




Get live Maths 1-on-1 Classs - Class 6 to 12
Ex 5.1
Ex 5.1 ,2
Ex 5.1, 3 (a)
Ex 5.1, 3 (b)
Ex 5.1, 3 (c) Important
Ex 5.1, 3 (d) Important
Ex 5.1 ,4
Ex 5.1 ,5 Important
Ex 5.1 ,6
Ex 5.1 ,7 Important You are here
Ex 5.1 ,8
Ex 5.1, 9 Important
Ex 5.1, 10
Ex 5.1, 11
Ex 5.1, 12 Important
Ex 5.1, 13
Ex 5.1, 14
Ex 5.1, 15 Important
Ex 5.1, 16
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 (i) Important
Ex 5.1, 22 (ii)
Ex 5.1, 22 (iii)
Ex 5.1, 22 (iv) Important
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
Last updated at March 22, 2023 by Teachoo
Ex 5.1, 7 Find all points of discontinuity of f, where f is defined by 𝑓(𝑥)={█(|𝑥|+3, 𝑖𝑓 𝑥≤−[email protected] −2𝑥, 𝑖𝑓−3<𝑥<[email protected] 6𝑥+2, 𝑖𝑓 𝑥≥3)┤ Since we need to find continuity at of the function We check continuity for different values of x When x < −3 When x = −3 When −3 < x < 3 When x = 3 When x > 3 Case 1 : When x < −3 For x < −3, f(x) = |𝑥|+3 f(x) = −x + 3 Since this a polynomial It is continuous ∴ f(x) is continuous for x < −3 (As x < −3, x is negative) Case 2 : When x = −3 f(x) is continuous at 𝑥 =−3 if L.H.L = R.H.L = 𝑓(−3) if lim┬(x→〖−3〗^− ) 𝑓(𝑥)=lim┬(x→〖−3〗^+ ) " " 𝑓(𝑥)= 𝑓(−3) Since there are two different functions on the left & right of −3, we take LHL & RHL . LHL at x → −3 lim┬(x→〖−3〗^− ) f(x) = lim┬(h→0) f(−3 − h) = lim┬(h→0) (|−3−ℎ|+3) = |−3−0|+3 = |−3|+3 = 3 + 3 = 6 RHL at x → −3 lim┬(x→〖−3〗^+ ) f(x) = lim┬(h→0) f(−3 + h) = lim┬(h→0) −2(−3+ℎ) = lim┬(h→0) 6−2ℎ = 6 − 0 = 6 & 𝒇(−𝟑) = |−3|+3 = 3+3 = 6 Hence, L.H.L = R.H.L = 𝑓(−3) ∴ f is continuous at x = −3 Case 3 : When −3 < x < 3 For −3 < x < 3, f(x) = −2x Since this a polynomial It is continuous ∴ f(x) is continuous for −3 < x < 3 Case 4 : When x = 3 f(x) is continuous at 𝑥 =3 if L.H.L = R.H.L = 𝑓(3) if lim┬(x→3^− ) 𝑓(𝑥)=lim┬(x→3^+ ) " " 𝑓(𝑥)= 𝑓(3) Since there are two different functions on the left & right of 3, we take LHL & RHL . LHL at x → 3 lim┬(x→3^− ) f(x) = lim┬(h→0) f(3 − h) = lim┬(h→0) −2(3−h) = lim┬(h→0) −6+2h = −6 + 0 = −6 RHL at x → 3 lim┬(x→3^+ ) f(x) = lim┬(h→0) f(3 + h) = lim┬(h→0) 6(3+ℎ)+2 = lim┬(h→0) 18+6ℎ+2 = lim┬(h→0) 20+6ℎ = 20 + 0 = 20 Since L.H.L ≠ R.H.L f(x) is not continuous at x=3 Case 5: When x > 3 For x > 3, f(x) = 6x + 2 Since this a polynomial It is continuous ∴ f(x) is continuous for x > 3 Hence, f is discontinuous at only 𝑥 =3 Thus, f is continuous at all real numbers except 3. f is continuous at x ∈𝐑−{𝟑}