LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
17 views
in Calculus by (54.3k points)
closed by
The equation of the tangent to the curve y = x3 at (1, 1) :
1. x - 10y + 50 = 0
2. 3x - y - 2 = 0
3. x + 3y - 4 = 0
4. x + 2y - 7 = 0
5. None of these

1 Answer

0 votes
by (30.0k points)
selected by
 
Best answer
Correct Answer - Option 2 : 3x - y - 2 = 0

Concept:

The equation of the tangent to a curve y = f(x) at a point (a, b) is given by (y - b) = m(x - a), where m = y'(b) = f'(a) [value of the derivative at point (a, b)].

Calculation:

y = f(x) = x3

⇒ y' = f'(x) = 3x2

m = f'(1) = 3 × 12 = 3

Equation of the tangent at (1, 1) will be:

(y - b) = m(x - a)

⇒ (y - 1) = 3(x - 1)

⇒ y - 1 = 3x - 3

3x - y - 2 = 0.

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

...