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, 16 Discuss the continuity of the function f, where f is defined by π(π₯)={β(β&2, ππ π₯β€β1@2&π₯, ππ β1<π₯β€1@2, ππ π₯>1 )β€ Since we need to find continuity at of the function We check continuity for different values of x When x < β1 When x = β1 When β1 < x < 1 When x = 1 When x > 1 Case 1 : When x < β1 For x < β1, f(x) = β2 Since this constant It is continuous β΄ f(x) is continuous for x < β1 Case 2 : When x = β1 f(x) is continuous at π₯ =β1 if L.H.L = R.H.L = π(β1) if limβ¬(xβγβ1γ^β ) π(π₯)=limβ¬(xβγβ1γ^+ ) " " π(π₯)= π(β1) Since there are two different functions on the left & right of β1, we take LHL & RHL . LHL at x β β1 limβ¬(xβγβ1γ^β ) f(x) = limβ¬(hβ0) f(β1 β h) = limβ¬(hβ0) 2(ββ) = 2 Γ 0 = 0 RHL at x β 0 limβ¬(xβγβ1γ^+ ) f(x) = limβ¬(hβ0) f(β1 + h) = limβ¬(hβ0) 0 = 0 & π(β1) = β2 Hence, L.H.L = R.H.L = π(β1) β΄ f is continuous at x=β1 Case 3 : When β1 < x < 1 For β1 < x < 1 f(x) = 2x Since this a polynomial It is continuous β΄ f(x) is continuous for β1 < x < 1 Case 4 : When x = 1 f(x) is continuous at π₯ =1 if L.H.L = R.H.L = π(1) if limβ¬(xβ1^β ) π(π₯)=limβ¬(xβ1^+ ) " " π(π₯)= π(1) Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β 1 limβ¬(xβ1^β ) f(x) = limβ¬(hβ0) f(1 β h) = limβ¬(hβ0) 0 = 0 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) 4(1+β) = 4(1 + 0) = 4 & π(1) = 2(1) = 2 Hence, L.H.L = R.H.L = π(1) β΄ f is continuous at x=1 Case 5: When x > 1 For x > 1, f(x) = 2 Since this constant It is continuous β΄ f(x) is continuous for x > 1 Thus, f(x) is continuous for all real numbers, i.e. f is continuous for all x β R limβ¬(xβγβ1γ^β ) π(π₯) = limβ¬(xββ1) (β2) = β2 limβ¬(xβγβ1γ^+ ) π(π₯) = limβ¬(xββ1) 2π₯ Putting x = β 1 = 2(β1) = β 2 And π(β2) = β2 Thus L.H.L = R.H.L = π(β2) = β2 Hence π(π) is continuous at x = βπ Case 1 At x = β1 A function is continuous at x = β 1 if A function is continuous at x = 1 if if L.H.L = R.H.L = π(1) i.e. limβ¬(xβ1^β ) π(π₯) = limβ¬(xβ1^+ ) π(π₯) = π(1) & π(1) = 2(1) = 2 Thus L.H.L = R.H.L = π(β2) Hence π(π) is continuous at x = 1 limβ¬(xβ1^β ) π(π₯) = limβ¬(xβ1^β ) 2π₯ Putting x = 1 = 2(1) = 2 limβ¬(xβ1^+ ) π(π₯) = limβ¬(xβ1^+ ) 2 = 2 Case 3 For π₯\<β1 π(π₯) = β2 Thus, π(π₯) is a constant function . & Every constant function is continuous for all real number. Hence π(π) is continuous at π\<βπ Case 4 For π₯>1 π(π₯) = 2 Thus, π(π₯) is a constant function . & Every constant function is continuous for all real number. Hence π(π) is continuous at π>π Case 5 For β1β€π₯β€1 π(π₯) = 2π₯ So, f(x) is a polynomial & Every polynomial is continuous. β π(π) is continuous at βπ<πβ€π Thus, f(x) is continuous for all real numbers, i.e. f is continuous for all x β R
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 You are here
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
Ex 5.1, 31
Ex 5.1, 32
Ex 5.1, 33
Ex 5.1, 34 Important
About the Author