Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.3

Question 1.
Solve the cubic equation: 2x³ – x² – 18x + 9 = 0 if sum of two of its roots vanishes.
Solution:
The given equation is 2x³ – x² – 18x + 9 = 0
\(x^{3}-\frac{x^{2}}{2}-9 x+\frac{9}{2}=0\)
Let the roots be α, -α, β
α – α + β = \(-\left(\frac{-1}{2}\right)\)
\(\Rightarrow \beta=\frac{1}{2}\)
(α) (-α) (β) = \(\frac{-9}{2}\)
\(\Rightarrow-\alpha^{2}\left(\frac{1}{2}\right)=\frac{-9}{2}\)
α² = 9
α = ±3
The roots are 3, -3, \(\frac { 1 }{ 2 }\)

Question 2.
Solve the equation 9x³ – 36x² + 44x – 16 = 0 if the roots form an arithmetic progression.
Solution:
Given the roots are in AP
Let the roots be a – d, a, a + d

Question 3.
Solve the equation 3x³ – 26x² + 52x – 24 = 0 if its roots form a geometric progression.
Solution:

3 + 3λ + 3λ² = 13λ
3λ² + 3λ – 13λ + 3 = 0
3λ² – 10λ + 3 = 0
(λ – 3) (3λ – 1) = 0
λ = 6 or λ = \(\frac{1}{3}\)

Question 4.
Determine k and solve the equation 2x³ – 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots.
Solution:
Given cubic equation
2x³ – 6x² + 3x + k = 0
Let the roots be α, β, γ
Given α = 2(β + γ)
β + γ = \(\frac{α}{2}\) ………. (1)
Sum of roots α(β + γ) = 3
From (1) α + \(\frac{α}{2}\) = 3
\(\frac{3α}{2}\) = 3 ⇒ α = 2
Again αβ + βγ + γα = \(\frac{3}{2}\)
α = 2 ⇒ 2β + βγ + 2γ = \(\frac{3}{2}\)
from (1) 2(\(\frac{α}{2}\)) + βγ = \(\frac{3}{2}\)
βγ = \(\frac{3}{2}\) – 2 = \(\frac{3-4}{2}\) = \(\frac{-1}{2}\)
βγ = \(\frac{-1}{2}\) ………… (2)
product of roots α β γ = –\(\frac{k}{2}\)
α = 2 ⇒ 2βγ = –\(\frac{k}{2}\)
βγ = –\(\frac{k}{4}\) ………… (3)

Question 5.
Find all zeros of the polynomial x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135, if it is known that 1 + 2i and √3 are two of its zeros.
Solution:
(i) Given that 1 + 2i, √3
Another roots be 1 – 2i, -√3
sum of roots = 1 + 2i + 1 – 2i
product roots = (1 + 2i)(1 – 2i)
1² + 2² = 1 + 4 = 5
x² – 2x + 5 = 0

(ii) sum of roots = √3 – √3
product roots = (√3)(-√3)
x² – 0x – 3 = 0
x² – 3 = 0
(x² – 2x + 5)(x² – 3) = x⁴ – 2x³ + 2x² + 6x – 15
x⁶– 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135
= (x⁴ – 2x³ + 2x² + 6x – 15)  (x² + px – 9)
Equate of co-efficient of x on both sides
-39 = -54 – 15 p
-39 + 54 = -15 p
15 = -15 p
p = -1
∴ x² – x – 9 = 0

Question 6.
Solve the cubic equations:
(i) 2x³ – 9x² + 10x = 3,
2x³ – 9x² + 10x – 3 = 0
Solution:
(i) 2x³ – 9x² + 10x = 3
2x³ – 9x² + 10x – 3 = 0
sum of the coefficients 2 – 9 + 10 – 3 = 0
∴ x = 1 is one of the roots.

2x² – 7x + 3 = 0
(x – 3)(2x – 1) = 0
x = 3 or x = \(\frac{1}{2}\)
roots are 3, \(\frac{1}{2}\), 1

(ii) 8x³ – 2x² – 7x + 3 = 0
sum of the alternative coefficients are equal
8 – 7 = -2 + 3
1 = 1
∴ (x + 1) is a factor.

8x² – 10x + 3 = 0
(4x – 3) (2x – 1) = 0
4x  = 3 (or) 2x = 1
x = \(\frac{3}{4}\) (or) \(\frac{1}{2}\)
∴ The roots are \(\frac{3}{4}\), \(\frac{1}{2}\), -1

Question 7.
Solve the equation:
x⁴ – 14x² + 45 = 0
Solution:
Put t = x² ⇒ (x²)² – 14x² + 45 = 0
t² – 14t + 45 = 0
t² – 9t – 5t + 45 = 0
t(t – 9) – 5(t – 9) = 0
(t – 9)(t – 5) = 0
t – 9 = 0 or t – 5 = 0
t = 9 or t = 5
x² = 9 or x² = 5
x = ± 3 or y = ±√5
The roots are ±3, ±√5

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.2

Question 1.
If k is real, discuss the nature of the roots of the polynomial equation 2x² + kx + k = 0, in terms of k.
Solution:
The given quadratic equation is 2x² + kx + k = 0
a = 2, b = k, c = k
∆ = b² – 4ac = k² – 4(2) k = k² – 8k
(i) If the roots are equal
k² – 8k = 0
⇒ k(k – 8) = 0
⇒ k = 0, k = 8
(ii) If the roots are real
k² – 8k > 0
k(k – 8) > 0
k ∈ (-∞, 0) ∪ (8, ∞)
(iii) If this roots are imaginary
k² – 8k < 0
⇒ k ∈ (0, 8)

Question 2.
Find a polynomial equation of minimum degree with rational coefficients, having 2 + √3 i as a root.
Solution:
Let the root be 2 + i √3
Another root be 2 – i √3
Sum of the roots = 2 + i √3 + 2 – i √3 = 4
Product of the roots = (2 + i √3) (2 – i √3) = 2² + √3² = 4 + 3 = 7
x² – (SR)x + PR = 0
x² – 4x + 7 = 0

Question 3.
Find a polynomial equation of minimum degree with rational coefficients, having 2i + 3 as a root.
Solution:
Given roots is (3 + 2i), the other root is (3 – 2i);
Since imaginary roots occur in with real co-efficient occurring conjugate pairs.
x² – x(S.O.R) + P.O.R = 0
⇒ x² – x(6) + (9 + 4) = 0
⇒ x² – 6x + 13 = 0

Question 4.
Find a polynomial equation of minimum degree with rational coefficients, having √5 – √3 as a root.
Solution:
Let the root be √5 – √3,
Another root is √5 + √3
Sum of the roots = √5 – √3 + √5 + √3 = 2√5
Product of roots = (√5 – √3) (√5 + √3)
√5² – √3² = 5 – 3 = 2
x² – (SR)x + PR = 0
x² – 2√5 x + 2 = 0 which is not rational co-efficient.
to make rational co-efficient
(x² + 2√5 x + 2) (x² + 2 + 2√5 x) = 0
(x² + 2)² – (2√5x)² = 0
x⁴ + 4 + 4x² – 20x² = 0
⇒ x⁴ – 16x² + 4 = 0 is a rational co-efficient polynomial equation.

Question 5.
Prove that a straight line and parabola cannot intersect at more than two points.
Solution:
Let the standard equation of parabola y² = 4ax …..(1)
Equation of line be y = mx + c …(2)
Solving (1) & (2)
(mx + c)² = 4ax
⇒ mx² + 2mcx + c² – 4ax = 0
⇒ mx² + 2x(mc – 2a) + c² = 0
This equation can not have more than two solutions and
hence a line and parabola cannot intersect at more than two points.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.1

Question 1.
If the sides of a cubic box are increased by 1, 2, 3 units respectively to form a cuboid, then the volume is increased by 52 cubic units. Find the volume of the cuboid.
Solution:
Let the side of the cube be ‘x’
Sides of cuboid are (x + 1) (x + 2) (x + 3)
∴ Volume of cuboid = x³ + 52
⇒ (x + 1) (x + 2) (x + 3) = x³ + 52
⇒ (x² + 3x + 2)(x + 3) = x³ + 52
⇒ x³ + 3x² + 3x² + 9x + 2x + 6 – x³ – 52 = 0
⇒ 6x² + 11x – 46 = 0 (÷2)
⇒ (x – 2) (6x + 23) = 0
⇒ x – 2 = 0 or 6x + 23 = 0
⇒ x = 2 or x = \(-\frac{23}{6}\) (not possible)
∴ x = 2
Volume of cube = 2³ = 8
Volume of cuboid = 52 + 8 = 60 cubic units

Question 2.
Construct a cubic equation with roots
Solution:
(i) 1, 2, and 3
α = 1, β = 2, γ = 3
α + β + γ = 6
αβ + βγ + γα = 2 + 6 + 3 = 11
αβγ = 6
x³ – (α + β + γ)x² + (αβ + βγ + γα)x – αβγ = 0
x³ – 6x² + 11x – 6 = 0

(ii) 1, 1, and -2
α = 1, β = 1, γ = -2
α + β + γ = 1 + 1 – 2 = 0
αβ + βγ + γα = 1 – 2 – 2 = -3
αβγ = 1(1)(-2) = -2
x³ – 0x² – 3x + 2 = 0
∴ x³ – 3x + 2 = 0

(iii) 2, \(\frac { 1 }{ 2 }\), and 1.

Multiplying by 2
2x³ – 7x² + 7x – 2 = 0

Question 3.
If α, β and γ are the roots of the cubic equation x³ + 2x² + 3x + 4 = 0, form a cubic equation whose roots are
(i) 2α, 2β, 2γ,
(ii) \(\frac{1}{α}\), \(\frac{1}{β}\), \(\frac{1}{γ}\)
(iii) – α, – β, – γ
Solution:
(i) Given that α, β, γ are the roots of x³ + 2x² + 3x + 4 = 0
Compare with x³ + bx² + cx + d = 0
b = 2, c = 3, d = 4
α + β + γ = -6 = -2
αβ + βγ + γα = c = 3
αβγ = -d = -4
Given roots are 2α, 2β, 2γ
2α + 2β + 2γ = 2 (α + β + γ)
= 2 (-2)
= -4
(2α) (2β) + (2β) (2γ) + (2γ) (2α) = (4αβ + 4βγ + 4γα)
= 4(αβ + βγ + γα)
= 4(3)
= 12
(2α) (2β) (2γ) = 8(αβγ)
= 8(-4)
= -32
The equation is
x³ – x² (2α + 2β + 2γ) + x (4αβ + 4βγ + 4γα) – 8 (αβγ) = 0
⇒ x³ – x² (-4) + x (12) – (-32) = 0
⇒ x³ + 4x² + 12x + 32 = 0

(ii) The new roots are \(\frac{1}{α}\), \(\frac{1}{β}\), \(\frac{1}{γ}\)

⇒ 4x³ + 3x² + 2x + 1 = 0

(iii) The given roots are -α, -β, -γ
The cubic equation is
x³ – x² (-α – β – γ) + x (αβ + βγ + γα) + (αβγ) = 0
⇒ x³ + x² (α + β + γ) + x (αβ + βγ + γα) + (αβγ) = 0
⇒ x³ + x² (-2) + x (3) – 4 = 0
⇒ x³ – 2x² + 3x – 4 = 0

Question 4.
Solve the equation 3x³ – 16x² + 23x – 6 = 0 if the product of two roots is 1.
Solution:
Let the roots be α, β, γ
Given αβ = 1, β = \(\frac{1}{α}\)

⇒ 3α² – 10α + 3 = 0
3α² – 9α – α + 3 = 0
3α(α – 3) -1(α – 3) = 0
(3α – 1) (α – 3) = 0
α = 3 or α = \(\frac{1}{3}\)
If α = 3, β = \(\frac{1}{3}\), γ = 2 (or)
α = \(\frac{1}{3}\), β = 3, γ = 2
⇒ [α, β, γ] = (\(\frac{1}{3}\), 3, 2)

Question 5.
Find the sum of squares of roots of the equation 2x⁴ – 8x³ + 6x² – 3 = 0.
Solution:
The given equation is 2x⁴ – 8x³ + 6x² – 3 = 0.
(÷ 2) ⇒ x⁴ – 4x³ + 3x² – \(\frac{3}{2}\) = 0
Let the roots be α, β, γ, δ
α + β + γ + δ = -b = 4
(αβ + βγ + γδ + αδ + αγ + βδ) = c = 3
αβγ + βγδ + γδα = -d = 0
αβγδ = \(\frac{-3}{2}\)
To Find α² + β² + γ² + δ² = (α + β + γ + δ)² – 2 (αβ + βγ + γδ + αδ + αγ + βδ)
= (4)² – 2(3)
= 16 – 6
= 10

Question 6.
Solve the equation x³ – 9x² + 14x + 24 = 0 if it is given that two of its roots are in the ratio 3 : 2.
Solution:
Let the roots are 3α, 2α, β
sum of the roots are
3α + 2α + β = 9
5α + β = 9 ………. (1)
Product of two roots
3α(2α) + 2α(β) + β(3α) = 14
6α² + 5αβ = 14 ……… (2)
Product of three roots
(3α) (2α)β = -24
α²β = -4 ………. (3)
(1) ⇒ β = 9 – 5 α
(2) ⇒ 6α² + 5α (9 – 5α) = 14
6α² + 45α – 25α² = 14
-19α² + 45α – 14 = 0
19α² – 45α + 14 = 0
(α – 2) (α – \(\frac{7}{19}\)) = 0
α = 2 or α = \(\frac{7}{19}\)
If α = 2, β = 9 – 5 (α) = 9 – 5(2) = 9 – 10 = -1
roots are 3α, 2α, β
3(2), 2(2), -1 (i,e.,) 6, 4, -1
If α = \(\frac{7}{19}\), β = 9 – 5(\(\frac{7}{19}\)) = (\(\frac{136}{19}\))
roots are 3α, 2α, β (i,e.,) \(\frac{21}{19}\), \(\frac{14}{19}\), \(\frac{136}{19}\)

Question 7.
If α, β, and γ are the roots of the polynomial equation ax³ + bx² + cx + d = 0, find the value of Σ\(\frac{α}{βγ}\) terms of the coefficients.
Solution:

Question 8.
If α, β, γ and δ are the roots of the polynomial equation 2x⁴ + 5x³ – 7x² + 8 = 0, find a quadratic equation with integer coefficients whose roots are α + β + γ + δ and αβγδ.
Solution:

pq = (\(\frac{-5}{2}\))(4) = -10
x² – (p + q)x + pq = 0
x² – \(\frac{3}{2}\)x – 10 = 0
2x² – 3x – 20 = 0

Question 9.
If p and q are the roots of the equation lx² + nx + n = 0,
show that,
Solution:
p and q are the roots of the equation lx² + nx + n = 0
p + q = –\(\frac{n}{l}\), pq = \(\frac{n}{l}\)
LHS

Question 10.
If the equations x² + px + q = 0 and x² + p’x + q’ = 0 have a common root, show
that it must be equal to \(\frac{pq’-p’q}{q-q’}\) and \(\frac{q-q’}{p’-p}\)
Solution:
Let it be α common roots of x² + px + q = 0
x² + p’x + q’ = 0
(i,e.,) α² + pα + q = 0 …… (1) and
α² + p’α + q’ = 0 ………. (2)
Solving 1 and 2 by cross multiplication method, we have

Hence proved.

Question 11.
A 12-meter tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away.
Solution:
Let the height of the tree = 12
length of the cut part = x³
Length of left out part = \(\sqrt[3]{x^{3}}\)
= x
Given x + x³ = 12
x³ + x – 12 = 0

Which is required mathematical problem

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.9

Choose the most suitable answer.

Question 1.
iⁿ + iⁿ⁺¹+ iⁿ⁺² + iⁿ⁺³ is:
(a) 0
(b) 1
(c) -1
(d) z
Solution:
(a) 0
Hint:
iⁿ + iⁿ⁺¹+ iⁿ⁺² + iⁿ⁺³
= iⁿ[1 + i + i² + i³]
= iⁿ[1 + i – 1 – i]
iⁿ (0) = 0

Question 2.
The value of \(\sum _{ i=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n-1 }) } \) is
(a) 1 + i
(b) i
(c) 1
(d) 0
Solution:
(a) 1 + i
Hint:
\(\sum _{ i=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n-1 }) } \)= (i¹ + i² + i³ + … + i¹³) + (i⁰ + i¹ + i² + … + i¹²)
= i⁰ + 2(i¹ + i² + i³  + ….. i¹²) + i¹³
= 1 + 2(i – 1 – i + 1 + … + 1) + 1
= 1 + 2(0) + i = 1 + i

Question 3.
The area of the triangle formed by the complex numbers z, iz, and z + iz in the Argand’s diagram is:
(a) \(\frac{1}{2}\) |z|²
(b) |z|²
(c) \(\frac{3}{2}\) |z|²
(d) 2|z|²
Solution:
(a) \(\frac{1}{2}\) |z|²
Hint:
Area of the triangle formed by the complex numbers z, iz and z + iz.
Let z = a + ib ⇒point (a, b)
iz = – b + ia ⇒ point (- b, a)
z + iz =(a – b) + i(a + b) point((a – b),(a + b))
Area of the triangle
= \(\frac {1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\) [a(a – a – b) -b(a + b – b) + (a – b)(b – a)]
= \(\frac {1}{2}\) [-ab – ab + ab – a² – b² + ab]

Question 4.
The conjugate of a complex number is \(\frac{1}{i-2}\) Then, the complex number is:
(a) \(\frac{1}{i+2}\)
(b) \(\frac{-1}{i+2}\)
(c) \(\frac{-1}{i-2}\)
(d) \(\frac{1}{i-2}\)
Solution:
(b) \(\frac{-1}{i+2}\)
Hint:
Conjugate of complex number is \(\frac{1}{i-2}\)
∴ the complex number is \(\frac{-1}{i+2}\)

Question 5.
If z = \(\frac{(√3+i)^3(3i+4)²}{(8+6i)²}\)
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(c) 2
Hint:

Question 6.
If z is a non zero complex number, such that 2iz² = \(\bar { z }\) then | z | is:
(a) \(\frac{1}{2}\)
(b) 0
(c) 1
(d) 2
Solution:
z is a non zero complex number
Given 2iz² = \(\bar { z }\)
let z = x + iy
2i(x + iy)² = x – iy
simplifying 2i(x² – y² + 2ixy) = x – iy
-4xy + 2i(x² – y²) = x – iy
Equating real and imaginary parts
-4xy = x, 2(x² – y²) = -y
solving x = \(\frac{√3}{4}\), y =-\(\frac{1}{4}\)
z = \(\frac{√3}{4}\) – \(\frac{i}{4}\)
|z| = \(\sqrt { \frac{3}{16}+\frac{1}{16}}\) = \(\sqrt { \frac{1}{4}}\)
= \(\frac {1}{2}\)

Question 7.
If |z – 2 + i | ≤ 2, then the greatest value of |z| is:
(a) √3 – 2
(b) √3 + 2
(c) √5 – 2
(d) √5 + 2
Solution:
(d) √5 + 2
Hint:
|z – 2 + i | ≤ 2
|z + (-2 + i)| ≤ |z| + |-2 + i|
|z| ± √5
Given |z – 2 + i| ≤ 2
∴ |z| ± √5 ≤ 2
|z| ≤ 2 – √5. |z| ≤ 2 + √5
∴ The greatest value is 2 + √5

Question 8.
If |z – \(\frac{3}{2}\)| = 2 then the least value of |z| is:
(a) 1
(b) 2
(c) 3
(d) 5
Solution:
(a) 1
Hint:

|z|² – 2 |z| + 1 ≤ 3 + 1
(|z| – 1)² ≤ 4
|z| – 1 ≤ ± 2
|z| ≤ 2 + 1 and |z| ≤ – 2 + 1
|z| = -1
But |z| = 1

Question 9.
If |z| = 1, then the value of \( \frac { 1+z }{ 1+\bar { z } } \) is
(a) z
(b) \( \bar { z } \)
(c) \(\frac{1}{z}\)
(d) 1
Solution:
(a) z
Hint:

Question 10.
The solution of the equation |z| – z = 1 + 2i is:
(a) \(\frac{3}{2}\) – 2i
(b) – \(\frac{3}{2}\) + 2i
(c) 2 – \(\frac{3}{2}\)i
(d) 2 + \(\frac{3}{2}\)i
Solution:
(a) \(\frac{3}{2}\) – 2i
Hint:
|z| – z = 1 + 2i
Let z = x + iy
|z| = x + iy + 1 + 2i
\(\sqrt{x^2+y^2}\) = (x + 1) + i(y + 2)
\(\sqrt{x^2+y^2}\) = x + 1 y + 2 = 0
x² + y² = (x + 1)² y = -2
y² = 2x + 1
2x = 3
x = \(\frac{3}{2}\)
∴z = x + iy = \(\frac{3}{2}\) – 2i

Question 11.
If |z₁| = 1,|z₂| = 2, |z₃| = 3 and |9z₁z₂ + 4z₁z₃ + z₂z₃| = 12, then the value of |z₁ + z₂ + z₃| is:
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2
Hint:

Question 12.
If z is a complex number such that z∈C\R and z + \(\frac{1}{z}\) ∈R, then |z| is:
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:
If z ∈ C\R and z + \(\frac{1}{z}\) ∈ R
Then |z| = 1

Question 13.
z₁, z₂ and z₃ are complex numbers such that z₁ + z₂ + z₃ = 0 and |z₁| = |z₂| = |z₃| = 1 then \({ z }_{ 1 }^{ 2 }+{ z }_{ 2 }^{ 2 }+{ z }_{ 3 }^{ 2 }\) is
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) 0

Question 14.
If \(\frac{z-1}{z+1}\)is purely imaginary, then |z| is
(a) \(\frac{1}{2}\)
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:

Question 15.
If z = x + iy is a complex number such that |z + 2| = |z – 2|, then the locus of z is:
(a) real axis
(b) imaginary axis
(c) ellipse
(d) circle
Solution:
(b) imaginary axis
Hint:
|z + 2| = |z – 2|
Let z = x + iy
|(x + 2) + iy| = |(x – 2) + iy|
(x + 2)² + y² = (x – 2)² + y²
x² + 4x + 4 = x² – 4x + 4
⇒ x = 0

Question 16.
The principal argument of \(\frac{3}{-1+i}\) is:
(a) \(\frac{-5π}{6}\)
(b) \(\frac{-2π}{3}\)
(c) \(\frac{-3π}{4}\)
(d) \(\frac{-π}{2}\)
Solution:
(c) \(\frac{-3π}{4}\)
Hint:

Since Real and imaginary parts are negative.
‘θ’ lies in 3rd quadrant.
∴ Principal argument = – \(\frac {3π}{4}\) [∵ \(\frac {π}{4}\) – π]

Question 17.
The principal argument of (sin 40° + i cos 40°)⁵ is:
(a) – 110°
(b) -70°
(c) 70°
(d) 110°
Solution:
(a) – 110°
Hint:
z = (sin 40° + i cos 40°)⁵
= (cos 50° + i sin 50°)⁵
= cos 250° + i sin 250°]
= cos (360° – 110°) + i sin (360° – 110°)
= cos 110° – i sin 110°
= cos (-110°) + i sin (-110°)
∴ Principal argument is – 110°

Question 18.
If (1 + i) (1 + 2i) (1 + 3i) ……. (l + ni) = x + iy, then 2.5.10 …… (1 + n²) is:
(a) 1
(b) i
(c) x² + y²
(d) 1 + n²
Solution:
(c) x² + y²
Hint:
(1 + i) (i + 2i) ….. (1 + ni) = x + iy
Taking of two modulli of each of squaring
2.5.10 … (1 +n²) = x²+ y²

Question 19.
If ω ≠ 1 is a cubic root of unity and (1 + ω)⁷ = A + Bω, then (A, B) equals:
(a) (1, 0)
(b) (-1, 1)
(c) (0, 1)
(d) (1, 1)
Solution:
(d) (1, 1)
Hint:
(1 + ω)⁷ = A + Bω
(- ω²)⁷ = A + Bω
– ω¹⁴ = A + Bω
– ω² = A + Bω
1 + ω² = A + Bω
∴ A = 1, B = 1

Question 20.
The principal argument of the complex number \(\frac{(1+i \sqrt{3})^{2}}{4 i(1-i \sqrt{3})}\) is:
(a) \(\frac{2π}{3}\)
(b) \(\frac{π}{6}\)
(c) \(\frac{5π}{6}\)
(d) \(\frac{π}{2}\)
Solution:
(d) \(\frac{π}{2}\)
Hint:

Question 21.
If α and β are the roots of x² + x + 1 = 0, then α²⁰²⁰ + β²⁰²⁰ is
(a) -2
(b) -1
(c) 1
(d) 2
Solution:
(b) -1
Hint:
x² + x + 1 = 0

Question 22.
The product of all four values of (cos\(\frac{π}{3}\) + i sin \(\frac{π}{3}\))^{\(\frac{3}{4}\)} is:
(a) -2
(b) -1
(c) 1
(d) 2
Solution:
(c) 1

Question 23.
If ω ≠ 1 is a cubic root of unity and

(a) 1
(b) -1
(c) √3 i
(d) -√3 i
Solution:
(d) -√3 i

= 1(ω² – ω⁴) – (ω – ω²) + 1(ω² – ω) = 3k
= ω² – ω – ω + ω² + ω² – ω = 3k
= 3ω² – 3ω = 3k
= 3(ω² – ω) = 3k
∴ k = ω² – ω

Question 24.
The value of (\(\frac{1+√3 i}{1-√3 i}\))¹⁰ is:
(a) cis\(\frac{2π}{3}\)
(b) cis\(\frac{4π}{3}\)
(c) -cis\(\frac{2π}{3}\)
(d) -cis\(\frac{2π}{3}\)
Solution:
(a) cis\(\frac{2π}{3}\)
Hint:

Question 25.
If ω = cis\(\frac{2π}{3}\), then the number of distinct roots of
(a) 1
(b) 2
(c) 3
(d) 4
Solution:


The expansion 1 becomes
z³ + (0) z² + (0) z + 0 = 0
⇒ z³ = 0
z = 0 is the only solution.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.8 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8

Question 1.
If to ω ≠ 1 is a cube root of unity, then show that

Solution:
L.H.S

Question 2.
Show that

Solution:

Question 3.

Solution:


Aliter method:

Question 4.
If 2cos α = x + \(\frac{1}{x}\) and 2 cos β = y + \(\frac{1}{x}\), show that
(i) \(\frac{x}{y}+\frac{y}{x}=2 \cos (\alpha-\beta)\)
Solution:
Given 2 cos α = x + \(\frac{1}{x}\) and cos β = y + \(\frac{1}{y}\)
simplifying x² – 2x cos α + 1 = 0

if x = cos α + i sin α, then \(\frac{1}{x}\) = cos α – i sin α
similarly y = cos β + i sin β and \(\frac{1}{y}\) = cos β – i sin β

Hence proved

(ii) xy = (cos α + i sin α)(cos β + i sin β )
Solution:
xy = (cos α + i sin α) (cos β + i sin β)
= cos (α + β) + i sin (α + β)
[∵ arg (z₁z₂) = arg z₁ + arg z₂
\(\frac{1}{xy}\) = cos (α + β) – i sin (α + β)
∴ xy – \(\frac{1}{xy}\) = cos (α + β) + i sin (α + β) – cos (α + β) + i sin (α + β)
= 2i sin (α + β)
Hence proved

(iii) \(\frac{x^{m}}{y^{n}}-\frac{y^{n}}{x^{m}}\) = 2 i sin (mα – nβ)
Solution:

Hence Proved

(iv) x^m yⁿ + \(\frac { 1 }{ x^m y^n }\) = 2 cos(mα – nβ)
Solution:
= (cos mα + sin mα) (cos nβ + i sin nβ)
= cos (mα + nβ) + i sin (mα + nβ)

Hence proved

Question 5.
Solve the equation z³ + 27 = 0.
Solution:
z³ + 27 = 0
z³ = – 27 = 27 (-1)
= 27 [cos(π + 2kπ) + i sin(π + 2kπ)], k ∈ z
∴ z = (27)^1/3[cos (2k + 1)π + i sin (2k+1)π]^1/3 k ∈ z

Question 6.
If ω ≠ 1 is a cube root of unity, show that the roots of the equation (z – 1)³ + 8 = 0 are -1, 1 – 2ω, 1 – 2ω².
Solution:
Given ω ≠ 1 is a active root of unity
(z – 1)³ + 8 = 0
(z- 1)³ = -8
z – 1 = (-8)^1/3 (1)^1/3
= (-2) (1, ω, ω²)
z – 1 = (-2, -2ω, -2ω²)
= z – 1 = -2
z = -2 + 1 = -1
z – 1 = -2ω
z = 1 – 2ω
z – 1 = -2ω²
z = 1 – 2ω²

Question 7.

Solution:
\(\sum_{k=1}^{8}\) (cos \(\frac { 2kπ }{ 9 }\) + i sin \(\frac { 2kπ }{ 9 }\))
The sum all nth root of unity is
1 + ω + ω² + …….. + ω^n-1 = 0
From the given polar from , it is clear that the complex number is 1 + i0 (unity)

Question 8.
If ω ≠ 1 is a cube root of unity, show that
(i) (1 – ω + ω²)⁶ + (1 + ω – ω²)⁶ = 128.
Solution:
ω is a cube root of unity ω³ = 1; 1 + ω + ω² = 0
(1 – ω + ω²)⁶ + (1 + ω – ω²)⁶
= (-ω – ω)⁶ + (-ω² – ω²)⁶
= (-2ω)⁶ + (-2ω²)⁶
= (-2)⁶(ω⁶ + ω¹²)
= (64)(1 + 1)
= 128

(ii) (1 + ω) (1 + ω²) (1 + ω⁴) (1 + ω⁸)….. (1 + ω²ⁿ) = 1
Solution:
(1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) …… (1 + ω²ⁿ)
= (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) ……. 2n factors
= (-ω²)(-ω)(-ω²)(-ω) …… 2n factors
= ω³. ω³
= 1
Hence proved.

Question 9.
If z = 2 – 2i, find the rotation of z by θ radians in the counterclockwise direction about the origin when
Solution:
Let 2 – 2i
Modules = |z| = \(\sqrt{2^2+2^2}\) = 2√2
Argument θ = tan^-1(\(\frac{-2}{2}\)) = tan^-1(-1) = –\(\frac{π}{4}\)
(i) when ‘z’ is rotated in the counter clockwise direction about the origin when θ = \(\frac{π}{3}\) i.,e argument of new position

(ii) θ = \(\frac{2π}{3}\)
Solution:

(iii) θ = \(\frac{3π}{3}\)
Solution:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.7

Question 1.
Write in polar form of the following complex numbers.
(i) 2 + i2 √3
Solution:
Let z + i2√3 = r (cos θ + i sin θ)
equating real and imaginary parts
r cos θ = 2 (+ve)
r sin θ = 2√3 (+ve)
r² cos² θ + r² sin² θ = (2)² + (2√3)²
r² = 4 + 12 = 16
|z| = r = 4
since cos cos θ and sin θ are positive ‘θ’ lies in 1st quadrant.
cos θ = \(\frac{1}{2}\), sin θ = \(\frac{√3}{2}\)
∴ θ = sin θ = \(\frac{π}{3}\) (or) θ = tan^-1 |\(\frac{y}{x}\)|
= tan^-1 |\(\frac{2√3}{2}\)|
= tan^-1 √3 = \(\frac{π}{3}\)
∴ argument = 2kπ + \(\frac{π}{3}\)
∴ Polar form is z = r (cos θ + i sin θ)
2 + 2i√3 = 4 (cos (2kπ + \(\frac{π}{3}\)) + i sin(2kπ +\(\frac{π}{3}\))) k ∈ z

(ii) 3 – i √3
Solution:
Let z = 3 – i √3 = r (cos θ + i sin θ)
equating real and imaginary parts
r cos θ = 3 (+ve)
r sin θ = -√3 (-ve)
r² cos² θ + r² sin² θ = (3)² + (-√3)²
r² = 9 + 3 = 12
|z| = r = 2√3
since cos cos θ positive and sin θ in -ve so lies in IV quadrant.
cos θ = \(\frac{√3}{2}\), sin θ = \(\frac{-1}{2}\), θ = \(\frac{-π}{6}\)
argument = 2kπ – \(\frac{π}{6}\), k ∈ Z
polar from z = r(cos θ + i sin θ)
3 – i√3 = 2√3 (cos (2kπ – \(\frac{π}{6}\)) + i sin(2kπ – \(\frac{π}{6}\))) k ∈ Z

(iii) -2 – i 2 = r (cos θ + i sin θ)
Solution:
Let z = -2 – i2 = r(cos θ + i sin θ)
equating real and imaginary parts
r cos θ = -2
r sin θ = -2
r² cos² θ + r² sin² θ = (-2)² + (-2)²
r² = 4 + 4 = 8
r² = 8
|z| = r = 2√2
cos θ = \(\frac{-2}{2√2}\) = \(\frac{-1}{√2}\), sin θ = \(\frac{-2}{2√2}\) = \(\frac{-1}{√2}\)
since cos θ and sin θ both are in -ve so lies in III quadrant.
argument = 2kπ – 3\(\frac{π}{4}\)
as θ = \(\frac{π}{4}\) – π = –\(\frac{3π}{4}\)
polar from z = r(cos θ + i sin θ)
-2 – i2 = 2√2 (cos (2kπ – \(\frac{3π}{4}\)) + i sin(2kπ – \(\frac{3}{4}\))) k ∈ Z

(iv) \(\frac{i-1}{cos{\frac{π}{3}}+isin{\frac{π}{3}}}\)
Solution:

Question 2.
Find the rectangular form of the complex numbers

Solution:

Solution:

Question 3.
\(\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right) \cdots\left(x_{n}+i y_{n}\right)=a+i b\), show that

Solution:
Let (x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …… (x_n + iy_n) = a + ib
Taking modulus
|(x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …… (x_n + iy_n)| = |a + ib|
|x₁ + iy₁| |x₂ + iy₂| |x₃ + iy₃| …… |x_n + iy_n| = |a + ib|
\(\sqrt{x_{1}^{2}+y_{1}^{2}} \sqrt{x_{2}^{2}+y_{2}^{2}} \sqrt{x_{3}^{2}+y_{3}^{2}} \ldots \sqrt{x_{n}^{2}+y_{n}^{2}}\) = \(\sqrt{a^2+b^2}\)
Squaring on both sides
\(\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)\left(x_{3}^{2}+y_{3}^{2}\right) \ldots\left(x_{n}^{2}+y_{n}^{2}\right)\) = a² + b²
Hence proved

(ii) \(\sum_{r=1}^{n}\) tan^-1 (\(\frac{y_r}{x_r}\)) = tan^-1 (\(\frac{b}{a}\)) + 2kπ, k ∈ Z
Solution:
Let (x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …… (x_n + iy_n) = a + ib
Taking arguments
arg [(x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …… (x_n + iy_n)] = arg (a + ib)
arg (x₁ + iy₁) + arg(x₂ + iy₂) + arg (x₃ + iy₃) …… + arg(x_n + iy_n) = arg(a + ib)

Question 4.
Given \(\frac{1+z}{1-z}\) = cos 2θ + i sin 2θ, show that To prove that z = i tan θ.
Solution:

Squaring on both sides
(1 + x)² + y² = (1 – x)² + y²
1 + 2x + x² + y² = 1 – 2x + x² +y²
x = 0
∴ z = 0 + iy = iy

∴ y = tan θ
hence z = iy
z = i tan θ

Question 5.
If cos α + cos β + cos γ = sin α + sin β + sin γ = 0. then show that
(i) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ) and
(ii) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ).
Solution:
Let a = cos α + i sin α = e^iα
b = cos β + i sin β = e^iβ
c = cos γ + i sin γ = e^iγ
a + b + c = (cos α + cos β + cos γ) + i (sin α + sin β + sin γ)
⇒ a + b + c = 0 + i 0
⇒ a + b + c = 0
If a + b + c = 0 then a³ + b³ + c³ = 3abc

(cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ) = 3 [cos (α + β + γ) + i sin (α + β + γ)]
(cos 3α + cos 3β + cos 3γ) + i (sin 3α + sin 3β + sin 3γ) = 3 cos (α + β + γ) + i 3sin(α + β + γ)
Equating real and Imaginary parts
(i) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
(ii) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)

Question 6.
If z = x + iy and arg \(\left(\frac{z-i}{z+2}\right)\) = \(\)\frac{π}{4}, then show that x² + y³ + 3x – 3y + 2 = 0.
Solution:

2y – x – 2 = x² + 2x + y² – y
x² + y² + 2x + x – y – 2y + 2 = 0
⇒ x² + y² + 3x – 3y + 2 = 0
Hence proved

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.6 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.6

Question 1.
If 2 = x + iy is a complex number such that \(\left|\frac{z-4 i}{z+4 i}\right|\) = 1 show that the locus of z is real axis.
Solution:
Let z = x + iy

Squaring on both sides
x² + (y – 4)² =  x² + (y + 4)²
simplifying
We get y = 0
Which is real axis

Question 2.
If z = x + iy is a complex number such that Im \(\left(\frac{2 z+1}{i z+1}\right)\) = 0, show that the locus of z is 2x² + 2y² + x – 2y = 0.
Solution:
Let z = x + iy

2y(1 – y) – x(2x + 1) = 0
⇒ 2y – 2y² – 2x² – x = 0
∴ The locus is 2x² + 2y² – 2y + x = 0
Hence proved

Question 3.
Obtain the Cartesian form of the locus of z = x + iy in each of the following cases:
(i) [Re (iz)]² = 3
Solution:
z = x + iy
[Re(iz)]² = 3
⇒ [Re[i(x + iy]]² = 3
⇒ [Re(ix – y)]² = 3
⇒ (-y)² = 3
⇒ y² = 3

(ii) Im[(1 – i)z + 1] = 0
⇒ Im [(1 – i)(z + iy) + 1] = 0
⇒ Im[x + iy – ix + y + 1] = 0
⇒ Im[(x + y + 1) + i(y – x)] = 0
Considering only the imaginary part
y – x = 0 ⇒ x = y

(iii) |z + i| = |z – 1|
⇒ |x + iy + i| = | x + iy – 1|
⇒ |x + i(y + 1)| = |(x – 1) + iy|
Squaring on both sides
|x + i(y + 1)|² = |(x – 1) + iy|²
⇒ x² + (y + 1)² = (x – 1)² + y²
⇒ x² + y² + 2y + 1 = x² – 2x + 1 + y²
⇒ 2y + 2x = 0
⇒ x + y = 0

(iv) \(\bar {z}\) = z^-1
Solution:

x² + y² = 1, x² + y² = -1 which cannot be true.
∴ x² + y² = 1

Question 4.
Show that the following equations represent a circle, and, find its centre and radius.
(i) |z – 2 – i| = 3
Solution:
Let z = x + iy
|z – 2 – i| = 3
⇒ |x + iy – 2 – i| = 3
⇒ |(x – 2) + i(y – 1)| = 3
⇒ \(\sqrt{(x-2)^{2}+(y-1)^{2}}=3\)
Squaring on both sides
(x – 2)² + (y – 1)² = 9
⇒ x² – 4x + 4 + y² – 2y + 1 – 9 = 0
⇒ x² + y² – 4x – 2y – 4 = 0 represents a circle
2g = -4 ⇒ g = -2
2f = -2 ⇒ f = -1
c = -4
(a) Centre (-g, -f) = (2, 1) = 2 + i
(b) Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+1+4}=3\)
Aliter: |z – (2 + i)| = 3
Centre = 2 + i
radius = 3

(ii) |2(x + iy) + 2 – 4i| = 2
⇒ |2x + i2y + 2 – 4i| =2
⇒ |(2x + 2) + i(2y – 4)| = 2
⇒ |2(x + 1) + 2i(y – 2)| = 2
⇒ |(x + 1) + i(y – 2)| = 1
⇒ \(\sqrt{(x+1)^{2}(y-2)^{2}}=1\)
Squaring on both sides,
x² + 2x + 1 + y² + 4 – 4y – 1 = 0
⇒ x² + y² + 2x – 4y + 4 = 0 represents a circle
2g = 2 ⇒ g = 1
2f = -4 ⇒ f = -2
c = 4
(a) Centre (-g, -f) = (-1, 2) = -1 + 2i
(b) Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{1+4-4}=1\)
Aliter: 2|(z + 1 – 2i)| = 2
|z – (-1 + 2i)| = 1
Centre = -1 + 2i
radius = 1

(iii) |3(x + iy) – 6 + 12i| = 8
⇒ |3x + i3y – 6 + 12i| = 8
⇒ |3(x – 2) + i3 (y + 4)| = 8
⇒ 3|(x – 2) + i (y + 4)| = 8
⇒ \(3 \sqrt{(x-2)^{2}+(y+4)^{2}}=8\)
Squaring on both sides,
9[(x – 2)² + (y + 4)²] = 64
⇒ x² – 4x + 4 + y² + 8y + 16 = \(\frac{64}{9}\)
⇒ x² + y² – 4x + 8y + 20 – \(\frac{64}{9}\) = 0
x² + y² – 4x + 8y + \(\frac{116}{9}\) = 0 represents a circle.
2g = -4 ⇒ g = -2
2f = 8 ⇒ f = 4
c = \(\frac{116}{9}\)
(a) Centre (-g, -f) = (2, -4) = 2 – 4i
(b) Radius = \(=\sqrt{g^{2}+f^{2}-c}=\sqrt{4+16-\frac{116}{9}}=\sqrt{\frac{180-116}{9}}=\frac{8}{3}\)
Aliter:
|z – 2 + 4i| = \(\frac{8}{3}\)
⇒ |z – (2 – 4i)| = \(\frac{8}{3}\)
Centre = 2 – 4i, Radius = \(\frac{8}{3}\)

Question 5.
Obtain the Cartesian equation for the locus of z = x + iy in each of the following cases:
(i) |z – 4| = 16
Solution:
Let z = x + iy
|x + iy – 4| – 16
|(x – 4) + iy| = 16
\(\sqrt{(x – 4)² + y²}\) = 16
∴ Squaring on both sides
(x – 4)² + y² = 256
x² – 8x + 16 + y² – 256 = 0
x² + y² – 8x – 240 = 0
The locus of the point is a circle.

(ii) |z – 4|² – |z – 1|² = 16.
Solution:
|x + iy – 4|² – |x + iy – 1|² = 16
⇒ |(x – 4) + iy|² – |(x – 1) + iy|² = 16
⇒ [(x – 4)² + y²] – [(x – 1)² + y²] = 16
⇒ (x² – 8x + 16 + y²) – (x² – 2x + 1 + y²) = 16
⇒ x² + y² – 8x + 16 – x² + 2x – 1 – y² = 16
⇒ -6x + 15 = 16
⇒ 6x + 1 = 0
The locus of the point is a straight line.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.5

Question 1.
Find the modulus of the following complex numbers.
(i) \(\frac{2i}{3+4i}\)
Solution:

(ii) \(\frac{2-i}{1+i}+\frac{1-2 i}{1-i}\)
Solution:

Modulus of z = |z| = \(\sqrt{4+4}\)
= √8
= 2√2

(iii) |(1 – i)¹⁰| = (|1 – i|)¹⁰
= \((\sqrt{1+1})^{10}=(\sqrt{2})^{10}=2^{5}=32\)
(iv) |2i(3 – 4i) (4 – 3i)|
= |2i| |3 – 4i| |4 – 3i|
= \(2 \sqrt{9+16} \sqrt{16+9}\)
= 2 × 5 × 5
= 50

Question 2.
For any two complex numbers z₁ and z₂, such that |z₁| = |z₂| = 1 and z₁ z₂ ≠ -1, then show that \(\frac{z_{1}+z_{2}}{1+z_{1} z_{2}}\) is real number.
Solution:
Given |z₁| = |z₂| = 1 and z₁ z₂ ≠ 1
|z₁|² = 1 |z₂|² = 1
z₁ \(\bar{z}_{1}\) = 1 similarly z₂ \(\bar{z}_{2}\) = 1

Since z = \(\bar{z}\), it is a real number.

Question 3.
Which one of the points 10 – 8i, 11 + 6i is closest to 1 + i.
Solution:
A (1 + i), B (10 – 8i), C (11 + 6i)
|AB| = |(10 – 8i) – (1 + i)|
= |10 – 8i – 1 – i|
= |9 – 9i|
= \(\sqrt{81+81}\)
= \(\sqrt{162}\)
= 9(1.414)
= 12.726
CA = |(11 + 6i) – (1 + i)|
= |11 + 6i – 1 – i|
= |10 + 5i|
= \(\sqrt{100+25}\)
= \(\sqrt{125}\)
C (11 + 6i) is closest to the point A (1 + i)

Question 4.
If |z| = 3, show that 7 ≤ |z + 6 – 8i| ≤ 13.
Solution:
given |z| = 3
|z + 6 – 8i| ≤ |z| + |6 – 8i|
= 3 + \(\sqrt{6^2+8^2}\)
= 3 + \(\sqrt{100}\)
= 3 + 10 = 13
∴ |z + 6 – 8i| ≤ 13 ……….. (1)
|z + 6 – 8i| ≥ ||z| – |-6 + 8i||
= |3 – 10|
= |-7| = 7
∴ |z + 6 – 8i| ≥ 7 ………… (2)
from 1 and 2
we get 7 ≤ |z + 6 – 8i| ≤ 13
hence proved.

Question 5.
If |z| = 1, show that 2 ≤ |z² – 3| ≤ 4.
Solution:
|z| = 1 ⇒ |z|² = 1
||z₁| – |z₂|| ≤ |z₁ + z₂| ≤ |z₁| + |z₂|
||z|² – |-3|| ≤ |z² – 3| ≤ |z|² + |-3|
|1 – 3| ≤ |z² – 3| ≤ 1 + 3
2 ≤ |z² – 3| ≤ 4

Question 6.
If |z| = 2 show that the 8 ≤ |z + 6 + 8i| ≤ 12
Solution:
Given |z| = 2
|z + 6 + 8i| = |z| + |6 + 8i|
= 2 + \(\sqrt{6^2+8^2}\)
= 2 + \(\sqrt{100}\)
= 2 + 10
= 12
∴ |z + 6 + 8i| ≤ 12 ……….. (1)
|z + 6 + 8i| ≥ ||z| – |-6 – 8i||
= |2 – 10|
= |-8|
= 8
|z + 6 + 8i| ≥ 8 ………… (2)
From 1 and 2 we get
8 ≤ |z + 6 + 8i| ≤ 12
Hence proved

Question 7.
If z₁ z₂ and z₃ are three complex numbers such that |z₁| = 1, |z₁| = 2, |z₃| = 3 and |z₁ + z₂ + z₃| = 1 show that |9z₁ z₂ + 4z₁ z₃ + z₂ z₃| = 6.
Solution:
|z₁| = 1, |z₁| = 2, |z₃| = 3
|z₁ + z₂ + z₃| = 1
Now |9z₁ z₂ + 4z₁ z₃ + z₂ z₃|

Hence proved.

Question 8.
If the area of the triangle formed by the vertices z, iz, and z + iz is 50 square units, find the value of |z|.
Solution:
The given vertices are z, iz, z + iz ⇒ z, iz are ⊥r to each other.
Area of triangle = \(\frac { 1 }{ 2 }\) bh = 50
⇒ \(\frac { 1 }{ 2 }\) |z| |iz| = 50
⇒ \(\frac { 1 }{ 2 }\) |z| |z| = 50
⇒ |z|² = 100
⇒ |z| = 10
Aliter:
Given the area of triangle = 50 sq. unit

x² + y² = 100
|z|² = 100
|z| = 10

Question 9.
Show that the equation z³ + 2 \(\bar {z}\) = 0 has five solutions.
Solution:
Given z³ + 2 \(\bar {z}\) = 0
z³ = -2 \(\bar {z}\)
|z³| = |-2| |\(\bar {z}\)|
|z|³ = 2|z| [∵ |z| = |\(\bar {z}\)|
|z|³ – 2 |z| = 0
|z| [|z|² – 2] = 0
|z| = 0 |z|² = 2
z\(\bar {z}\) = 2
z = \(\frac{2}{\bar {z}}\) = ± √2 [∵ \(\bar {z}\) = \(\frac{-z^3}{2}\) ]
z = \(\frac{2}{(\frac{z^3}{-2})}\)
z⁴ = 4
It has 4 non zero solutions.
∴ Including z = 0 we have 5 solutions.

Question 10.
Find the square roots of
(i) 4 + 3i
Solution:
|4 + 3i| = \(\sqrt {4^2+3^2}\) = \(\sqrt {16+9}\)
\(\sqrt {25}\) = 5
Let \(\sqrt {4+3i}\) = a + ib
squaring on both sides
4 + 3i = (a + ib)²
4 + 3i = (a² – b²) + 2 jab
Equating real and imaginary parts
a² – b² = 4, 2ab = 3
(a² + b²)² = (a² – b²)² + 4a² b²
= (4)² + (3)²
= 16 + 9 = 25
∴ a² + b² = 5
Solving a² – b² = 4 and a² + b² = 5.
we get a² = \(\frac {9}{2}\) , b² = \(\frac {1}{2}\)
a = ±\(\frac {3}{√2}\) and b = ±\(\frac {1}{√2}\)
∴ \(\sqrt {4 + 3i}\) = a + ib
= ±(\(\frac {3}{√2}\) + ±\(\frac {i}{√2}\))
Aliter:
Square root of 4 + 3i
formula method

(ii) -6 + 8i
Solution:
Let \(\sqrt {-6 + 8i}\) = a + ib
Squaring on both sides
-6 + 8i = (a + ib)²
-6 + 8i = a² – b² + 2iab
Equating real and imaginary parts
a² – b² = -6 and 2ab = 8
Now (a² + b²)² = (a² – b²)² + 4a²b²
= (-6)² + (8)²
= 36 + 64 = 100
∴ a + b² = 10
Solving a² – b² = -6 and a² + b² = 10
we get 2a² = 4, b² = 8
a² = 2, b² = ±2√2
a = ±√2
∴ \(\sqrt {-6 + 8i}\) = ±√2 ± i 2√2
= ±(√2 + i 2√2)
Aliter:
square root of -6 + 8i
let a + ib = -6 + 8i
a = -6, b = 8

(iii) -5 – 12i
Solution:
Let \(\sqrt{-5-12 i}\) = a + ib
Squaring on both sides
-5 – 12i = (a+ib)²
-5 – 12i = a² – b² + 2iab
Equating real and imaginary parts
a² – b² = -5, 2ab = -12
(a² + b²)² = (a² – b²)² + 4a²b²
= (-5)² + (-12)² = 169
∴ a² + b² = 13
Solving a²- b² = -5 and a² + b² = 13
we get a² = 4, b² = 9
a = ±2, b = ±3
Since 2ab = -12 < 0, a, b are of opposite signs.
∴ When a = ±2, b = ±3
Now \(\sqrt{-5-12 i}\) = ± (2 – 3i)
Aliter
Square root of -5 – 12i
Let a + ib = -5 – 12i
a = -5, b = -12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.4

Question 1.
Write the following in the rectangular form:
(i) \(\overline { (5+9i)+(2-4i) } \)
Solution:
\(\overline { (5+9i)+(2-4i) } \)
= \(\overline {(5+9i)} \) + \(\overline {(2-4i)} \)
= 5 – 9i + 2 + 4i
= 7 – 5i

(ii) \(\frac {10-5i}{6+2i} \)
Solution:

(iii) \(\overline {3i} + \frac{2}{2-i}\)
Solution:

Question 2.
If z = x + iy, find the following in rectangular form.
(i) Re(\(\frac {1}{z} \))
Answer:

(ii) Re(i\(\bar{z}\)) = Re[i(\(\overline{x+i y}\))]
= Re(ix + y)
= y
(iii) Im(3z + 4\(\bar{z}\) – 4i)
= Im (3(x + iy) + 4(x – iy) – 4i)
= Im (3x + 3iy + 4x – 4iy – 4i)
= Im (3x + 4 + i (3y – 4y – 4)
= Im (3x + 4x + i(-y – 4))
= Im [7x + i(-y – 4)]
= -y – 4
= -(y + 4)

Question 3.
If z₁ = 2 – i and z₂ = -4 + 3i, find the inverse of z₁, z₂ and \(\frac {z_1}{z_2} \)
Solution:
z₁ = 2 – i, z₂ = -4 + 3i
z₁ z₂ = (2 – i) (-4 + 3i)
= -8 + 3 + 4i + 6i
= -5 + 10i

Question 4.
The complex numbers u, v, and w are related by \(\frac {1}{u}\) = \(\frac {1}{v}\) + \(\frac {1}{w}\). If v = 3 – 4i and w = 4 + 3i, find u in rectangular form.
Solution:
v = 3 – 4i, w = 4 + 3i

Question 5.
Prove the following properties:
(i) z is real if and only if z = \(\overline {z}\)
Solution:
z is real iff z = \(\bar{z}\)
Let z = x + iy
z = \(\bar{z}\)
⇒ x + iy = x – iy
⇒ 2iy = 0
⇒ y = 0
⇒ z is real.
z is real iff z = \(\bar{z}\)

(ii) Re(z) = \(\frac{z+\bar{z}}{2}\) and Im(z) = \(\frac{z-\bar{z}}{2i}\)
Solution:
let z = x + iy
\(\overline {z}\) = x – iy

Hence proved

Question 6.
Find the least value of the positive integer n for which (√3 + i)ⁿ (i) real, (ii) purely imaginary.
Solution:
Given (√3 + i)ⁿ
= (√3)² + 2i √3 + (i)²
= 3 + 2i √3 – 1
= 2 + 2i √3
= 2(1 + i√3)
put n = 3 or 4 or 5
then real part is not possible

which is purely real ∴ n = 6

(ii) (√3 + i)ⁿ

which is purely imaginary
∴ n = 3

Question 7.
Show that
(i) (2 + i√3)¹⁰ – (2 – i√3)¹⁰ is purely imaginary.
Solution:
Let z = (2 + i√3)¹⁰ – (2 – i√3)¹⁰
Let Z

= (2 – i√3)¹⁰ – (2 + i√3)¹⁰
= -[(z + i√3)¹⁰ – (2 – i√3)¹⁰]
= -z
(2 + i√3)¹⁰ – (2 – i√3)¹⁰ is purely imaginary

(ii) \(\left(\frac{19-7 i}{9+i}\right)^{12}\) + \(\left(\frac{20-5 i}{7-6 i}\right)^{12}\) is real
Solution:

∴ z is real.

Students can Download Tamil Nadu 12th Maths Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Maths Model Question Paper 5 English Medium

Instructions:

1.  The question paper comprises of four parts.
2.  You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Part I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

I. Choose the correct answer. Answer all the questions. [20 × 1 = 20]

Question 1.
If | adj(adj A) | = |A|⁹, then the order of the square matrix A is _______.
(a) 3
(b) 4
(c) 2
(d) 5
Answer:
(b) 4

Question 2.
If |z₁| = 1, |z₂| = 2, |z₃| = 3 and |9z₁z₂ + 4z₁z₂ + z₂z₃| = 12, then the value of |z₁ + z₂+ z₃| is ________.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 3.
The value of \(\left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10}\) is ________.

Answer:
(a) cis \(\frac{2 \pi}{3}\)

Question 4.
If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})=u\), then cos 2u is equal to _____.
(a) tan² α
(b) 0
(c) -1
(d) tan 2α
Answer:
(c) -1

Question 5.
If \(\cot ^{-1} x=\frac{2 \pi}{5}\) for some x∈R, the value of tan^-1 x is _______.

Answer:
(c) \(\frac{\pi}{10}\)

Question 6.
The radius of the circle passing through the point (6, 2) two of whose diameter are x + y = 6 and x + 2y = 4 is ____.
(a) 10
(b) \(2 \sqrt{5}\)
(c) 6
(d) 4
Answer:
(b) \(2 \sqrt{5}\)

Question 7.
The length of the L.R. of x² = -4y is _______.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 8.
Distance from the origin to the plane 3x – 6y + 2z + 7 = 0 is ______.
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 9.
The distance from the origin to the plane \(\vec{r} \cdot(2 \vec{i}-\vec{j}+5 \vec{k})=7\) is _____.

Answer:
(a) \(\frac{7}{\sqrt{30}}\)

Question 10.
The number given by the Mean value theorem for the function \(\frac{1}{x}\), x ∈ [1, 9] is ______.
(a) 2
(b) 2.5
(c) 3
(d) 3.5
Answer:
(c) 3

Question 11.
f is a differentiable function defined on an interval I with positive derivative. Then f is ______.
(a) increasing on I
(b) decreasing on I
(c) strictly increasing on I
(d) strictly decreasing on I
Answer:
(c) strictly increasing on I

Question 12.
If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in our calculation of the volume is ________.
(a) 0.4 cu.cm
(b) 0.45 cu.cm
(c) 2 cu.cm
(d) 4.8 cu.cm
Answer:
(d) 4.8 cu.cm

Question 13.
If u(x, y) = \(e^{x^{2}+y^{2}}\), then \(\frac{\partial u}{\partial x}\) is equal to _______.
(a) \(e^{x^{2}+y^{2}}\)
(b) 2xu
(c) x²u
(d) y²u
Answer:
(b) 2xu

Question 14.
The value of \(\int_{0}^{\infty} e^{-3 x} x^{2} d x\) is _______.
(a) \(\frac{7}{27}\)
(b) \(\frac{5}{27}\)
(c) \(\frac{4}{27}\)
(d) \(\frac{2}{27}\)
Answer:
(d) \(\frac{2}{27}\)

Question 15.
\(\int_{0}^{a} f(x) d x\) is _____.

Answer:
(b) \(\int_{0}^{a} f(a-x) d x\)

Question 16.
The integrating factor of the differential equation \(\frac{d y}{d x}\) + P(x) y = Q (x) is x, then P(x) ______.
(a) x
(b) \(\frac{x^{2}}{2}\)
(c) \(\frac{1}{x}\)
(d) \(\frac{1}{x^{2}}\frac{1}{x^{2}}\)
Answer:
(c) \(\frac{1}{x}\)

Question 17.
The order and degree of the differential equation p\(\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{1 / 3}+x^{1 / 4}=0\) are respectively _______.
(a) 2, 3
(b) 3, 3
(c) 2, 6
(d) 2, 4
Answer:
(a) 2, 3

Question 18 .
Which of the following is a discrete random variable?
I. The number of cars crossing a particular signal in a day.
II. The number of customers in a queue to buy train tickets at a moment.
III. The time taken to complete a telephone call.
(a) I and II
(b) II only
(c) III only
(d) II and III
Answer:
(a) I and II

Question 19.
If p is true and q is false then which of the following is not true?
(a) p → q is false
(b)p ∨ q is true
(c)p ∧ q is false
(d) p ↔ q is true
Answer:
(d) p ↔ q is true

Question 20.
The operation * defined by a*b = \(\frac{a b}{7}\) is not a binary operation on ______.
(a) Q⁺
(b) Z
(c) R
(c) C
Answer:
(b) Z

Part – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
Using elementary transformations find the inverse of the following matrix \(\left[\begin{array}{ll}
4 & 7 \\
3 & 0
\end{array}\right]\)
Answer:

Question 22.
If Z₁ = 1 – 3i, z₂ = -4i, and z₃ = 5, show that (z₁ + z₂) + z₃ = Z₁ + (z₂ + z₃)
Answer:
z₁ = 1 – 3i, z₂ = 4i, z₃ = 5
(z₁ + z₂) + z₃ = (1 – 3i – 4i) + 5(1 – 7i) + 5
= 6 – 7i …..(1)
z₁ + (z₂ + z₃) = (1 – 3i) + (-4i + 5)
= 6 – 7i …..(2)
from(1)& (2)we get
∴ (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)

Question 23.
Find a polynomial equation of minimum degree with rational coefficients, having 2+ \(\sqrt{3}\) i as
a root.
Answer:
Given roots is (2 + \(\sqrt{3}\) i)
∴ The other root is (2 – \(\sqrt{3}\) i), since the imaginary roots with real co-efficient occur as conjugate
pairs.
x² – x(S.O.R) + P.O.R = 0 ⇒ x² – x(4) + (4 + 3) = 0
x² – 4x + 7 = 0.

Question 24.
Evaluate: \(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+17 x+29}{x^{4}}\right)\)
Answer:
This is an indeterminate of the form (\(\frac{\infty}{\infty}\)). To evaluate this limit, we apply l’Hôpital Rule.

Question 25.
Let g(x, y) = 2y + x², x = 2r – s, y = r² + 2s, r, s ∈R. Find \(\frac{\partial g}{\partial r}, \frac{\partial g}{\partial s}\).
Answer:

Question 26.
Evaluate: \(\int_{0}^{\frac{\pi}{2}}\left(\sin ^{2} x+\cos ^{4} x\right) d x\)
Answer:

Question 27.
Find the differential equation corresponding to the family of curves represented by the equation y = Ae^8x + Be^-8x, where A and B are arbitrary constants.
Answer:
y = Ae^8x + Be^-8x Where A and B are arbitrary constants.
Differentiate with respect to ‘x’

Question 28.
If F(x) = \(\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right)\) – ∞ < x < ∞ is a distribution function of a continuous variable X, find P (0 ≤ x ≤ 1).
Answer:

Question 29.
Show that p → q and q → p are not equivalent.
Answer:
Truth table for p → q

Truth table for q → p

The entries in the column corresponding to p → q and q → p are not identical, hence they are not equivalent.

Question 30.
Show that the lines \(\frac{x-1}{4}=\frac{2-y}{6}=\frac{z-4}{12}\) and \(\frac{x-3}{-2}=\frac{y-3}{3}=\frac{5-z}{6}\) are parallel.
Answer:
We observe that the straight line \(\frac{x-1}{4}=\frac{2-y}{6}=\frac{z-4}{12}\) is parallel to the vector \(4 \hat{i}-6 \hat{j}+12 \hat{k}\) and the straight line \(\frac{x-3}{-2}=\frac{y-3}{3}=\frac{5-z}{6}\) is parallel to the vector \(-2 \hat{i}+3 \hat{j}-6 \hat{k}\).
Since \(4 \hat{i}-6 \hat{j}+12 \hat{k}=-2(-2 \hat{i}+3 \hat{j}-6 \hat{k})\) the two vectors are parallel, and hence the two straight lines are parallel.

Part – III

II. Answer any seven questions. Question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
Using elementary transformations find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\)

Question 32.
Find the square roots of – 15 – 8i

Question 33.
Find the sum of the squares of the roots of ax⁴ + bx³ + cx² + dx + e = 0, a ≠ 0

Question 34.
For what value of x, the inequality \(\frac{\pi}{2}\) < cos^-1 (3x – 1) < π holds?

Question 35.
Find the foot of the perpendicular drawn from the point (5, 4, 2) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}\). Also, find the equation of the perpendicular.

Question 36.
Evaluate \(\int_{0}^{\infty} \frac{x^{n}}{n^{x}} d x\), where n is a positive integer ≥ 2.

Question 37.
The engine of a motor boat moving at 10 m/s is shut off. Given that the retardation at any subsequent time (after shutting off the engine) equal to die velocity at that time. Find the velocity after 2 seconds of switching off the engine.

Question 38.
The probability that Mr.Q hits a target at any trial is \(\frac{1}{4}\). Suppose he tries at the target 10 times. Find the probability that he hits the target (i) exactly 4 times (ii) at least one time.

Question 39.
Consider the binary operation * defined on the set A = {a, b, c, d} by the following table:

Is it commutative and associative?

Question 40.
Evaluate the following limit, if necessary use l’Hopital Rule. \(\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\)

Part – IV

IV. Answer all the questions. [7 × 5 = 35]

Question 41.
(a) Find the inverse of A = \(\left[\begin{array}{lll}
2 & 1 & 1 \\
3 & 2 & 1 \\
2 & 1 & 2
\end{array}\right]\) by Gauss-Jordan method
[OR]
(b) Find the point of intersection of the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-4}{5}=\frac{y-1}{2}=z\).

Question 42.
(a) Suppose z₁, z₂ and z₃ are the vertices of an equilateral triangle inscribed in the circle.
|z| = 2. If z₁ = 1 + i\(\sqrt{3}\), then find z₂ and z₃.
[OR]
(b) If A = \(\left[\begin{array}{cc}
5 & 3 \\
-1 & -2
\end{array}\right]\) show that A² – 3A – 7I₂ = O₂. Hence Find A^-1.

Question 43.
(a) Solve the equation 3x³ – 16x² + 23x – 6 = 0 if the product of two roots is 1.
[OR]
(b) The mean and variance of a binomial variate X are respectively 2 and 1.5.
Find (i) P(X = 0) (ii) P(X = 1) (iii) P(X ≥ 1)

Question 44.
(a) Find the value of \(\cos \left(\sin ^{-1}\left(\frac{4}{5}\right)-\tan ^{-1}\left(\frac{3}{4}\right)\right)\)
[OR]
(b) Find, by integration, the volume of the solid generated by revolving about y-axis the region bounded between the curve y = \(\frac{3}{4} \sqrt{x^{2}-16}\), x ≥ 4, the y-axis, and the lines y = 1 and y = 6.

Question 45.
(a) A semielliptical archway over a one-way road has a height of 3m and a width of 12m. The truck has a width of 3m and a height of 2.7m. Will the truck clear the opening of the archway?
[OR]
(b) For the function f(x, y) = \(\frac{3 x}{y+\sin x}\) find the f_x, f_y, and show that f_xy = f_yx.

Question 46.
(a) Derive the equation of the plane in the intercept form.
[OR]
(b) Let A be Q\{1}. Define * on A by x * y = x +y – xy. Is * binary on A ? If so, examine the existence of identity, existence of inverse properties for the operation * on A .

Question 47.
(a) Find the angle between the rectangular hyperbola xy = 2 and the parabola x² + 4y = 0.
[OR]
(b) A pot of boiling water at 100° C is removed from a stove at time t = 0 and left to cool in the kitchen. After 5 minutes, the water temperature has decreased to 80° C, and another 5 minutes later it has dropped to 65° C. Determine the temperature of the kitchen.