NCERT Solutions Class 11th Mathematics. FREE Flip-BOOK by Study Innovations. 614 Pages

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NCERT Solutions for Class 11 Maths Maths Chapter 4

Principle of Mathematical Induction Class 11

Chapter 4 Principle of Mathematical Induction Exercise 4.1 Solutions
Exercise 4.1 : Solutions of Questions on Page Number : 94
Q1 :
Prove the following by using the principle of mathematical induction for all n ∈ N:

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

P(n): 1 + 3 + 32 + …+ 3n–1 =
For n = 1, we have

P(1): 1 = , which is true.

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

We shall now prove that P(k + 1) is true.
Consider
1 + 3 + 32 + … + 3k–1 + 3(k+1) – 1
= (1 + 3 + 32 +… + 3 )k–1 + 3k

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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.

Q2 :
Prove the following by using the principle of mathematical induction for

all n ∈ N:
Answer :
Let the given statement be P(n), i.e.,

P(n):
For n = 1, we have

P(1): 13 = 1 = , which is true.

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

We shall now prove that P(k + 1) is true.
Consider
13 + 23 + 33 + … + k3 + (k + 1)3

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= (13 + 23 + 33 + …. + k3) + (k + 1)3
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.

Q3 :
Prove the following by using the principle of mathematical induction for

all n ∈ N:

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

P(n):
For n = 1, we have

P(1): 1 = which is true.

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

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

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.
Q4 :
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3 + 2.3.4 + … + n(n + 1)
(n + 2) =

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

P(n): 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =
For n = 1, we have

P(1): 1.2.3 = 6 = , which is true.
Let P(k) be true for some positive integer k, i.e.,

1.2.3 + 2.3.4 + … + k(k + 1) (k + 2)
We shall now prove that P(k + 1) is true.
Consider
1.2.3 + 2.3.4 + … + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)
= {1.2.3 + 2.3.4 + … + k(k + 1) (k + 2)} + (k + 1) (k + 2) (k + 3)

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.

Q5 :
Prove the following by using the principle of mathematical induction for

all n ∈ N:
Answer :
Let the given statement be P(n), i.e.,

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P(n) :
For n = 1, we have

P(1): 1.3 = 3 , which is true.
Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.
Consider
1.3 + 2.32 + 3.33 + … + k3k+ (k + 1) 3k+1
= (1.3 + 2.32 + 3.33 + …+ k.3k) + (k + 1) 3k+1

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.

Q6 :
Prove the following by using the principle of mathematical induction for

all n ∈ N:

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

P(n):
For n = 1, we have

P(1): , which is true.

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

We shall now prove that P(k + 1) is true.
Consider
1.2 + 2.3 + 3.4 + … + k.(k + 1) + (k + 1).(k + 2)
= [1.2 + 2.3 + 3.4 + … + k.(k + 1)] + (k + 1).(k + 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.

Q7 :
Prove the following by using the principle of mathematical induction for

all n ∈ N:
Answer :
Let the given statement be P(n), i.e.,

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P(n): , which is true.
For n = 1, we have

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

We shall now prove that P(k + 1) is true.
Consider
(1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + {2(k + 1) – 1}{2(k + 1) + 1}

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www.ncrtsolutions.in

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.

Q8 :
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n=
(n - 1) 2n+1 + 2
Answer :
Let the given statement be P(n), i.e.,
P(n): 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2
For n = 1, we have
P(1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2, which is true.
Let P(k) be true for some positive integer k, i.e.,
1.2 + 2.22 + 3.22 + … + k.2k = (k – 1) 2k + 1 + 2 … (i)
We shall now prove that P(k + 1) is true.
Consider

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.

Q9 :

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Prove the following by using the principle of mathematical induction for

all n ∈ N:

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

P(n):
For n = 1, we have

P(1): , which is true.

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

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

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.

Q10 :
Prove the following by using the principle of mathematical induction for

all n ∈ N:

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www.ncrtsolutions.in

Answer :
Let the given statement be P(n), i.e.,
P(n):
For n = 1, we have

, which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider

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www.ncrtsolutions.in

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.

Q11 :
Prove the following by using the principle of mathematical induction for

all n ∈ N:
Answer :
Let the given statement be P(n), i.e.,

P(n):
For n = 1, we have

, which is true.

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

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

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www.ncrtsolutions.in

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.

Q12 :
Prove the following by using the principle of mathematical induction for

all n ∈ N:
Answer :

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

For n = 1, we have

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

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

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.

Q13 :
Prove the following by using the principle of mathematical induction for

all n ∈ N:
Answer :

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www.ncrtsolutions.in

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

For n = 1, we have
Let P(k) be true for some positive integer k, i.e.,

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

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.

Q14 :
Prove the following by using the principle of mathematical induction for
all n ∈ N:
Answer :
Let the given statement be P(n), i.e.,

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For n = 1, we have

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

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

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.
Q15 :
Prove the following by using the principle of mathematical induction for
all n ∈ N:
Answer :
Let the given statement be P(n), i.e.,

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

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www.ncrtsolutions.in

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

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.
Q16 :
Prove the following by using the principle of mathematical induction for
all n ∈ N:
Answer :
Let the given statement be P(n), i.e.,

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www.ncrtsolutions.in

Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider

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.
Q17 :

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Prove the following by using the principle of mathematical induction for
all n ∈ N:
Answer :
Let the given statement be P(n), i.e.,
For n = 1, we have

, which is true.
Let P(k) be true for some positive integer k, i.e.,
We shall now prove that P(k + 1) is true.
Consider

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www.ncrtsolutions.in

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.

Q18 :
Prove the following by using the principle of mathematical induction for
all n ∈ N:
Answer :
Let the given statement be P(n), i.e.,

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

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

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

Hence,
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.

Q19 :
Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a
multiple of 3.
Answer :
Let the given statement be P(n), i.e.,
P(n): n (n + 1) (n + 5), which is a multiple of 3.
It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3.
Let P(k) be true for some positive integer k, i.e.,
k (k + 1) (k + 5) is a multiple of 3.
∴k (k + 1) (k + 5) = 3m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider

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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.
Q20 :
Prove the following by using the principle of mathematical induction for all n ∈ N: 102n - 1 + 1 is divisible by 11.
Answer :
Let the given statement be P(n), i.e.,
P(n): 102n – 1 + 1 is divisible by 11.
It can be observed that P(n) is true for n = 1 since P(1) = 102.1 – 1 + 1 = 11, which is divisible by 11.
Let P(k) be true for some positive integer k, i.e.,
102k– 1 + 1 is divisible by 11.
∴102k– 1 + 1 = 11m, where m ∈ N … (1)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Consider

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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.
Q21 :
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

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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.
Q22 :
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.,
P(n): 32n + 2 – 8n – 9 is divisible by 8.
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

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www.ncrtsolutions.in

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.

Q23 :
Prove the following by using the principle of mathematical induction for all n ∈ N: 41n - 14n is a multiple of 27.

Answer : , which is a multiple of 27.
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
Let P(k) be true for some positive integer k, i.e.,
41k – 14kis 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

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www.ncrtsolutions.in

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.
Q24 :
Prove the following by using the principle of mathematical induction for all
(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

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www.ncrtsolutions.in

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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NCERT Solutions for Class 11 Maths Chapter 5

Complex Numbers and Quadratic Equations Class 11

Chapter 5 Complex Numbers and Quadratic Equations Exercise 5.1, 5.2, 5.3, miscellaneous Solutions
Exercise 5.1 : Solutions of Questions on Page Number : 103
Q1 :
Express the given complex number in the form a + ib:
Answer :

Q2 :
Express the given complex number in the form a + ib: i9 + i19
Answer :

Q3 :
Express the given complex number in the form a + ib: i-39
Answer :

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www.ncrtsolutions.in

Q4 :
Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7)
Answer :

Q5 :
Express the given complex number in the form a + ib: (1 - i) - (-1 + i6)
Answer :

Q6 :
Express the given complex number in the form a + ib:
Answer :

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www.ncrtsolutions.in

Q7 :
Express the given complex number in the form a + ib:
Answer :

Q8 :
Express the given complex number in the form a + ib: (1 - i)4
Answer :

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www.ncrtsolutions.in

Q9 :
Express the given complex number in the form a + ib:
Answer :

Q10 :
Express the given complex number in the form a + ib:
Answer :

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www.ncrtsolutions.in

Q11 :
Find the multiplicative inverse of the complex number 4 - 3i
Answer :
Let z = 4 – 3i
Then, = 4 + 3i and
Therefore, the multiplicative inverse of 4 – 3i is given by

Q12 :
Find the multiplicative inverse of the complex number
Answer :
Let z =

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www.ncrtsolutions.in

Therefore, the multiplicative inverse of is given by

Q13 :
Find the multiplicative inverse of the complex number -i
Answer :
Let z = –i
Therefore, the multiplicative inverse of –i is given by

Q14 :
Express the following expression in the form of a + ib.

Answer :

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www.ncrtsolutions.in

Exercise 5.2 : Solutions of Questions on Page Number : 108
Q1 :
Find the modulus and the argument of the complex number
Answer :
On squaring and adding, we obtain

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www.ncrtsolutions.in

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

Thus, the modulus and argument of the complex number are 2 and respectively.

Q2 :
Find the modulus and the argument of the complex number
Answer :

On squaring and adding, we obtain

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www.ncrtsolutions.in

Thus, the modulus and argument of the complex number are 2 and respectively.

Q3 :
Convert the given complex number in polar form: 1 - i

Answer :
1 – i
Let rcos θ = 1 and rsin θ = –1
On squaring and adding, we obtain

This is the

required polar form.

Q4 :
Convert the given complex number in polar form: - 1 + i
Answer :
– 1 + i
Let rcos θ = –1 and rsin θ = 1
On squaring and adding, we obtain

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www.ncrtsolutions.in

It can be written,
This is the required polar form.
Q5 :
Convert the given complex number in polar form: - 1 - i
Answer :
– 1 – i
Let rcos θ = –1 and rsin θ = –1
On squaring and adding, we obtain

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www.ncrtsolutions.in This is the

required polar form.

Q6 :
Convert the given complex number in polar form: -3
Answer :
–3
Let rcos θ = –3 and rsin θ = 0
On squaring and adding, we obtain

This is the required polar form.
Q7 :
Convert the given complex number in polar form:
Answer :
Let rcos θ = and rsin θ = 1
On squaring and adding, we obtain

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www.ncrtsolutions.in

This is the required polar form.
Q8 :
Convert the given complex number in polar form: i
Answer :
i
Let rcosθ = 0 and rsin θ = 1
On squaring and adding, we obtain

This is the required polar form.
Exercise 5.3 : Solutions of Questions on Page Number : 109

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www.ncrtsolutions.in

Q1 :
Solve the equation x2 + 3 = 0
Answer :
The given quadratic equation is x2 + 3 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 1, b = 0, and c = 3
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 02 – 4 × 1 × 3 = –12
Therefore, the required solutions are

Q2 :
Solve the equation 2x2 + x + 1 = 0
Answer :
The given quadratic equation is 2x2 + x + 1 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 2, b = 1, andc = 1
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = –7
Therefore, the required solutions are

Q3 :
Solve the equation x2 + 3x + 9 = 0
Answer :
The given quadratic equation is x2 + 3x + 9 = 0

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On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 1, b = 3, and c = 9
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 32 – 4 × 1 × 9 = 9 – 36 = –27
Therefore, the required solutions are

Q4 :
Solve the equation -x2 + x - 2 = 0
Answer :
The given quadratic equation is –x2 + x – 2 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = –1, b = 1, and c = –2
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 12 – 4 × (–1) × (–2) = 1 – 8 = –7
Therefore, the required solutions are

Q5 :
Solve the equation x2 + 3x + 5 = 0
Answer :
The given quadratic equation is x2 + 3x + 5 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 1, b = 3, and c = 5
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 32 – 4 × 1 × 5 =9 – 20 = –11
Therefore, the required solutions are

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www.ncrtsolutions.in

Q6 :

Solve the equation x2 - x + 2 = 0

Answer :
The given quadratic equation is x2 – x + 2 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 1, b = –1, and c = 2
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–1)2 – 4 × 1 × 2 = 1 – 8 = –7
Therefore, the required solutions are

Q7 :
Solve the equation

Answer :

The given quadratic equation is
On comparing the given equation with ax2 + bx + c = 0, we obtain

a = , b = 1, and c =
Therefore, the discriminant of the given equation is

D = b2 – 4ac = 12 – = 1 – 8 = –7

Therefore, the required solutions are

Q8 :

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Solve the equation

Answer :

The given quadratic equation is
On comparing the given equation with ax2 + bx + c = 0, we obtain

a= ,b= , and c =

Therefore, the discriminant of the given equation is

D = b2 – 4ac =
Therefore, the required solutions are

Q9 :

Solve the equation
Answer :

The given quadratic equation is
This equation can also be written as
On comparing this equation with ax2 + bx + c = 0, we obtain
a = , b = , and c = 1

Therefore, the required solutions are

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www.ncrtsolutions.in

Q10 :
Solve the equation
Answer :
The given quadratic equation is
This equation can also be written as
On comparing this equation with ax2 + bx + c = 0, we obtain
a = , b = 1, and c =
Therefore, the required solutions are

Exercise Miscellaneous : Solutions of Questions on Page Number : 112
Q1 :
Evaluate:
Answer :

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www.ncrtsolutions.in

Q2 :
For any two complex numbers z1 and z2, prove that
Re (z1z2) = Re z1 Re z2 - Im z1 Im z2
Answer :

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www.ncrtsolutions.in

Q3 : to the standard form.

Reduce
Answer :

Q4 :

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If x – iy = prove that .
Answer :

Q5 :

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