Question

The points at which the tangents to the curve y = x³ - 12x + 18 are parallel to x-axis are
(a) (2, -2)(-2, -34) (b) (2, 34)(-2, 0)
(c) (0, 34)(-2, 0) (d) (2, 2)(-2, 34)

Answer 1

Let (x₁, y₁) be the point of contact of tangent with the given curve y = x³ − 12x + 18.

∴ $y_1=x_1^3-12x+18$ .....(1)

y = x³ − 12x + 18

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=3x^2-12$

∴ Slope of tangent at (x₁, y₁) = ${\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=3x_1^2-12$

It is given that, the tangent is parallel to the x-axis.

$\therefore {\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=0$

$\Rightarrow 3x_1^2-12=0$

$\Rightarrow x_1^2=4$

⇒ x₁ = −2 or x₁ = 2

Putting x₁ = −2 in (1), we get

$y_1={\left(-2\right)}^3-12\times \left(-2\right)+18=-8+24+18=34$

Putting x₁ = 2 in (1), we get

$y_1={\left(2\right)}^3-12\times 2+18=8-24+18=2$

So, the coordinates of the point of contact are (−2, 34) and (2, 2).

Thus, the points at which the tangents to the given curve are parallel to x-axis are (−2, 34) and (2, 2).

Hence, the correct answer is option (d).