Question 11
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).

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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