The points at which the tangents to the curve y = x 3 – 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)

 

This question is similar to Ex 6.3, 7 - Chapter 6 Class 12 - Application of Derivatives

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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).

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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 and Computer Science at Teachoo