Question

The abscissa of the points of the curve $y=x^3$ in the interval $[–2,2]$, where the slope of the tangents can be obtained by mean value theorem for the interval $[–2,2]$, are

• A. $\pm \frac{2}{\sqrt{3}}$
• B. $\pm \sqrt{3}$
• C. $\pm \frac{\sqrt{3}}{2}$
• D. $0$

Answer 1

The correct option is A

$\pm \frac{2}{\sqrt{3}}$

Explanation for the correct option:

Finding the abscissa of the points of the curve.

The given curve is $y=x^3...\left(i\right)$

Considering $f\left(x\right)=x^3$

Therefore, $f\left(2\right)=8\&f(-2)=–8$

We know that coordinates of every point on the curve be of the form $(x,y)$

And in $(x,y)$, $x$ is called abscissa.

Now differentiating $\left(i\right)$ with respect to $x$ to find the slope of the tangent

$\Rightarrow \frac{dy}{dx}=f\text{'}\left(x\right)=3x^2$

We know according to mean value theorem that slope of tangent

$=\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}$

Therefore,

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{\left[f\left(2\right)–f(-2)\right]}{[2–(-2\left)\right]} \\ =\frac{\left[8–(-8)\right]}{4} \\ \Rightarrow 3x^2=4\left[\because f\text{'}\left(x\right)=3x^2\right] \\ \Rightarrow x=\pm \frac{2}{\sqrt{3}} \end{gathered}$$

Hence, option (A) is the correct answer.