Inequalities Test - 1

We have,
3^x + 2^2x ≥5^x
⇒(3/5)^x + (4/5)^x ≥1
⇒(3/5)^x + (4/5)^x ≥(3/5)² + (4/5)²
⇒x ≤2 ⇒x ∈(-∞, 2]
[If a^x + b^x ≥1 and a² + b² = 1], then x ∈(-∞, 2)

x² ―3|x| + 2 <0
⇒|x|
2 ―3|x| + 2 <0
⇒(|x| ―1)(|x| ―2) <0
⇒1 <|x| <2
⇒―2 ∴X E ( ―2, ―1) ∪(1,2)

We have, 2^x + 2|x| ≥2 √2
If x ≥0, then 2^x + 2^x ≥2 √2
⇒2^x ≥√2 ⇒x ≥1/2
And if x <0, then 2^x + 2^-x ≥2 √2
⇒t + 1/t ≥√2 (where t = 2^x )
⇒t² - 2 √2t + 1 ≥0
⇒(t - √2 - 1)(t - √2 + 1) ≥0
But t >0
⇒0 <2^x ≤√2 - 1 or 2^x ≥√2 + 1
⇒-∞ Which is not possible because x >0
∴x ∈(-∞, log2(√2 - 1)) ∪[1/2, ∞)

Since,
AM ≥GM
⇒(a + b) / 2 ≥√ab
⇒(a + b) / 2 ≥√4 (∴ab = 4, given)
⇒a + b ≥4

(x - 1)(x ²- 5x + 7) <(x - 1)
⇒(x - 1)(x ²- 5x + 6) <0
⇒(x - 1)(x - 2)(x - 3) <0
∴x ∈(-∞, 1) ∪(2, 3)

We have,
(1/2) |a + 11/a| ≥√11, equality holding iff a = ±√11
⇒|x| ≥√11, |y| ≥√11, |z| ≥√11, |t| ≥√11
Let x ≥0, then y ≥√11, z ≥√11, t ≥√11
Now,
y - √11 = (1/2) (11/x + x) - √11
⇒y - √11 = ((x - √11)²) / (2x) = ((x - √11) / (2x)) (x - √11)|
⇒y - √11 = (1/2) (1 - √11/x) (x - √11) <(x - √11)
⇒x Similarly, we have y >z, z >t, t >x ⇒x >y
Thus, x = y = z = t = √11 is the only solution for x >0.
We observe that (x, y, z, t) is a solution iff (-x, -y, -z, -t) is a solution.
Thus, x = y = z = t = √11 is the only other solution.

Given, 3^x + 3^1-x - 4 <0
⇒3^2x + 3 - 4.3^x <0
⇒(3^x - 1)(3^x - 3) <0
⇒1 <3^x <3 ⇒0 Thus, the solution set is (0, 1).

Given, inequality can be rewritten as (5/13)^x + (12/13)^x ≥1
⇒cos^x α+ sin^x α≥1
Where, cos α= 5/13
If x = 2, the above equality holds.
If x <2, both cos αand sin αincrease in positive fraction.
Hence, the above inequality holds for x ∈(-∞, 2].

4x + 3 <5x + 7
subtract 4 both sides,
4x + 3 - 3 <5x + 7 - 3
⇒4x <5x + 4
subtract '5x 'both sides ,
[ equal number may be subtracted from both sides of an inequality without affecting the sign of inequality]
4x - 5x <5x + 4 - 5
-x <4
now, multiple with (-1) then, sign of inequality change .
-x.(-1) >4(-1)
x >-4
hence, x €( -4 , ∞)

Let f(x) = loge((x - 2) / (x - 3))
f(x) is defined either if (x - 2) >0, (x - 3) >0 or (x - 2) <0, (x - 3) <0
⇒f(x) is defined either if x >3 or x <2 or x ≠2, 3
i.e., x ∈(-∞, 2) ∪(3, ∞)

3 ≤3t - 18 ≤18
⇒21 ≤3t ≤36
⇒7 ≤t ≤12
⇒8 ≤t + 1 ≤13

To solve the inequality 6x - 7 >5, we need to isolate x on one side of the inequality sign. Here are the steps:
1. Add 7 to both sides of the inequality:
6x - 7 + 7 >5 + 7
6x >12
2. Divide both sides of the inequality by 6:
6x/6 >12/6
x >2
Therefore, the solution to the inequality 6x - 7 >5 is x >2. This means that x is greater than 2. Therefore, the correct answer to the question is D: x >2

f(-1) <1 ⇒a - b + c <1 ...(i)
and f(1) >-1, f(3) <-4, then
a + b + c >-1 ...(ii)
9a + 3b + c <-4 ...(iii)
From Eq. (ii),
-a - b - c <1 ...(iv)

24x <100
⇒(24x) / 24 <100 / 24
⇒x <25/6 or x <4 1/6
When x ∈N, x = 4, 3, 2, 1
Thus, the solution set is {1, 2, 3, 4}

Given, (2x + 3) / (2x - 9) <0
⇒2x + 3 <0 and 2x - 9 >0
Or 2x + 3 >0 and 2x - 9 <0 and x ≠9/2
⇒x <-3/2 and x >9/2
or x >-3/2 and x <9/2
⇒x ∈(-3/2, 9/2)