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

f ( x ) = − x ln x and lim x → 0 + f ( x ) = 0 . f ′ ( x ) = − ( 1 + ln x ) ⎩ ⎨ ⎧ > 0 = 0 < 0 if if if 0 < x < e 1 x = e 1 e 1 < x < 1 f has a 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 DerivativeMiscellaneous/MixedMedium2 minQB

MathematicsMediumsingle choice

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

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Answer:

Solution:

$f (x) = - x ln x$ and $lim_{x \to 0 +} f (x) = 0$ .

$f^{'} (x) = - (1 + ln x) ⎩ ⎨ ⎧ > 0 = 0 < 0 if if if 0 < x < \frac{1}{e} x = \frac{1}{e} \frac{1}{e} < x < 1$

$f$ has a maximum value at $x = \frac{1}{e}$ and $f (\frac{1}{e}) = \frac{1}{e}$ .

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: QB

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