For what values of a and b is the function

`f(x) = (x^2 - 4)/(x - 2)` for x < 2

= ax^{2} − bx + 3 for 2 ≤ x < 3

= 2x – a + b for x ≥ 3

continuous in its domain.

#### Solution

Function f is continuous for every x on R.

∴ Function f is continuous at x = 2 and x = 3.

As f is continuous at x = 2.

∴ `lim_(x→2^-) "f"(x) = lim_(x→2^+) "f"(x)`

∴ `lim_(x→2^-) (x^2 - 4)/(x - 2) = lim_(x→2^+) ("a"x^2 - "b"x + 3)`

∴ `lim_(x→2^-) ((x - 2)(x + 2))/(x - 2) = lim_(x→2^+) ("a"x^2 - "b"x + 3)`

∴ `lim_(x→2^-) (x + 2) = lim_(x→2^+) ("a"x^2 - "b"x + 3) ...[(because x → 2 therefore x ≠ 2),(therefore x - 2 ≠ 0)]`

∴ 2 + 2 = a(2)^{2} – b(2) + 3

∴ 4 = 4a – 2b + 3

∴ 4a – 2b = 1 ...(i)

Also function f is continuous at x = 3

∴ `lim_(x→3^-) "f"(x) = lim_(x→3^+) "f"(x)`

∴ `lim_(x→3^-) ("a"x^2 - "b"x + 3) = lim_(x→3^+) (2x - "a + b")`

∴ a(3)^{2} – b(3) + 3 = 2(3) – a + b

∴ 9a – 3b + 3 = 6 – a + b

∴ 10a – 4b = 3 ...(ii)

Multiplying (i) by 2, we get

8a – 4b = 2 ...(iii)

Subtracting (iii) from (ii), we get

2a = 1

∴ a = `1/2`

Substituting a = `1/2` in (i), we get

`4(1/2) - 2"b" = 1`

∴ 2 – 2b = 1

∴ 1 = 2b

∴ b = `1/2`

∴ a = `1/2` and b = `1/2`