Q. The equation of the tangent to the curve $y = x + \frac{4}{x^{2}}$ that is parallel to the x-axis, is
A) Y=0
B) Y=1
C) Y=2
D) Y=3
Explanation
(d)
$We\ have\ ,\ y = x + \frac{4}{x^{2}}$
On differencing with respect to x, we get
$\frac{dy}{dx} = 1 - \frac{8}{x^{3}}$
Since, the tangent is parallel to $x - axis,$ therefore
$\frac{dy}{dx} = 0$
$\Rightarrow \ \ \ \ \ \ \ x^{3} = 8\ \ \ \ \Rightarrow x = 2\ and\ y = 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.