PE 1 - Inequalities & Modulus | Algebra - Inequalities & Modulus
For two real numbers a and b, let x = |a| + |b|, y = |a| − |b|, and z = |a − b|. Which of the following is true?
- A.
z ≤ x ≤ y
- B.
x ≤ y ≤ z
- C.
y ≤ z ≤ x
- D.
y ≤ x ≤ z
Answer: Option C
Explanation :
|a| and |b| are two non-negative numbers, hence we can be sure that |a| + |b| ≥ |a| - |b|, i.e., x ≥ y.
z = |a - b|, hence z will always lie between x and y.
⇒ x ≥ z ≥ y
Hence, option (c).
Workspace:
The range of f(x) = |x| + |x – 3| + |x + 5| is?
- A.
[–5, 3]
- B.
[0, 5]
- C.
[11, ∞)
- D.
[8, ∞)
Answer: Option D
Explanation :
Here the critical points are x = -5, 0, 3
∴ f(x) will be minimum when x = 0
⇒ Minimum value of f(x) = f(0) = 0 + 3 + 5 = 8
There is no limit on maximum value of f(x).
∴ f(x) ∈ [8, ∞)
Hence, option (d).
Workspace:
The solution set for the inequation |x − 2| + |x + 3| < 2 is _______.
- A.
(−∞,−2) ∪ (1, ∞)
- B.
[2, 1)
- C.
(−1, 2)
- D.
an empty set
Answer: Option D
Explanation :
Here, the critical points are x = -3 and 2
Case 1: x > 2
∴ x – 2 + x + 3 < 2
⇒ 2x < 1
⇒ x < ½ (this contradicts our assumption that x > 2)
∴ No solution
Case 2: -3 < x < 2
∴ -(x - 2) + x + 3 < 2
⇒ 5 < 2 (not true)
∴ No solution
Case 3: x < -3
∴ -(x - 2) – (x + 3) < 2
⇒ -2x < 3
⇒ x > -3/2 (this contradicts our assumption that x < -3/2)
∴ No solution
∴ There is no real value of x for which |x − 2| + |x + 3| < 2.
Hence, option (d).
Workspace:
If x, y and z are positive real numbers such that x + y + z = 5, then which of the following is always true?
- A.
xyz ≤ 5/3
- B.
xyz ≥ 5/3
- C.
xyz ≤ 125/27
- D.
xyz ≥ 27/125
Answer: Option C
Explanation :
We know AM ≥ GM.
⇒ ≥
⇒ ≥
⇒ xyz ≤
Hence, option (c).
Workspace:
What is minimum value of , , , where x, y and z are positive real numbers.
- A.
3
- B.
6
- C.
0
- D.
1
Answer: Option B
Explanation :
Let S = + + = 1 + + 1 + + 1 + = 3 + + +
For S to be minimum + + should be minimized.
We know, AM ≥ GM
⇒ ≥
⇒ + + ≥ 3
∴ S ≥ 3 + 3 = 6
Hence, option (b).
Workspace:
What is minimum value of + + , where x, y and z are positive real numbers
- A.
3
- B.
6
- C.
0
- D.
1
Answer: Option B
Explanation :
Let S = + + = + +
⇒ S = + +
We know sum of a positive number and its reciprocal is always greater than or equal to 2.
∴ ≥ 2, ≥ 2 and ≥ 2
∴ S ≥ 2 + 2 + 2 = 6
Hence, option (b).
Workspace:
If x and y are positive real numbers and x3y2 = 48, then the minimum value of 2x + 3y is?
- A.
10
- B.
14
- C.
24
- D.
Cannot be determined
Answer: Option A
Explanation :
Given, x3y2 = 24
We need to calculate minimum value of A = 2x + 3y
Since the power of x is 3 and y is 2 in x3y2 = 24, we rewrite write A as + + + +
We know, AM ≥ GM
⇒ ≥
⇒ 2x + 3y ≥ 5 ×
⇒ 2x + 3y ≥ 5 ×
⇒ 2x + 3y ≥ 5 ×
⇒ 2x + 3y ≥ 10
∴ The least possible value of 2x + 3y = 10
Hence, option (a).
Workspace:
Find the number of solutions of the equation 2x2 + |4x - 7| = 9
Answer: 2
Explanation :
Case 1: 4x – 7 > 0 i.e., x > 7/4
∴ 2x2 + 4x - 7 = 9
⇒ 2x2 + 4x - 16 = 0
⇒ x2 + 2x - 8 = 0
⇒ (x - 2)(x + 4) = 0
∴ x = -4 or 2
[x > 7/4, hence x = 4 is rejected]
∴ x = 2
Case 2: 4x – 7 < 0 i.e., x < 7/4
∴ 2x2 - 4x + 7 = 9
⇒ 2x2 - 4x - 2 = 0
⇒ x2 - 2x - 1 = 0
⇒ x = = 1 ± 2√2
x = 1 + 2√2 will be rejected as it is greater than 7/4
∴ x = 1 - 2√2
⇒ Total number of solutions = 1 + 1 = 2.
Hence, 2.
Workspace:
If + < 0, then the range of x is?
- A.
x ∈ (-∞, -1/2) ∪ [1/2, 1)
- B.
x ∈ (-1, 1/2) ∪ (1, ∞)
- C.
x ∈ (-∞, -1) ∪ (1/2, 1)
- D.
None of these
Answer: Option C
Explanation :
Given, + < 0
⇒ < 0
⇒ < 0
The range of this inequality is same as the range of inequality (2x - 1)(x + 1)x(x - 1) < 0
Critical points are -1, 0, ½ and 1
∴ x ∈ (-1, 0) ∪ (1/2, 1)
Hence, option (c).
Workspace:
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