Check sibling questions

Ex 6.2, 13 On which intervals f (x) = x100 + sin ⁑x -1 strictly

Ex 6.2,13 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2,13 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.2,13 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.2,13 - Chapter 6 Class 12 Application of Derivatives - Part 5

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.