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# Solution for C = 3 2 ∈ ( 1 , 2 ) the Value of C in Rolle'S Theorem for the Function F (X) = X3 − 3x in the Interval [0, √ 3 ] is (A) 1 (B) −1 (C) 3/2 (D) 1/3 - CBSE (Science) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

$c = \frac{3}{2} \in \left( 1, 2 \right)$The value of c in Rolle's theorem for the function f (x) = x3 − 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.

Is there an error in this question or solution?

#### APPEARS IN

Solution C = 3 2 ∈ ( 1 , 2 ) the Value of C in Rolle'S Theorem for the Function F (X) = X3 − 3x in the Interval [0, √ 3 ] is (A) 1 (B) −1 (C) 3/2 (D) 1/3 Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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