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Test: Linear Inequalities One Variable - JEE MCQ


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15 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Linear Inequalities One Variable

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Test: Linear Inequalities One Variable - Question 1

If 3x + 22x ≥ 5x, then the solution set for x is:

Detailed Solution for Test: Linear Inequalities One Variable - Question 1

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, ∞)

Test: Linear Inequalities One Variable - Question 2

x2 ―3|x| + 2 < 0, then x belongs to

Detailed Solution for Test: Linear Inequalities One Variable - Question 2

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)

Test: Linear Inequalities One Variable - Question 3

Solution of 2x + 2|x| ≥ 2√2

Detailed Solution for Test: Linear Inequalities One Variable - Question 3

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

Test: Linear Inequalities One Variable - Question 4

If ab = 4 (a, b ∈ ℝ⁺), then

Detailed Solution for Test: Linear Inequalities One Variable - Question 4

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

Test: Linear Inequalities One Variable - Question 5

x - 1)(x² - 5x + 7) < (x - 1), then x belongs to

Detailed Solution for Test: Linear Inequalities One Variable - Question 5

(x - 1)(x² - 5x + 7) < (x - 1)
⇒ (x - 1)(x² - 5x + 6) < 0
⇒ (x - 1)(x - 2)(x - 3) < 0
∴ x ∈ (-∞, 1) ∪ (2, 3)

Test: Linear Inequalities One Variable - Question 6

The number of real solutions (x, y, z, t) of simultaneous equations: 2y = 11/x, 2z = 11/y, 2x = 11/z, 2x = 11/t

Detailed Solution for Test: Linear Inequalities One Variable - Question 6

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.

Test: Linear Inequalities One Variable - Question 7

The solution set contained in R of the inequality 3x + 31-x - 4 < 0 is:

Detailed Solution for Test: Linear Inequalities One Variable - Question 7

Given, 3x + 31-x - 4 < 0
⇒ 32x + 3 - 4.3x < 0
⇒ (3x - 1)(3x - 3) < 0
⇒ 1 < 3x < 3 ⇒ 0 < x < 1
Thus, the solution set is (0, 1).

Test: Linear Inequalities One Variable - Question 8

If 5x + (2√3) 2x ≥ 13x, then the solution set for x is:

Detailed Solution for Test: Linear Inequalities One Variable - Question 8

Given, inequality can be rewritten as (5/13)x + (12/13)x ≥ 1
⇒ cosx α + sinx 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].

Test: Linear Inequalities One Variable - Question 9

The solution of inequality 4x + 3 < 5x + 7 when x is a real number is

Detailed Solution for Test: Linear Inequalities One Variable - Question 9

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 , ∞)

Test: Linear Inequalities One Variable - Question 10

Solution set of inequality loge((x - 2) / (x - 3)) is:

Detailed Solution for Test: Linear Inequalities One Variable - Question 10
  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, ∞)

 

Test: Linear Inequalities One Variable - Question 11

By solving the inequality 6x - 7 > 5, the answer will be

Detailed Solution for Test: Linear Inequalities One Variable - Question 11

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

Test: Linear Inequalities One Variable - Question 12

Let f(x) = ax² + bx + c and f(-1) < 1, f(1) > -1, f(3) < -4 and a ≠ 0, then:

Detailed Solution for Test: Linear Inequalities One Variable - Question 12

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)

Test: Linear Inequalities One Variable - Question 13

The shaded region shown in the figure is given by the inequations

Detailed Solution for Test: Linear Inequalities One Variable - Question 13

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

Test: Linear Inequalities One Variable - Question 14

The set of admissible values of x such that (2x + 3) / (2x - 9) < 0 is:

Detailed Solution for Test: Linear Inequalities One Variable - Question 14

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)

Test: Linear Inequalities One Variable - Question 15

The number of real values of parameter k for which (log16x)2−log16x+log16k=0 will have exactly one solution is

Detailed Solution for Test: Linear Inequalities One Variable - Question 15

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.

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