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Question
For what values of a and b is the function
f(x) `{:(= (x^2 - 4)/(x - 2)",", "for" x < 2),(= "a"x^2 - "b"x + 3",", "for" 2 ≤ x < 3),(= 2x - "a" + "b"",", "for" x ≥ 3):}}` continuous for every x on R?
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Solution
f(x) is continuous for every x on R.
∴ f(x) is continuous at x = 2 and x = 3.
f(x) 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
∴ 4a – 2b + 3 = 4
∴ 4a – 2b = 1 ...(i)
Also f(x) 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)
Multiply (i) by 2 and (ii) by 1, we get
8a – 4b = 2 ...(iii)
10a – 4b = 3 ...(iv)
Subtracting (iv) from (iii)
– 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`
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