#### Question

The value of *c* in Rolle's theorem when

f (x) = 2x^{3} − 5x^{2} − 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 .