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

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

Transcript

Ex 5.1, 30 Find the values of a and b such that the function defined by 𝑓(π‘₯)={β–ˆ(5, 𝑖𝑓 π‘₯≀2@π‘Žπ‘₯+𝑏, 𝑖𝑓 2<π‘₯<10@21, 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
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.