Question

Let $f(x)$ be a quadratic expression which is positive for all real values of $x$. If $g(x)=f(x)+f^{\prime }(x)+f"(x),$ then for any real $x$

• A. $g(x)<0$
• B. $g(x)>0$
• C. $g(x)=0$
• D. $g(x)\ge 0$

Answer 1

Answer: (B)

Let $f(x)=a{x}^2+bx+c>0,\forall x\in R$
$\Rightarrow a>0$
and $b^2-4ac<0$ ...(i)
$\therefore g(x)=f(x)+f^{\prime }(x)+f^{\prime \prime }(x)$
$\Rightarrow g(x)=a{x}^2+bx+c+2ax+b+2a$
$\Rightarrow g(x)=a{x}^2+x(b+2a)+(c+b+2a)$
whose discriminant
= $(b+2a{)}^2-4a(c+b+2a)$
= $b^2+4a^2+4ab-4ac-4ab-8a^2$
= $b^2-4a^2-4ac=(b^2-4ac)<0$ [ from Eq. (i) ]
$\therefore g(x)>0,\forall x,$ as $a>0$ and discriminant $<0$.
Thus, $g(x)>0,\forall x\in R$.