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Ex 5.1, 15 - Ex 5.1

Ex 5.1, 15 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 15 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.1, 15 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.1, 15 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.1, 15 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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Ex 5.1, 15 Discuss the continuity of the function f, where f is defined by 𝑓(π‘₯)={ β–ˆ(2π‘₯, 𝑖𝑓 π‘₯<0@ 0, 𝑖𝑓 0≀π‘₯≀1@ 4π‘₯, 𝑖𝑓 π‘₯>1 )─ Since we need to find continuity at of the function We check continuity for different values of x When x < 0 When x = 0 When 0 < x < 1 When x = 1 When x > 1 Case 1 : When x < 0 For x < 0, f(x) = 2x Since this a polynomial It is continuous ∴ f(x) is continuous for x < 0 Case 2 : When x = 0 f(x) is continuous at π‘₯ =0 if L.H.L = R.H.L = 𝑓(0) if lim┬(xβ†’0^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’0^+ ) " " 𝑓(π‘₯)= 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x β†’ 0 lim┬(xβ†’0^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = lim┬(hβ†’0) f(βˆ’h) = lim┬(hβ†’0) 2(βˆ’β„Ž) = 2 Γ— 0 = 0 RHL at x β†’ 0 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) 0 = 0 & 𝑓(0) = 0 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x = 0 Case 3 : When 0 < x < 1 For 0 < x < 1 f(x) = 0 Since this constant It is continuous ∴ f(x) is continuous for 0 < x < 1 Case 4 : 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) Since L.H.L β‰  R.H.L f(x) is not continuous at x = 1 Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β†’ 1 lim┬(xβ†’1^βˆ’ ) f(x) = lim┬(hβ†’0) f(1 βˆ’ h) = lim┬(hβ†’0) 0 = 0 RHL at x β†’ 1 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) 4(1+β„Ž) = 4(1 + 0) = 4 Case 5: When x > 1 For x > 1, f(x) = 4x Since this a polynomial It is continuous ∴ f(x) is continuous for x > 1 Hence points of discontinuity are x = 1 Thus, f is continuous for all x ∈ R βˆ’ {𝟏}

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.