‌f(x)=x3+x2f′(1)+xf′′(2)+6 ∴‌‌f′(x)=3x2+2xf′(1)+f′′(2)...(i) ∴f′′(x)=6x+2f′(1)....(ii) Substituting x=1 in (i), we get ‌f′(1)=3(1)2+2(1)f′(1)+f′′(2) ‌⇒f′(1)+f′′(2)=−3...(iii) Substituting x=2 in (ii), we get ‌f′′(2)=6(2)+2f′(1) ‌⇒f′′(2)=12+2f′(1)....(iv) From (iii) and (iv), we get ‌f′(1)+12+2f′(1)=−3 ‌⇒3f′(1)=−15 ‌⇒f′(1)=−5 ‌‌ From (iii), ‌−5+f′′(2)=−3 ‌⇒f′′(2)=2 ‌∴‌‌f(2)=23+22(−5)+2(2)+6 ‌=8−20+4+6 ‌=−2