Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12




Last updated at Jan. 3, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12
Transcript
Ex 5.1, 30 Find the values of a and b such that the function defined by π(π₯)={β(5, ππ π₯β€2@ππ₯+π, ππ 2<π₯<10@21, ππ π₯β₯10)β€ is a continuous function Since f(x) is a continuous function, It will be continuous for all values of At x = 2 A function is continuous at x = 2 if L.H.L = R.H.L = π(2) i.e. limβ¬(xβ2^β ) π(π₯)=limβ¬(xβ2^+ ) " " π(π₯)= π(2) LHL at x β 2 (πππ)β¬(π₯β2^β ) f(x) = (πππ)β¬(ββ0) f(2 β h) = limβ¬(hβ0) 5 = 5 RHL at x β 2 (πππ)β¬(π₯β2^+ ) f(x) = (πππ)β¬(ββ0) f(2 + h) = limβ¬(hβ0) a(2 + h) + b = a(2 + 0) + b = 2a + b Since, LHL = RHL 2a + b = 5 At x = 10 π is continuous at x = 10 if L.H.L = R.H.L = π(10) i.e. limβ¬(xβγ10γ^β ) π(π₯)=limβ¬(xβγ10γ^+ ) " " π(π₯)= π(10) LHL at x β 10 (πππ)β¬(π₯βγ10γ^β ) f(x) = (πππ)β¬(ββ0) f(10 β h) = limβ¬(hβ0) a(10 β h) + b = a(10 β 0) + b = 10a + b RHL at x β 10 (πππ)β¬(π₯βγ10γ^+ ) f(x) = (πππ)β¬(ββ0) f(10 + h) = limβ¬(hβ0) 21 = 21 Since, L.H.L = R.H.L 10a + b = 21 Now, our equations are 2a + b = 5 β¦(1) 10a + b = 21 β¦(2) From (1) 2a + b = 5 b = 5 β 2a Putting value of b in (2) 10π+(5β2π) = 21 10π+5β2π = 21 8π = 21β5 8π = 16 π = 16/8 π = 2 Putting value of a in (1) 2π+π=5 2(2)+π=5 4+π=5 π=5β4 π=1 Hence, a = 2 & b = 1
Ex 5.1
Ex 5.1 ,2
Ex 5.1 ,3 Important
Ex 5.1 ,4
Ex 5.1 ,5 Important
Ex 5.1 ,6
Ex 5.1 ,7 Important
Ex 5.1 ,8
Ex 5.1, 9 Important
Ex 5.1, 10
Ex 5.1, 11
Ex 5.1, 12
Ex 5.1, 13
Ex 5.1, 14
Ex 5.1, 15 Important
Ex 5.1, 16
Ex 5.1, 17 Important
Ex 5.1, 18 Important
Ex 5.1, 19 Important
Ex 5.1, 20
Ex 5.1, 21
Ex 5.1, 22
Ex 5.1, 23
Ex 5.1, 24 Important
Ex 5.1, 25
Ex 5.1, 26 Important
Ex 5.1, 27
Ex 5.1, 28 Important
Ex 5.1, 29
Ex 5.1, 30 Important You are here
Ex 5.1, 31
Ex 5.1, 32
Ex 5.1, 33
Ex 5.1, 34 Important
About the Author