Q46 / 82 · NTA Abhyas (JEE Main) 2020 · Mathematics · Quadratic Equation

If both roots of $x^{2} - 2 a x + a^{2} + a - 3 = 0$ are less than $3,$ then

Solution

Both roots are less than

3

\Rightarrow D \geq 0, - \frac{B}{2 A} < 3, f (3) > 0

D = 4 a^{2} - 4 (a^{2} + a - 3) = - 4 (a - 3) \geq 0

\Rightarrow a \leq 3 \dots . (i)

\frac{- B}{2 A} = \frac{2 a}{2} = a < 3 \dots . (i i)

f (3) = 9 - 6 a + a^{2} + a - 3 > 0

a^{2} - 5 a + 6 > 0

(a - 2) (a - 3) > 0 \Rightarrow a < 2

a > 3 \dots . . (i i i)

Taking the intersection of

(i), (i i), (i i i)

, we get,

a < 2

Hence, the correct answer is option: 'A'.