JEE Main 2023 — Mathematics Algebra

Let

$$f(x)={x}^{3}-a{x}^{2}+bx-c$$f(x)=x3−ax2+bx−c

$${f}^{'}(x)=3{x}^{2}-2ax+b$$f′(x)=3x2−2ax+b

$${f}^{"}(x)=6x-2a$$f"(x)=6x−2a

$${f}^{'''}(x)=6$$f′′′(x)=6

Also,

$${f}^{'}(1)=a$$f′(1)=a

$$\Rightarrow 3-2a+b=a$$⇒3−2a+b=a

$$\Rightarrow 3a=b+3$$⇒3a=b+3

Also,

$${f}^{"}(2)=12-2a=b$$f"(2)=12−2a=b

$$\Rightarrow 12-2a=3a-3$$⇒12−2a=3a−3

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

$$\Rightarrow b=6$$⇒b=6

And,

$${f}^{'''}(3)=c\Rightarrow c=6$$f′′′(3)=c⇒c=6

Hence,

$$f(x)={x}^{3}-3{x}^{2}+6x-6$$f(x)=x3−3x2+6x−6

Now,

$$f(3)+2f(0)=27-27+18-6-12=0$$f(3)+2f(0)=27−27+18−6−12=0

$$f(2)+f(1)=8-12+12-6+1-3+6-6=0$$f(2)+f(1)=8−12+12−6+1−3+6−6=0

So,

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

$\Rightarrow$⇒$2f(0)-f(1)+f(3)=f(2)$2f(0)−f(1)+f(3)=f(2)