Question

I. lf $f\left(x\right)$ be a quadratic expression which is positive for all real $x$ and $g\left(x\right)=f\left(x\right)+f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)$ for any real $x$, then $g\left(x\right)>0$.

II. lf the equations $p{x}^2-7x+3p=0$ and $2x^2+qx+6=0$ have the same roots, then $pq=-14$.Which of the above statement(s) is/are true?

• A. Only I
• B. Only II
• C. Both I and II
• D. Neither I nor II

Answer 1

The correct option is C Both I and II

(I) Let $f\left(x\right)={ax}^2+bx+c>0$ for all $x\in R$

$\Rightarrow a>0$

and $b^2-4ac<0$ ...(1)

$\therefore g\left(x\right)=f\left(x\right)+f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)$

$\Rightarrow g\left(x\right)={ax}^2+bx+c+2ax+b+2a$

$\Rightarrow g\left(x\right)={ax}^2+x(b+2a)+(c+b+2a)$

whose discriminant

$={(b+2a)}^2-4a(c+b+2a)$

$=b^2+4a^2+4ab-4ac-4ab-8a^2$

$=b^2-4a^2-4ac$

$=(b^2-4ac)-4a^2<0$ [from eq. (1)]

$\therefore g\left(x\right)>0$ for all $x,$ as $a>0$ and discriminant $<0.$

Thus, $g\left(x\right)>0$ for all $x\in R.$

(II) As $P{x}^2-7x+3p=0$ and $2x^2+qx+6=0$ have same roots, then

$\frac{P}{2}=\frac{-7}{q}=\frac{3P}{6}$

$\Rightarrow Pq=-7\times 2=-14$