Ans.

(i)

Sol. Let t = 3 x then 9 x – a.3 x + 3–a = t 2 – at – (a – 3)

f(t) = t 2 – at – (a – 3)

f(t) ≤ 0 where t = 3 x > a ∀ x ∈ R D = a

2 + 4(a – 3) ⇒ f(t) = 0 has real roots if

a ≤ – 6 or a ≥ 2

for at least one negative solution 3 x < 1 ⇒ t < 1

⇒ t ∈ (0,1)

Exactly one f(0) f(1) < 0 ⇒ (3 – a) (2 – a) < 0

⇒ a ∈ (2,3)

for both ⇒ a ∈ (0, 2) ⇒ not possible ⇒ a ∈ φ

Hence for at least one negative solution a ∈ (2, 3)

(ii)

Sol. Positive solution ⇒ t > 1

for exactly one f(1) < 0 ⇒ a > 2

for both D ≥ 0, f(1) > 0 & > 1

⇒ a < 2 ⇒ not possible

Hence a > 2

(iii)

Sol. x ∈ (1, 2) ⇒ t ∈ (3, 9)

exactly one in (3, 9); f(3) f(9) < 0

⇒ a ∈

Both in (3, 9) 3 < a/2 < 9 & f(3) > 0, f(9) > 0

⇒ 6 < a < 18 ⇒ a < & a < 3

⇒ a ∈ φ hence a ∈ (3, 84/10)