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Question
Find the relationship between a and b so that the function f defined by `f(x)= {(ax + 1, if x<= 3),(bx + 3, if x > 3):}` is continuous at x = 3.
Solution
`f(x)= {(ax + 1, if x<= 3),(bx + 3, if x > 3):}`
At x = 3
f(x) = ax + 1
When x = 3
L.H.L = `lim_(x -> 0)` + f(x)
= `lim_(x -> 0)` + (ax + 1)
= 3a + 1
f(3) = 3a + 1
f(x) = bx + 1 when x > 3
R.H.L = `lim_(x -> 0)` + f(x)
= `lim_(x -> 0)` + (bx + 3)
= 3b + 3
3a + 1 = 3b + 3
a = b + `2/3`
The value of a can be found for any arbitrary value of b.
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