The set of all values of k > -1, for which the equation (3x2 + 4x + 3)2 - (k +1) (3x2 + 4x + 3) (3x2 + 4x + 2) + k(3x2 + 4x + 2)2 = 0 has real roots, is
(1) \(\Big(1, \frac{5}{2}\Big]\)
(2) [2, 3)
(3) \(\Big[-\frac{1}{2}, 1\Big)\)
(4) \(\Big(\frac{1}{2}, \frac{3}{2}\Big]\) - {1}
Correct answer: (1) \(\Big(1, \frac{5}{2}\Big]\)
Explanation:
(3x2 + 4x + 3)2 – (k + 1) (3x2 + 4x + 3) (3x2 + 4x + 2) + k ( 3x2 + 4x + 2)2 = 0
Let 3x2 + 4x + 3 = a
and 3x2 + 4x + 2 = b ⇒ b = a – 1
Given equation becomes
⇒ a2 – ( k +1) ab + k b2 = 0
⇒ a (a – kb) – b ( a – kb) = 0
⇒ (a – kb) (a – b) = 0
⇒ a = kb or a = b (reject)
∴ a = kb
⇒ 3x2 + 4x + 3 = k (3x2 + 4x + 2)
⇒ 3 (k – 1)x2 + 4 (k – 1) x + (2k – 3) = 0
for real roots D ≥ 0
⇒ 16 ( k –1)2 – 4 (3(k–1)) (2k – 3) ≥ 0
⇒ 4 (k –1) {4 (k –1) – 3 (2k – 3)} ≥ 0
⇒ 4 ( k –1) (-2k + 5) ≥ 0
⇒ –4 (k –1) (2k – 5) ≥ 0
⇒ (k – 1) (2k – 5) ≤ 0
∴ k ∈ \(\Big[1, \frac{5}{2}\Big]\)
∴ k ≠ 1
∴ k ∈ \(\Big(1, \frac{5}{2}\Big]\)