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

Transcript

Ex 6.2,13 On which of the following intervals is the function f given by f (π‘₯) = π‘₯^100 + sin⁑π‘₯ –1 strictly decreasing ? (A) (0, 1) (B) (πœ‹/2, πœ‹) (C) (0, πœ‹/2) (D) None of these f (x) = π‘₯100 + sin x βˆ’ 1 f’(x) = 100π‘₯99 + cos x (A) (0, 1) For 0 < x < 1 0 < x < 1.57 099 < x99 < (1.57)99 0 < x99 < 2.47 Γ— 1099 0 < 100x99 < 247 Γ— 1019 0 < x < 1 ∴ 0 < x < πœ‹/2 So x is in 1st quadrant ∴ cos x is positive (As πœ‹/2 = 3.14/2 = 1.57 > 1) So, 100 x99 + cos x is positive. ∴ f’(x) > 0 f (x) is increasing on (0, 1) (B) (𝝅/𝟐, 𝝅) πœ‹/2 < x < πœ‹ 1.57 < x < 3.14 Since 100 x99 is much greater than –1 So 100 x99 + cos x is positive ∴ f’ (x) > 0 1.57 < x < 3.14 (1.57)99 < x99 < (3.14)99 2.47 Γ— 1099 < x99 < 3.14 Γ— 1099 247 Γ— 1099 < 100 x99 < 314 Γ— 1099 Since πœ‹/2 < x < πœ‹ so x is in 2nd quadrant ∴ cos x is negative. Minimum value of cos x = βˆ’1 (π‘ͺ)(𝟎, 𝝅/𝟐) 0<π‘₯< πœ‹/2 So 100 x99 + cos x is positive ∴ f’ (x) > 0 f(x) is increasing on (0,πœ‹/2) 0<π‘₯< πœ‹/2 0 < x < 1.57 099 < x99 < (1.57)99 0 < x99 < 2.47 Γ— 1099 0 < cos x < πœ‹/2 So x is in 1st quadrant ∴ cos x is positive Thus, f(x) is strictly decreasing for none of the intervals. So, (D) is the correct answer

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.