1. Class 11
2. Important Question for exams Class 11

Transcript

Ex 13.1, 28 (Method 1) Suppose f(x) = a+bx, x<1﷮4, x=1﷮b−ax, x>1﷯﷯ and if lim﷮x→1﷯ f(x) = f(1) what are possible values of a and b? Given lim﷮x→1﷯ f(x) = f (1) Thus, limit exists at x = 1 ⇒ Left hand limit = Right hand limit ⇒ lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) So, lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) = lim﷮x→1﷯ f(x) lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) = f (1) = 4 Thus , lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) = 4 f(x) = a+bx, x<1﷮4, x=1﷮b−ax, x>1﷯﷯ From (1) lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) = 4 Putting values b – a = a + b = 4 Thus, we have equations b – a = 4 …(1) & a + b = 4 …(2) Adding (1) & (2) b – a + a + b = 4 + 4 2b = 8 b = 8﷮2﷯ = 4 Putting b = 4 in (2) a + b = 4 a + 4 = 4 a = 4 – 4 a = 0 Hence a = 0 & b = 4 Ex13.1, 28 (Method 2) Suppose f(x) = a+bx, x<1﷮4, x=1﷮b−ax, x>1﷯﷯ and if lim﷮x→1﷯ f(x) = f(1) what are possible values of a and b? Given lim﷮x→1﷯ f(x) = f (1) Thus, limit exists at x = 1 ⇒ Left hand limit = Right hand limit ⇒ lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) So, lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) = lim﷮x→1﷯ f(x) lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) = f (1) = 4 Thus , lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) = 4 Finding 𝒍𝒊𝒎﷮ 𝐱→𝟏﷮+﷯﷯f(x) So, f(x) = b – ax So, as x tends towards 1, f(x) tends towards b – (1) a Thus, lim﷮ x→1﷮+﷯﷯ f(x) = b – (1)a lim﷮ x→1﷮+﷯﷯ f(x) = b – a Finding 𝒍𝒊𝒎﷮ 𝐱→𝟏﷮−﷯﷯f(x) So, f(x) = a + bx So, as x tends towards 1, f(x) tends towards a + b(1) Thus, lim﷮ x→1﷮−﷯﷯ f(x) = a + b(1) = a + b From (1) lim﷮ x→1﷮+﷯﷯f(x) = lim﷮ x→1﷮−﷯﷯f(x) = 4 Putting values b – a = a + b = 4 Thus, we have equations b – a = 4 …(1) & a + b = 4 …(2) Adding (1) & (2) b – a + a + b = 4 + 4 2b = 8 b = 8﷮2﷯ = 4 Putting b = 4 in (2) a + b = 4 a + 4 = 4 a = 4 – 4 a = 0 Hence a = 0 & b = 4

Class 11
Important Question for exams Class 11