Ex 6.2

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

Ex 6.2, 13 On which of the following intervals is the function f given by f (π₯) = π₯^100 + sinβ‘π₯ β1 strictly decreasing ? (A) (0, 1) (B) (π/2, π) (C) (0, π/2) (D) None of these f(x) = π₯100 + sin x β 1 fβ(x) = 100π₯99 + cos x Letβs check sign of fβ(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 β΄ 100x99 is positive Since 0 < x < 1 As π/2 = 3.14/2 = 1.57 > 1 Therefore, 0 < x < π/2 So, x is in 1st quadrant β΄ cos x is positive So, 100 x99 + cos x is positive. β΄ fβ(x) > 0 f(x) is increasing on (0, 1) (B) (π/π, π) π/2 < x < π 1.57 < x < 3.14 Since 100 x99 is much greater than β1 So 100 x99 + cos x is positive β΄ fβ(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/2 < x < π So x is in 2nd quadrant β΄ cos x is negative. Minimum value of cos x = β1 (C) (π,π/π) 0 < x < π/2 0 < x < 1.57 So, 100 x99 + cos x is positive. β΄ fβ(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/2 So x is in 1st quadrant β΄ cos x is positive Thus, f(x) is strictly decreasing for none of the intervals. So, (D) is the correct answer