First slide

Theory of expressions

Question

The set of values of a for which each one of the roots of $x^{2} - 4 a x + 2 a^{2} - 3 a + 5 = 0$ is greater than $2,$ is

Moderate

A

$a \in (1, \infty)$
B

$a = 1$
C

$a \in (- \infty, 1)$
D

$a \in (9 / 2, \infty)$

Solution

Let $f (x) = x^{2} - 4 a x + 2 a^{2} - 3 a + 5 .$
Clearly, $y = f (x)$ represents a parabola opening upward. So, its both roots will be greater than $2,$ if
(i) Discriminant $\geq 0$
(ii) $x -$ coordinate of vertex $> 2$
(iii) $2$ lies outside the roots i.e. $f (2) > 0$
Now,
(i) Discriminant $\geq 0$
$\begin{array}{l} \Rightarrow 16 a^{2} - 4 (2 a^{2} - 3 a + 5) \geq 0 \\ \Rightarrow 2 a^{2} + 3 a - 5 \geq 0 \end{array}$
$\Rightarrow (2 a + 5) (a - 1) \geq 0 \Rightarrow a \leq - \frac{5}{2}$ or, $a \geq 1$ …(i)
(ii) $x -$ coordinate of vertex $> 2$
$\Rightarrow 2 a > 2 \Rightarrow a > 1$ …(ii)
(iii) $f (2) > 0$
$\Rightarrow 2 a^{2} - 11 a + 9 > 0$
$\Rightarrow (a - 1) (2 a - 9) > 0 \Rightarrow a < 1$ or, $a > \frac{9}{2}$ …(iii)
From (i), (ii) and (iii), we get $a > \frac{9}{2}$ .

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