Answer
Verified
482.1k+ views
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE