Question

Suppose that $f\left(x\right)$ is a quadratic expression, positive for all real $x$. If $g\left(x\right)=f\left(x\right)+f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)$, then
(where $f^{\prime }\left(x\right)$ and $f^{\prime \prime }\left(x\right)$ represent first and second derivative respectively)

• A. $g\left(x\right)<0$ for all real values of $x$
• B. $g\left(x\right)>0$ for all real values of $x$
• C. $g\left(x\right)=0$ for some real values of $x$
• D. $g\left(x\right)\le 0$ for some real values of $x$

Answer 1

The correct option is B $g\left(x\right)>0$ for all real values of $x$

Let $f\left(x\right)=a{x}^2+bx+c,a\ne 0$ such that $f\left(x\right)>0$ for all $x\in R$.
Then, $a>0$ and $b^2-4ac<0$.

Now, $g\left(x\right)=f\left(x\right)+f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)$
$\Rightarrow g\left(x\right)=a{x}^2+(b+2a)x+(b+2a+c)$
Discriminant of $g\left(x\right)$ is
$D=(b+2a{)}^2-4a(b+2a+c)$
$=b^2-4a^2-4ac$
$=(b^2-4ac)-4a^2$
$<0(\because b^2-4ac<0)$
Therefore, $g\left(x\right)>0$ for all $x\in R$