The numbers A, B and C such that a function of the form f (x) = A x2 + Bx + C satisfies the conditions
f′(1)=8,f(2)+f′′(2)=33 and ∫01 f(x)dx=7/3 are
A = 1, B = – 4, C = 2
A=7,B=–6, C=3
A=8,B=−6,C=3
none of these.
8=f′(1)=2A+B33=f(2)+f′′(2)=6A+2B+C Also, 73=13A+12B+C⇒14=2A+3B+6C
Solving we obtain
A=7,B=–6,C=3