Answer :

It is given that f(x) is differentiable at each x є R


For x ≤ 1,


f(x) = x2 + 3x + a i.e. a polynomial


for x > 1,


f(x) = bx + 2, which is also a polynomial


Since, a polynomial function is everywhere differentiable. Therefore, f(x) is differentiable for all x > 1 and for all x < 1.


f(x) is continuous at x = 1





4 + a = b + 2


a – b + 2 = 0 …(1)


As function is differentiable, therefore, LHD = RHD


LHD at x = 1:




RHD at x = 1:




As, LHD = RHD


Therefore,


5 = b


Putting b in (1), we get,


a – b + 2 = 0


a - 5 + 2 =0


a = 3


Hence,


a = 3 and b = 5


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