FunctionHard
Question
Find the values of ′a′ in the domain of the definition of the function , f(a) = 
for which the roots of the equation, x2 + (a + 1) x + (a - 1) = 0 lie between - 2 & 1.
for which the roots of the equation, x2 + (a + 1) x + (a - 1) = 0 lie between - 2 & 1.Options
A.(- 1/2, 0] ∪ [1/2, 1)
B.(- 2/3, 0] ∪ [1/3, 1)
C.(- 1/2, - 1/4] ∪ [0, 1)
D.none of these
Solution
f(a) = 
for domain of f(x)
2a2 - a ≥ 0 ⇒ a(2a - 1) ≥ 0
∴ a ∈ (-∞ 0] ∪
Let g(x) ≡ x2 + (a + 1)x + (a - 1) = 0
(i) D ≥ 0
(a + 1)2 - 4(a - 1) ≥ 0 ⇒ a ∈ R ......(i)
(ii) - 2 < -
< 1
⇒ - 2 < -
< 1
⇒ a ∈ (-3, 3) ....(ii)
(iii) g(- 2) > 0
⇒ 4 - 2(a + 1) + (a - 1) > 0 ⇒ a < 1
(iv) g(1) > 0
⇒ 4 - 2(a + 1) + (a - 1) > 0 ⇒ a < 1
(iv) g(1) > 0 ⇒ 1 + a + 1 + a - 1 > 0
⇒ a > - 1/2
Now (i) ∩ (ii) ∩ (iii) ∩ (iv) we get
Ans. : a ∈
for domain of f(x)2a2 - a ≥ 0 ⇒ a(2a - 1) ≥ 0
∴ a ∈ (-∞ 0] ∪
Let g(x) ≡ x2 + (a + 1)x + (a - 1) = 0
(i) D ≥ 0
(a + 1)2 - 4(a - 1) ≥ 0 ⇒ a ∈ R ......(i)
(ii) - 2 < -
< 1 ⇒ - 2 < -
< 1⇒ a ∈ (-3, 3) ....(ii)
(iii) g(- 2) > 0
⇒ 4 - 2(a + 1) + (a - 1) > 0 ⇒ a < 1
(iv) g(1) > 0
⇒ 4 - 2(a + 1) + (a - 1) > 0 ⇒ a < 1
(iv) g(1) > 0 ⇒ 1 + a + 1 + a - 1 > 0
⇒ a > - 1/2
Now (i) ∩ (ii) ∩ (iii) ∩ (iv) we get
Ans. : a ∈
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