Question

Let $Z$ be the set of all integers and $f:Z\to Z$ be a function given by $f\left(x\right)=a{x}^2+bx+c$, such that $f\left(0\right)$ is $2,$ $f\left(1\right)$ is $6,$ $f\left(2\right)$ is $12.$ Then $f(-1)$ is

• A. 0
• B. 000
• C. 00

Answer 1

Answer: (A), (B), (C)

We have, $f:Z\to Z,f\left(x\right)=a{x}^2+bx+c$

$\because f\left(0\right)=2$

$\Rightarrow a.0+b.0+c=2\Rightarrow c=2$

$\because f\left(1\right)=6$

$\Rightarrow a.(1{)}^2+b.(1)+c=6\Rightarrow a+b+c=6$

$\Rightarrow a+b+2=6\Rightarrow a+b=4$ .............(1)

$\because f\left(2\right)=12$

$\Rightarrow a.(2{)}^2+b.(2)+c=12$

$\Rightarrow 4a+2b+c=12\Rightarrow 4a+2b=12-2=10$

$\Rightarrow 2(2a+b)=10\Rightarrow 2a+b=5$ .........(2)

Now, from $e{q}^n\left(2\right)-\left(1\right)$, we get

$(2a+b)-(a+b)=5-4\Rightarrow a=1$

Put $a=1$ in equation (1), we get

$1+b=4\Rightarrow b=4-1=3$

Putting value of $a,b,c$ in $f\left(x\right)=a{x}^2+bx+c$, we get

$\Rightarrow f\left(x\right)=1.x^2+3.x+2$

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

Now, $f(-1)=(-1{)}^2+3.(-1)+2$

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

Hence, $f(-1)=0$