The function f (x) = |x| + |x – 1| is

(A) continuous at x = 0 as well as at x = 1.

(B) continuous at x = 1 but not at x = 0.

(C) discontinuous at x = 0 as well as at x = 1.

(D) continuous at x = 0 but not at x = 1.

This question is similar to Ex 5.1, 34 - Chapter 5 Class 12 - Continuity and Differentiability

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

Transcript

Question 6 The function f (x) = |x| + |x – 1| is (A) continuous at x = 0 as well as at x = 1. (B) continuous at x = 1 but not at x = 0. (C) discontinuous at x = 0 as well as at x = 1. (D) continuous at x = 0 but not at x = 1. Given 𝑓(π‘₯)= |π‘₯|+|π‘₯βˆ’1| Here, we have 2 critical points i.e. π‘₯ = 0, and 𝒙 = 1 So, our intervals will be When π’™β‰€πŸŽ When 0<𝒙<𝟏 When 𝒙β‰₯𝟏 When π’™β‰€πŸŽ 𝑓(π‘₯)= |π‘₯|+ |π‘₯βˆ’1|. Here, both will be negative 𝑓(π‘₯)=(βˆ’π‘₯)+(βˆ’(π‘₯βˆ’1)) 𝑓(π‘₯)=βˆ’π‘₯βˆ’(π‘₯βˆ’1) 𝑓(π‘₯)=βˆ’π‘₯βˆ’π‘₯+1 " " 𝒇(𝒙)=βˆ’πŸπ’™+𝟏 When 0<𝒙<𝟏 𝑓(π‘₯)= |π‘₯| +|π‘₯βˆ’1|. Here, x will be positive, but (x - 1) will be negative 𝑓(π‘₯)=(π‘₯)+(βˆ’(π‘₯βˆ’1)) 𝑓(π‘₯)=π‘₯βˆ’(π‘₯βˆ’1) 𝑓(π‘₯)=π‘₯βˆ’π‘₯+1 " " 𝒇(𝒙)=𝟏 When 𝒙β‰₯𝟏 𝑓(π‘₯)= |π‘₯|+|π‘₯βˆ’1|. Here, both will be positive 𝑓(π‘₯)=π‘₯+(π‘₯βˆ’1) 𝑓(π‘₯)=π‘₯+π‘₯βˆ’1 " " 𝒇(𝒙)=πŸπ’™βˆ’πŸ Thus, our function becomes 𝒇(𝒙)={β–ˆ(βˆ’πŸπ’™+𝟏" " π’Šπ’‡ π’™β‰€πŸŽ@ 𝟏 π’Šπ’‡ 𝟎<𝒙<𝟏@πŸπ’™βˆ’πŸ π’Šπ’‡ 𝒙β‰₯𝟏)─ Since, from options we need to check continuity of the function when x = 0 and x = 1 Let’s check continuity Case 1 : When x = 0 f(x) is continuous at π‘₯ =0 if L.H.L = R.H.L = 𝑓(0) if (π₯𝐒𝐦)┬(π±β†’πŸŽ^βˆ’ ) 𝒇(𝒙)=(π₯𝐒𝐦)┬(π±β†’πŸŽ^+ ) " " 𝒇(𝒙)= 𝒇(𝟎) LHL at x β†’ 0 (π’π’Šπ’Ž)┬(π±β†’πŸŽ^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) f (βˆ’ h) = lim┬(hβ†’0) (βˆ’2(βˆ’β„Ž))+1 = lim┬(hβ†’0) (2β„Ž)+1 = 2(0) + 1 = 0 + 1 = 1 RHL at x β†’ 0 (π’π’Šπ’Ž)┬(π±β†’γ€–βˆ’πŸγ€—^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) f(h) = lim┬(hβ†’0) 1 = 1 Case 2 : When x = 1 f(x) is continuous at π‘₯=1 if L.H.L = R.H.L = 𝑓(1) if lim┬(xβ†’1^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’1^+ ) " " 𝑓(π‘₯)=𝑓(1) LHL at x β†’ 1 lim┬(xβ†’1^βˆ’ ) f(x) = lim┬(hβ†’0) f(1 βˆ’ h) = lim┬(hβ†’0) 1 = 1 RHL at x β†’ 0 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) 2(1+β„Ž)βˆ’1 = lim┬(hβ†’0) 2+2β„Žβˆ’1 = 2 + 2(0) – 1 =2 βˆ’ 1 = 1 & 𝑓(1) = 2(1)βˆ’1 𝒇(𝟏) = 1 Hence, L.H.L = R.H.L = 𝑓(1) ∴ f is continuous at x = 1 Therefore, f is continuous at x = 0 as well as at x = 1. So, the correct answer is (A)

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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.