**Question 4:**

Solve 3x + 8 > 2, when

(i) x is an integer (ii) x is a real number

**Answer**

The given inequality is 3x + 8 > 2.

3x + 8 > 2

⇒ 3x + 8 – 8 > 2 – 8

⇒ 3x > −6

**Question 4:**

Solve 3x + 8 > 2, when

(i) x is an integer (ii) x is a real number

**Answer**

The given inequality is 3x + 8 > 2.

3x + 8 > 2

⇒ 3x + 8 – 8 > 2 – 8

⇒ 3x > −6

**Question 3:**

Solve 5x– 3 < 7, when

(i) x is an integer (ii) x is a real number

**Answer**

The given inequality is 5x– 3 < 7.

−12x > 30

⇒ 5x – 3 + 3 < 7 + 3

⇒ 5x < 10

(i) The integers less than 2 are …, –4, –3, –2, –1, 0, 1.

Thus, when x is an integer, the solutions of the given inequality are …, –4, –3, –2, –1, 0, 1.

Hence, in this case, the solution set is {…, –4, –3, –2, –1, 0, 1}.

(ii) When x is a real number, the solutions of the given inequality are given by x < 2,

that is, all real numbers x which are less than 2.

Thus, the solution set of the given inequality is x ∈ (–∞, 2).

**Question 2:**

Solve –12x > 30, when

(i) x is a natural number (ii) x is an integer

**Answer**

The given inequality is –12x > 30.

−12x > 30

(i) There is no natural number less than (−5/2).

Thus, when x is a natural number, there is no solution of the given inequality.

(ii) The integers less than (−5/2) are …, –5, –4, –3.

Thus, when x is an integer, the solutions of the given inequality are …, –5, –4, –3.

Hence, in this case, the solution set is {…, –5, –4, –3}.

**Question 1:**

Solve 24*x *< 100, when (i) *x *is a natural number (ii) *x *is an integer

**Answer**

The given inequality is 24*x *< 100

24x < 100

(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than 25/6.

Thus, when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.

Hence, in this case, the solution set is {1, 2, 3, 4}.

(ii) The integers less than 25/6 are … are …–3, –2, –1, 0, 1, 2, 3, 4.

Thus, when x is an integer, the solutions of the given inequality are …–3, –2, –1, 0, 1, 2, 3, 4.

Hence, in this case, the solution set is {…–3, –2, –1, 0, 1, 2, 3, 4}.

**Question 20:**

**Answer**

∴ m = 4k, where k is some integer.

Therefore, the least positive integer is 1.

Thus, least positive integral value of m is 4(= 4 ×1).

**Question 19:**

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} + B^{2}.

**Answer**

(a + ib) (c + id) (e + if) (g + ih) =A + IB’

∴ |(a + ib) (c + id) (e + if) (g + ih)| = |A + iB|

⇒ |(a + ib) ×|(c + id)| × |(e + if | × |(g + ih)| = |A + iB| [|z_{1}z_{2}| = |z_{1}||z_{2}|]

On squaring both sides, we obtain

(a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} + B^{2}

Hence, proved.

**Question 18:**

Find the number of non-zero integral solutions of the equation |1 –i|^{x} = 2^{x} It is given that, |β| = 1

**Answer**

|1 – i|^{x} = 2^{x}

Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0.

**Question 17:**

If α and β are different complex numbers with |β| = 1, then find

**Answer**

Let a = a + ib and β = x + iy

It is given that, |β| = 1

**Question 16:**

**Answer**

(x + iy)^{3} = u + iv

⇒ x^{3} + (iy)^{3} + 3 ∙ x ∙ iy(x + iy) = u + iv

⇒ x^{3} + i^{3}y^{3} + 3x^{2}yi + 3xy^{2}i^{2} = u + iv

⇒ x^{3} – iy^{3} + 3x^{2}yi− 3xy^{2} = u + iv

⇒ (x^{3} – 3xy^{2}) + i(3x^{2}y – y^{3}) = u + iv

On equating real and imaginary parts, we obtain

u = x^{3} – 3xy^{2}, v = 3x^{2}y – y^{3}