ln x\) and \(f'(x)=\ln x+1>0\), so f is increasing.
Hence f is decreasing only on \(\left[\frac1e,1\right]\).
Answer: (2) \(\left[\frac{1}{e}, 1\right]\)
\(f'(x)>0\) when \(\ln x<-1\), i.e. \(0<x<\frac1e\) (increasing); \(f'(x)<0\) when \(\frac1e<x<1\) (decreasing).
For \(x>1\), \(f(x)=x\ln x\) and \(f'(x)=\ln x+1>0\), so f is increasing.
Hence f is decreasing only on \(\left[\frac1e,1\right]\).
Answer: (2) \(\left[\frac{1}{e}, 1\right]\)
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