Let $P\left(x\right)=x^2+bx+c$ be a quadratic polynomial with real coefficients such that ${\int }_0^{1}P\left(x\right)dx=1$ and $P\left(x\right)$ leaves remainder $5$ when it is divided by $(x–2)$. Then the value of $9(b+c)$ is equal to :

The correct option is A

$7$

Explanation for the correct option.

Finding the value of $9(b+c)$ :

Given that,

${\int }_0^{1}P\left(x\right)dx=1$

So,

$$\begin{gathered} {\int }_0^1\left(x^2+bx+c\right)dx=1\mathbf{}\left[\because GivenP\left(x\right)=\left(x^2+bx+c\right)\right] \\ \Rightarrow \frac{1}{3}+\frac{b}{2}+c=1 \\ \Rightarrow \frac{b}{2}+c=\frac{2}{3} \\ \Rightarrow 3b+6c=4...\left(1\right) \end{gathered}$$

Now, given that $P\left(2\right)=5$

$$\begin{gathered} P\left(2\right)=5 \\ \Rightarrow 2^2+2b+c=5 \\ \Rightarrow 4+2b+c=5 \\ \Rightarrow 2b+c=1...\left(2\right) \end{gathered}$$

Now from the equation $\left(1\right)$ and $\left(2\right)$ we get,

$b=\frac{2}{9}\mathrm{and}c=\frac{5}{9}$

Now substituting these values in $9(b+c)$ we get,

$$\begin{gathered} 9(b+c) \\ =9\left(\frac{2}{9}+\frac{5}{9}\right) \\ =9\left(\frac{7}{9}\right) \\ =7 \end{gathered}$$

Therefore, the correct answer is option (A).