Question

The abscissa of the points of the curve \[y={{x}^{3}}\] in the interval \[\left[ -2,2 \right]\], where the slope of the tangents can be obtained by mean value theorem for the interval \[\left[ -2,2 \right]\] are
(1) \[\pm \dfrac{2}{\sqrt{3}}\]
(2) \[\pm \sqrt{3}\]
(3) \[\pm \dfrac{\sqrt{3}}{2}\]
(4) \[0\]

Answer 1

Hint: In this type of question we have to use the Mean Value Theorem. We know that the Mean value Theorem states that, if \[f\left( x \right)\] is a function which is continuous on the closed interval \[\left[ a,b \right]\] and differentiable on the open interval \[\left( a,b \right)\] then there exists a number \[c\] such that \[a < c < b\] and \[f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{\left( b-a \right)}\].

Complete step-by-step solution:
Now we have to find the slope of the tangents of the abscissa of the points of the curve \[y={{x}^{3}}\] in the interval \[\left[ -2,2 \right]\] by using Mean Value Theorem.
Given that the equation of the curve is \[y={{x}^{3}}\] so let us consider
\[\Rightarrow f\left( x \right)={{x}^{3}}\]
Now the Mean value Theorem states that, if \[f\left( x \right)\] is a function which is continuous on the closed interval \[\left[ a,b \right]\] and differentiable on the open interval \[\left( a,b \right)\] then there exists a number \[c\] such that \[a < c < b\] and \[f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{\left( b-a \right)}\].
Here, we can clearly observe that the function \[f\left( x \right)={{x}^{3}}\] is a continuous function on \[\left[ -2,2 \right]\] and differentiable on \[\left( -2,2 \right)\]. Also for the given function we have
\[\Rightarrow f\left( x \right)={{x}^{3}}\text{ and }a=-2,b=2\]
\[\Rightarrow f\left( a \right)=f\left( -2 \right)={{\left( -2 \right)}^{3}}=-8\text{ and }f\left( b \right)\text{=}f\left( 2 \right)={{\left( 2 \right)}^{3}}=8\]
By differentiating the given function with respect to \[x\] we get,
\[\Rightarrow f'\left( x \right)=3{{x}^{2}}\]
Hence, by Mean value Theorem there exists a number \[c\] such that \[a < c < b\] and \[f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{\left( b-a \right)}\]
\[\Rightarrow f'\left( c \right)=\dfrac{f\left( b \right)-f\left( a \right)}{\left( b-a \right)}\]
By substituting the values and on simplification we get
\[\begin{align}
  & \Rightarrow 3{{c}^{2}}=\dfrac{8-\left( -8 \right)}{2-\left( -2 \right)} \\
 & \Rightarrow 3{{c}^{2}}=\dfrac{16}{4} \\
 & \Rightarrow 3{{c}^{2}}=4 \\
 & \Rightarrow {{c}^{2}}=\dfrac{4}{3} \\
 & \Rightarrow c=\pm \dfrac{2}{\sqrt{3}} \\
\end{align}\]
Hence, option (1) is the correct option.

Note: In this type of question students have to remember the statement of Mean Value Theorem as well as they must have an idea about the implementation of the same. Also after obtaining the values of \[c\] students have to accept only those value of \[c\] which satisfies the condition \[a < c < b\] in the given problem as both values of \[c\] satisfies the condition \[a < c < b\] we accept both values.