Q. If the roots of the equation $ x^2 - 2ax + a^2 + a - 3 = 0 $ are real and less than 3, then

Solution:

Let $ f (x) = x^2 - 2ax + a^2 + a - 3 $
Since, both root are less than 3.
$\Rightarrow \alpha < 3, \beta < 3 $
$\Rightarrow $ Sum , S = $ \alpha + \beta < 6 $
$\Rightarrow \frac{ \alpha + \beta }{ 2} < 3 $
$\Rightarrow \frac{ 2a}{ 2} < 3 $
$\Rightarrow a < 3 $
Again, product, P = $ \alpha \ \beta $
$\Rightarrow P < 9 \Rightarrow \alpha \beta < 9 $
$\Rightarrow \alpha^2 + \alpha - 12 = 0 $
$\Rightarrow ( \alpha - 3 ) \ (\alpha + 4 ) < 0 $ $\Rightarrow - 4 < a < 3 $ $ \hspace16mm$ ...(iii)
Again, D = $ B^2 - 4 AC \ge 0 $
$\Rightarrow ( - 2a)^2 - 4 . 1 (a^2 + a - 3 ) \ge 0 $
$\Rightarrow 4a^2 - 4a^2 - 4a + 12 \ge 0 $
$\Rightarrow - 4 a + 12 \ge 0 \Rightarrow a \le 3 $ $ \hspace19mm$ ...(ii)
Again a f (3) > 0
$\Rightarrow 1 [ ( 3)^2 - 2a \ (3) + a^2 + a - 3 ] > 0 $
$\Rightarrow 9 - 6 a + a^2 + a - 3 > 0 \Rightarrow a \le 3 $ $ \hspace19mm$ ...(iii)
$\Rightarrow (a - 2) \ ( a - 3) > 0 $
$\therefore$ $ a \ \in \ ( - \infty, 2 ) \cup (3, \infty) $ $ \hspace19mm$ ...(iv)
From Eqs. (i), (ii), (iii) and (iv), we get
$ a \ \in (4, 2)$
Therefore, (a) is the answer
NOTE There is correction in answer a < 2 should be - 4 < a < 2

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