Let g(x) = {(ax^2-b) if x<2, (bx-a) if x>2}
Find a relationship between a and b (that is, solve for a in terms of b or vice versa) So that g(x) is continuous for all of x.
Since, g(x) is continuous for all of x.
Therefore, "(a.2^2-b)=b.2-a" [LHL = RHL]
"4a-b=2b-a\\Rightarrow 5a=3b \\Rightarrow a=\\frac{3}{5}b"
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