f(x)=Pex+Qe2x+ReJxf(0)=P+Q+R=6 ....(1)
f′(x)=Pex+2Qe2x+3Re3xf′(ln3)=Peln3+2Qe2ln3+3Re3ln3⇒3P+18Q+81R=282⇒P+6Q+27R=94 ....(2)
Since
∫f(x)dx=Pex+2Qe2x+3Re3x⇒∫0lnf(x)dx=11(PeIn2+2Qe2In2+3Re3In2−P−2Q−3R)=11⇒2P+2Q+38R−P−2Q−3R=11⇒P+23Q+37R=11 ....(3)
(2) - (1) and (3) - (1),
5Q+26R=88 ....(4)
and,
2Q+34R=53Q+8R=30 ....(5)
From (5),
R=830−3Q
From (4),
5Q+26(830−3Q)=88⇒20Q+390−39Q=352⇒19Q=38⇒Q=2