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
365.7k+ 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.
Last updated date: 28th Sep 2023
•
Total views: 365.7k
•
Views today: 8.65k
Recently Updated Pages
What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE
