Check sibling questions

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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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.