Question

Let \[f\left( x \right) = 8{x^3} - 6{x^2} - 2x + 1\] , then
A.\[f\left( x \right) = 0\] has no root in \[\left( {0,1} \right)\]
B.\[f\left( x \right) = 0\] has at least one root in \[\left( {0,1} \right)\]
C.\[f'\left( c \right)\] vanishes for some \[c \in \left( {0,1} \right)\]
D.None

Answer 1

Hint: Here, we will find the root of the equation and the condition for differentiability. We will consider the function and find the root. Then by applying the Rolle’s theorem we will find the conditions of differentiability of the given equation.

Complete step-by-step answer:
Let the given polynomial \[f\left( x \right) = 8{x^3} - 6{x^2} - 2x + 1\].
By substituting \[x = 0\] in the polynomial \[f\left( x \right)\], we get
\[ \Rightarrow f\left( 0 \right) = 8{\left( 0 \right)^3} - 6{\left( 0 \right)^2} - 2\left( 0 \right) + 1\].
By subtracting the terms, we get
\[ \Rightarrow f\left( 0 \right) = 1\].
Now substituting \[x = 1\] in the polynomial \[f\left( x \right)\],we get
\[f\left( 1 \right) = 8{\left( 1 \right)^3} - 6{\left( 1 \right)^2} - 2\left( 1 \right) + 1\]
By subtracting the terms, we get
\[ \Rightarrow f\left( 1 \right) = 8 - 6 - 2 + 1\]
\[ \Rightarrow f\left( 1 \right) = 1\]
Since \[f\left( 0 \right) = f\left( 1 \right) = 1\] i.e., \[f\left( a \right) = f\left( b \right)\]
Now differentiating \[f\left( x \right) = 8{x^3} - 6{x^2} - 2x + 1\], we get
\[f'\left( x \right) = 24{x^2} - 12x - 2\]

Applying Rolle’s theorem in the interval \[\left[ {0,1} \right]\] , we get
\[f'\left( x \right) = 24{x^2} - 12x - 2\]
\[ \Rightarrow f'\left( c \right) = 12{c^2} - 6c - 1\]
Now, \[f'\left( c \right) = 0\] vanishes at \[c \in \left( {0,1} \right)\]
So, the function vanishes for some \[c \in \left( {0,1} \right)\].
Therefore, the given polynomial \[f\left( x \right) = 8{x^3} - 6{x^2} - 2x + 1\], \[f\left( x \right) = 0\] has at least one root in \[\left( {0,1} \right)\] and \[f'\left( c \right)\] vanishes for some \[c \in \left( {0,1} \right)\].
Thus, option (B) and (C) are the correct answers.

Note: We know that Rolle’s theorem states that suppose that a function \[f\left( x \right)\] is continuous on the closed interval \[\left[ {a,b} \right]\] and differentiable on the open interval \[\left( {a,b} \right)\] , then if \[f\left( a \right) = f\left( b \right)\] , then there exists at least one point \[c\] in the open interval \[\left( {a,b} \right)\]for which \[f'\left( c \right) = 0\] . The necessary conditions of Rolle’s theorem to be true is \[f\left( x \right)\] is continuous on the closed interval \[\left[ {a,b} \right]\] , \[f\left( x \right)\] is differentiable on the open interval and \[f\left( a \right) = f\left( b \right)\]. A continuous function is defined as a function when there is no abrupt change in the values of a function and always lies in the limits. A differentiable function is defined as a function whose derivative exists at all points of the interval.