Least value of f = min (Range of f ). f ( − 1 ) = 0 ; x → 2 1 − lim f ( x ) ≡ − ∞ as g ( x ) → 0 + as x → 2 1 − f has no minima.

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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

$f (x) = ln (1 - x^{2} - x)$

Least value of $f$ is

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Answer:

Solution:

Least value of $f = min$ (Range of $f$ ). $f (- 1) = 0; x \to \frac{1}{2} - lim f (x) \equiv - \infty as g (x) \to 0 + as x \to \frac{1}{2} -$ $f$ has no minima.

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

⏱ 2mℹ️ Source: QB

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