The value of c in Rolle's theorem when


The value of c in Rolle's theorem when

$f(x)=2 x^{3}-5 x^{2}-4 x+3, x \in[1 / 3,3]$ is

(a) 2

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

(c) $-2$

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


(a) 2


$f(x)=2 x^{3}-5 x^{2}-4 x+3$

Differentiating the given function with respect to x, we get 

$f^{\prime}(x)=6 x^{2}-10 x-4$

$\Rightarrow f^{\prime}(c)=6 c^{2}-10 c-4$

$\therefore f^{\prime}(c)=0$

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

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

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

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

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

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

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

Hence, the required value of c is 2.

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