We have,
f(x)=(x2+3x+a, if x≤1bx+2, if x>1It is given that f(x) is differentiableat x=1 and every differentiablefunction is continuous.So, f(x) is continuous at x=1∴limx→1−f(x)=limx→1+f(x)=f(1)limx→1−(x2+3x+a)=limx→1+bx+2=4+a⇒4+a=b+2=4+a⇒a=b−2Now, f(x) is differentiable at x=1⇒(LHD at x=1)=(RHD at x=1)⇒limx→1−f(x)−f(1)x−1=limx→1+f(x)−f(1)x−1⇒limx→1−(x2+3x+a)−4−ax−1=limx→1+bx+2−4−ax−1⇒limx→1−x2+3x−4x−1=limx→1+bx+2−4−ax−1⇒limx→1−x2+4x−x−4x−1=limx→1+bx−2−(b−2)x−1⇒limx→1−(x+4)(x−1)x−1=limx→1+bx−bx−1⇒limx→1−x+4=limx→1+b(x−1)x−1⇒5=b∴a=3