Since, function f (x) is continuous at x = 1, 3 ∴f(1)=f(1+) ⇒ae+be−1=c ......(i) f(3)=f(3+) ⇒9c=9a+6c⇒c=3a .......(ii) From (i) and (ii), b=ae(3−e) ........(iii)
f′(x)=[
aex−be−x
−1<x<1
2cx
1<x<3
2ax+2c
3<x<4
f′(0)=a−b,f′(2)=4c Given, f′(0)+f′(2)=e a−b+4c=e ......(iv) From eqs. (i), (ii), (iii) and (iv), a−3ae+ae2+12a=e ⇒13a−3ae+ae2=e ⇒a=