Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
32.9k views
in Derivatives by (50.4k points)
closed by

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)

1 Answer

+1 vote
by (49.0k points)
selected by
 
Best answer

Given the equation of the curve is

y = x3 – 12x + 18

Differentiating on both sides with respect to x, we get

We know derivative of a constant is 0, so above equation becomes

So equating equation (i) to 0, we get

3x2-12=0

⇒ 3x2=12

⇒ x2=4

⇒ x=±2

When x=2, the given equation of curve becomes,

y = x3 – 12x + 18

⇒ y = (2)3 – 12(2) + 18

⇒ y = 8– 24 + 18

⇒ y = 2

When x=-2, the given equation of curve becomes,

y = x3 – 12x + 18

⇒ y = (-2)3 – 12(-2) + 18

⇒ y = -8+24 + 18

⇒ y = 34

Hence the points at which the tangents to the curve y = x3 – 12x + 18 are parallel to x-axis are (2, 2) and (-2, 34).

So the correct option is option D.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...