Question

Let $Q\left(x\right)$ be a quadratic polynomial with real coefficients satisfying the inequality
$x^2-4x+6\le Q\left(x\right)\le 3x^2-12x+14,\forall x\in R.$ If $Q\left(12\right)=162$, then the value of $Q\left(7\right)$ is

• A. $37$
• B. $42$
• C. $48$
• D. $52$

Answer 1

The correct option is B $42$

Let $f\left(x\right)=x^2-4x+6=(x-2{)}^2+2g(x)=3x^2-12x+14=3(x-2{)}^2+2$
Both the polynomial $f\left(x\right)$ and $g\left(x\right)$ have a common vertex $(2,2)$, and $g\left(x\right)$ inclines more sharply than $f\left(x\right)$.

Thus, the graph of $Q\left(x\right)$ must lie between these two curves and must also have $(2,2)$ as its vertex in order to hold the inequality $f\left(x\right)\le Q\left(x\right)\le g\left(x\right)$.

$\therefore Q\left(x\right)=k(x-2{)}^2+2,(1\le k\le 3)$
$Q\left(12\right)=162\Rightarrow k(10{)}^2+2=162\Rightarrow k=\frac{8}{5}$
$\Rightarrow Q\left(x\right)=\frac{8}{5}(x-2{)}^2+2\therefore Q(7)=\frac{8}{5}\times 5^2+2=42$