Mathematics - Functions Question with Solution | TestHub

Let

$f\left(x\right)=x^3-a{x}^2+bx-c$

$f^{\text{'}}\left(x\right)=3x^2-2ax+b$

$f^{"}\left(x\right)=6x-2a$

$f^{\text{'}\text{'}\text{'}}\left(x\right)=6$

Also,

$f^{\text{'}}\left(1\right)=a$

$\Rightarrow 3-2a+b=a$

$\Rightarrow 3a=b+3$

Also,

$f^{"}\left(2\right)=12-2a=b$

$\Rightarrow 12-2a=3a-3$

$\Rightarrow 5a=15\Rightarrow a=3$

$\Rightarrow b=6$

And,

$f^{\text{'}\text{'}\text{'}}\left(3\right)=c\Rightarrow c=6$

Hence,

$f\left(x\right)=x^3-3x^2+6x-6$

Now,

$f\left(3\right)+2f\left(0\right)=27-27+18-6-12=0$

$f\left(2\right)+f\left(1\right)=8-12+12-6+1-3+6-6=0$

So,

$f\left(3\right)+2f\left(0\right)=f\left(2\right)+f\left(1\right)$

$\Rightarrow$$2f\left(0\right)-f\left(1\right)+f\left(3\right)=f\left(2\right)$