The function f ( x ) = x 3 + a x 2 + b x + c , a 2 ≤ 3 b has

Given, f ( x ) = x 3 + a x 2 + b x + c , a 2 ≤ 3 b ⇒ f ′ ( x ) = 3 x 2 + 2 a x + b Put f ′ ( x ) = 0 ⇒ 3 x 2 + 2 a x + b = 0 ⇒ x = 2 × 3 − 2 a ± 4 a 2 − 12 b = 3 − 2 a ± 2 a 2 − 3 b Since, a 2 ≤ 3 b , ∴ x has an imaginary value. Hence, no extreme value of x exists.

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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaEasy2 minPYQ_2009

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The function $f (x) = x^{3} + a x^{2} + b x + c, a^{2} \leq 3 b$ has

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Given, $f (x) = x^{3} + a x^{2} + b x + c, a^{2} \leq 3 b$ $\Rightarrow f^{'} (x) = 3 x^{2} + 2 a x + b$ Put $f^{'} (x) = 0$ $\Rightarrow 3 x^{2} + 2 a x + b = 0 \Rightarrow x = \frac{- 2 a \pm 4 a ^{2} - 12 b}{2 \times 3} = \frac{- 2 a \pm 2 a ^{2} - 3 b}{3}$ Since, $a^{2} \leq 3 b$ , $∴ x$ has an imaginary value. Hence, no extreme value of $x$ exists.

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

⏱ 2mℹ️ Source: PYQ_2009

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