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
[MP PET 1993]

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

Answer 1

The correct option is A $\pm \frac{2}{\sqrt{3}}$

According to the mean value theorem,
f'(c) = $\frac{f\left(b\right)-f\left(a\right)}{b-a}$ in the interval [a,b]
f'(c) = $\frac{f\left(8\right)-f(-8)}{8+8}$
f'(c) = 4
Also, f'(c) = $3c^2$
so, $3c^2$ = 4
or c = $\pm \frac{2}{\sqrt{3}}$