Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
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Ex 5.1, 30 Important You are here
Ex 5.1, 34 Important
Ex 5.2, 5
Ex 5.2, 9 Important
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Ex 5.3, 10 Important
Ex 5.3, 14
Example 29 Important
Example 30 Important
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Ex 5.6, 7 Important
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Example 38 Important
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Question 4 Important
Question 5 Important
Example 39 (i)
Example 40 (i)
Example 42 Important
Misc 6 Important
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Misc 16 Important
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Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
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 x 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¬(xβ10^β ) π(π₯)=limβ¬(xβ10^+ ) " " π(π₯)= π(10) LHL at x β 10 (πππ)β¬(π₯β10^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₯β10^+ ) 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 = π Putting value of a in (1) 2π+π=5 2(2)+π=5 4+π=5 π=5β4 π=π Hence, a = 2 & b = 1