CRE 2 - Solving Inequalities | Algebra - Inequalities & Modulus

Answer: Option A

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Explanation :

Given, 2x + 5 ≥ 1
⇒ x ≥ -2   …(1)

Also, x + 7 > 2x + 3
⇒ x < 4   …(2)

For both (1) and (2) to be true

-2 ≤ x < 4

Hence, option (a).

Answer: Option C

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Explanation :

Given, (x + 3)(x – 5) > 0

The given inequality is possible when 

Case 1: Both (x + 3) and (x – 5) are positive
i.e., x + 3 > 0 ⇒ x > -3   …(1)
x – 5 > 0 ⇒ x > 5   …(2)

From (1) and (2)
x > 5   …(3)

Case 2: Both (x + 3) and (x – 5) are negative
i.e., x + 3 < 0 ⇒ x < -3   …(4)
x – 5 < 0 ⇒ x < 5   …(5)

From (1) and (2)
x < -3   …(6)

From (3) and (6), we get
-3 > x > 5

i.e., x ∈ (-∞, -3) ∪ (5, ∞)

Hence, option (c).

Answer: 6

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Explanation :

Given, x² – 5x – 6 > 0

⇒ (x - 6)(x + 1) > 0

This is only possible when (x - 6) is negative and (x + 1) is positive.
∴ -1 < x < 6

Integral values x can take are 0, 1, 2, 3 and 4, i.e., 5 values

Hence, 6.

Answer: Option D

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Explanation :

In such questions we need to identify the critical points, i.e., values of x when LHS = 0.

∴ critical points are -5, 3 and 7.

Now, for x > 7, the given inequality will be always positive (range not accepted)
Now, for -3 < x < 7, the given inequality will be always negative (range accepted)
Now, for -5 < x < -3, the given inequality will be always positive (range not accepted)

∴ x ∈ (-3, 7)

Also, we also want (x-3)(x+5)(x-7) = 0, hence all critical points will be included in our final answer.

∴ x ∈ [-3, 7] + {-5}

Hence, option (d.)

Answer: Option C

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Explanation :

Here we only need to consider the signs of each of these 4 expressions given and not their value.

Sign of (x + 3)³ will be same as the sign of (x + 3)
Sign of (x + 5)⁵ will be same as the sign of (x + 5)
(x - 4)⁴ is always positive

∴ The range of given inequality will be same as the range for (x - 1)(x + 3)(x + 5) ≥ 0

Here the critical points are -5, -3 and 1.

∴ (x - 1)(x + 3)(x + 5) is

> 0 when x > 1 (range accepted)
< 0 when -3 < x < 1 (range not accepted)
> 0 when -5 < x < -3 (range accepted)
< 0 when x < -5 (range not accepted)

∴ For (x - 1)(x + 3)(x + 5) > 0

x ∈ (-5, -3) ∪ (1, ∞)

and for (x - 1)(x + 3)(x + 5) ≥ 0, we will also include the critical points.

∴ x ∈ [-5, -3] ∪ [1, ∞)

Now, the original inequality is (x - 1)(x + 3)³(x - 4)⁴(x + 5)⁵ ≥ 0

∴ The range of x should also include the critical point x = 4.

⇒ x ∈ [-5, -3] ∪ [1, ∞)

Hence option (c).

Answer: Option A

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Explanation :

Here we only need to consider the signs of each of these 3 expressions given and not their value.

∴ The range of given inequality will be same as the range for (x + 3)(x - 7)(x + 1) ≤ 0

The critical points here are -3, -1 and 7.

∴ (x + 3)(x - 7)(x + 1) is

> 0 when x > 7 (range not accepted)
< 0 when -1 < x < 7 (range accepted)
> 0 when -3 < x < -1 (range not accepted)
< 0 when x < -3 (range accepted)

∴ For (x + 3)(x - 7)(x + 1) ≤ 0, the acceptable range of x ∈ (-∞, -3] ∪ [-1, 7]

Here we need to be careful about critical points. For x = -1, the original expression is invalid, hence x = -1 should not be included in our range of x.

∴ For $\frac{{(x+3)}^3(x-7)}{(x+1)}$ ≤ 0

x ∈ (-∞, -3] ∪ (-1, 7]

Hence, option (a).

Answer: Option D

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Explanation :

Given inequality is x + 2√x – 24 ≤ 0

Substituting √x with a, ∴ x = a²

The given inequality becomes:

a² + 2a – 24 ≤ 0

⇒ (a + 6)(a - 4) ≤ 0

⇒ a ∈ [-6, 4]

a cannot be negative, hence a ∈ [0, 4]

⇒ √x ∈ [0, 4]

⇒ x ∈ [0, 16]

Hence, option (d).

Answer: Option A

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Explanation :

In a quadratic expression ax² + bx + c, if a > 0 and (b² – 4ac) < 0, then the expression is always positive for all real values of x.

Here, b² – 4ac = 36 – 4 × 1 × 19 = - 40 < 0 and a > 0

∴ x² + 6x + 19 is always positive for all possible values of x.

Hence, option (a).

Answer: Option D

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Explanation :

In the given inequality RHS is not equal to 0, hence we first bring all the terms to LHS and then simplify

⇒ $\frac{x+2}{x-1}$ - 2 ≥ 0

⇒ $\frac{x+2-2(x-1)}{x-1}$ ≥ 0

⇒ $\frac{-x+4}{x-1}$ ≥ 0

⇒ $\frac{x-4}{x-1}$ ≤ 0

Critical points are 1 and 5.

∴ x ∈ (1, 4]

Note: we will not include x = 1 since the expression becomes invalid at x = 1.

Hence, option (d).

Answer: Option A

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Explanation :

$\frac{x^2-7x+12}{2x^2+9x-5}$ < 0

⇒ $\frac{(x-3)(x-4)}{(2x-1)(x+5)}$ < 0

Critical points are -5, ½, 3 and 4

∴ x ∈ (-5, ½) ∪ (3, 4)

Hence, option (a).