Question

If f(x) is quadratic expression which is positive for all real value of x and $g(x) = f(x) + f^{'}(x) + f^{'^{'}}$. Then for any real value of x

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

Answer 1

Answer: (B)

$g(x) > 0$

Answer 2

Let $f(x) = ax^{2} + bx + c$, then

$g(x) = ax^{2} + bx + c + 2ax + b + 2a = ax^{2} + (b + 2a)x + (b + c + 2a)$ ∵ $f(x) > 0$. Therefore $b^{2} - 4ac < 0$ and $a > 0$

Now for g(x),

Discriminant $= (b + 2a)^{2} - 4a(b + c + 2a) = b^{2} + 4a^{2} + 4ab$ $$- 4ab - 4ac - 8a^{2} = (b^{2} - 4ac) - 4a^{2} < 0$$

as $b^{2} - 4ac < 0$

Therefore sign of g(x) and a are same i.e. $g(x) > 0$.