Q. The point on the curve $y = 2x^{2} - 6x - 4$ at which the tangent is parallel to the $x$-axis, is
A) $\left( \frac{3}{2},\frac{13}{2} \right)$
B) $\left( - \frac{5}{2},\ - \frac{17}{2} \right)$
C) $\left( \frac{3}{2},\frac{17}{2} \right)$
D) $\left( \frac{3}{2},\ - \frac{17}{2} \right)$
Explanation
(d)
Given, $y = 2x^{2} - 6x - 4\ \ \ \Rightarrow \ \ \ \ \frac{dy}{dx} = 4x - 6$
Since, $\frac{dy}{dx} = 0\ \ \ \ \ \ \Rightarrow \ \ \ \ \frac{dy}{dx} = 4x - 6 = 0\ \ \ \ \ \ \Rightarrow \ \ \ \ \ x = \frac{3}{2}$
$\Rightarrow \ \ \ \ \ \ \ y = 2.\frac{9}{4} - 6.\frac{3}{2} - 4 = - \frac{17}{2}\$
$\therefore$ Required point is $\left( \frac{3}{2},\ - \frac{17}{2} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.