First slide

Theory of expressions

Question

Let $f (x)$ be a quadratic expression which is positive for all x. If $g (x) = f (x) + f^{'} (x) + f^{''} (x)$ then for all
real x,

Moderate

A

$g (x) < 0$
B

$g (x) > 0$
C

$g (x) = 0$
D

$g (x) \geq 0$

Solution

Let $f (x) = a x^{2} + b x + c$
As $f (x) > 0 \forall x \in R$ , we must have $a > 0$ and $b^{2} - 4 a c < 0$ .
Also, $f^{'} (x) = 2 a x + b$ and $f^{''} (x) = 2 a$
Thus, $g (x) = a x^{2} + (2 a + b) x + 2 a + b + c$
Since $a > 0$ and discriminant
$\begin{array}{l} (2 a + b)^{2} - 4 a (2 a + b + c) \\ = 4 a^{2} + 4 a b + b^{2} - 8 a^{2} - 4 a b - 4 a c \\ = - 4 a^{2} + (b^{2} - 4 a c) < 0 [∵ b^{2} - 4 a c < 0] \end{array}$
We get $g (x) > 0 \forall x \in R$

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