# Find a and b if following function is continuous at the point or on the interval indicated against them: f(x) =ax2+bx+1, for |2x-3|≥2=3x+2, for 12<x<52 - Mathematics and Statistics

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Find a and b if following function is continuous at the point or on the interval indicated against them:

f(x) {:(= "a"x^2 + "b"x + 1",", "for"  |2x - 3| ≥ 2),(= 3x + 2",", "for"  1/2 < x < 5/2):}

#### Solution

|2x – 3| = ± (2x – 3)

∴ 2x – 3 ≥ 2 or 3 – 2x ≥ 2

∴ 2x ≥ 5 or 1 ≥ 2x

∴ x ≥ 5/2 or x ≤ 1/2

∴ f can be expressed as

f(x) {:(= "a"x^2 + "b"x + 1",", "for"  x ≤ 1/2),(= 3x + 2",", "for"  1/2 < x < 5/2), (= "a"x^2 + "b"x + 1",", "for"  x ≥ 5/2):}

∵ f is continuous on its domain

∴ f is continuous at x = 1/2 and x = 5/2

Since f is continuous at x = 1/2,

"f"(1/2) = lim_(x -> (1/2)^+) "f"(x)

∴ ("a"x^2 + "b"x + 1)_("at"  x = 1/2) =  lim_(x -> 1/2) (3x + 2)

∴ "a"/4 + "b"/2 + 1 = 3 xx 1/2 + 2 = 7/2

∴ a + 2b + 4 = 14

∴ a + 2b = 10      ...(1)

Also f is continuous at x = 5/2

∴ "f"(5/2) = lim_(x -> (5/2)^(-)) "f"(x)

∴ ("a"x^2 + "b"x + 1)_("at"  x = 5/2) =  lim_(x -> 5/2) (3x + 2)

∴ (25"a")/4 + (5"b")/2 + 1 = 3 xx 5/2 + 2 = 19/2

∴ 25a + 10b + 4 = 38

∴  25a + 10b = 34

∴ 25 (10 – 2b) + 10b = 34    ...[By (1)]

∴ – 40b = – 216

∴ b = 216/40 = 27/5

∴ from (1),

∴ "a" + 54/5 = 10

∴ a = 10 - 54/5 = - 4/5

Hence, a =  -4/5, b = 27/5

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 8 Continuity
Miscellaneous Exercise 8 | Q VI. (2) | Page 178