f ( x ) = ∣ x ln x ∣ : x ∈ ( 0 , 1 ) , then f ( x ) has maximum value=

f ( x ) = − x ln x x → 0 + lim f ( x ) = 0 f ′ ( x ) = − ( 1 + ln x ) { > 0 if 0 < x < e 1 = 0 if x = e 1 < 0 if e 1 < x < 1 f has maximum value at x = e 1 and f ( e 1 ) = e 1

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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaEasy2 minQB

MathematicsEasysingle choice

$f (x) = ∣ x ln x ∣ : x \in (0, 1)$ , then $f (x)$ has maximum value=

Options:

Answer:

Solution:

$f (x) = - x ln x$

$x \to 0 + lim f (x) = 0$

$f^{'} (x) = - (1 + ln x) {> 0 if 0 < x < \frac{1}{e} = 0 if x = \frac{1}{e} < 0 if \frac{1}{e} < x < 1$ f has maximum value at $x = \frac{1}{e}$ and $f (\frac{1}{e}) = \frac{1}{e}$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima

⏱ 2mℹ️ Source: QB

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