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