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`