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For what values of a and b is the function f(x)=x2-4x-2 for x < 2 = ax2 − bx + 3 for 2 ≤ x < 3= 2x – a + b for x ≥ 3continuous in its domain. - Mathematics and Statistics

Sum

For what values of a and b is the function

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

= ax2 − bx + 3         for 2 ≤ x < 3
= 2x – a + b             for x ≥ 3
continuous in its domain.

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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`

Concept: Continuity in the Domain of the Function
  Is there an error in this question or solution?
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