# What are the values of \[c\] for which Rolle`s 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

Answer

Verified

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Hint: Rolle`s 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 Rolle`s theorem

Conditions for Rolle`s 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 Rolle`s 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 Rolle`s 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 Rolle`s theorem. Otherwise the Rolle`s theorem cannot be verified.

First of all, we should know the conditions of Rolle`s theorem

Conditions for Rolle`s 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 Rolle`s 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 Rolle`s 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 Rolle`s theorem. Otherwise the Rolle`s theorem cannot be verified.

Last updated date: 28th Sep 2023

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