A point moves along the curve 12y = x3 in such a way that the rate of increase of its ordinate is more than the rate of increase of abscissa. The abscissa of the point lies in the interval

A
$(- 2, 2)$
B
$(- \infty, - 2) \cup (2, \infty)$
C
$[- 2, 2]$
D
none of these

Solution:

It is given that

$\frac{d y}{d t} > \frac{d x}{d t}, \frac{d y}{d t} > 0$ and $\frac{d x}{d t} > 0$

Now, $12 y = x^{3}$

$\Rightarrow 12 \frac{d y}{d t} = 3 x^{2} \frac{d x}{d t} \Rightarrow 3 (x^{2} \frac{d x}{d t} - 4 \frac{d y}{d t}) = 0 \Rightarrow x^{2} \frac{d x}{d t} = 4 \frac{d y}{d t}$

We have,

$\begin{array}{l} \frac{d y}{d t} > \frac{d x}{d t} \\ \Rightarrow \frac{x^{2}}{4} \frac{d x}{d t} > \frac{d x}{d t} \\ \Rightarrow (x^{2} - 4) \frac{d x}{d t} > 0 \Rightarrow x^{2} - 4 > 0 \Rightarrow x \in (- \infty, - 2) \cup (2, \infty) \end{array}$

A point moves along the curve $12 y = x^{3}$ in such a way that the rate of increase of its ordinate is more than the rate of increase of abscissa. The abscissa of the point lies in the interval

Solution:

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