Question
On the curve $${x^3} = 12y$$ the abscissa changes at a faster rate than the ordinate. Then $$x$$ belongs to the interval :
A.
$$\left( { - 2,\,2} \right)$$
B.
$$\left( { - 1,\,1} \right)$$
C.
$$\left( {0,\,2} \right)$$
D.
none of these
Answer :
$$\left( { - 2,\,2} \right)$$
Solution :
From the question, $$\left| {\frac{{dx}}{{dy}}} \right| > 1.$$ Differentiating $${x^3} = 12y$$ w.r.t. $$y$$
$$\eqalign{
& 3{x^2}\frac{{dx}}{{dy}} = 12\,\,\,\,\,\,\,\,\,\therefore \frac{{dx}}{{dy}} = \frac{4}{{{x^2}}} \cr
& \therefore \frac{4}{{{x^2}}} > 1\,\,\,\,\, \Rightarrow {x^2} - 4 < 0\,\,\,\,\, \Rightarrow - 2 < x < 2 \cr} $$