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

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


Transcript

Ex 5.1, 30 Find the values of a and b such that the function defined by 𝑓(𝑥)={█(5, 𝑖𝑓 𝑥≤2@𝑎𝑥+𝑏, 𝑖𝑓 2<𝑥<10@21, 𝑖𝑓 𝑥≥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┬(x→2^− ) 𝑓(𝑥)=lim┬(x→2^+ ) " " 𝑓(𝑥)= 𝑓(2) LHL at x → 2 (𝑙𝑖𝑚)┬(𝑥→2^− ) f(x) = (𝑙𝑖𝑚)┬(ℎ→0) f(2 − h) = lim┬(h→0) 5 = 5 RHL at x → 2 (𝑙𝑖𝑚)┬(𝑥→2^+ ) f(x) = (𝑙𝑖𝑚)┬(ℎ→0) f(2 + h) = lim┬(h→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┬(x→10^− ) 𝑓(𝑥)=lim┬(x→10^+ ) " " 𝑓(𝑥)= 𝑓(10) LHL at x → 10 (𝑙𝑖𝑚)┬(𝑥→10^− ) f(x) = (𝑙𝑖𝑚)┬(ℎ→0) f(10 − h) = lim┬(h→0) a(10 − h) + b = a(10 − 0) + b = 10a + b RHL at x → 10 (𝑙𝑖𝑚)┬(𝑥→10^+ ) f(x) = (𝑙𝑖𝑚)┬(ℎ→0) f(10 + h) = lim┬(h→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 − 2a Putting value of b in (2) 10𝑎+(5−2𝑎) = 21 10𝑎+5−2𝑎 = 21 8𝑎 = 21−5 8𝑎 = 16 𝑎 = 16/8 𝒂 = 𝟐 Putting value of a in (1) 2𝑎+𝑏=5 2(2)+𝑏=5 4+𝑏=5 𝑏=5−4 𝒃=𝟏 Hence, a = 2 & b = 1

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.