Ex 5.1, 12 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Transcript

Ex 5.1, 12 Find all points of discontinuity of f, where f is defined by 𝑓(𝑥)={ █(𝑥10−1, 𝑖𝑓 𝑥≤1@&𝑥2 , 𝑖𝑓 𝑥>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 𝑥 =1 if L.H.L = R.H.L = 𝑓(1) if lim┬(x→1^− ) 𝑓(𝑥)=lim┬(x→1^+ ) " " 𝑓(𝑥)= 𝑓(1) Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x → 1 lim┬(x→1^− ) f(x) = lim┬(h→0) f(1 − h) = lim┬(h→0) ((1−ℎ)^10−1) = (1−0)^10+1 = 1^10−1 = 1 − 1 = 0 RHL at x → 1 lim┬(x→1^+ ) f(x) = lim┬(h→0) f(1 + h) = lim┬(h→0) (1+ℎ)^2 = (1 + 0)2 = 12 = 1 Since L.H.L ≠ R.H.L ∴ f is not continuous at x = 1 Case 2 : When x < 1 For x < 1, f(x) = 𝑥^10−1 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 𝑥=1 is point of discontinuity. ∴ f is continuous at all real numbers except 1 Thus, f is continuous for 𝒙∈ R − {1}