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Question 8 (MCQ) - Finding approximate value of function - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Finding approximate value of function
Chapter 6 Class 12 Application of Derivatives
Concept wise
- Finding rate of change
- To show increasing/decreasing in whole domain
- To show increasing/decreasing in intervals
- Find intervals of increasing/decreasing
- Finding slope of tangent/normal
- Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
- Finding equation of tangent/normal when slope and curve are given
- Finding approximate value of numbers
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Finding approximate value of function
- Finding approximate value- Statement questions
- Finding minimum and maximum values from graph
- Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
- Absolute minima/maxima
Transcript
Question 8 If f(x) = 3x2 + 15x + 5, then the approximate value of f (3.02) is (A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66Let x = 3 and ∆x = 0.02 f’(x) = 6x + 15 Now, ∆y = f’(x) ∆x = (6x + 15) 0.02 Also, ∆y = f (x + ∆x) − f(x) f (x + ∆x) = f (x) + ∆y f (3.02) = 3x2 + 15x + 5 + (6x + 15) (0.02) Since x = 3 f (3.02) = 3(3)2 + 15(3) + 5 + (0.02)[6(3)+15] = (27 + 45 + 5) + 33(0.02) = 77 + 0.66 = 77.66 Hence, part (D) is the correct answer.
