PE 1 - Inequalities & Modulus | Algebra - Inequalities & Modulus

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).

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).

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).

Answer: Option C

Explanation :

We know AM ≥ GM.

⇒ $\frac{x+y+z}{3}$ ≥ $\sqrt[3]{xyz}$

⇒ $\frac{5}{3}$ ≥ $\sqrt[3]{xyz}$

⇒ xyz ≤ $\frac{125}{27}$

Hence, option (c).

Answer: Option B

Explanation :

Let S = $\frac{x+y}{x}$ + $\frac{y+z}{y}$ + $\frac{z+x}{z}$ = 1 + $\frac{y}{x}$ + 1 + $\frac{z}{y}$ + 1 + $\frac{x}{z}$ = 3 + $\frac{y}{x}$ + $\frac{z}{y}$ + $\frac{x}{z}$

For S to be minimum $\frac{y}{x}$ + $\frac{z}{y}$ + $\frac{x}{z}$ should be minimized.

We know, AM ≥ GM

⇒ $\frac{{\displaystyle \frac{y}{x}}+{\displaystyle \frac{z}{y}}+{\displaystyle \frac{x}{z}} }{3}$ ≥ $\sqrt[3]{\frac{y}{x}\times \frac{z}{y}\times \frac{x}{z}}$

⇒ $\frac{y}{x}$ + $\frac{z}{y}$ + $\frac{x}{z}$ ≥ 3

∴ S ≥ 3 + 3 = 6

Hence, option (b).

Answer: Option B

Explanation :

Let S = $\frac{x+y}{z}$+$\frac{y+z}{x}$ + $\frac{z+x}{y}$ = $\left(\frac{x}{z}+\frac{y}{z}\right)$ + $\left(\frac{y}{x}+\frac{z}{x}\right)$ + $\left(\frac{z}{y}+\frac{x}{y}\right)$

⇒ S = $\left(\frac{x}{z}+\frac{z}{x}\right)$ + $\left(\frac{y}{z}+\frac{z}{y}\right)$ + $\left(\frac{y}{x}+\frac{x}{y}\right)$

We know sum of a positive number and its reciprocal is always greater than or equal to 2.

∴ $\left(\frac{x}{z}+\frac{z}{x}\right)$ ≥ 2, $\left(\frac{y}{z}+\frac{z}{y}\right)$ ≥ 2 and $\left(\frac{y}{x}+\frac{x}{y}\right)$ ≥ 2

∴ S ≥ 2 + 2 + 2 = 6

Hence, option (b).

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 $\frac{2x}{3}$+ $\frac{2x}{3}$+ $\frac{2x}{3}$+ $\frac{3y}{2}$ + $\frac{3y}{2}$

We know, AM ≥ GM

⇒ $\frac{{\displaystyle \frac{2x}{3}}+\frac{2x}{3}+\frac{2x}{3}+{\displaystyle \frac{3y}{2}}+\frac{3y}{2}}{5}$ ≥  $\sqrt[5]{\frac{2x}{3}\times \frac{2x}{3}\times \frac{2x}{3}\times \frac{3y}{2}\times \frac{3y}{2}}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{\frac{2}{3}\times x^3{y}^2}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{\frac{2}{3}\times 48}$

⇒ 2x + 3y ≥ 5 × $\sqrt[5]{32}$

⇒ 2x + 3y ≥ 10

∴ The least possible value of 2x + 3y = 10

Hence, option (a).

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 = $\frac{2\pm \sqrt{4–4\times 1\times -1}}{2}$ = 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.

Answer: Option C

Explanation :

Given, $\frac{x}{x-1}$ + $\frac{1}{2x}$ < 0

⇒ $\frac{2x^2+x-1}{2x(x-1)}$ < 0

⇒ $\frac{(2x-1)(x+1)}{2x(x-1)}$ < 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).