Ex 6.4, 8 - If f(x) = 3x2 + 15x + 5, then approx value f(3.02)

Ex 6.4,8 - Chapter 6 Class 12 Application of Derivatives - Part 2

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.4, 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.

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