First slide

Theory of expressions

Question

If $(x^{2} + ax + 3) / (x^{2} + x + a)$ takes all real values for possible real values of x, then

Moderate

A

$4 a^{3} + 39 < 0$
B

$4 a^{3} + 39 \geq 0$
C

$a \geq \frac{1}{4}$
D

$a < \frac{1}{4}$

Solution

Let $\frac{x^{2} + ax + 3}{x^{2} + x + a} = y$
$\begin{array}{ll} \Rightarrow x^{2} (1 - y) - x (y - a) + 3 - ay = 0 \\ ∵ x \in R \\ (y - a)^{2} - 4 (1 - y) (3 - ay) \geq 0 \end{array}$
$\Rightarrow (1 - 4 a) y^{2} + (2 a + 12) y + a^{2} - 12 \geq 0 (1)$
Now, (1) is true for all $y \in R$ .if 1 - 4a > 0 and D < 0. Hence,
$a < \frac{1}{4} and 4 (a + 6)^{2} - 4 (a^{2} - 12) (1 - 4 a) \leq 0$
$\begin{array}{l} \Rightarrow a < \frac{1}{4} and 4 a^{3} - 36 a + 48 \leq 0 \\ \Rightarrow a < \frac{1}{4} and 4 a^{3} \leq 36 a - 48 \\ \Rightarrow 4 a^{3} < 36 (\frac{1}{4}) - 48 \\ \Rightarrow 4 a^{3} + 39 < 0 [∵ a < \frac{1}{4}] \end{array}$

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