Question

If $f\left(x\right)$ is a quadratic polynomial with leading coefficient $1$ such that $\underset{x\to 0}{lim}f\left(x\right)=3=\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)$, then $\underset{x\to 2}{lim}f\left(x\right)=$
$\left(f^{\prime }\right(x)=\frac{df\left(x\right)}{dx})$

• A. $\underset{x\to 1}{lim}f\left(2x\right)$
• B. $\underset{x\to -5}{lim}f\left(x\right)$
• C. $\underset{2x\to -5}{lim}f\left(2x\right)$
• D. $13$

Answer 1

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

$13$

$f\left(x\right)=x^2+ax+b$
For the polynomial function, we can directly substitute the value of $x$
$\underset{x\to 0}{lim}f\left(x\right)=b=3$ and $\underset{x\to 0}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 0}{lim}2x+a=a=3$
$\therefore a=b=3\Rightarrow f\left(x\right)=x^2+3x+3$
$\underset{x\to 2}{lim}f\left(x\right)=13$ also $\underset{x\to 1}{lim}f\left(2x\right)=f\left(2\right)=13$
$\underset{x\to -5}{lim}f\left(x\right)=\underset{x\to -5}{lim}{x}^2+3x+3=13$ and
$\underset{2x\to -5}{lim}f\left(2x\right)=f(-5)=13$