Q. The slope at any point of a curve $y = f(x)$ is given by $\frac{dy}{dx} = 3x^{2}$ and it passes through (-1,1). The equation of the curve is
A) $y = x^{3} + 2$
B) $y = - x^{3} - 2$
C) $y = 3x^{3} + 4$
D) $y = - x^{3} + 2$
Explanation
(a)
Given, $\frac{dy}{dx} = 3x^{3}$
$\Rightarrow$ $dy = 3x^{3}\ \ dx$
On integrating, we get
$y = \frac{3x^{3}}{3} + c$
$\Rightarrow$ $y = x^{3} + c$
It passes through (-1, 1).
$\therefore$ $1 = ( - 1)^{3} + c$
$\Rightarrow$ $c = 2$
$\therefore$ $y = x^{3} + 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.