Find all points on the curve y = 4x^3-2x^5 at which the tangents pass…

The correct answer is Option (3) → (0, 0), (1, 2), and (−1, −2)

The given curve is $y = 4x^3-2x^5$   ...(i)

Let $P(x_1,y_1)$ be a point on the curve (i) at which the tangent to it passes through origin.

Differentiating (i) w.r.t. x, we get

$\frac{dy}{dx}=4.3x^2-2.5x^4=12x^2 - 10x^4$.

The slope of tangent to (i) at $P(x_1,y_1) = 12{x_1}^2 - 10{x_1}^4$.

The equation of tangent at $P(x_1,y_1)$ is

$y-y_1 = (12{x_1}^2 - 10{x_1}^4) (x − x_1)$.

As it passes through origin (0, 0),

$0-y_1 = (12{x_1}^2 - 10{x_1}^4) (0-x_1) ⇒ y_1 = x_1 (12{x_1}^2 - 10{x_1}^4)$   ...(ii)

As the given curve passes through $P(x_1,y_1), y_1 = 4{x_1}^3-2{x_1}^5$   ...(iii)

From (ii) and (iii), we get

$4{x_1}^3-2{x_1}^5 = 12{x_1}^3-10{x_1}^5⇒8{x_1}^5-8{x_1}^3 = 0$

$⇒{x_1}^3 ({x_1}^2-1)=0⇒{x_1} = 0, 1, -1$.

When $x_1 = 0, y_1 = 0$; when $x_1 = 1, y_1 =4-2=2;$

when $x_1 = -1, y_1 = -4+2=-2$.

Hence, the required points are (0, 0), (1, 2), (-1,-2).