#### Question

\[c = \frac{3}{2} \in \left( 1, 2 \right)\]The value of *c* in Rolle's theorem for the function f (x) = x^{3} − 3x in the interval [0,\[\sqrt{3}\]] is

(a) 1

(b) −1

(c) 3/2

(d) 1/3

#### Solution

(a) 1

The given function is \[f\left( x \right) = x^3 - 3x\] .

This is a polynomial function, which is continuous and derivable in *R.*

Therefore, the function is continuous on [0,\[\sqrt{3}\]] and derivable on (0,\[\sqrt{3}\])

Differentiating the given function with respect to *x,* we get

\[f'\left( x \right) = 3 x^2 - 3\]

\[ \Rightarrow f'\left( c \right) = 3 c^2 - 3\]

\[ \therefore f'\left( c \right) = 0 \]

\[ \Rightarrow 3 c^2 - 3 = 0\]

\[ \Rightarrow c^2 = 1\]

\[ \Rightarrow c = \pm 1\]

Thus,

\[c = 1 \in \left[ 0, \sqrt{3} \right]\] for which Rolle's theorem holds.

Hence, the required value of *c* is 1.