Test: Linear Inequalities One Variable - JEE MCQ

Step 1: Simplify the inequality
The given inequality:
3x + 22x ≥ 5x
Combine like terms on the left-hand side:
(3x + 22x) = 25x
So, the inequality becomes:

Step 2: Subtract 5x from both sides
Subtract 5x from both sides to simplify further.
This simplifies to:
20x ≥ 0

Step 3: Solve for x
Divide through by 20 (a positive number, so the inequality direction remains unchanged):
x ≥ 0

Step 4: Interpret the solution
The solution to the inequality is:
x ≥ 0
This means that x can take any value in the interval [0, ∞).

Final Answer:
The solution set for x is:
[0, ∞)

x² ― 3|x| + 2 < 0
⇒|x|
2 ― 3|x| + 2 < 0
⇒(|x| ― 1)(|x| ― 2) < 0
⇒1 < |x| < 2
⇒ ― 2 < x < ―1 or 1 < x < 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
⇒ -∞ < x ≤ log2(√2 - 1) or x ≥ log2(√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)(x² - 5x + 7) < (x - 1)
⇒ (x - 1)(x² - 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 < x, i.e., x < y
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 < x < 1
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 α and sin α increase 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 , ∞)

1. The expression (x - 2)(x - 3) equals zero when x = 2 or x = 3. These points divide the number line into three intervals:

   • x < 2
   • 2 < x < 3
   • x > 3
2. Test each interval:
   • For x < 2: Choose x = 1. Then, (1 - 2)(1 - 3) = (-1)(-2) = 2 > 0. So, the inequality holds in this interval.
   • For 2 < x < 3: Choose x = 2.5. Then, (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0. So, the inequality does not hold in this interval.
   • For x > 3: Choose x = 4. Then, (4 - 2)(4 - 3) = (2)(1) = 2 > 0. So, the inequality holds in this interval.

• Combine the intervals where the inequality holds:

  The inequality (x - 2)(x - 3) > 0 holds for x < 2 and x > 3. Therefore, the solution set is:

  (-∞, 2) ∪ (3, ∞)

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)

Consider the equation of the line joining (0,14) and (5,0). Applying intercept form gives us

Now it has been shaded away from the origin, hence the inequality has to be false at x,y=(0,0). At x,y=(0,0) we get 14x+5y=0.
Therefore for false inequality, 14x+5y≥70
Similarly the equation of the line joining (0,14) and (19,14), is given by y=14.
Since it is shaded towards the origin, hence y≤14.
The equation of the line joining (5,0) and (19,14) is given by

xy−14x−5y+70=xy−19y
14x−14y=70 or x−y=5. Since it is shaded towards the origin, x−y≤5
Hence the shaded area is given by 14x+5y≥70,y≤14,x−y≤5

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)

Let log₁₆x=t
Hence, t²−t+log₁₆k=0 and change k>0
The equation has exactly one solution. Hence, the discriminant must be zero.
Δ=1−4log₁₆k
⇒0=1−4log₁₆k
⇒log₁₆k=1/4
⇒k=2
So, number of real values of k is 1 when k=2.