Ex 5.1, 12 - Ex 5.1

Ex 5.1, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.1, 12 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Ex 5.1, 12 Find all points of discontinuity of f, where f is defined by 𝑓(π‘₯)={ β–ˆ(π‘₯10βˆ’1, 𝑖𝑓 π‘₯≀1@&π‘₯2 , 𝑖𝑓 π‘₯>1)─ Since we need to find continuity at of the function We check continuity for different values of x When x = 1 When x < 1 When x > 1 Case 1 : When x = 1 f(x) is continuous at π‘₯ =1 if L.H.L = R.H.L = 𝑓(1) if lim┬(xβ†’1^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’1^+ ) " " 𝑓(π‘₯)= 𝑓(1) Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β†’ 1 lim┬(xβ†’1^βˆ’ ) f(x) = lim┬(hβ†’0) f(1 βˆ’ h) = lim┬(hβ†’0) ((1βˆ’β„Ž)^10βˆ’1) = (1βˆ’0)^10+1 = 1^10βˆ’1 = 1 βˆ’ 1 = 0 RHL at x β†’ 1 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) (1+β„Ž)^2 = (1 + 0)2 = 12 = 1 Since L.H.L β‰  R.H.L ∴ f is not continuous at x = 1 Case 2 : When x < 1 For x < 1, f(x) = π‘₯^10βˆ’1 Since this a polynomial It is continuous ∴ f(x) is continuous for x < 1 Case 3 : When x > 1 For x > 1, f(x) = x2 Since this a polynomial It is continuous ∴ f(x) is continuous for x > 1 Hence, only π‘₯=1 is point of discontinuity. ∴ f is continuous at all real numbers except 1 Thus, f is continuous for π’™βˆˆ R βˆ’ {1}

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