Question

Let $f\left(x\right)$ is a cubic polynomial such that $f\left(1\right)=1,f\left(2\right)=2,f\left(3\right)=3$ and $f\left(4\right)=16$, then $f\left(0\right)$ is equal to

• A. $6$
• B. $-6$
• C. $12$
• D. $-12$

Answer 1

The correct option is D $-12$

Let us assume:

$f\left(x\right)=a{x}^3+b{x}^2+cx+d$

$f\left(1\right)=a+b+c+d=1$ ………….$\left(1\right)$

$f\left(2\right)=8a+4b+2c+d=2$ ………..$\left(2\right)$

$f\left(3\right)=27a+9b+3c+d=3$ …………$\left(3\right)$

$f\left(4\right)=64a+16b+4c+d=16$ ……….$\left(4\right)$

Solving equation $\left(1\right)$ & $\left(2\right)$, we get

$7a+3b+c=1$ ………..$\left(5\right)$

Solving equation $\left(1\right)$ & $\left(3\right)$, we get

$26a+8b+2c=2$ …….$\left(6\right)$

Solving equation $\left(1\right)$ & $\left(4\right)$, we get

$63a+15b+3c=15$ ……..$\left(7\right)$

Solving $\left(5\right)$ and $\left(6\right)$,

$6a+b=0$

Solving $\left(5\right)$ and $\left(7\right)$

$7a+b=2$

This gives $a=2,b=-12,c=23,d=12$

$f\left(x\right)=2x^3-12x^2+23x-12$

$f\left(0\right)=2(0{)}^3-12(0{)}^3+23\left(0\right)-12$

$f\left(0\right)=-12$

The option [D] is right.