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In which of the following interval, the function f(x)=xlogxf(x) = \frac{x}{\log x} is decreasing?

Solution

Correct Option: 3

The function f(x)=xlogxf(x) = \frac{x}{\log x} is decreasing when its derivative is negative.

The domain requires x>0x > 0 and logx0\log x \neq 0, so x1x \neq 1.

Domain: x(0,){1}x \in (0, \infty) - \{1\}


Using the quotient rule with u=xu = x and v=logxv = \log x:

f(x)=(1)(logx)(x)(1x)(logx)2f'(x) = \frac{(1) \cdot (\log x) - (x) \cdot \left(\frac{1}{x}\right)}{(\log x)^2}

f(x)=logx1(logx)2f'(x) = \frac{\log x - 1}{(\log x)^2}


For the function to be decreasing, f(x)<0f'(x) < 0:

logx1(logx)2<0\frac{\log x - 1}{(\log x)^2} < 0

Since (logx)2(\log x)^2 is always positive, the sign depends on the numerator:

logx1<0\log x - 1 < 0

logx<1\log x < 1

x<ex < e


Combining with the domain:

x<ex < e and x>0x > 0 and x1x \neq 1

Therefore, the function is decreasing on (0,e){1}(0, e) - \{1\}.

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