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Ex 5.1, 30 - Find a and b such that f(x) = {5, ax + b, 21

Ex 5.1, 30 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 30 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.1, 30 - Chapter 5 Class 12 Continuity and Differentiability - Part 4 Ex 5.1, 30 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

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Ex 5.1, 30 Find the values of a and b such that the function defined by 𝑓(π‘₯)={β–ˆ(5, 𝑖𝑓 π‘₯≀[email protected]π‘Žπ‘₯+𝑏, 𝑖𝑓 2<π‘₯<[email protected], 𝑖𝑓 π‘₯β‰₯10)─ is a continuous function Since f(x) is a continuous function, It will be continuous for all values of x At x = 2 A function is continuous at x = 2 if L.H.L = R.H.L = 𝑓(2) i.e. lim┬(xβ†’2^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’2^+ ) " " 𝑓(π‘₯)= 𝑓(2) LHL at x β†’ 2 (π‘™π‘–π‘š)┬(π‘₯β†’2^βˆ’ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(2 βˆ’ h) = lim┬(hβ†’0) 5 = 5 RHL at x β†’ 2 (π‘™π‘–π‘š)┬(π‘₯β†’2^+ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(2 + h) = lim┬(hβ†’0) a(2 + h) + b = a(2 + 0) + b = 2a + b Since, LHL = RHL 2a + b = 5 At x = 10 𝑓 is continuous at x = 10 if L.H.L = R.H.L = 𝑓(10) i.e. lim┬(xβ†’10^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’10^+ ) " " 𝑓(π‘₯)= 𝑓(10) LHL at x β†’ 10 (π‘™π‘–π‘š)┬(π‘₯β†’10^βˆ’ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(10 βˆ’ h) = lim┬(hβ†’0) a(10 βˆ’ h) + b = a(10 βˆ’ 0) + b = 10a + b RHL at x β†’ 10 (π‘™π‘–π‘š)┬(π‘₯β†’10^+ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(10 + h) = lim┬(hβ†’0) 21 = 21 Since, L.H.L = R.H.L 10a + b = 21 Now, our equations are 2a + b = 5 …(1) 10a + b = 21 …(2) From (1) 2a + b = 5 b = 5 βˆ’ 2a Putting value of b in (2) 10π‘Ž+(5βˆ’2π‘Ž) = 21 10π‘Ž+5βˆ’2π‘Ž = 21 8π‘Ž = 21βˆ’5 8π‘Ž = 16 π‘Ž = 16/8 𝒂 = 𝟐 Putting value of a in (1) 2π‘Ž+𝑏=5 2(2)+𝑏=5 4+𝑏=5 𝑏=5βˆ’4 𝒃=𝟏 Hence, a = 2 & b = 1

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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.