Expansions — Multiple Choice Questions

If $x + \dfrac{1}{x} = 2, \text{ then } x^2 + \dfrac{1}{x^2}$ is equal to

1. 4

2. 2

3. 0

4. none of these

Answer

We know that,

$x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 -2$.

Substituting values we get,

$x^2 + \dfrac{1}{x^2} = (2)^2 - 2 = 4 - 2 = 2$.

Hence, Option 2 is the correct option.

If x² + y² = 9 and xy = 8, then x + y is equal to

1. 25

2. 5

3. -5

4. ±5

Answer

We know that,

$(x + y) = \sqrt{x^2 + y^2 + 2xy}$.

Substituting values we get,

$(x + y) = \sqrt{9 + 2 \times 8} = \sqrt{9 + 16} = \sqrt{25} = \pm 5$.

Hence, Option 4 is the correct option.

(102)² - (98)² is equal to

1. 200

2. 400

3. 600

4. 800

Answer

Applying a² - b² = (a - b)(a + b) formula to (102)² - (98)² we get,

(102)² - (98)² = (102 - 98)(102 + 98) = 4 × 200 = 800.

Hence, Option 4 is the correct option.

96 × 104 is equal to

1. 9984

2. 9974

3. 9964

4. none of these

Answer

We can write 96 × 104 as (100 - 4) × (100 + 4).

By a² - b² = (a - b)(a + b) formula,

(100 - 4) × (100 + 4) = (100)² - (4)² = 10000 - 16 = 9984.

Hence, Option 1 is the correct option.

$\dfrac{103^2 - 97^2}{200}$ is equal to

1. 3

2. 4

3. 5

4. 6

Answer

By a² - b² = (a - b)(a + b) formula,

$\dfrac{103^2 - 97^2}{200} = \dfrac{(103 - 97)(103 + 97)}{200} \\[1em] = \dfrac{6 \times 200}{200} \\[1em] = 6.$

Hence, Option 4 is the correct option.

If x + y = 11 and xy = 24, then x² + y² is equal to

1. 121

2. 73

3. 48

4. 169

Answer

We know that,

x² + y² = (x + y)² - 2xy.

Substituting values we get,

x² + y² = (11)² - 2 × 24 = 121 - 48 = 73.

Hence, Option 2 is the correct option.

The value of 249² - 248² is

1. 1²

2. 477

3. 487

4. 497

Answer

By a² - b² = (a - b)(a + b) formula,

249² - 248² = (249 - 248)(249 + 248) = 1 × 497 = 497.

Hence, Option 4 is the correct option.

If $\dfrac{x}{y} + \dfrac{y}{x}$ = -1 (x, y ≠ 0) then the value of x³ - y³ is

1. 1

2. -1

3. 0

4. $\dfrac{1}{2}$

Answer

Given,

$\Rightarrow \dfrac{x}{y} + \dfrac{y}{x} = -1 \\[1em] \Rightarrow \dfrac{x^2 + y^2}{xy} = -1 \\[1em] \Rightarrow x^2 + y^2 = -xy \\[1em] \Rightarrow x^2 + y^2 + xy = 0.$

We know that,

x³ - y³ = (x - y)(x² + y² + xy) = (x - y) × 0 = 0.

Hence, Option 3 is the correct option.

If a + b + c = 0, then the value of a³ + b³ + c³ is

1. 0

2. abc

3. 2abc

4. 3abc

Answer

We know that,

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - bc - ca - ab).

Substituting values we get,

a³ + b³ + c³ - 3abc = 0 × (a² + b² + c² - bc - ca - ab)

a³ + b³ + c³ - 3abc = 0

a³ + b³ + c³ = 3abc.

Hence, Option 4 is the correct option.

If $x - \dfrac{2}{x} = 3, \text{ then } x^3 - \dfrac{8}{x^3}$ is equal to

1. 27

2. 36

3. 45

4. 54

Answer

We know that,

(a - b)³ = a³ -3ab(a - b) - b³

⇒ a³ - b³ = (a - b)³ + 3ab(a - b)

$\therefore x^3 - \dfrac{8}{x^3} = \Big(x - \dfrac{2}{x}\Big)^3 + 6\Big(x - \dfrac{2}{x}\Big)$

Substituting values we get,

$x^3 - \dfrac{8}{x^3} = 3^3 + 6 \times 3 \\[1em] = 27 + 18 \\[1em] = 45.$

Hence, Option 3 is the correct option.

Consider the following two statements :

Statement 1: If a + b = 0, then a² + b² = 0

Statement 2: a² + b² = (a + b)².

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

Given,

⇒ a + b = 0

Squaring both sides, we get

⇒ (a + b)² = 0²

⇒ a² + b² + 2ab = 0

⇒ a² + b² = -2ab

∴ Statement 1 is false.

(a + b)² = a² + b² + 2ab

∴ Statement 2 is false.

∴ Both statements are false.

Hence, option 2 is the correct option.