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

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Ex 5.1, 3 Examine the following functions for continuity. (a) f(x) = x – 5 f(x) = x – 5 Since x – 5 is a polynomial. ∴ f(x) is defined for every real number c. Let us check continuity at x = c f(x) is is continuous at x = c if lim┬(x→𝑐) 𝑓(𝑥) = 𝑓(𝑐) lim┬(x→𝑐) 𝑓(𝑥) = lim┬(x→𝑐) 𝑥 − 5 = c − 5 f (c) = c − 5 Since lim┬(x→𝑐) 𝑓(𝑥) = 𝑓(𝑐) So, f is continuous for x = c, where c is a real number ∴ f is continuous for all real numbers Hence, f is continuous for each x ∈ R Ex 5.1, 3 Examine the following functions for continuity. (b) f (x) = 1/(𝑥 − 5) , x ≠ 5 If x = 5 f (x) = 1/(5 − 5) = 1/0 = ∞ Hence at x = 5, f (x) is not defined. Let us check continuity at x = c, where x ≠ 5 f(x) is continuous at x = c, x ≠ 5 if lim┬(x→𝑐) 𝑓(𝑥) = 𝑓(𝑐) lim┬(x→𝑐) 𝑓(𝑥) = lim┬(𝑥⟶𝑐)⁡〖1/(𝑥 − 5)〗 = 1/(𝑐 − 5) f (c) = 1/(𝐶 − 5) Since lim┬(x→𝑐) 𝑓(𝑥) = 𝑓(𝑐) f is continuous for all real numbers except 5 ∴ f is continuous at each 𝐱 ∈ R − {𝟓} Ex 5.1, 3 Examine the following functions for continuity. (c) f (x) = (𝑥^(2 )− 25 )/(𝑥 + 5), x ≠ –5 f (x) = (𝑥^(2 )− 25 )/(𝑥 + 5) Putting x = –5 f (−5) = (〖(−5)〗^(2 )− 25 )/(−5 + 5) = (25− 25 )/(−5 + 5) = 0/0 = undefined Hence at x = −5, f (x) is not defined. Let us check continuity at x = c, where x ≠ −5 Given function is continuous at x = c , where c is a real number except −5 if lim┬(x→𝑐) 𝑓(𝑥) = 𝑓(𝑐) f(x) is continuous at x = c, x ≠ −5 if lim┬(x→𝑐) 𝑓(𝑥) = 𝑓(𝑐) Since lim┬(x→𝑐) 𝑓(𝑥) = f (c) f is continuous for all real numbers except −5 ∴ f is continuous at each 𝐱 ∈ R − {−𝟓} lim┬(x→𝑐) 𝑓(𝑥) = lim┬(x→𝑐) (𝑥^2− 25)/(𝑥 + 5) = lim┬(x→𝑐) ((𝑥 − 5) (𝑥 + 5))/(𝑥 + 5) = lim┬(x→𝑐) 𝑥−5 Putting x = c = c − 5 f (x) = (𝑥^(2 )− 25 )/(𝑥 + 5) f (c) = (𝑐^(2 )− 25 )/(𝑐 + 5) f (c) = ((𝑐 − 5)(𝑐 + 5))/((𝑐 + 5)) f (c) = c − 5 Ex 5.1, 3 Examine the following functions for continuity. (d) f (x) = |x – 5| f(x) = |𝑥−5| = {█((𝑥−5), 𝑥−5≥0@−(𝑥−5), 𝑥−5<0)┤ = {█((𝑥−5), 𝑥≥5@−(𝑥−5), 𝑥<5)┤ Since we need to find continuity at of the function We check continuity for different values of x When x = 5 When x < 5 When x > 5 Case 1 : When x = 5 f(x) is continuous at 𝑥 = 5 if L.H.L = R.H.L = 𝑓(5) if lim┬(x→5^− ) 𝑓(𝑥)=lim┬(x→5^+ ) " " 𝑓(𝑥)= 𝑓(5) Since there are two different functions on the left & right of 5, we take LHL & RHL . LHL at x → 5 lim┬(x→5^− ) f(x) = lim┬(h→0) f(5 − h) = lim┬(h→0) − ((5−ℎ)−5) = lim┬(h→0) −(−ℎ) = 0 RHL at x → 5 lim┬(x→5^+ ) f(x) = lim┬(h→0) f(5 + h) = lim┬(h→0) ((5+ℎ)−5) = lim┬(h→0) ℎ = 0 & 𝑓(5) = (𝑥−5) = 5−5 = 0 Hence, L.H.L = R.H.L = 𝑓(5) ∴ f is continuous at x=5 Case 2 : When x < 5 For x < 5, f(x) = − (x − 5) Since this a polynomial It is continuous ∴ f(x) is continuous for x < 5 Case 3 : When x > 5 For x > 5, f(x) = (x − 5) Since this a polynomial It is continuous ∴ f(x) is continuous for x > 5 Hence, 𝑓(𝑥)= |𝑥−5| is continuous at all points. i.e. f is continuous at 𝒙 ∈ R.

Ex 5.1 