Slide13.JPG

Slide14.JPG
Slide15.JPG Slide16.JPG Slide17.JPG

  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 Let’s check sign of f’(x) in different intervals (A) (0, 1) For 0 < x < 1 Checking sign of 100x99 and cos x Since 0 < x < 1 099 < x99 < (1)99 0 < x99 < 1 0 < 100x99 < 100 ∴ 100x99 is positive Since 0 < x < 1 As πœ‹/2 = 3.14/2 = 1.57 > 1 Therefore, 0 < x < πœ‹/2 So, x is in 1st quadrant ∴ cos x is positive 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 Since 1.57 < x < 3.14 (1.57)99 < x99 < (3.14)99 (1.57)99 Γ— 100 < 100x99 < (3.14)99 Γ— 100 Since πœ‹/2 < x < πœ‹ So x is in 2nd quadrant ∴ cos x is negative. Minimum value of cos x = βˆ’1 (C) (𝟎,𝝅/𝟐) 0 < x < πœ‹/2 0 < x < 1.57 So, 100 x99 + cos x is positive. ∴ f’(x) > 0 f(x) is increasing on (0, 1) Since 0 < x < 1.57 099 < x99 < (1.57)99 0 Γ— 100 < 100x99 < (1.57)99 Γ— 100 0 < 100x99 < (1.57)99 Γ— 100 Since 0 < 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

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.