Question

A particle is in motion along a curve $12y=x^3$. The rate of change of its ordinate exceeds that of abscissa in

• A. $–2<x<2$
• B. $x=\pm 2$
• C. $x<–2$
• D. $x>2$

Answer 1

Answer: (C), (D)

The correct options are
C $x<–2$
D $x>2$

Given $\frac{dy}{dt}>\frac{dx}{dt}...\left(1\right)$

$12y=x^3$
Differentiate w.r.t $t$
$\Rightarrow 12\frac{dy}{dt}=3x^2\frac{dx}{dt}\Rightarrow \frac{dy}{dt}=\frac{x^2}{4}\cdot \frac{dx}{dt}...\left(2\right)$

From eqn(1) and (2)
$\frac{x^2}{4}\cdot \frac{dx}{dt}>\frac{dx}{dt}$
$\Rightarrow x^2>4$
$\Rightarrow x\in (-\infty ,-2)\cup (2,\infty )$