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Tamilnadu Samacheer Kalvi 12th Maths Book Solutions Chapter 10 Ordinary Differential Equations Ex 10.5

Question 1.
If F is the constant force generated by the motor of an automobile of mass M, its velocity V is given by M\(\frac{d \mathbf{V}}{d t}\) = F – kV, where A is a constant. Express V in terms of t given that V = 0 when t = 0.
Solution:

Substituting in (1)

Question 2.
The velocity v, of a parachute falling vertically satisfies the equation, where g and k are constants. If v and x are both initially zero, find v in terms of x.
Solution:


Squaring on both sides

Question 3.
Find the equation of the curve whose slope is \(\frac{y-1}{x^{2}+x}\) and which passes through the point (1, 0).
Solution:


This passes through (1, 0)

Question 4.
Solve the following differential equations:

Solution:

(ii) ydx + (1 + x²) tan^-1 x dy = 0
Solution:
ydx + (1 + x²) tan^-1 xdy = 0
(1 + x²) tan^-1x dy = -y dx
Separating the variables

log y = – log (tan^-1 x) + log c
log y + log (tan^-1 x) = log c
log (y tan^-1 x) = log c
y tan^-1 x = c

(iii) sin \(\frac{d y}{d x}\) = a, y (0) = 1
Solution:

Given y(0) = 1
i.e., When x = 0, y = 1
∴ 1 = 0(sin^-1 a) + c ⇒ c = 1
∴ The solution is y = xsin^-1 a + 1

(iv)
Solution:

(v) (e^y + 1) cos x dx + e^y sin x dy = 0
Solution:
(e^y + 1) cos x dx + e^y sin x dy = 0
e^y sin x dy = – (e^y + 1) cos x dx

log (e^y + 1) = – log sin x + log c
log [(e^y + 1) + log sin x = log c
log (e^y +1) sin x] = log c
(e^y+ 1) sin x = c

(vi)
Solution:

(vii)
Solution:

(viii) x cos y dy = e^x(x log x + 1)dx
Solution:

(ix)
Solution:

(x)
Solution:
\(\frac{d y}{d x}\) = tan² (x +y)
Let u = x + y
Differentiating with respect to ‘x’

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.5 Additional Problems

Question 1.

Solution:
The given equation can be written as

Question 2.

Solution:
Put x +y = z. Differentiating with respect to x we get

Question 3.
Find the cubic polynomial in x which attains its maximum value 4 and minimum value 0 at x = -1 and 1 respectively.
Solution:
Let the cubic polynomial bey = f(x). Since it attains a maximum atx = -1 and a minimum at x = 1.

Separating the variables we have dy = k (x² – 1) dx


When x = – 1, y = 4 and when x =1,7 = 0
Substituting the equation (1) we have
2k + 3c = 12; – 2k + 3c =0
On solving we have k = 3 and c = 2. Substituting these values in (1) we get the required cubic polynomial y = x³ – 3x + 2.

Question 4.
The normal lines to a given curve at each point (x, y) on the curve pass through the point (2, 0). The curve passes through the point (2, 3). Formulate the differential equation representing the problem and hence find the equation of the curve.
Solution:
Slope of the normal at any point P(x, y) = \(-\frac{d x}{d y}\)

Since the curve passes through (2, 3)

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.3

Question 1.
Find the differential equation of the family of
(i) all non-vertical lines in a plane
Solution:
Equation of family of all non-vertical lines is
y = mx + c (m ≠ 0)
Differentiate with respect to ‘x’

(ii) all non-horizontal lines in a plane.
Solution:
Equation of family of all non-horizontal lines is

Question 2.
Form the differential equation of all straight lines touching the circle x² + y² = r²
Solution:
Equation of circle x² + y² = r² of the line y = mx + c is to be a tangent to the circle, then the equation of the tangent is

Differentiating with respect to V dy

Question 3.
Find the differential equation of the family of circles passing through the origin and having their centres on the x -axis.
Solution:
All circles passing through the origin and having their centre on the x -axis say at (a, 0) will have radius ‘a’ units.
∴ Equation of circle is (x – a²) + y² = a² ….(1) [ ∵ ‘a’ arbitrary constant]
Differentiate with respect to ‘x’

Question 4.
Find the differential equation of the family of all the parabolas with latus rectum 4a and whose axes are parallel to the x -axis.
Solution:
Equation of all parabolas whose axis is parallel to X – axis is
(y – k)² = 4a (x – h)
Where (h, k) is the vertex

Question 5.
Find the differential equation of the family of parabolas with vertex at (0, -1) and having axis along the y – axis.
Solution:
Given, vertex (0, -1) and axis along y-axis
Equation of Parabola, (x + 1)² = – 4ay …… (1) [∵ a is the perameter]
Differentiate with respect to ‘x’

Question 6.
Find the differential equations of the family of all the ellipses having foci on the y – axis and centre at the origin.
Solution:
Equations of the family of all the Ellipses having foci on the y – axis and centre at the origin is

Differentiate with respect to ’x’

Question 7.
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.
Solution:

Question 8.
Find the differential equation of the curve represented by xy = ae^x + be^-x + x².
Solution:
xy = ae^x + be^-x + x²
xy – x² = ae^x + be^-x …… (1) [∵ a’, b’ are arbitrary constants]
Differentiate with respect to ‘x’
xy’ +y – 2x = ae^x – be^-x
Again, Differentiate with respect to ‘x’

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.3 Additional Problems

Question 1.
Find the differential equation of the family of straight lines y = mx + \(\frac{a}{m}\) when
(i) m is the parameter,
(ii) a is the parameter,
(iii) a, m both are parameters.
Solution:

Question 2.
Find the differential equation that will represent family of all circles having centres on the x-axis and the radius is unity.
Solution:
Equation of a circle with centre on x-axis and radius 1 unit is
(x – a)² + y² = 1 ….. (1)
Differentiating with respect to x,
2 (x – a) + 2yy’ – 0
⇒ 2 (x – a) = – 2yy’
(or) x – a = -yy’ ……(2)
Substituting (2) in (1), we get,

Question 3.
From the differential equation from the following equations.
(i) y = e^2x (A + Bx)
Solution:
ye^-2x = A + Bx ……. (1)
Since the above equation contains two arbitrary constants, differentiating twice,

(ii) y = e^x(A cos 3x + B sin 3x)
Solution:

(iii) Ax² + By² = 1
Solution:

Eliminating A and B between (1), (2) and (3) we get

(iv) y² = 4a(x – a)
Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.2

Question 1.
Express each of the following physical statements in the form of differential equation.
(i) Radium decays at a rate proportional to the amount Q present.
Solution:
Radium decays at a rate proportional to the amount Q present.

(ii) The population P of a city increases at a rate proportional to the product of population and to the difference between 5,00,000 and the population.
Solution:
Rate of change of P with respect to ‘t’ is \(\frac{d \mathrm{P}}{d t}\)
Product of population and the difference between 50,000 and the population is P (50,000 – P)

(iii) For a certain substance, the rate of change of vapor pressure P with respect to temperature T is proportional to the vapor pressure and inversely proportional to the square of the temperature.
Solution:

(iv) A saving amount pays 8% interest per year, compounded continuously. In addition, the income from another investment is credited to the amount continuously at the rate of ₹ 400 per year.
Solution:
Let ‘x’ be the amount invested.

Question 2.
Assume that a spherical rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.
Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.1

Question 1.
For each of the following differential equations, determine its order, degree (if exists)

Solution:
Order = 1,
Degree = 1

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

(vii)
Solution:

(viii)
Solution:
Order = 2

(ix)
Solution:

(x)
Solution:

Order = 1,
degree = Not exist

Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.1 Additional Problems

Question 1.
Find the order and degree of the following differential equations:

Solution:
Order = 1,
Degree = 1

(ii) y’ + y² + y³ = 0
Solution:
Order = 1,
Degree = 1

(iii)
Solution:

(iv)
Solution:

∴ Order = 2
Degree = 2

(v)
Solution:

(vi)
Solution:

(vii)
Solution:

(viii)
Solution:

(ix)
Solution:

(x) sin x(dx + dy) = cos x(dx – dy)
Solution:
Order = 1;
Degree = 1

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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 \(\frac{a+b \omega+c \omega^{2}}{b+c \omega+a \omega^{2}}+\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}=1\)
Solution:
Since ω is a cube root of unity, we have ω³ = 1 and 1 + ω + ω² = 0

Question 2.
Show that \(\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}=-\sqrt{3}\)
Solution:

Question 3.
Find the value of \(\left(\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right)^{10}\)
Solution:

Question 4.
If 2 cos α = x + \(\frac{1}{x}\) and 2 cos β = y + \(\frac{1}{y}\), show that

Solution:
(i) 2 cos α = x + \(\frac{1}{x}\)
⇒ 2 cos α = \(\frac{x^{2}+1}{x}\)
⇒ 2x cos α = x² + 1
⇒ x² – 2x cos α + 1 = 0

Question 5.
Solve the equation z³ + 27 = 0
Solution:
z³ + 27 = 0
⇒ z³ = -27
⇒ z³ = 3³(-1)
⇒ z = 3(-1)^1/3

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:
(z – 1)³ + 8 = 0
⇒ (z – 1 )³ = -8

The roots are -1, 1 – 2ω, 1 – 2ω²

Question 7.
Find the value of \(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\)
Solution:
\(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\)
We know that 9th roots of unit are 1, ω, ω², ……., ω⁸
Sum of the roots:
1 + ω + ω² + …. + ω⁸ = 0 ⇒ ω + ω² + ω³ + …… + ω⁸ = -1

The sum of all the terms \(\sum_{k=1}^{8}\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)\) = -1

Question 8.
If ω ≠ 1 is a cube root of unity, show that
(i) (1 – ω + ω²)⁶ + (1 + ω – ω²)⁶ = 128
(ii) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸)……(1 + ω²ⁿ) = 1
Solution:
(i) 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 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) ……. 2n factors
= (-ω²)(-ω)(-ω²)(-ω) …… 2n factors
= ω³. ω³
= 1

Question 9.
If z = 2 – 2i, find the rotation of z by θ radians in the counter clockwise direction about the origin when
(i) θ = \(\frac{\pi}{3}\)
(ii) θ = \(\frac{2 \pi}{3}\)
(iii) θ = \(\frac{3 \pi}{2}\)
Solution:
(i) z = 2 – 2i = 2 (1 – i) = r(cos θ + i sin θ)
\(r=\sqrt{x^{2}+y^{2}}=2 \sqrt{1+1}=2 \sqrt{2}\)
\(\alpha=\tan ^{-1}=\left|\frac{y}{x}\right|=\tan ^{-1}|1|=\frac{\pi}{4}\)
(1 – i) lies in IV quadrant
θ = -α = \(-\frac{\pi}{4}\)
\(\Rightarrow z=2 \sqrt{2}\left[\cos \left(\frac{-\pi}{4}\right)+i \sin \left(\frac{-\pi}{4}\right)\right]\)
z is rotated by θ = \(\frac{\pi}{3}\) in the counter clock wise direction.

(ii) z is rotated by θ = \(\frac{2 \pi}{3}\) in the counter clockwise direction.

(iii) z is rotated by θ = \(\frac{3 \pi}{2}\) in the counter clockwise direction.

Question 10.
Prove that the values of \(\sqrt[4]{-1} \text { are } \pm \frac{1}{\sqrt{2}}(1 \pm i)\)
Solution:

The roots are \(\pm \frac{1}{\sqrt{2}}(1 \pm i)\)

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

Question 1.

Solution:

Question 2.

Solution:

Question 3.
Prove that: (1 + i)⁴ⁿ and (1 + i)^{4n + 2} are real and purely imaginary respectively.
Solution:
Let z = 1 + i

= 2i (-1)ⁿ which is purely imaginary.

Question 4.

Solution:

Question 5.
If a = cos 2α + i sin 2α, b = cos 2β + i sin 2β and c = cos 2γ + i sin 2γ, prove that

Solution:

Question 6.
Solve: x⁴ + 4 = 0
Solution:

Question 7.
If x = a + b, y = aω + bω², z = aω² + bω, show that
(i) xyz = a³ + b³
(ii) x³ + y³ + z³ = 3(a³ + b³)
Solution:
(i) x = a + A; y = aω + bω², z = aω² + bω
Now xyz = (a + b) (aω + bω²) (abω² + bω) = (a+A) [aω³ + abω² + abω + b²ω³]
= (a + b) (a² – ab + b²) = a³ + b³
xyz = a³ + b³

(ii) x = a + b, y = aω + bω², z = aω² + bω
x + y + z = (a + aω + aω²) + (b + bω² + bω)
= a (1 + ω + ω²) + b (1 + ω + ω²) = a(0) + b(0) = 0
Now x + y + z = 0 ⇒ x³ + y³ + z³ = 3xyz
Here xyz = a³ + b³
∴ x³ + y³ + z³ = 3 (a³ + b³)

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.3

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
A binary operation on a set S is a function from …….
(a) S ➝ S
(b) (S × S) ➝ S
(c)S ➝ (S × S)
(d) (S × S) ➝ (S × S)
Solution:
(b) (S × S) ➝ S

Question 2.
Subtraction is not a binary operation in
(a) R
(b) Z
(c) N
(d) Q
Solution:
(c) N
Hint:
For example 2, 5 ∈ N but 2 – 5 = 3 ∉ N

Question 3.
Which one of the following is a binary operation on N ?
(a) Subtraction
(b) Multiplication
(c) Division
(c) All the above
Solution:
(b) Multiplication

Question 4.
In the set R of real numbers ‘*’ is defined as follows. Which one of the following is not a binary operation on R ?
(a) a * b = min (a.b)
(b) a * b = max (a, b)
(c) a * b = a
(d) a * b = a^b
Solution:
(d) a * b = a^b
Hint:
Since -2, 1/2 ∈ R , but (-2)^1/2 ∉ R.

Question 5.
The operation * defined by a * b = \(\frac{a b}{7}\) is not a binary operation on ……….
(a) Q⁺
(b) Z
(c) R
(c) C
Solution:
(b) Z
Hint:
Since 3, 5 ∈ Z, but \(\frac{3 \times 5}{7} \notin\) Z.

Question 6.



Solution:
(b) y = \(\frac{-2}{3}\)

Question 7.
If a * b = \(\sqrt{a^{2}+b^{2}}\) on the real numbers then * is ……..
(a) commutative but not associative
(b) associative but not commutative
(c) both commutative and associative
(d) neither commutative nor associative
Let, a, b ∈ R

(1) = (2) = * is associative
So * is both commutative and associative
Solution:
(c) both commutative and associative

Question 8.
Which one of the following statements has the truth value T ?
(a) sin x is an even function.
(b) Every square matrix is non-singular
(c) The product of complex number and its conjugate is purely imaginary
(d) \(\sqrt{5}\) is an irrational number
Solution:
(d) \(\sqrt{5}\) is an irrational number

Question 9.
Which one of the following statements has truth value F ?
(a) Chennai is in India or \(\sqrt{2}\) is an integer
(b) Chennai is in India or \(\sqrt{2}\) is an irrational number
(c) Chennai is in China or \(\sqrt{2}\) is an integer
(d) Chennai is in China or \(\sqrt{2}\) is an irrational number
Solution:
(c) Chennai is in China or \(\sqrt{2}\) is an integer

Question 10.
If a compound statement involves 3 simple statements, then the number of rows in the truth table is ……….
(a) 9
(b) 8
(c) 6
(d) 3
Solution:
(b) 8
Hint:
(i.e.) 2³ = 8

Question 11.
Which one is the inverse of the statement

Solution:
(a) \((\neg p \wedge \neg q) \rightarrow(\neg p \vee \neg q)\)

Question 12.
Which one is the contrapositive of the statement \((p \vee q) \rightarrow r\)?

Solution:
(a) \(\neg r \rightarrow(\neg p \wedge \neg q)\)

Question 13.
The truth table for \((p \wedge q) \vee \neg q\) is given below

Which one of the following is true?

Hint: The truth table for \((p \wedge q) \vee \neg q\)

Solution:
(3) T T F T

Question 14.
In the last column of the truth table for \(\neg(p \vee \neg q)\) the number of final outcomes of the truth value ‘F’ are
(a) 1
(b) 2
(c) 3
(d) 4
Hint:
The truth table for \(\neg(p \vee \neg q)\)

Solution:
(c) 3

Question 15.
Which one of the following is incorrect? For any two propositions p and q, we have …….

Solution:
(c) \(\neg(p \vee q) \equiv \neg p \vee \neg q\)

Question 16.

Which of the following is correct for the truth \((p \wedge q) \rightarrow \neg p\) ?

Hint:

Solution:
(2) F T T T

Question 17.


Solution:
(d) \(\neg(p \wedge q) \wedge | p \wedge(p \vee \neg r)]\)

Question 18.
The proposition \(p \wedge(\neg p \vee q)]\) is ……..
(a) a tautology
(b) a contradiction
(c) logically equivalent to \(p \wedge q\)
(d) logically equivalent to \(p \vee q\)
Solution:
(c) logically equivalent to \(p \wedge q\)

Question 19.
Determine the truth value of each of the following statements:
(a) 4 + 2 = 5 and 6+ 3 = 9
(b) 3 + 2 = 5 and 6 + 1 = 7
(c) 4 + 5 = 9 and 1 + 2 = 4
(d) 3 + 2 = 5 and 4 + 7 = 11

Solution:
(1) F T F T

Question 20.
Which one of the following is not true?
(a) Negation of a negation of a statement is the statement itself.
(b) If the last column of the truth table contains only T then it is a tautology.
(c) If the last column of its truth table contains only F then it is a contradiction
(d) If p and q are any two statements then p ⟷ q is a tautology.
Solution:
(d) If p and q are any two statements then p ⟷ q is a tautology.

Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.3 Additional Problems

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
Which of the following are statements?
(i) May God bless you
(ii) Rose is a flower
(iii) milk is white
(iv) 1 is a prime number
(a) (i), (ii), (iii)
(b) (i), (ii), (iv)
(c) (i), (iii), (iv)
(d) (ii), (iii), (iv)
Hint:
Sentence (ii), (iii) and (iv) are statements
(ii) Rose is a flower – True
(iii) Milk is white – True
(iv) 1 is a prime number
∴ (ii), (iii), (iv) are statements
(i) May god bless you. This statement can not be assigned True or False.
∴ (i) is not a statements
Solution:
(d) (ii), (iii), (iv)

Question 2.
If a compound statement is made up of the three simple statements, then the number of rows in the truth table is …….
(a) 8
(b) 6
(c) 4
(d) 2
Hint:
The number of rows in truth table = 2ⁿ = 2³ = 8
Solution:
(a) 8

Question 3.
If p is T and q is F, then which of the following have the truth value T? ………

(a) (i), (ii), (iii)
(b) (i), (ii), (iv)
(c) (i), (iii), (iv)
(d) (ii), (iii), (iv)
Hint:
p is T then ~p is F
q is F then ~ q is T

Solution:
(c) (i), (iii), (iv)

Question 4.
The number of rows in the truth offimg6 is ……..
(a) 2
(b) 4
(c) 6
(d) 8
Hint:
Number of simple statements given is 2. i.e., p and q.
Number of rows in the truth table of \(\sim[p \wedge(\sim q)]\) = 2² = 4
Solution:
(b) 4

Question 5.
The conditional statement p ➝ q is equivalent to ……..


The truth table for p ➝ q and \((\sim p \vee q)\) having the last column identical.
∴ p ➝ q is equivalent to \((\sim p \vee q)\)
Solution:
(3) \(\sim p \vee q\)

Question 6.
Which of the following is a tautology?

Solution:
(c) \(\boldsymbol{p} \vee \sim \boldsymbol{p}\)
Hint:
A statement is said to be a tautology if the last column of its truth table contains only T.

Question 7.
In the set of integers with operation * defined by a * b = a + b – ab, the value of 3 * (4 * 5) is ……..
(a) 25
(b) 15
(c) 10
(d) 5
Hint:
a * b = a + b – ab
3 * (4 * 5) = 3 * (4 + 5 – 4(5))
= 3 * (9 – 20)
= 3 * (-11)
= 3 + (-11) – 3(-11)
= 3 – 11 + 33
= -8 + 33 = 25
Solution:
(a) 25

Question 8.
In the multiplicative group of cube root of unity, the order of \(\omega^{2}\) is ……
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(b) 3
Hint:

Question 10.

(a) 5
(b) 5\(\sqrt{2}\)
(c) 25
(d) 50
Hint:

Solution:
(b) 5\(\sqrt{2}\)

Question 11.
The order of -i in the multiplicative group of 4^th roots of unity is ……..
(a) 4
(b) 3
(c) 2
(d) 1
Hint:
The roots of fourth roots of unity are 1, -1, i, -i
The identity element is 1
(-i)⁴ = i⁴ = 1
Order of (-i) = 4.
Solution:
(a) 4

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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
(ii) 3 – i√3
(iii) -2 – i2
(iv) \(\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}\)
Solution:

Question 2.
Find the rectangular form of the following complex numbers.

Solution:

Question 3.

Solution:
(i) (x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …….. (x_n + iy_n) = a + ib …… (1)
Taking modulus on both sides,
|(x₁ + iy₁) (x₂ + iy₂) (x₃ + iy₃) …….. (x_n + iy_n)| = |a + ib|
|x₁ + iy₁| |x₂ + iy₂| |x₃ + iy₃| ….. |x_n + iy_n| = |a + ib|

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

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)
(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 \(\left(\frac{z-i}{z+2}\right)=\frac{\pi}{4}\), then show that x² + y² + 3x – 3y + 2 = 0.
Solution:
arg \(\left(\frac{z-i}{z+2}\right)=\frac{\pi}{4}\)
We have arg (\(\frac{z_{1}}{z_{2}}\)) = arg(z₁) – arg(z₂)
arg (z – i) – arg (z + 2) = \(\frac{\pi}{4}\)
Let z = x + iy
arg (x + iy – i) – arg (x + iy + 2) = \(\frac{\pi}{4}\)
arg(x + i(y – 1)) – arg(x + 2 + iy) = \(\frac{\pi}{4}\)

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

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

Question 1.
Write the following complex numbers in the polar form:

Solution:

Question 2.
Find the modulus and principal argument of (1 + i) and hence express it in the polar form.
Solution:

Question 3.
Express the following complex numbers in the polar form.

Solution:

Question 4.
Express the following complex numbers in the polar form: \(2+2 \sqrt{3} i\)
Solution:

Question 5.
Express the following complex numbers in the polar form: \(-1+i \sqrt{3}\)
Solution:

Question 6.
Express the following complex numbers in the polar form: -1 – i
Solution:

Question 7.
Express the following complex numbers in the polar form: 1 – i
Solution:

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.
If z = 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:
\(\left|\frac{z-4 i}{z+4 i}\right|=1\)
⇒ |z – 4i| = |z + 4i|
let z = x + iy
⇒ |x + iy – 4i| = |x + iy + 4i|
⇒ |x + i(y – 4)| = |x +(y + 4)|
⇒ \(\sqrt{x^{2}+(y-4)^{2}}=\sqrt{x^{2}+(y+4)^{2}}\)
Squaring on both sides, we get
x² + y² – 8y + 16 = x² + y² + 16 + 8y
⇒ -16y = 0
⇒ y = 0 in two equation of 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

2x² + 2y² + x – 2y = 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
(ii) Im[(1 – i)z + 1] = 0
(iii) |z + i| = |z – 1|
(iv) \(\bar{z}=z^{-1}\)
Solution:
(i) 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) \(\bar{z}=z^{-1}\)
⇒ \(\bar{z}=\frac{1}{z}\)
⇒ \(z \bar{z}=1\)
⇒ |z|² = 1
⇒ |x + iy|² = 1
⇒ x² + y² = 1

Question 4.
Show that the following equations represent a circle, and, find its centre and radius.
(i) |z – 2 – i| = 3
(ii) |2z + 2 – 4i| = 2
(iii) |3z – 6 + 12i| = 8
Solution:
(i) 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
(ii) |z – 4|² – |z – 1|² = 16
Solution:
(i) z = x + iy
|z – 4| = 16
⇒ |x + iy – 4| = 16
⇒ |(x – 4) + iy| = 16
⇒ \(\sqrt{(x-4)^{2}+y^{2}}=16\)
Squaring on both sides
(x – 4)² + y² = 256
⇒ x² – 8x + 16 + y² – 256 = 0
⇒ x² + y² – 8x – 240 = 0 represents the equation of circle

(ii) |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

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

Question 1.
If the imaginary part of is -2, then show that the locus of the point representing z in the argand plane is a straight line.
Solution:
Let z = x + iy. Then,


Hence, the locus of z is a straight line

Question 2.
If the real part of is 4, then show that locus of the point representing z in the complex plane is a circle.
Solution:

It is given that the real part of is 4.

Question 3.

Solution:

Question 4.
If arg (z – 1) = \(\frac{\pi}{6}\) and arg (z + 1) = 2 \(\frac{\pi}{3}\) , then prove that |z| = 1.
Solution:

Question 5.
P represents the variable complex number z. Find the locus of P, if
Solution:

Question 6.
P represents the variable complex number z. Find the locus of P, if
Solution:
Let z = x + iy

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.2

Question 1.
Let p : Jupiter is a planet and q : India is an island be any two simple statements. Give verbal sentence describing each of the following statements.

Solution:
(i) \(\neg p\) : Jupiter is not a planet
(ii) \(p \wedge \neg q\) : Jupiter is not a planet and India is not an island
(iii) \(\neg p \vee q\) : Jupiter is not a planet or India is an island.
(iv) \(p \rightarrow \neg q\) : If Jupiter is a planet then India is not an island
(v) \(p \leftrightarrow q\) : If Jupiter is a planet if and only if India is an island

Question 2.
Write each of the following sentences in symbolic form using statement variables p and q.
(i) 19 is not a prime number and all the angles of a triangle are equal.
(ii) 19 is a prime number or all the angles of a triangle are not equal
(iii) 19 is a prime number and all the angles of a triangle are equal
(iv) 19 is not a prime number
Solution:
p : 19 is a prime number
q : All the angles of a triangle are equal

Question 3.
Determine the truth value of each of the following statements
(i) If 6 + 2 = 5 , then the milk is white.
(ii) China is in Europe or \(\sqrt{3}\) is an integer
(iii) It is not true that 5 + 5 = 9 or Earth is a planet
(iv) 11 is a prime number and all the sides of a rectangle are equal
Solution:

Question 4.
Which one of the following sentences is a proposition?
(i) 4 + 7 = 12
(ii) What are you doing?
(iii) 3ⁿ ≤ 81, n ∈ N
(iv) Peacock is our national bird
(v) How tall this mountain is!
Solution:
(i) is a proposition
(ii) not a proposition
(iii) is a proposition
(iv) is a proposition
(v) not a proposition

Question 5.
Write the converse, inverse, and contrapositive of each of the following implication.
(i) If x and y are numbers such that x = y, then x² = y²
(ii) If a quadrilateral is a square then it is a rectangle
Solution:
(i) Converse: If x and y are numbers such that x² = y² then x = y.
Inverse: If x and y are numbers such that x ≠ y then x² ≠ y².
Contrapositive : If x and v are numbers such that x² ≠ y² then x ≠ y.

(ii) Converse: If a quadrilateral is a rectangle then it is a square.
Inverse: If a quadrilateral is not a square then it is not a rectangle.
Contrapositive : If a quadrilateral is not a rectangle then it is not a square.

Question 6.
Construct the truth table for the following statements.

Solution:

Question 7.
Verify whether the following compound propositions are tautologies or contradictions or contingency

Solution:

In the above Truth table the last column entries are ‘F’. So the given propositions is a contradiction.

In the above truth table the last column entries are ‘T’. So the given propositions is a tautology.

In the above truth table the entries in the last column are a combination of’ T ‘ and ‘ F ‘. So the given statement is neither propositions is neither tautology nor a contradiction. It is a contingency.

The last column entires are ‘T’. So the given proposition is a tautology.

Question 8.

Solution:

Question 9.

Solution:

The entries in the column corresponding to q ➝ p and \(\neg p \rightarrow \neg q\) are identical and hence they are equivalent.

Question 10.
Show that p ➝ q and q ➝ p are not equivalent
Solution:

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

Question 11.

Solution:

Question 12.
Check whether the statement p ➝ (q ➝ p) is a tautology or a contradiction without using the truth table.
Solution:

Question 13.
Using truth table check whether the statements are logically equivalent.
Solution:

Question 14.
without using truth table
Solution:

Question 15.

Solution:

The entries in the column corresponding to are identical.
Hence they are equivalent.

Samacheer Kalvi 12th Maths Solutions Chapter 12 Discrete Mathematics Ex 12.2 Additional Problems

Question 1.
Show that is a tautology.
Solution:

Question 2.
Show that is a contradiction.
Solution:

Question 3.
Use the truth table to determine whether the statement is a tautology.
Solution:

The last column contains only T. ∴ The given statement is a tautology.

Question 4.
Show that
Solution:
(i) Truth table for p \(\leftrightarrow\) q

Question 5.
Show that .
Solution:


The last columns of statements (i) and (ii) are identical.