Solve the following inequalities
(i)
≥ 1
(ii) 
(iii) 
(iv) log 1/4 (2 – x) > log 1/4 
(v) log 1/3 (2 x+2 – 4 x ) ≥ – 2
(vi) log x (4x – 3) ≥ 2
Text Solution
Verified by Experts(ii) (i); (ii) (i); (ii) (vi)
(i)
⇒ 2x 2 – x –
≤
⇒ 16x 2 – 8x – 8 ≤ 0
⇒ 2x 2 – x – 1 ≤ 0 ⇒ (2x + 1)(x – 1) ≤ 0
x ∈
......(i)
also 2x 2 – x –
> 0 ⇒ 16x 2 – 8x – 3 > 0
⇒ 16x 2 – 12x + 4x – 3 > 0 ⇒ (4x – 3)(4x + 1) > 0
⇒ x ∈
.......(ii)
(i) ∩ (ii) ⇒ 
(ii) x 2 – 5x + 6 > 0 ⇒ (x – 3)(x – 2) > 0
⇒ x ∈ (– ∞ , 2) ∪ (3, ∞ ) ......(i)
and x 2 – 5x + 6 < 
⇒ x 2 – 5x + 4 < 0 ⇒ (x – 1)(x – 4) < 0 ⇒ x ∈ (1, 4) .....(ii)
(i) ∩ (ii) ⇒ x ∈ (1, 2) ∪ (3, 4)
(iii) log 7
⇒
⇒ 
⇒
⇒
⇒ 
and
⇒ 
Ans. 
(iv) 2 – x <
⇒
+ x – 2 > 0 ⇒ 
⇒
⇒ 
and 2 – x > 0 ⇒ x < 2 and
⇒ x > –1
Ans. (–1, 0) ∪ (1, 2)
(v) 2 2 .2 x – 4 x ≤
⇒ 4.2 x – (2 x ) 2 ≤ 9
Let 2 x = t ⇒ t 2 – 4t + 9 ≥ 0 always true
2 x+2 – 4 x > 0 ⇒ 2 x .(4 – 2 x ) > 0
4 – 2 x > 0 ⇒ 2 x < 2 x
⇒ x < 2 ⇒ x ∈ (– ∞ , 2)
(vi) log x (4x – 3) ≥ 2
Case -I : When x > 1 ⇒ 4x – 3 ≥ x 2 ⇒ x 2 – 4x + 3 ≤ 0
⇒ (x – 1)(x – 3) ≤ 0
x ∈ (1, 3] .......(i)
Case-II : When 0< x < 1 ⇒ 4x – 3 ≤ x 2
and 4x – 3 > 0 ⇒ (x – 1)(x – 3) ≥ 0
x > 3/4 x ∈
......(ii)
Ans. (i) ∪ (ii)
(vi) log x (4x – 3) ≥ 2
Case- Ι 0 < x < 1 and 4x – 3 > 0 ⇒ x > 
then 4x – 3 ≤ x 2 ⇒ x 2 – 4x + 3 ≥ 0 ⇒ (x – 3)(x – 1) ≥ 0
⇒ x ∈ 
Case - ΙΙ x > 1, 4x – 3 > 0 ⇒ x > 
then 4x – 3 ≥ x 2 ⇒ x 2 – 4x + 3 ≤ 0
⇒ (x – 3)(x – 1) ≤ 0 ⇒ x ∈ (1,3]
Ans. 
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