If f(x)=x^3-x^2 f^{prime}(1)+x f^{prime prime | vedclass

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(C) Given $f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3)$.Let $f^{\prime}(1)=a$,$f^{\prime \prime}(2)=b$,and $f^{\prim

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$2 f(0)-f(1)+f(3)=f(2)$

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(C) Given $f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3)$.Let $f^{\prime}(1)=a$,$f^{\prime \prime}(2)=b$,and $f^{\prime \prime \prime}(3)=c$.Then $f(x)=x^3-ax^2+bx-c$.Calculating derivatives:$f^{\prime}(x)=3x^2-2ax+b$$f^{\prime \prime}(x)=6x-2a$$f^{\prime \prime \prime}(x)=6$Now,substitute the values:$f^{\prime \prime \prime}(3)=6 \implies c=6$.$f^{\prime \prime}(2)=6(2)-2a=12-2a=b \implies 2a+b=12$.$f^{\prime}(1)=3(1)^2-2a(1)+b=3-2a+b=a \implies 3a-b=3$.Adding the two equations: $(2a+b)+(3a-b)=12+3 \implies 5a=15 \implies a=3$.Substituting $a=3$ into $2a+b=12$: $2(3)+b=12 \implies b=6$.Thus,$f(x)=x^3-3x^2+6x-6$.Calculating values:$f(0)=-6$$f(1)=1-3+6-6=-2$$f(2)=8-12+12-6=2$$f(3)=27-27+18-6=12$Checking option $C$: $2f(0)-f(1)+f(3) = 2(-6)-(-2)+12 = -12+2+12 = 2 = f(2)$.Therefore,the correct option is $C$.

If $f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R$,then

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