Check sibling questions

The set of points where the function f given by f (x) = |2x−1| sin x is differentiable is

(A)R     

(B) R − {1/2} 

(C) (0, ∞)    

(D) none of these

This question is similar to Ex 5.2, 9 - Chapter 5 Class 12 - Continuity and Differentiability


Transcript

Question 13 The set of points where the function f given by f (x) = |2x−1| sin x is differentiable is R (B) R − {1/2} (C) (0, ∞) (D) none of these f(x) = |2𝑥−1| sin⁡𝑥 = {█((2𝑥−1) sin⁡𝑥, 2𝑥−1≥0@−(2𝑥−1) sin⁡𝑥, 2𝑥−1<0)┤ = {█((2𝑥−1) sin⁡𝑥, 𝑥≥1/2@−(2𝑥−1) sin⁡〖𝑥 ,〗 𝑥<1/2)┤ Now, f(x) is a differentiable at x = 1/2 if LHD = RHD (𝒍𝒊𝒎)┬(𝐡→𝟎) (𝒇(𝒙) − 𝒇(𝒙 − 𝒉))/𝒉 = (𝑙𝑖𝑚)┬(h→0) (𝑓(1/2) − 𝑓(1/2 − ℎ))/ℎ =(𝑙𝑖𝑚)┬(h→0) (|2(1/2)−1| sin⁡(1/2)−|2(1/2 − ℎ)− 1| sin⁡(1/2 −ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) (0 −|1 − ℎ − 1| 〖sin 〗⁡(1/2 − ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) (0 −|− ℎ| sin⁡(1/2 − ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) (−ℎ 〖sin 〗⁡(1/2 − ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) −sin⁡(1/2 − ℎ) = −sin⁡(1/2−0) = −𝑠𝑖𝑛 1/2 = − 𝝅/𝟔 (𝒍𝒊𝒎)┬(𝐡→𝟎) (𝒇(𝒙+𝒉) − 𝒇(𝒙 ))/𝒉 = (𝑙𝑖𝑚)┬(h→0) (𝑓(1/2+ℎ) − 𝑓(1/2))/ℎ = (𝑙𝑖𝑚)┬(h→0) (|(2(1/2+ℎ)−1| 𝑠𝑖𝑛⁡(1/2+ℎ)−|2(1/2)−1| 𝑠𝑖𝑛⁡(1/2))/ℎ = (𝑙𝑖𝑚)┬(h→0) (|1+ ℎ −1| 〖sin 〗⁡〖(1/2 + ℎ) 〗− 0 )/ℎ = (𝑙𝑖𝑚)┬(h→0) (|ℎ| 〖sin 〗⁡(1/2+ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) (ℎ 〖sin 〗⁡(1/2+ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) sin⁡(1/2+ℎ) = 〖sin 〗⁡(1/2+0) = sin 1/2 = 𝝅/𝟔 Since LHD ≠ RHD ∴ f(x) is not differentiable at x = 1/2 Hence, we can say that f(x) is differentiable on R − {𝟏/𝟐} So, the correct answer is (B)

  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo