Question

Let $f\left(x\right)$ be a quadratic expression which is 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 for any real $x$

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

Answer 1

The correct option is B $g\left(x\right)>0$

Given, $g\left(x\right)=f\left(x\right)+f^{{}^{\prime }}\left(x\right)+f^{\prime \prime }\left(x\right)$ ....(1)
Let $f\left(x\right)=a{x}^2+bx+ca\ne 0$
$f^{\prime }\left(x\right)=2ax+b$
$f^{\prime \prime }\left(x\right)=2a$
Substitute these values in (1)
$g\left(x\right)=a{x}^2+(2a+b)x+(c+b+2a)$ ....(2)
Since given, $f\left(x\right)>0$
$\Rightarrow a>0$ and $D<0$
or, $a>0$ and $b^2-4ac<0$
Now, we will find D for $g\left(x\right)$,
$D=(2a+b{)}^2-4a(c+b+2a)$
$\Rightarrow D=(b^2-4ac)-4a^2$
$\Rightarrow D<0(\because b^2-4ac<0$ and $4a^2>0)$
So, for $g\left(x\right)$, $a>0$ and $D<0$
Hence, $g\left(x\right)>0\forall x$