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) $\frac{2}{3}$

Answer 1

(a) 2

Given:
$f\left(x\right)=2x^3-5x^2-4x+3$

Differentiating the given function with respect to x, we get

$$\begin{gathered} f\text{'}\left(x\right)=6x^2-10x-4 \\ \Rightarrow f\text{'}\left(c\right)=6c^2-10c-4 \\ \therefore f\text{'}\left(c\right)=0 \\ \Rightarrow 3c^2-5c-2=0 \\ \Rightarrow 3c^2-6c+c-2=0 \\ \Rightarrow 3c\left(c-2\right)+c-2=0 \\ \Rightarrow \left(3c+1\right)\left(c-2\right)=0 \\ \Rightarrow c=2,\frac{-1}{3} \\ \therefore c=2\in \left(\frac{1}{3},3\right) \end{gathered}$$

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

Hence, the required value of c is 2.