         1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise
3. Ex 5.1

Transcript

Ex 5.1, 34 Find all the points of discontinuity of f defined by 𝑓(𝑥)= |𝑥| – |𝑥+1|. Given 𝑓(𝑥)= |𝑥| – |𝑥+1|. Here, we have 2 critical points x = 0 and x + 1 = 0 i.e. x = 0, and x = −1 So, our intervals will be When 𝑥≤−1 When −1<𝑥<0 When 𝑥≥0 When 𝒙≤−𝟏 𝑓(𝑥)= |𝑥| – |𝑥+1|. Here, both will be negative 𝑓(𝑥)=(−𝑥) –(−(𝑥+1)) 𝑓(𝑥)=−𝑥+(𝑥+1) " " 𝑓(𝑥)=1 When −𝟏<𝒙≤𝟎 𝑓(𝑥)= |𝑥| – |𝑥+1|. Here, x will be negative, but (x + 1) will be positive 𝑓(𝑥)=(−𝑥) –(𝑥+1) 𝑓(𝑥)=−𝑥−𝑥−1 " " 𝑓(𝑥)=−2𝑥−1 |𝑥| = {█(𝑥, 𝑥 ≥0@−𝑥, 𝑥<0)┤ |𝑥+1| = {█((𝑥+1) , 𝑥+1≥0@−(𝑥+1) 𝑥+1<0)┤ = {█((𝑥+1) , 𝑥≥−1@−(𝑥+1) 𝑥<1)┤ When 𝒙≥𝟎 𝑓(𝑥)= |𝑥| – |𝑥+1|. Here, both will be positive 𝑓(𝑥)=𝑥 –(𝑥+1) 𝑓(𝑥)=𝑥−𝑥−1 " " 𝑓(𝑥)=−1 Thus, our function becomes 𝑓(𝑥)={█(1 𝑖𝑓 𝑥≤−1@−2𝑥−1 𝑖𝑓 −1<𝑥<0@−1 𝑖𝑓 𝑥≥0)┤ 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 −1 < x < 0 When x = 0 When x > 0 Checking continuity Case 1 : When x < −1 For x < −1, f(x) = 1 Since this constant It is continuous ∴ f(x) is continuous for x < −1 Case 2 : 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 = 1 RHL at x → 0 lim┬(x→〖−1〗^+ ) f(x) = lim┬(h→0) f(−1 + h) = lim┬(h→0) (−2(−1+ℎ))−1 = lim┬(h→0) (2−2ℎ)−1 = (2 − 2(0)) − 1 = 2 − 0 − 1 = 1 & 𝑓(−1) = 1 Hence, L.H.L = R.H.L = 𝑓(−1) ∴ f is continuous at x=−1 Case 3 : When −1 < x < 0 For −1 < x < 0 f(x) = −2x − 1 Since this a polynomial It is continuous ∴ f(x) is continuous for −1 < x < 0 Case 4 : When x = 0 f(x) is continuous at 𝑥 =0 if L.H.L = R.H.L = 𝑓(0) if lim┬(x→0^− ) 𝑓(𝑥)=lim┬(x→0^+ ) " " 𝑓(𝑥)= 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x → 0 lim┬(x→0^− ) f(x) = lim┬(h→0) f(0 − h) = lim┬(h→0) f(−h) = lim┬(h→0) −2(−ℎ)−1 = lim┬(h→0) 2ℎ−1 = 2(0) − 1 = −1 RHL at x → 0 lim┬(x→0^+ ) f(x) = lim┬(h→0) f(0 + h) = lim┬(h→0) f(h) = lim┬(h→0) −1 = −1 & 𝑓(0) = −1 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x=0 Case 5: When x > 0 For x > 0, f(x) = −1 Since this constant It is continuous ∴ f(x) is continuous for x > 0 Since there is no point of discontinuity Therefore, f is continuous for all x ∈ R

Ex 5.1 