Question

If $f\left(0\right)=1,f\left(1\right)=5$ and $f\left(2\right)=11$$4a+2b+1=11$, then the equation of polynomial of degree two is

• A. $x^2+1=0$
• B. $x^2+3x+1=0$
• C. $x^2-2x+1=0$
• D. None of these

Answer 1

The correct option is B

$x^2+3x+1=0$

Explanation for the correct option:

Step1. Assume the function:

Given that $f\left(0\right)=1,f\left(1\right)=5$ and $f\left(2\right)=11$

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

Now, Put $x=0$ we get

$$\begin{gathered} \Rightarrow f\left(0\right)=c \\ \Rightarrow 1=c \\ \Rightarrow c=1 \end{gathered}$$

Now, Put $x=2$we get

$\Rightarrow$ $f\left(2\right)=4a+2b+1$

$\Rightarrow$$4a+2b+1=11.....\left(1\right)$

Now, Put $x=1$we get

$\Rightarrow$ $f\left(1\right)=a+b+1$

$\Rightarrow$ $a+b+1=5......\left(2\right)$

Step2. Find the equation of polynomial of degree two:

Multiply equation $\left(2\right)$ with $2$

$$\begin{gathered} \Rightarrow 2a+2b+2=10 \\ \Rightarrow 4a+2b+1=11 \\ \Rightarrow -2a+1=-1 \\ \Rightarrow a=1 \end{gathered}$$

From $\left(2\right)$ ,we get

$$\begin{gathered} \Rightarrow 1+b+1=5 \\ \Rightarrow b=3 \end{gathered}$$

So polynomial of degree two is $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}={\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{1}$

Hence, Option ‘B’ is Correct.