Prasanna

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 2.9

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

Question 1.
iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ is ______
(a) 0
(b) 1
(c) -1
(d) i
Answer:
(a) 0

Question 2.
The value of \(\sum_{i=1}^{13}\left(i^{n}+i^{n-1}\right)\) is ______
(a) 1 + i
(b) i
(c) 1
(d) 0
Answer:
(a) 1 + i

Question 3.
The area of the triangle formed by the complex numbers z, iz, and z + iz in the Argand’s diagrams is ______
(a) \(\frac { 1 }{ 2 }\) |z|
(b) |z|²
(c) \(\frac { 3 }{ 2 }\)|z|²
(d) 2|z|²
Answer:
(a) \(\frac { 1 }{ 2 }\) |z|
Hint: Area of triangle = \(\frac { 1 }{ 2 }\) bh
= \(\frac { 1 }{ 2 }\) |z| |iz|
= \(\frac { 1 }{ 2 }\) |z|²

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) \(\frac{1}{i-2}\)
Answer:
(b) \(\frac{-1}{i+2}\)
Hint:

Question 5.
If \(z=\frac{(\sqrt{3}+i)^{3}(3 i+4)^{2}}{(8+6 i)^{2}}\) then |z| is equal to ______
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(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) 1
(c) 2
(d) 3
Answer:
(a) \(\frac { 1 }{ 2 }\)
Hint:

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
Answer:
(d) √5 + 2
Hint:

Question 8.
If \(\left|z-\frac{3}{z}\right|\), then the least value of |z| is ______
(a) 1
(b) 2
(c) 3
(d) 5
Answer:
(a) 1
Hint:

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
Answer:
(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
Answer:
(a) \(\frac { 3 }{ 2 }\) – 2i

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

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
Answer:
(b) 1
Hint: We have

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
Answer:
(d) 0
Hint:

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

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
Answer:
(b) imaginary axis
Hint:
|z + 2| = | z – 2|
⇒ |x + iy + 2| = |x + iy – 2|
⇒ |x + 2 + iy|² = |x – 2 + iy|²
⇒ (x + 2)² + y² = (x – 2)² + y²
⇒ x² + 4 + 4x = x² + 4 – 4x
⇒ 8x = 0
⇒ x = 0

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

The complex number lies in III quadrant. \(\theta=-(\pi-\alpha)=-\left(\pi-\frac{\pi}{4}\right)=\frac{-3 \pi}{4}\)

Question 17.
The principal argument of (sin 40° + i cos 40°)⁵ is ______
(a) -110°
(b) -70°
(c) 70°
(d) 110°
Answer:
(a) -110°
Hint:
(sin 40° + i cos 40°)⁵
= (cos 50° + i sin 50°)⁵
= (cos 250° + i sin 250°)
250° lies in III quadrant.
To find the principal argument the rotation must be in a clockwise direction which coincides with 250°
θ = -110°

Question 18.
If (1 + i) (1 + 2i) (1 + 3i)…(1 + ni) = x + iy, then 2.5.10……(1 + n² ) is _____
(a) 1
(b) i
(c) x² + y²
(d) 1 + n²
Answer:
(c) x² + y²

Question 19.
If ω ≠ 1 is a cubic root of unity and (1 + ω)⁷ = A + B ω, then (A, B) equal to ______
(a) (1, 0)
(b) (-1, 1)
(c) (0, 1)
(d) (1, 1)
Answer:
(d) (1, 1)
Hint:

Question 20.
The principal argument of the complex number \(\frac{(1+i \sqrt{3})^{2}}{4 i(1-i \sqrt{3})}\) is _____
(a) \(\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{6}\)
(c) \(\frac{5 \pi}{6}\)
(d) \(\frac{\pi}{2}\)
Answer:
(d) \(\frac{\pi}{2}\)
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
Answer:
(b) -1
Hint:
x² + x + 1 = 0
α and β are the roots of the equation.
There are the two roots of cube roots of unity except 1.

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

Question 23.
If ω ≠ 1 is a cubic root of unity and \(\left|\begin{array}{cccc}{1} & {1} & {1} \\ {1} & {-\omega^{2}-1} & {\omega^{2}} \\ {1} & {\omega^{2}} & {\omega^{7}}\end{array}\right|\) = 3k, then k is equal to ______
(a) 1
(b) -1
(c) i√3
(d) -i√3
Answer:
(d) -i√3
Hint:

Question 24.
The value of \(\left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10}\) is ______
(a) cis \(\frac{2 \pi}{3}\)
(b) cis \(\frac{4 \pi}{3}\)
(c) -cis \(\frac{2 \pi}{3}\)
(d) -cis \(\frac{4 \pi}{3}\)
Answer:
(a) cis \(\frac{2 \pi}{3}\)
Hint:

Question 25.
If ω = cis \(\frac{2 \pi}{3}\), then the number of distinct roots of \(\left|\begin{array}{ccc}{z+1} & {\omega} & {\omega^{2}} \\ {\omega} & {z+\omega^{2}} & {1} \\ {\omega^{2}} & {1} & {z+\omega}\end{array}\right|=0\) are _____
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1
Hint:

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

Question 1.

(a) (2, -1)
(b) (1, 0)
(c) (0, 1)
(d) (-1, 2)
Solution:
(b) (1, 0)

Question 2.

(a) (0, 2)
(b) (-2, 0)
(c) (0, -2)
(d) (-2, 0)
Solution:
(c) (0, -2)

Question 3.
The point represented by the complex number 2 – i is rotated about origin through an angle
of \(\frac{\pi}{2}\). in the clockwise direction, the new position of point is ……
(a) 1 + 2 i
(b) -1 – 2 i
(c) 2 + i
(d) -1 + 2 i
Solution:
(b) -1 – 2i
Hint.
Let z = 2 – i. Let the new position of point when the point represented by the complex number z = 2 – i is rotated about origin through an angle of \(\frac{\pi}{2}\) in the clockwise direction be denoted by z₁.

Question 4.
Fill in the blanks of the following:
For any two complex numbers z₁, z₂ and any real number a, b,
Solution:

Question 5.
Multiplicative inverse of 1 + i is …..
Solution:

Question 6.

Solution:

Question 7.
If |z| = 4 and arg (z) = \(\frac{5 \pi}{6}\), then z = ……
Solution:

Question 8.
State true or false for the following:
(i) The order relation is defined on the set of complex numbers.
Solution:
The given statement is false because the order relation “greater than” and “less than” are not defined on the set of complex numbers.

(ii) For any complex number 2, the minimum value of |z| + |z – 1| is 1.
Solution:
Let z = x + iy be any complex number.

When x = 0, y = 0, the value of |z| + |z – 1| is minimum and the minimum value.

Hence, the given statement is true.

(iii) 2 is not a complex number.
Solution:
It is a false statement because 2 = 2 + i(0), which is of the form α + ib, and the number of the form a + ib, where a and b are real numbers is called a complex number.

(iv) The locus represented by |z – 1| = |z – i| is a line perpendicular to the join of (1, 0) and (0, 1).
Solution:
Equation of a straight line passing through (1, 0) and (0, 1) is

Now, the lines y = -x + 1 and y = x are perpendicular to each other, so the given statement is true.

Question 9.

Solution:
1

Question 10.
If – \(\bar{z}\) lies in the third quadrant, then z lies in the ……..
(a) first quadrant
(b) second quadrant
(c) third quadrant
(d) fourth quadrant
Solution:
(d) fourth quadrant
Hint:

Question 11.

(a) cos 2(α – β + γ) + i sin 2(α – β + γ)
(b) – 2cos(α – β + γ)
(c) – 2 i sin(α – β + γ)
(d) 2 cos(α – β + γ)
Solution:
(c) -2 i sin (α – β + γ)
Hint: α β γ

Question 12.


Solution:

Hint:

Question 13.

(a) 1
(b) -1
(c) 0
(d)-i
Solution:
(c) 0
Hint:

Question 14.
If -i + 2 is one root of the equation ax² – bx + c = 0, then the other root is
(a) – i – 2
(b) i – 2
(c) 2 + i
(d) 2i + i
Solution:
(c) 2 + i
Hint:
Complex roots occur in pairs when -i + 2 is one root, i.e., when 2 – i is a root, the other root is 2 + i

Question 15.
The equation having 4 – 3i and 4 + 3i as roots is …….
(a) x² + 8x + 25 = 0
(b) x² + 8x – 25 = 0
(c) x² – 8x + 25 = 0
(d) x² – 8x – 25 = 0
Solution:
(c) x² – 8x + 25 = 0
Hint.

Question 16.

(a) (1, 1)
(b) (1, -1)
(c) (0, 1)
(d) (1, 0)
Solution:
(d) (1, 0)
Hint:

Question 17.
If – i + 3 is a root of x² – 6x + k = 0, then the value of k is

Solution:
(d) 10
Hint:
α = -i + 3 = 3 – 7 ⇒ The other root is β = 3 + i
Product of the roots = αβ = (3 – i)(3 + i) = 10
Product of the roots of x² – 6x + k = 0 is k ⇒ k = 10

Question 18.
If ω is a cube root of unity, then the value of (1 – ω + ω²)⁴ + (1 + ω – ω²)⁴ is ……
(a) 0
(b) 32
(c) -16
(d) -32
Solution:
(c) -16
Hint.

Question 19.
If ω is a cube root of unity, then the value of (1 – ω) (1 – ω²) (1 – ω⁴)(1 – ω⁸) is …….
(a) 9
(b) -9
(c) 16
(d) 32
Solution:
(a) 9
Hint.

Question 20.

(a) | z | = 1
(b) | z | > 1
(c) | z | < 1
(d) None of these.
Solution:
(a) | z | = 1

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 3 Theory of Equations Ex 3.4

Question 1.
Solve:
(i) (x – 5) (x – 7) (x + 6) (x + 4) = 504
(ii) (x – 4) (x – 7) (x – 2) (x + 1) = 16
Solution:
(i) (x – 5) (x + 4) (x – 7) (x + 6) = 504
(x² – x – 20) (x² – x – 42) = 504
Let y = (x² – x)
(y – 20) (y – 42) = 504
⇒ y² – 42y – 20y + 840 = 504
⇒ y² – 62y + 336 = 0
⇒ (y – 56) (y – 6) = 0
⇒ (y – 56) = 0 or (y – 6) = 0
⇒ x² – x – 56 = 0 or x² – x – 6 = 0
⇒ (x – 8) (x + 7) = 0 or (x – 3) (x + 2) = 0
⇒ x = 8, -7 or x = 3, -2
The roots are 8, -7, 3, -2

(ii) (x – 4) (x – 7) (x – 2) (x + 1) = 16
⇒ (x – 4) (x – 2) (x – 7) (x + 1) = 16
⇒ (x² – 6x + 8) (x² – 6x – 7) = 16
Let x² – 6x = y
(y + 8)(y – 7) = 16
⇒ y² – 7y + 8y – 56 – 16 = 0
⇒ y² + y – 72 = 0
⇒ (y + 9) (y – 8) = 0
y + 9 = 0
x² – 6x + 9 = 0
(x – 3)² = 0
x = 3, 3
or
y – 8 = 0
x² – 6x – 8 = 0

The roots are 3, 3, 3 ±√17

Question 2.
Solve : (2x – 1) (x + 3) (x – 2) (2x + 3) + 20 = 0.
Solution:
(2x – 1) (2x + 3) (x + 3) (x – 2) + 20 = 0
⇒ (4x² + 6x – 2x – 3) (x² – 2x + 3x – 6) + 20 = 0
⇒ (4x² + 4x – 3) (x² + x – 6) + 20 = 0
⇒ [4(x² + x) – 3] [x² + x – 6] + 20 = 0
Let y = x² + x
⇒ (4y – 3) (y – 6) + 20 = 0
⇒ 4y² – 24y – 3y + 18 + 20 = 0
⇒ 4y² – 27y + 38 = 0
⇒ (4y – 19) (y – 2) = 0
(4y – 19) = 0
4(x² + x) – 19 = 0
4x² + 4x – 19 = 0

or
(y – 2) = 0
x² + x – 2 = 0
(x + 2) (x – 1) = 0
x = -2, +1
The roots are -2, 1, \(\frac{-1 \pm 2 \sqrt{5}}{2}\)

Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.4 Additional Questions

Questions 1.
Solve (x – 3) (x – 6) (x – 1) (x + 2) + 54 = 0.
Solution:
(x – 3) (x – 1) (x – 6) (x + 2) + 54 = 0
(x² – 4x + 3) (x² – 4x – 12) + 54 = 0
Put x² – 4x = y
(y + 3)(y – 12) + 54 = 0
y² – 9y – 36 + 54 = 0
y² – 9y + 18 = 0
(y – 3)(y – 6) = 0

Question 2.
Solve the equation (x – 4) (x – 2) (x – 1) (x + 1) + 8 = 0
Solution:
The equation can be rewritten as {(x – 4) (x + 1)} {(x – 2) (x – 1)} + 8 = 0
(x² – 3x – 4)(x² – 3x + 2) + 8 = 0

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 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:
The given equation is 9x³ – 36x² + 44x – 16 = 0
\(x^{3}-4 x^{2}+\frac{44}{9} x-\frac{16}{9}=0\)
Let the roots be α – d, α, α + d
As they are in A.P

Question 3.
Solve the equation 3x³ – 26x² + 52x – 24 = 0 if its roots form a geometric progression.
Solution:
The given equation is 3x³– 26x² + 52x – 24 = 0
\(x^{3}-\frac{26}{3} x^{2}+\frac{52}{3} x-8=0\)
Given that the root are GP

Question 4.
Determine ft and solve the equation 2x³ – 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots.
Solution:
The given equation is 2x³ – 6x² + 3x + k = 0

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:
The given equation is x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135 = 0
The given roots are 1 + 2i, √3
The other roots are 1 – 2i, -√3
The factors are
= {x² – x(2) + (1 + 4)}{(x + √3)(x – √3)}
= (x² – 2x + 5) (x² – 3)
= x⁴ – 3x² – 2x³ + 6x + 5x² – 15
= x⁴ – 2x³ + 2x² + 6x – 15
To find this roots,

Question 6.
Solve the cubic equations:
(i) 2x³ – 9x² + 10x = 3
(ii) 8x³ – 2x² – 7x + 3 = 0
Solution:
(i) Given equation is 2x³ – 9x² + 10x = 3
Sum of the co-efficients = 0
(x – 1) is a factor.
The other factor is 2x² – 7x + 3
2x² – 7x + 3 = 0
(x – 3)(2x – 1) = 0
The roots are 1, 3, \(\frac { 1 }{ 2 }\)

(ii) Given equation is 8x³ – 2x² – 7x + 3 = 0
Sum of odd co-efficients = Sum of even co-efficients
(x + 1) is a factor.
The other factor is 8x² – 10x + 3
8x² – 10x + 3 = 0
(4x – 3) (2x – 1) = 0
The roots are \(\frac{3}{4}, \frac{1}{2},-1\)

Question 7.
Solve the equation: x⁴ – 14x² + 45 = 0.
Solution:
The given equation is x⁴ – 14x² + 45 = 0
Let x² = y
y² – 14y + 45 = 0
(y – 9)(y – 5) = 0
y = 9, 5
x² = 9, x² = 5
x = ± 3, x = ± √5
The roots are ± 3, ± √5

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

Question 1.
If one root of x³ + 2x² + 3x + k = 0 is sum of the other two roots then find the value of k.
Solution:
Let α, β, γ be the roots of given equation.
But, α = β + γ
α + β + γ = -2 … (1) ⇒ 2α = -2 ⇒ α = -1
αβ + βγ + γα = 3 … (2) This gives β + γ = – 1
αβγ = -k …(3)

Question 2.
If sum of the roots of the equation x³ – 3x² – 16x + k = 0 is zero then find the value of k.
Solution:
Let α, β, γ be the roots of given equation.
But, α + β = 0

Question 3.
Find all zeros of the polynomial x³ – 5x² + 9x – 5 = 0, If 2 + i is a root.
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 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)

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 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:

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 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:

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 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

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 2.8

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)

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 12.3

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