Quadratic EquationHard

Question

If the roots of the equation x2 - 2ax + a2 + a - 3 = 0 are real and less than 3, then

Options

A.
- 4 < a < 2 
B.2 ≤ a ≤ 3
C.3 ≤ a ≤ 4
D.a > 4

Solution


       
Let f (x) = x2 - 2ax = a2 + a - 3
Since, both root are less than 3.
⇒     α < 3, β < 3
⇒     Sum, S = α + β < 6
⇒     < 3
⇒     < 3
⇒     α < 3      .......(i)
Again, product, P = αβ
⇒     P < 9
⇒     αβ < 9
⇒     α2 + a - 3 < 9
⇒     α2 + a - 12 < 0
⇒    (α - 3)(a + 4) < 0
⇒     -4 < a < 3      .......(ii) Agani,   D = B2 - 4AC ≥ 0
⇒     (-2a)2 - 4.1 (a2 + a - 3) ≥ 0
⇒     4a2 - 4a2 - 4a + 12 ≥ 0
⇒     - 4a + 12 ≥ 0
⇒     a ≤ 3      .......(iii)
Agani,   a f (3) > 0
⇒     1[(3)2 - 2a(3) + a2 + a - 3] > 0
⇒     9 - 6a + a2 + a - 3 > 0
⇒     a2 - 5a + 6 > 0
⇒     (a - 2)(a - 3) > 0
∴     a ∈ ( - ∞, 2) (3, ∞)      .......(iv)
From Eqs. (i),(ii),(iii) and (iv), we get
    a ∈ (-4, 2)
Thereform, (a) is the answer.

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