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Question

Let two functions are defined as g(x)={x2,1x2x+2,2x3, and f(x)={x+1x12x+11<x2, then find gof(1)

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Solution

From the given functions we get
f(1)=2 since f(x)=x+1 for all x1...(i)
Now
limx2g(x)
=limx2(x2)
=4
limx2+g(x)
=limx2+(x+2)
=4
Hence
LHL=RHL.
Thus the function g(x) is continuous at x=2.
Now
g(f(1))
=g(2) since f(1)=2
=4.... from above limit calculation.
Hence
g(f(1))=4

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