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What are the values of $c$ for which Rolles Theorem for the function $f\left( x \right) = {x^3} - 3{x^2} + 2x$ in the interval $\left[ {0,2} \right]$is verified?A. $c = \pm 1$B. $c = 1 \pm \dfrac{1}{{\sqrt 3 }}$C. $c = \pm 2$D. None of these

Last updated date: 23rd Jul 2024
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Hint: Rolles Theorem must satisfy all the three conditions and if $f\left( x \right)$ is a polynomial function then the function is continuous in the interval.
First of all, we should know the conditions of Rolles theorem
Conditions for Rolles theorem is
(1) $f\left( x \right)$ is a continuous at $\left[ {a,b} \right]$
(2) $f\left( x \right)$is derivable at $\left( {a,b} \right)$
(3) $f\left( a \right) = f\left( b \right)$
If all the three conditions satisfied then there exist some $c$ in $f\left( a \right) = f\left( b \right)$ such that $f'\left( c \right) = 0$
Here we can clearly observe that
(a) $f\left( x \right)$ is a polynomial, so it is continuous in the interval $\left[ {0,2} \right]$
(b) $f'\left( x \right) = 3{x^2} - 6x + 2$ exists for all $x \in \left( {0,2} \right)$
So, $f\left( x \right)$ is differentiable for all $x \in \left( {0,2} \right)$ and
(c) $f\left( 0 \right) = f\left( 2 \right)$ since,
$f\left( 0 \right) = {\left( 0 \right)^3} - 3{\left( 0 \right)^2} + 2\left( 0 \right) = 0 \\ f\left( 2 \right) = {\left( 2 \right)^3} - 3{\left( 2 \right)^2} + 2\left( 2 \right) \\ {\text{ = }}8 - 3 \times 4 + 4 = 0 \\$
Thus, all the three conditions of Rolles theorem are satisfied.
So, there must be exist $c \in \left[ {0,2} \right]$ such that $f'\left( c \right) = 0$
$f'\left( c \right) = 3{\left( c \right)^2} - 6c + 2 = 0$
solving the equation $3{\left( c \right)^2} - 6c + 2 = 0$ we get
$\Rightarrow 3{c^2} - 6c + 2 = 0 \\ c = \dfrac{{6 \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( {3 \times 2} \right)} }}{{2\left( 3 \right)}} \\ c = \dfrac{{6 \pm \sqrt {36 - 24} }}{6} \\ c = \dfrac{{6 \pm \sqrt {12} }}{6} \\ c = \dfrac{{6 \pm 2\sqrt 3 }}{6} \\$
Separating the positive terms, we get
$c = \dfrac{{6 + 2\sqrt 3 }}{6} \\ c = 1 + \dfrac{{\sqrt 3 }}{3} \\ c = 1 + \dfrac{1}{{\sqrt 3 }} \\$
Separating the negative terms, we get
$c = \dfrac{{6 - 2\sqrt 3 }}{6} \\ c = 1 - \dfrac{{\sqrt 3 }}{3} \\ c = 1 - \dfrac{1}{{\sqrt 3 }} \\$
Thus, $c = 1 \pm \dfrac{1}{{\sqrt 3 }} \in \left[ {0,2} \right]$ and Rolles theorem is verified.
Therefore, the value of $c$ is $1 \pm \dfrac{1}{{\sqrt 3 }}$.
So, option B. $1 \pm \dfrac{1}{{\sqrt 3 }}$
Note: All the three conditions must be satisfied to obtain the value of $c$ in Rolles theorem. Otherwise the Rolles theorem cannot be verified.