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MathematicsApplication of DerivativeMonotonicity-Rolle's-theroemMedium2 minQB

MathematicsMediumsingle choice

The equation $2 x^{3} - 3 x^{2} - 12 x + 1 = 0$ has in the interval $(- 2, 1)$

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Solution:

Let $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 1$ $f (- 2) < 0; f (0) > 0; f (1) < 0$ $∴ f (x) = 0$ has atleast two roots in the interval (-2, 1). Suppose all the real roots of $f (x) \in (- 2, 1)$ . Then by Rolle's theorm, both the roots of the equation $f^{1} (x) = 0$ should belong to $(- 2, 1)$ $f^{1} (x) = 6 x^{2} - 6 x - 12 = 0 \Rightarrow x^{2} - x - 2 = 0 \Rightarrow (x - 2) (x + 1) = 0 \Rightarrow x = 2, - 1$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Monotonicity-Rolle's-theroem

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