Question 13
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)
Given Curve
"y = x3 – 12x + 18"
Finding slope of tangent
Differentiating w.r.t. 𝑥
𝑑𝑦/𝑑𝑥=𝑑(𝑥^3 − 12𝑥 + 18)/𝑑𝑥
𝑑𝑦/𝑑𝑥=3𝑥^2−12+0
𝑑𝑦/𝑑𝑥=3𝑥^2−12
𝑑𝑦/𝑑𝑥=3(𝑥^2−4)
𝒅𝒚/𝒅𝒙=𝟑(𝒙−𝟐)(𝒙+𝟐)
Given tangent to the curve is parallel to x-axis
Therefore,
Slope of tangent = Slope of x-axis
𝒅𝒚/𝒅𝒙=𝟎
3(𝑥−2)(𝑥+2)=0
(𝑥−2)(𝑥+2)=0
Thus, 𝒙=−𝟐 & 𝒙=𝟐
When 𝒙=−𝟐
"y = " 𝑥^3−12𝑥+18
=(−2)^3−12(−2)+18
=−8+24+18
=−8+42
=𝟑𝟒
Point is (−𝟐 , 𝟑𝟒)
When 𝒙=𝟐
"y = " 𝑥^3−12𝑥+18
=(2)^3−12(2)+18
=8−24+18
=26−24
=𝟐
Point is (𝟐 , 𝟐)
Hence , required points are (−𝟐 , 𝟑𝟒) & (𝟐 , 𝟐)
So, the correct answer is (D).

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.