Question

lf $f\left(x\right)$ is a quadratic expression which is positive for all real vaues of $x$ and $g\left(x\right)=f\left(x\right)+f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)$ then for any real value of $x$

• A. $g\left(x\right)<0$
• B. $g\left(x\right)>0$
• C. $g\left(x\right)=0$
• D. $g\left(x\right)\underset{–}{>}0$

Answer 1

Answer: (B)

$g\left(x\right)>0$

Let $f\left(x\right)=a{x}^2+bx+c$
According to the given condition $a>0,b^2-4ac<0$.........(i)
$\therefore g\left(x\right)=a{x}^2+bx+c+2ax+b+2a=a{x}^2+(b+2a)x+(b+c+2a)$
Now discriminant of $g\left(x\right)$ is $D=(b+2a{)}^2-4a(2a+c+b)=b^2+4a^2+4ab-4ab-4ac-8a^2=(b^2-4ac)-4a^2<0$ using (i)
Hence $g\left(x\right)>0$ for any $x\in R$