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# Solution for The Value of C in Rolle'S Theorem When F (X) = 2x3 − 5x2 − 4x + 3, X ∈ [1/3, 3] is (A) 2 (B) − 1 3 (C) −2 (D) 2 3 - CBSE (Science) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

The value of c in Rolle's theorem when
f (x) = 2x3 − 5x2 − 4x + 3, x ∈ [1/3, 3] is

(a) 2

(b)$- \frac{1}{3}$

(c) −2

(d) $\frac{2}{3}$

#### Solution

(a) 2

Given: $f\left( x \right) = 2 x^3 - 5 x^2 - 4x + 3$ Differentiating the given function with respect to x, we get

$f'\left( x \right) = 6 x^2 - 10x - 4$

$\Rightarrow f'\left( c \right) = 6 c^2 - 10c - 4$

$\therefore f'\left( c \right) = 0$

$\Rightarrow 3 c^2 - 5c - 2 = 0$

$\Rightarrow 3 c^2 - 6c + c - 2 = 0$

$\Rightarrow 3c\left( c - 2 \right) + c - 2 = 0$

$\Rightarrow \left( 3c + 1 \right)\left( c - 2 \right) = 0$

$\Rightarrow c = 2, \frac{- 1}{3}$

$\therefore c = 2 \in \left( \frac{1}{3}, 3 \right)$

Thus,

$c = 2 \in \left( \frac{1}{3}, 3 \right)$ for which Rolle's theorem holds.

Hence, the required value of c is 2 .

Is there an error in this question or solution?

#### APPEARS IN

Solution The Value of C in Rolle'S Theorem When F (X) = 2x3 − 5x2 − 4x + 3, X ∈ [1/3, 3] is (A) 2 (B) − 1 3 (C) −2 (D) 2 3 Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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