Maths 5 Marks Questions

1. 2y(3x² + y²)

Solution:

We know the identity

a³ - b³ = (a - b)(a² + b² + ab)

Let x + y = a and x - y = b

Then,

a³ - b³

= (x + y)³ - (x - y)³

= [(x + y) - (x - y)][(x + y)² + (x - y)² + (x + y)(x - y)]

= 2y[x² + y² + 2xy + x² + y² - 2xy + x² - y²]

= 2y(3x² + y²)

Hence, correct option is (a).

4. (x - 1 + y)(x² + 1 + y² + x + y - xy)

Solution:

By using identity

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

We can write,

x³ - 1 + y³ + 3xy

= (x³) + (-1)³ + (y³) - 3(-1)(x)(y)

= [x + (-1) + y][x² + (-1)² + y² - x(-1) - y(-1) - xy]

= (x - 1 + y)(x² + 1 + y² + x + y - xy)

Hence, correct option is (a).

1. (x² + 3x + 5)(x² - 3x + 5)

Solution:

For making perfect square to x⁴ + x² + 25

We add +10x² and -10x² to it.

= x⁴ + x² + 25

= x⁴ + x² + 25 + 10x² - 10x²

= [x⁴ + 10x² + 25] - 9x²

= (x² + 5)² + (3x)²

= [(x² + 5) + 3x][(x² + 5) - 3x]

= (x² + 3x + 5)(x² - 3x + 5)

Hence, correct option is (a).

4. (x - y)²(x + y)

Solution:

x³ - x²y - xy² + y³ = x³ + y³ - xy(x + y)

Now by identity x³ + y³ = (x + y)(x² + y² - xy), we have

x³ - x²y - xy² + y³ = (x + y)(x² + y² - xy) - xy(x + y)

= (x + y)(x² + y² - xy - xy)

= (x + y)(x² + y² - 2xy)

= (x + y)(x - y)²

Hence, correct option is (d).

4. (x - 1)(x + 3)(x - 2)

Solution:

x³ - 7x + 6 = x³ - 7x + 6 + 1 - 1 (by adding +1 & -1 to R.H.S)

= x³ - 7x + 7 - 1

= (x³ - 1) - 7(x - 1)

Now by identity a³ - b³ = (a - b)(a² + b² + ab), we get

x³ - 7x + 6 = (x³ - 1) - 7(x - 1)

= (x - 1)(x² + x + 1) - 7(x - 1)

= (x - 1)(x² + x + 1 - 7)

= (x - 1)(x² + x - 6)

= (x - 1)(x + 3)(x - 2)

Hence, correct option is (d).

1. (x - 2y - 4)(x - 2y + 2)

Solution:

x² + 4y² + 4y - 4xy - 2x - 8

= x² +(2y)² - 2 × x(2y) + 4y - 2x - 8

= (x - 2y)² + 4y - 2x - 8 ...(1)

Now making eq(1) a perfect square by adding 1 and -1

(x - 2y)² + 4y - 2x - 8 = (x - 2y)² + 4y - 2x - 8 + 1 - 1

= (x - 2y)² + (1)² - 2 × (1) × (x - 2y) - 9

= (x - 2y - 1)² - (3)²

= [(x - 2y - 1) - 3][x - 2y - 1 + 3]

= (x - 2y - 4)(x - 2y + 2)

Hence, correct option is (a).

3. (a + x +1)(a - x + 1)

Solution:

a² - 1 - 2x - x²

= a² - (1 + 2x + x²)

= a² - (1 + x)²

= [a - (1 + x)][a + (1 + x)]

= (a - x - 1)(a + x + 1)

Hence, correct option is (c).

3. (2a + b + 1)(4a² + b² + 1 -2ab - b - 2a)

Solution:

We know the identity

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

So by using identity, we can write given expression as

(2a)³ + (b)³ + (1)³ - 3(2a)(b)(1)

= (2a + b + 1)[(2a)² + b² + 1² -2a × b - b × 1 - 2a × 1]

= (2a + b + 1)(4a² + b² + 1 -2ab - b - 2a)

Hence, correct option is (c).

1. (x² + 2x + 2)(x² - 2x + 2)

Solution:

x⁴ + 4

= x⁴ + 4 + 4x² - 4x²

= (x⁴ + 4x² + 4) - 4x²

= (x² + 2)² - (2x)²

= (x² + 2 - 2x)(x² + 2 + 2x)

= (x² + 2x + 2)(x² - 2x + 2)

Hence, correct option is (a).

2. 3(a - b)(b - c)(c - a)

Solution:

By we know that a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

If a + b + c = 0, then

a³ + b³ + c³ = 3abc

In given expression,

Let a - b = A, b - c = B, c - a = C

Now, a - b + b - c + c - a = 0

i.e. A + B + C = 0

⇒ A³ + B³ + C³ = 3ABC

⇒ (a - b)³ + (b - c)³ + (c - a)³ = 3(a - b)(b - c)(c - a)

Hence, correct option is (b).

4. 8

Solution:

Let x + y = A and x - y = B

Now, (A - B)³ = A³ - B³ - 3AB(A - B)

⇒ [(x + y) - (x - y)]³ = (x + y)³ - (x - y)³ - 3(x + y)(x - y)[(x + y) - (x - y)]

= (x + y)³ - (x - y)³ - 3(x² - y²)(2y)

= (x + y)³ - (x - y)³ - 6y(x² - y²)

But, (x + y)³ - (x - y)³ - 6y(x² - y²) = ky³

⇒ [(x + y) - (x - y)]³ = (2y)³ = k8y³

⇒ (2y)³ = ky³

⇒ 8y³ = ky³

⇒ k = 8

Hence, correct option is (d).

3. -10

Solution:

The given equation is 

x³ - 3x² + 3x - 7 = (x + 1)(ax² + bx + c)

This can be written as

x³ - 3x² + 3x - 7 = (x + 1)(ax² + bx + c)

= x³ - 3x² + 3x - 7 = ax³ + bx² + cx + ax² + bx + c

= x³ - 3x² + 3x - 7 = ax³ + (a + b)x² + (b + c)x + c

Comparing the cofficients on both sides of the equation.

We get,

a = 1 ...(1)

a + b = 3 ...(2)

b + c = 3 ...(3)

c = -7 ...(4)

Putting the value of a form (1) in (2)

We get,

1 + b = 3,

b = -3 - 1

b= -4

So the value of a, b and c is 1, -4 and -7 respectively.

Therefore,

a + b + c = 1 - 4 - 7 = -10

Hence, correct option is (c).

3. 0

Solution:

3x = a + b + c

⇒ a + b + c - 3x = 0

⇒ 3x - (a + b + c) = 0

⇒ (x - a) + (x - b) + (x - c) = 0 ...(1)

Using identity if a + b + c = 0 then, a³ + b³ + c³ - 3abc = 0

If we take x - a = A, x - b = B, x - c = C in equation (1), we get

A + B + C = 0

⇒ A³ + B³ + C³ - 3ABC= 0

⇒ (x - a)³ + (x - b)³ + (x - c)³ - 3(x - a) (x - b) (x - c) = 0

Hence, correct option is (c).