Q. The set $\{ x^{3} - 12x: - 3 \leq x \leq 3\}$ is equal to
A) $\{ x:\ - 16 \leq x \leq 16\}$
B) $\{ x:\ - 12 \leq x \leq 12\}$
C) $\{ x:\ - 9 \leq x \leq 9\}$
D) $\{ x:\ 0 \leq x \leq 10\}$
Explanation
(a)
Let $y = x^{3} - 12x\ \ \ \ \ \ \ \Rightarrow \ \ \ \frac{dy}{dx} = 3x^{2} - 12$
Put $\frac{dy}{dx} = 0,\ \ \ \ \ 3x^{2} - 12 = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ x = \pm 2$
At $x = 2,\ \ \ \ \ y = 2^{3} - 12(2) = - 16$
At $x = - 2,\ \ \ \ \ y = ( - 2)^{3} - 12( - 2) = 16$
Hence, option (a) is correct
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.