Question

Let $f$ be a monic polynomial with real coefficients. It is given that $\underset{x\to \infty }{lim}{f}^{\prime \prime }\left(x\right)=\underset{x\to \infty }{lim}f\left(\frac{1}{x}\right)$ and $f\left(x\right)\ge f\left(5\right)$ for all $x\in R.$ Then the value of $f\left(3\right)$ is

Note: a monic polynomial is a single-variable polynomial in which the leading coefficient (the nonzero coefficient of highest degree) is equal to $1$

• A. $-15$
• B. $-19$
• C. $9$
• D. $15$

Answer 1

The correct option is B $-19$

$\underset{x\to \infty }{lim}{f}^{\prime \prime }\left(x\right)=\underset{x\to \infty }{lim}f\left(\frac{1}{x}\right)\cdots \left(1\right)$
$\Rightarrow f\left(x\right)$ is a polynomial of degree $2.$
Also, $f$ is a monic polynomial.
$\therefore f\left(x\right)=x^2+bx+c$

$f\left(\frac{1}{x}\right)=\frac{1+bx+c{x}^2}{x^2}$
From $\left(1\right),$ we get $c=2$


$f\left(x\right)\ge f\left(5\right)$
$\Rightarrow f$ has minimum at $x=5.$
$\Rightarrow f^{\prime }\left(5\right)=0$
$\Rightarrow b=-10$

$\therefore f\left(x\right)=x^2-10x+2$
$f\left(3\right)=-19$