Expansions — Multiple Choice Questions

If a + b = 12 and ab = 35, then (a − b)² =

1. 2

2. 4

3. 16

4. 20

Answer

Given,

a + b = 12 and ab = 35

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

⇒ (a − b)² = (a + b)² - 4ab

⇒ (a − b)² = (12)² - 4(35)

⇒ (a − b)² = 144 - 140

⇒ (a − b)² = 4

Hence, Option 2 is the correct option.

If a + b = 16 and a − b = 2, then ab =

1. 48

2. 56

3. 63

4. 65

Answer

Given,

a + b = 16 and a - b = 2

Using identity,

⇒ (a + b)² - (a - b)² = 4ab

⇒ 16² - 2² = 4ab

⇒ 4ab = 256 - 4

⇒ 4ab = 252

⇒ ab = $\dfrac{252}{4}$

⇒ ab = 63.

Hence, Option 3 is the correct option.

If $p + \dfrac{1}{p} = \dfrac{5}{2} \text{ then }p^2 + \dfrac{1}{p^2}$ =

1. $\dfrac{25}{4}$

2. $\dfrac{19}{4}$

3. $\dfrac{9}{4}$

4. $\dfrac{17}{4}$

Answer

Given,

$p + \dfrac{1}{p} = \dfrac{5}{2}$

Using identity,

$\Rightarrow \Big(p + \dfrac{1}{p}\Big)^2 = p^2 + \dfrac{1}{p^2} + 2 \\[1em] \Rightarrow \Big(\dfrac{5}{2}\Big)^2 = \Big(p^2 + \dfrac{1}{p^2}\Big) + 2 \\[1em] \Rightarrow \dfrac{25}{4} = \Big(p^2 + \dfrac{1}{p^2}\Big) + 2 \\[1em] \Rightarrow \dfrac{25}{4} - 2 = \Big(p^2 + \dfrac{1}{p^2}\Big) \\[1em] \Rightarrow \Big(\dfrac{25 - 8}{4}\Big) = \Big(p^2 + \dfrac{1}{p^2}\Big) \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2}\Big) = \dfrac{17}{4} \\[1em]$

Hence, Option 4 is the correct option.

If $x + \dfrac{1}{x} = 5$, then $x - \dfrac{1}{x}$ =

1. $\pm \sqrt{21}$

2. $\pm \sqrt{29}$

3. $\pm 3$

4. $\pm 2$

Answer

Given,

$x + \dfrac{1}{x} = 5$

Using identity,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow (5)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow 25 - 2 = x^2 + \dfrac{1}{x^2} \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 23\\[1em]$

We know that,

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 23 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 21 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{21} \\[1em]$

Hence, Option 1 is the correct option.

If $y + \dfrac{1}{y} = 2$, then $y^5 + \dfrac{1}{y^5}$ =

1. 5

2. 3

3. 2

4. 1

Answer

Given,

$y + \dfrac{1}{y} = 2$

$\Rightarrow \dfrac{y^2 + 1}{y} = 2 \\[1em] \Rightarrow y^2 + 1 = 2y \\[1em] \Rightarrow y^2 - 2y + 1 = 0 \\[1em] \Rightarrow (y - 1)^2 = 0 \\[1em] \Rightarrow y-1 = 0 \\[1em] \Rightarrow y = 1$

Substituting value of y, we get :

$\Rightarrow y^5 + \dfrac{1}{y^5} = (1)^5 + \dfrac{1}{(1)^5} = 1 + 1 = 2$.

Hence, option 3 is correct option.

If $a^2 - \dfrac{1}{a^2} = 7$, then $a^4 + \dfrac{1}{a^4}$ =

1. 49

2. 51

3. 53

4. 60

Answer

Given,

$a^2 - \dfrac{1}{a^2} = 7$

Upon squaring both sides,

$\Rightarrow \Big(a^2 - \dfrac{1}{a^2}\Big)^2 = a^4 + \dfrac{1}{a^4} - 2 \\[1em] \Rightarrow (7)^2 = \Big(a^4 + \dfrac{1}{a^4}\Big) - 2 \\[1em] \Rightarrow 49 = \Big(a^4 + \dfrac{1}{a^4}\Big) - 2 \\[1em] \Rightarrow \Big(a^4 + \dfrac{1}{a^4}\Big) = 49 + 2 \\[1em] \Rightarrow \Big(a^4 + \dfrac{1}{a^4}\Big) = 51 \\[1em]$

Hence, Option 2 is the correct option.

If x + y + z = 0, then x³ + y³ + z³ =

1. xyz

2. 3xyz

3. 27xyz

4. 0

Answer

Given,

x + y + z = 0

Using identity,

⇒ x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

⇒ x³ + y³ + z³ - 3xyz = (0) × (x² + y² + z² - xy - yz - zx)

⇒ x³ + y³ + z³ - 3xyz = 0

⇒ x³ + y³ + z³ = 3xyz.

Hence, Option 2 is the correct option.

If $\dfrac{a}{b} + \dfrac{b}{a} = 1$ (a, b ≠ 0), then a³ + b³ =

1. 0

2. 1

3. 2

4. 8

Answer

Given,

$\Rightarrow \dfrac{a}{b} + \dfrac{b}{a} = 1 \\[1em] \Rightarrow \dfrac{a^2 + b^2}{ab} = 1 \\[1em] \Rightarrow a^2 + b^2 = ab \\[1em] \Rightarrow a^2 + b^2 - ab = 0.$

We know that,

⇒ a³ + b³ = (a + b)(a² - ab + b²)

⇒ a³ + b³ = (a + b)(0)

⇒ a³ + b³ = 0.

Hence, Option 1 is the correct option.

If $p + \dfrac{1}{p} = x\text{ and }p - \dfrac{1}{p} = y$, then the relation between x and y is :

1. x² = y²

2. x² + y² = 4

3. x² - y² = 4

4. xy = 2

Answer

Given,

Upon squaring $p + \dfrac{1}{p} = x$ we get,

$\Rightarrow \Big(p + \dfrac{1}{p}\Big)^2 = x^2 \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2} + 2\Big) = x^2 \text{ ....(1)}$

Upon squaring $p - \dfrac{1}{p} = y$ we get,

$\Rightarrow \Big(p - \dfrac{1}{p}\Big)^2 = y^2 \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2} - 2\Big) = y^2 \text{ ....(2)}$

Subtracting (2) from (1) we get,

$\Rightarrow \Big(p^2 + \dfrac{1}{p^2} + 2\Big) - \Big(p^2 + \dfrac{1}{p^2} - 2\Big) = x^2 - y^2 \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2} + 2 - p^2 - \dfrac{1}{p^2} + 2\Big) = x^2 - y^2 \\[1em] \Rightarrow x^2 - y^2 = 4.$

Hence, Option 3 is the correct option.

If $a = \dfrac{1}{a - 7}$, then $a - \dfrac{1}{a}$ =

1. 0

2. 1

3. 7

4. $\dfrac{1}{7}$

Answer

Given,

$\Rightarrow a = \dfrac{1}{a - 7} \\[1em] \Rightarrow a \times (a-7) = 1 \\[1em] \Rightarrow a - 7 = \dfrac{1}{a} \\[1em] \Rightarrow a - \dfrac{1}{a} = 7$

Hence, option 3 is correct option.

2.51 × 2.51 + 1.31 × 1.31 − 2.62 × 2.51 =

1. 1.44

2. 0.44

3. 1

4. 0

Answer

Given,

⇒ 2.51 × 2.51 + 1.31 × 1.31 − 2.62 × 2.51

⇒ 2.51² + 1.31² - 2(1.31)(2.51)

⇒ 2.51² + 1.31² - 2(2.51)(1.31)

⇒ (2.51 - 1.31)²     [(a - b)² = a² + b² - 2ab]

⇒ (1.2)²

⇒ 1.44

Hence, Option 1 is the correct option.

If $\Big(x + \dfrac{1}{x}\Big)^2 = 3$, then $x^3 + \dfrac{1}{x^3}$ =

1. 9

2. 3

3. 2

4. 0

Answer

Given,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 3 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm \sqrt{3}$

Case 1:

$\Rightarrow \Big(x + \dfrac{1}{x}\Big) = \sqrt{3}$

Using identity,

$\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (\sqrt{3})^3 - 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (3\sqrt{3}) - 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = 0$

Case 2:

$\Rightarrow \Big(x + \dfrac{1}{x}\Big) = -\sqrt{3}$

Using identity,

$\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (-\sqrt{3})^3 - 3(-\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (-3\sqrt{3}) + 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = 0$

Hence, Option 4 is the correct option.

The value of ab if 3a + 5b = 15 and 9a² + 25b² = 75 is :

1. 4

2. 5

3. 6

4. 8

Answer

Given,

3a + 5b = 15 and 9a² + 25b² = 75

⇒ (3a + 5b)² = (3a)² + (5b)² + 2 × (3a) × (5b)

⇒ (3a + 5b)² = 9a² + 25b² + 30ab

⇒ 15² = 75 + 30ab

⇒ 225 = 75 + 30ab

⇒ 30ab = 225 - 75

⇒ 30ab = 150

⇒ ab = $\dfrac{150}{30}$

⇒ ab = 5.

Hence, Option 2 is the correct option.

If l + m − n = 9 and l² + m² + n² = 31, then mn + nl − lm is :

1. -25

2. 25

3. -2

4. -5

Answer

Given,

l + m − n = 9

l² + m² + n² = 31

Solving,

⇒ [(l + m) − (n)]² = (l + m)² + n² - 2 × (l + m) × n

⇒ [(l + m) − (n)]² = l² + m² + 2 × l × m + n² - 2ln - 2mn

⇒ (9)² = l² + m² + 2lm + n² - 2ln - 2mn

⇒ 81 = l² + m² + n² + 2lm - 2ln - 2mn

⇒ 81 = 31 + 2(lm - ln - mn)

⇒ 2(lm - ln - mn) = 81 - 31

⇒ (lm - ln - mn) = $\dfrac{50}{2}$

⇒ (lm - ln - mn) = 25

⇒ -(lm - ln - mn) = -25

⇒ (mn + nl - lm) = -25

Hence, Option 1 is the correct option.

If a − b + c = 6 and a² + b² + c² = 38, then ab + bc − ca is :

1. 0

2. 1

3. -1

4. not possible

Answer

Given,

a - b + c = 6

a² + b² + c² = 38

We know that,

⇒ [(a - b) + (c)]² = (a - b)² + c² + 2 × (a - b) × c

⇒ [(a - b) + (c)]² = a² + b² - 2 × a × b + c² + 2ac - 2bc

⇒ (6)² = a² + b² + c² - 2(ab - ac + bc)

⇒ 36 = 38 - 2(ab - ac + bc)

⇒ 2(ab + bc − ca) = 38 - 36

⇒ 2(ab + bc − ca) = 2

⇒ (ab + bc − ca) = $\dfrac{2}{2}$

⇒ (ab + bc − ca) = 1.

Hence, Option 2 is the correct option.