NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 2 (Ex 5.2)

Question 2:

Find the modulus and the argument of the complex number z = −√3 + i

Answer

z = −√3 + i

Let r cos θ = −√3 and r sin θ = 1

On squaring and adding, we obtain

r2 cos2 θ + r2 sin2 θ =(−√3)2 + 12

⇒ r2 = 3 + 1 = 4                           [cos2 θ + sin2 θ = 1]

⇒ r = √4 = 2                                 [Conventionally, r > 0]

∴ Modulus = 2

∴ 2 cos θ = −√3 and 2 sin θ = 1

⇒ cos θ = −√3/2 and sin θ = 1/2

 

Thus, the modulus and argument of the complex number −√3 + i are 2 and 5π/6 respectively.

NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 1 (Ex 5.2)

Question 1:

Find the modulus and the argument of the complex number z = −1 – i√3

Answer

z = −1 – i√3

Let r cos θ = −1 and r sin θ = −√3

On squaring and adding, we obtain

(r cos θ)2 + (r sin θ)2 = (−1)2 + (−√3)2

⇒ r2(cos2 θ + sin2 θ) = 1 + 3

⇒ r2 = 4                            [cos2 θ + sin2 θ = 1]

⇒ r = √4 = 2                   [Conventionally, r >0]

∴ Modulus = 2 ∴ 2 cos θ = −1 and 2 sin θ = −√3

⇒ cos θ = −1/2 and sin θ = −√3/2

Since both the values of sin θ and cos θ are negative and sin θ and cos θ are negative in III quadrant,

Thus, the modulus and argument of the complex number −1 – √3i are 2 and −2π/3 respectively.

NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 8 (Ex 5.1)

Question 8:

Express the given complex number in the form a + ib: (1 – i)4

Answer

(1 – i)4 = [(1 – i)2]2            

              = [12 + i2 – 2i]2            

              = [1 – 1 – 2i]2            

              = (−2i)2            

              = (−2i) × (−2i)

              = 4i2 = −4          [i2 = −1]

NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 2 (Ex 5.1)

Question 2:

Express the given complex number in the form a + ib: i9 + i19

Answer

i9 + i19 = i4×2+1 + i4×4+3            

              =(i4)2 ∙ i + (i4)4 ∙ i3              

              = 1 × i +1×(−i)              [i4 = 1, i3 = −i]              

              = i + (−i)

              = 0

NCERT Solution Class XI Mathematics Principle of Mathematical Induction Question 24 (Ex 4.1)

Question 24:

Prove the following by using the principle of mathematical induction for all n ∈ N : (2n +7) < (n + 3)2

Answer

Let the given statement be P(n), i.e.,

P(n): (2n +7) < (n + 3)2

It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)2 = 16, which is true.

Let P(k) be true for some positive integer k,

i.e., (2k + 7) < (k + 3)2 … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

{2(k +1) + 7} = (2k + 7) + 2

∴{2(k + 1) + 7} = (2k + 7) + 2 < (k + 3)2 + 2                 [using (1)]

2(k + 1) + 7 < k2 + 6k + 9 + 2

2(k + 1) + 7 < k2 + 6k + 11

Now, k2 + 6k + 11 < k2 +8k + 16

∴2(K + 1) + 7 < (k + 4)2

2(k + 1) + 7<{(k + 1) + 3}2

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

NCERT Solution Class XI Mathematics Principle of Mathematical Induction Question 23 (Ex 4.1)

Question 23:

Prove the following by using the principle of mathematical induction for all n ∈N : 4n – 14n is a multiple of 27.

Answer

Let the given statement be P(n), i.e.,

P(n):41n – 14nis a multiple of 27.

It can be observed that P(n) is true for n = 1 since 411 – 141 = 27, which is multiple of 27.

Let P(k) be true for some positive integer k,

41k – 14k is a multiple of 27

∴41k − 14k = 27m, where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

41k+1 – 14k+1

=41k ∙ 41 – 14k ∙ 14

=41(41k – 14k + 14k) – 14k ∙ 14

=41(41k – 14k) + 41∙14k – 14k ∙ 14

=41.27m + 14k(41 – 14)

=41.27m + 27.14k

=27(41m – 14k)

= 27 × r, where r = (41m – 14k) is a natural number

Therefore, 41k+1 – 14k+1 is a multiple of 27.

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

NCERT Solution Class XI Mathematics Principle of Mathematical Induction Question 22 (Ex 4.1)

Question 22:

Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2 – 8n – 9 is divisible by 8.

Answer

Let the given statement be P(n), i.e.,

It can be observed that P(n) is true for n = 1 since 32 × 1 + 2 − 8 × 1 – 9 = 64, which is divisible by 8.

Let P(k) be true for some positive integer k,

i.e., 32k + 2 − 8k − 9 is divisible by 8.

∴32k + 2 − 8k − 9 = 8m; where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

32(k+1)+2 – 8(k + 1) – 9

=32k+2 ∙ 32 – 8k – 8 – 9

=32(32k+2 – 8k – 9 + 8k + 9) – 8k – 17

= 32(32k+2 – 8k – 9) + 32(8k + 9) – 8k – 17

= 9.8 m + 9(8k + 9) – 8k – 17

= 9.8m + 72k + 81 – 8k – 17

= 9.8m + 72k + 81 – 8k – 17

= 9.8m + 64k + 64

=8(9m + 8k + 8)

=8r, where r =(9m + 8k + 8) is a natural number

Therefore, 32k+2 – 8(k + 1) – 9 is divisible by 8.

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

NCERT Solution Class XI Mathematics Principle of Mathematical Induction Question 21 (Ex 4.1)

Question 21:

Prove the following by using the principle of mathematical induction for all n ∈N : x2n – y2n is divisible by x + y.

Answer

Let the given statement be P(n), i.e.,

P(n): x2n − y2n is divisible by x + y.

It can be observed that P(n) is true for n = 1.

This is so because x2 × 1 – y2 × 1 = x2 – y2 = (x + y) (x – y) is divisible by (x +y)

Let P(k) be true for some positive integer k, i.e.,

x2k – y2k is divisible by x + y.

∴x2k – y2k = m (x + y), where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

x2(k+1) – y2(k + 1)

=x2k ∙ x2 – y2k ∙ y2

=x2(x2k – y2k + y2k) – y2k ∙y2

=x2{m(x + y) + y2k} – y2k ∙ y2              [Using (1)]

=m(x + y)x2 + y2k ∙ x2 – y2k ∙ y2

=m(x + y)x2 + y2k (x2 – y2)

=m(x + y)x2 + y2k(x + y) (x – y)

=(x + y) {mx2 + y2k(x – y)}, which is factor of (x + y).

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

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