TS Inter 2nd Year

Students must practice this TS Intermediate Maths 2A Solutions Chapter 1 Functions Ex 1(d) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Exercise 1(d)

I.
Question 1.
i) Find the equation of the perpendicular bisector of the line segment joining the points 7 + 7i, 7 – 7i.
ii) Find the equation of the straight line joining the points – 9 + 6i, 11 – 4i in the Argand plane.
Solution:
i) z₁ = 7 + 7i
z₂ = 7 – 71
A (7, 7) B (7, – 7)

M(7, 0)

Slope of AB = \(\frac{7+7}{7-7}\) → ∞
Line ⊥ to AB slope is zero,
y = 0 is line.

ii) A (- 9, 6) B (11, – 4)
Slope of AB = \(\frac{6+4}{-9-11}\)
= \(\frac{10}{-20}=\frac{-1}{2}\)
Equation of line AB,
y – 6 = \(\frac{- 1}{2}\) (x + 9)
2y – 12 = – x – 9
x + 2y = 3.

Question 2.
If z = x + ily and if the point Pin the Argand plane represents z, then describe geometrically the locus of z satisfying the equations
i) |z – 2 – 3i| = 5
ii) 2|z – 2| = |z – 1|
iii) im z² = 4
iv) Arg \(\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)
Solution:
i) |z – 2 – 3i| = 5
|(x – 2) + (y – 3)i| = 5
(x – 2)² + (y – 3)² = 25
x² + y² – 4x – 6y + 4 + 9 – 25 = 0
x² + y² – 4x – 6y – 12 = 0

ii) 2|z – 2| = |z – 1|
4(z – 2) (- 2) = (z – 1) (\(\overline{\mathbf{z}}\) – 1)
4z\(\overline{\mathbf{z}}\) – 8z – 8\(\overline{\mathbf{z}}\) + 16 = z\(\overline{\mathbf{z}}\) – z – \(\overline{\mathbf{z}}\) + 1
3z\(\overline{\mathbf{z}}\) – 7z – 7\(\overline{\mathbf{z}}\) + 15 = 0
3(x² + y²) – 7(2x) + 15 = 0.

iii) Im (z²) = 4
Im (z²) = 4
z = x + iy
z² = (x + iy)²
z² = x² – y² + 2xyi
Im(z²) = 2xy
2xy = 4
xy = 2 rectangualr hyperbola

iv) Arg \(\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)

Question 3.
Show that the points in the Argand diagram represented by the complex numbers 2 + 2i, – 2 – 2i, – 2√3 + 2√3i are the vertices of an equilateral triangle..
Solution:
A(2, 2), B(- 2, – 2), C (- 2√3 + 2√3)
AB = \(\sqrt{(2+2)^2+(2+2)^2}\) = 4√2
BC = \(\sqrt{(-2+2 \sqrt{3})^2+(-2-2 \sqrt{3})^2}\)
BC = \(\sqrt{4+12-8 \sqrt{3}+4+12+8 \sqrt{3}}\)= 4√2
AC = \(\sqrt{\left(2+2 \sqrt{3}^2\right)+(2-2 \sqrt{3})^2}\) = 4√2
AB = AC = BC
∆ ABC is equilateral.

Question 4.
Find the eccentricity of the ellipse whose equtaion is | z – 4 | + |z – \(\frac{12}{5}\)| = 10
Solution:
SP + S’P = 2a
S (4, 0) S’(\(\frac{12}{5}\), 0)
2a = 10
a = 5
SS’ = 2ae
4 – \(\frac{12}{5}\) = 2 × 5e
\(\frac{8}{5}\) = 10e
e = \(\frac{4}{25}\).

II.
Question 1.
If \(\frac{z_3-z_1}{z_2-z_1}\) is a real number, show that the points represented by the complex numbers z₁, z₂, z₃ are collinear.
Solution:

Arg \(\left(\frac{z_1-z_3}{z_1-z_2}\right)\) = 0 then \(\frac{z_1-z_3}{z_1-z_2}\) is real θ = 0.
∴ z₁, z₂, z₃ are collinear.

Question 2.
Show that the four points in the Argand plane represented by the complex numbers 2 + i, 4 + 3i, 2 + 5i, 3i are vertices of a square.
Solution:
A (2, 1), B (4, 3), C (2, 5), D (0, 3)
AB = \(\sqrt{(4-2)^2+(3-1)^2}\) = 2√2
BC = \(\sqrt{(4-2)^2+(3-5)^2}\) = 2√2
CD = \(\sqrt{(2-0)^2+(5-3)^2}\) = 2√2
AD = \(\sqrt{(2-0)^2+(1-3)^2}\) = 2√2
Slope of AB = \(\frac{5-3}{2-4}\)= 1
Slope of BC = \(\frac{3-1}{4-2}\) = 1
AB ⊥ BC
BC ⊥ CD
⇒ ABCD is a square.

Question 3.
Show that die points in the Argand plane represented by the complex numbers – 2 + 7i, – \(\frac{-3}{2}\) + \(\frac{1}{2}\)i, 4 – 3i, \(\frac{7}{2}\) (1 + i) are the vertices of a rhombus.
Solution:

AC ⊥ BD
∴ ABCD is rhombus.

Question 4.
Show that the points in the Argand diagram represented by the complex numbers z₁, z₂, z₃ are collhitear if and only if there exists three real numbers p, q, r not all zero satisfying p + qz₂ + rz₃ = 0 and p + q + r = 0.
Solution:
pz₁ + qz₂ + rz₃ = 0
pz₁ + qz₂ = – rz₃
\(\left(\frac{p z_1+q z_2}{p+q}\right)\) (p + q) = – rz₃
Now p + q = – r
\(\left(\frac{p z_1+q z_2}{p+q}\right)\) = z₃
⇒ z₃ divides z₁ and z₂ is q : p ratio.
∴ z₁, z₂, z₃ are collinear.

Question 5.
The points P, Q denote the complex numbers z₁, z₂ in the Argand diagram. O is
origin. If z₁\(\overline{\mathbf{z}}_2\) + \(\overline{\mathbf{z}}_1\)z₂ = 0 tlien show that ∠POQ = 90°.
Solution:
z₁\(\overline{\mathbf{z}}_2\) + \(\overline{\mathbf{z}}_1\)z₂ = 0
\(\frac{\mathbf{z}_1 \overline{\mathbf{z}}_2+\overline{\mathbf{z}}_1 \mathbf{z}_2}{\mathbf{z}_2 \overline{\mathrm{z}}_2}\) = 0
⇒ Real of \(\frac{\mathrm{z}_1}{\mathrm{z}_2}\) = 0
\(\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}+\frac{\overline{\mathrm{z}}_1}{\mathrm{z}_2}\right)\)
Imaginary part of (\(\frac{\mathrm{z}_1}{\mathrm{z}_2}\)) is k.
\(\left(\frac{\mathrm{z}_1}{\mathrm{z}_2}\right)+\left(\frac{\overline{\mathrm{z}}_1}{\mathrm{z}_2}\right)\) = 0 or
\(\frac{\mathrm{z}_1}{\mathrm{z}_2}\) is purely imaginary.
\(\frac{\mathrm{z}_1}{\mathrm{z}_2}\) = ki
⇒ Arg\(\frac{\mathrm{z}_1}{\mathrm{z}_2}\) = \(\frac{\pi}{2}\).

Question 6.
The complex number z has argument θ 0 < θ < \(\frac{\pi}{2}\) and satisfy the equation |z – 3i| = 3. Then prove that (cot θ – \(\frac{6}{z}\)) = 1.
Solution:
(x²) + (y – 3)² = 9
x + y = 0
x + y = 6y

Students must practice this TS Intermediate Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(d)

I.
Question 1.
Find the algebraic equation whose roots are 3 times the roots of x³ + 2x² – 4x + 1 = 0.
Solution:
Given equation is x³ – 2x² – 4x + 1 = 0 ……………(1)
Let f(x) = x³ + 2x² – 4x + 1
The equation whose roots are 3 times the roots of f(x) = 0 is given by f(\(\frac{x}{3}\)) = 0.
i.e., \(\left(\frac{x}{3}\right)^3+2\left(\frac{x}{3}\right)^2-4\left(\frac{x}{3}\right)\) + 1 = 0
⇒ \(\frac{x^3}{27}+\frac{2 x^2}{9}-\frac{4 x}{3}\) + 1 = 0
⇒ x³ + 6x² – 36x + 27 = 0.

Question 2.
Find the algebraic equation whose roots are 2 times the roots of x5 – 2×4 + 3×3 – 2×2 + 4x + 3 = 0.
Solution:
Given equation is x⁵ – 2x⁴ + 3x³ – 2x² + 4x + 3 = 0 ……..(1)
Let f(x) = x⁵ – 2x⁴ + 3x³ – 2x² + 4x + 3
The equation whose roots are 2 times the roots of f(x) = 0 is given by f(\(\frac{x}{2}\)) = 0
i.e., \(\left(\frac{x}{2}\right)^5-2\left(\frac{x}{2}\right)^4+3\left(\frac{x}{2}\right)^3-2\left(\frac{x}{2}\right)^2+4\left(\frac{x}{2}\right)+3\) = 0
⇒ x⁵ – 4x⁴ + 12x³ – 16x² + 64x + 96 = 0.

Question 3.
Find the transformed equation whose roots are the negatives of the roots of x4 + 5×3 + lix + = 0.
Solution:
Given equation is x⁴ + 5x³ + 11x + 3 = 0
Let f(x) = x⁴ + 5x³ + 11x + 3
The transformed equation whose roots are the negatives of the roots of f(x) = 0 is
f(- x) = 0.
i.e., (- x)⁴ + 5(- x)³ + 11 (- x) + 3 = 0
⇒ x⁴ – 5x³ – 11x + 3 = 0.

Question 4.
Find the transformed equation whose roots are the negatives of the root of x⁷ + 3x⁵ + x³ – x² + 7x + 2 = 0.
Solution:
Given equation is
x⁷ + 3x⁵ + x³ – x² + 7x + 2 = 0
Let f(x) = x⁷ + 3x⁵ + x³ – x² + 7x + 2
The transformed equation whose roots are the negatives of the roots of f(x) = 0 is
f(- x) = 0.
i.e., (- x)² + 3(- x)⁵ + (- x)³ – (- x)² + 7(- x) + 2 = 0
x⁷ + 3x⁵ + x³ + x² + 7x – 2 = 0.

Question 5.
Find the polynomial equation whose roots are the reciprocals of the roots of x⁴ – 3x³ + 7x² + 5x – 2 = 0.
Solution:
Given equation is x⁴ – 3x³ + 7x² + 5x – 2 = 0 ………….(1)
Let f(x) = x⁴ – 3x³ + 7x² + 5x – 2
The polynomial equation whose roots are the reciprocals of the roots of (1) is given by
f(\(\frac{1}{x}\)) = 0
i.e., \(\left(\frac{1}{x}\right)^4-3\left(\frac{1}{x}\right)^3+7\left(\frac{1}{x}\right)^2+5\left(\frac{1}{x}\right)\) – 2 = 0
⇒ 1 – 3x + 7x² + 5x³ – 2x⁴ = 0
⇒ 2x⁴ – 5x³ – 7x² + 3x – 1 = 0.

Question 6.
Find the polynomial equation whose roots are the reciprocals of the roots of x⁵ + 11x⁴ + x³ + 4x² – 13x + 6 = 0.
Solution:
Given equation is x⁵ + 11x⁴ + x³ + 4x² – 13x + 6 = 0 …………(1)
Let f(x) = x⁵ + 11x⁴ + x³ + 4x² – 13x + 6
The polynomial equation whose roots are the reciprocals of the roots of f(x) = 0 is
f(\(\frac{1}{x}\)) = 0
i.e., \(\left(\frac{1}{x}\right)^5+11\left(\frac{1}{x}\right)^4+\left(\frac{1}{x}\right)^3\) + 6 = 0
⇒ 6x⁵ – 13x⁴ + 4x³ + x² + 11x + 1 = 0.

II.
Question 1.
Find the polynomial equation whose roots are the squares of the roots of x⁴ + x³ + 2x² + x + 1 = 0.
Solution:
Given equation is x⁴ + x³ + 2x² + x + 1 = 0
Let f(x) = x⁴ + x³ + 2x² + x + 1
The polynomial equation whose roots are squares of the roots of f(x) = 0 is f (√x) = 0.
i.e.. (4√x)⁴ + (√x)³ + 2(√x)² + √x + 1 = 0
⇒ x² + 2x + 1 = √x (x + 1)
⇒ (x + 1)² = √x (x + 1)
⇒ (x + 1)⁴ = x (x + 1)²
⇒ x⁴ + 4x³ + 6x² + 4x + 1 = x(x² +2x + 1)
⇒ x⁴ + 3x³ + 4x² + 3x + 1 = 0 is the required equation.

Question 2.
Form the polynomial equation whose roots are the squares of the roots of x³ + 3x² – 7x + 6 = 0.
Solution:
Given equation is x³ + 3x² – 7x + 6 = 0 …… (1)
Let f(x) = x³ + 3x² – 7x + 6
The polynomial equation whose roots are the squares of the roots of f(x) = 0 is f(√x) = 0.
i.e., (√x)³ + 3(√x)² – 7(√x) + 6 = 0
⇒ x√x – 7√x = – 3x – 6
⇒ √x (x – 7) = – (3x + 6)
⇒ x (x² + 49 – 14x) = 9x² + 36 + 36x
⇒ x³ – 23x² + 13x – 36 = 0 is the required equation.

Question 3.
Form the polynomial equation whose roots are the cubes of the roots of x³ + 3x² + 2 = 0.
Solution:
Given equation is x³ + 3x² + 2 = 0 …………. (1)
Let f(x) = x³ + 3x² + 2
The polynomial equation whose roots are the cubes of the roots of f(x) = 0 is f(\(\sqrt[3]{x}\)) = 0.
i.e., \((\sqrt[3]{x})^3\) + 3 (\((\sqrt[3]{x})^2\)) + 2 = 0
⇒ x + 2 = – 3x^2/3
⇒ (x + 2)³ = – 27 (x²)
⇒ x³ + 6x² + 12x + 8 = – 27x²
⇒ x³ + 33x² + 12x + 8 = 0 is the required equation.

III.
Question 1.
Find the polynomial equation whose roots are the translates of those of the equation – 5x³ + 7x² – 17x + 11 = 0 by – 2.
Solution:
Given equation is
x⁴ – 5x³ + 7x² – 17x + 11 = 0 ………(1)
Let f(x) = x⁴ – 5x³ + 7x² – 17x + 11
The polynomial equation, whose roots are
the translates of those of the f(x) = 0 by – 2 is f(x + 2) = 0.
Suppose that
f(x + 2) = A₀x⁴ + A₁x³ + A₂x²+ A₃x + A₄
By synthetic division, the coefficients A₀, A₁, A₂, A₃ and A₄ are obtained as follows.

∴ The roots ol the equation x⁴ – 3x³ + x² – 17x + 19 = 0 are the translates of the roots of the given equation by – 2.

Question 2.
Find the polynomial equation whose roots are the translates of those of the equation x₅ – 4x₄ + 3x₂ – 4x + 6 = 0 by – 3.
Solution:
Given equation is x₅ – 4x₄ + 3x₂ – 4x 4 6 = 0 ………….(1)
Let f(x) = x₅ – 4x₄ + 3x₂ – 4x + 6
The polynomial equation, whose roots are the translates of those of the f(x) = 0 by – 3 is f(x + 3) = 0.
Suppose that f(x + 3) = A₀x⁵ + A₁x⁴ + A₂x³ + A₃x² + A₄x + A₅
By synthetic division, the coefficients A₀, A₁, A₂, A₃, A₄ and A₅ are obtained as follows.

∴ The roots of the equation x⁵ + 11x⁴ + 42x³ + 57x² – 13x – 60 = 0 are the translates of the roots of the given equation by – 3.

Question 3.
Find the polynomial equation whose roots are the translates of the roots of the equation x⁴ – x³ – 10x² + 4x + 24 = 0 by 2.
Solution:
Given equation is x⁴ – x³ – 10x² + 4x + 24 = 0
Let f(x) = x⁴ – x³ – 10x² + 4x + 24
The polynomial equation whose roots are trans-lates of those of the f(x) = 0 by 2 is f(x – 2) = 0.
Suppose that f(x – 2) = A₀x⁴ + A₁x³ + A₂x² + A₃x + A₄
By synthetic division, the coefficients A₀, A₁, A₂, A₃, A₄ are obtained as follows – 2.

∴ The roots of the equation x⁴ – 9x³ + 20x² = 0 are the translates of the roots of the given equati on by 2.

Question 4.
Find the polynomial equation whose roots are the translates of the roots of the equation 3x⁵ – 5x³ + 7 = 0 by 4.
Solution:
Given equation is 3x⁵ – 5x³ + 7 = 0
Let f(x) = 3x⁵ – 5x³ + 7
The polynomial equation whose roots are the translates of those of the f(x) = 0 by 4 is f(x – 4) = 0.
Suppose that
f(x – 4) = A₀x⁵ + A₁x⁴ + A₂x³ + A₃x² + A₄x + A₅
By synthetic division, the coefficients A₀, A₁, A₂, A₃, A₄, A₅ are obtained as follows.

∴ The roots of the equation 3x⁵ – 60x⁴ + 475x² – 1860x³ + 3600x – 2745 = 0 are the translates of the roots of the given equation by 4.

Question 5.
Transform e jach of the following equations into ones i n which the coefficients of the second hig >hest power of x is zero and also find their transformed equations.
i) x³ – 6x² + 10x – 3 = 0
ii) x⁴ + 4x³ + 2x² – 4x – 2 = 0
iii) x³ – 6x² + 4x – 7 = 0
iv) x³ + 6x²+ 4x + 4 = 0
Solution:
i) Given equation is x³ – 6x + 10x – 3 = 0
Let f(x) = x³ – 6x² + 10x – 3
We have to find ’h’ so that the coefficient of the Second highest power of x in f(x + h) is zero.
i.e., Coefficient of x² in f(x + h) is zero,
f (x + h) = (x + h)³ – 6 (x + h)² + 10 (x + h) – 3 Coefficients of x² in f(x + h) is 3h – 6.
We choose ‘h‘ such that 3h – 6 = 0 i.e., h = 2
∴ f (x + 2) = (x + 2)³ – 6 (x + 2)² + 10 (x + 2) – 3
= x³ + 6x² + 12x + 8 – 6 (x² + x + 4) + 10x + 20 – 3
= x³ – 2x + 1
∴ x³ – 2x + 1 = 0 is the required equation.

ii) Given equation is
x⁴ + 4x³ + 2x² – 4x – 2 = 0
Let f(x) = x⁴ + 4x³ + 2x² – 4x – 2
We have to find ‘h’ so that the coefficient of the second highest power of x in f(x + h) is zero.
i.e., coefficient of x² in f(x + h) is zero,
f (x + h) = (x + h)⁴ + 4 (x + h)³ + 2 (x + h)² , – 4 (x + h) – 2
Coefficient of x² in f(x + h) is 4h + 4.
We choose ‘h‘ such that 4h + 4 = 0 i.e., h = – 1
∴ f(x – 1) = (x – 1)⁴ + 4 (x – 1)³ + 2 (x – 1)² – 4 (x – 1) – 2
= (x⁴ – 4x³ + 6x² – 4x + 1) + 4 (x³ – 3x² + 3x – 1) + 2 (x² – 2x + 1) – 4 (x – 1) – 2
= x⁴ – 4x² + 1
∴ x⁴ – 4x² + 1 = 0 is the required equation.

iii) Given equation is x³ – 6x² + 4x – 7 = 0
Let f(x) = x³ – 6x² + 4x – 7
We have to find ‘h’ so that the coefficient of the second highest power of x in f(x + h) is zero.
i.e., coefficient of x² in f(x + h) is zero.
f(x + h) = (x + h)³ – 6 (x + h)² + 4 (x + h) – 7
Coefficient of x² in f(x + h) is 3h – 6
We choose ‘h’ such that 3h – 6 = 0 i.e., h = 2
∴ f(x + 2) = (x + 2)³ – 6 (x + 2)² + 4 (x + 2) – 7
= (x³ + 6x² + 12x + 8) – 6 (x² + 4x + 4) + 4 (x + 2) – 7
= x³ – 8x – 15
∴ x3 – 8x – 15 = 0 is the required equation.

iv) Given equation is x³ + 6x² + 4x + 4 = 0
Let f(x) = x³ + 6x² + 4x + 4
We have to find ’h’ so that the coefficient of the second highest power of x in f(x + h) is zero, i.e., coefficient of x² in f(x + h) is zero.
f(x + h) = (x + h)³ + 6(x + h)² + 4(x + h) + 4
Coefficient of x² in 1(x + h) is 3h + 6.
We have to choose ‘h’ such that 3h + 6 = 0 i.e., h = – 2
∴ f(x – 2)= (x – 2)³ + 6(x – 2)² + 4(x – 2) + 4
= (x³ – 6x² + 12x – 8) + 6 (x² – 4x + 4) + 4 (x – 2) + 4
= x³ – 8x + 12
∴ x³ – 8x + 12 = is the required equation.

Question 6.
Transform each of the following equations into ones in which the coefficients of the third highest power of x is zero.
i) x⁴ + 2x<sup3 – 12x² + 2x – 1 = 0
ii) x³ + 2x² + x + 1 = 0
Solution:
i) Given equation is
x⁴ + 2x³ – 12x² + 2x – 1 = 0
Let f(x) = x⁴ + 2x³ – 12x² + 2x – 1
We have to find h’ so that the coefficient of the third highest power of ‘x’ in f(x + h) is zero.
i.e., Coefficient of x² in [(x + h) is zero.
f(x + h) = (x + h)⁴ + 2 (x + h)³ – 12 (x + h)² + 2 (x + h) – 1
Coefficient of x³ in 1(x + h) is 6h² + 6h – 12
We have to choose ‘h’ such that
6h² + 6h – 12 = 0
⇒ (h + 2) (h – 1) = 0
⇒ h = – 2 or 1.

Case – (I):
When h = – 2
f(x – 2) = (x – 2)⁴ + 2 (x – 2)³ – 12(x – 2)² + 2 (x – 2) – 1
= x⁴ – 8x³ + 24x² – 32x + 16 + 2(x³ – 6x² + 12x – 8) – 12(x² – 4x + 4) + 2(x – 2) – 1
= x⁴ – 6x³ + 42x – 53
∴ Tranformed equation is x⁴ – 6x³ + 42x – 53 = 0.

Case-(ii):
When h = 1
f(x + 1) = (x + 1)⁴ + 2(x + 1)³ – 12(x + 1)² + 2 (x + 1) – 1
= (x⁴ + 4x³ + 6x² + 4x + 1) + 2(x³ + 3x² + 3x + 1) – 12 (x² + 2x + 1) + 2(x + 1) – 1
= x⁴ + 6x³ – 12x – 8
∴ Tranformed equation is x⁴ + 6x³ – 12x – 8 = 0
∴ Required equation is x⁴ – 6x³ + 4x – 53 = 0
or x⁴ + 6x³ – 12x – 8 = 0.

ii) Given equation is x³ + 2x² + x + 1 = 0
Let f(x) = x³ + 2x² + x + 1
We have o find ‘h’ so that the coefficient of the third highest power of ‘x’ in f(x + h) is zero.
i.e., Coefficient of x in f(x + h) is zero.
f(x + h) = (x + h)³ + 2(x + h)² + (x + h) + 1
Coefficient of ‘x³’ in f(x + h) is 3h² + 4h + 1
We have to Choose ‘h’ such that 3h² + 4h + 1 = 0
i.e., h = – 1 or h = – \(\frac{1}{3}\).

Case – (I):
When h = – 1
f(x – 1) = (x – 1)³ + 2(x – 1)² +(x – 1) + 1
= (x³ – 3x² + 3x – 1)² + 2 (x² – 2x + 1) + x – 1 + 1
= x³ – x² + 1
∴ Transformed euluation is x³ – x² + 1 = 0.

Case – (ii):
When h = – \(\frac{1}{3}\)
\(f\left(x-\frac{1}{3}\right)=\left(x-\frac{1}{3}\right)^3+2\left(x-\frac{1}{3}\right)^2+\left(x-\frac{1}{3}\right)\)
= \(\left(x^3-x^2+\frac{x}{3}-\frac{1}{27}\right)+2\left(x^2-\frac{2}{3} x+\frac{1}{9}\right)\) + x – \(\frac{1}{3}\) + 1
= x³ + x² + \(\frac{23}{27}\)
∴ Transformed equation is x³ + x² + \(\frac{23}{27}\)
⇒ 27x³ + 27x² + 23 = 0
∴ Required equation is x³ – x² + 1 = 0 or 27x³ + 27x² + 23 = 0.

Question 7.
Solve the following equations.
i) x⁴ – 10x³ + 26x² – 10x + 1 = 0.
ii) 2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0
Solution:
i) Given equatIon is
x⁴ – 10x³ + 26x² – 10x + 1 = 0 (1)
This is even degree reciprocal equation of class one.
On dividing (1) by x²,
x² – 10x + 26 – \(\frac{10}{x}+\frac{1}{x^2}\) = 0
⇒ \(\left(x^2+\frac{1}{x^2}\right)-10\left(x+\frac{1}{x}\right)\) + 26 = 0
⇒ \(\left(x^2+\frac{1}{x^2}\right)-10\left(x+\frac{1}{x}\right)\) + 24 = 0
[Put x + \(\frac{1}{x}\) = y)
⇒ y² – 10y + 24 = 0
⇒ (y – 4) (y – 6) = 0
⇒ y = 4 or y = 6.

Case – (i):
When y = 4
⇒ x + \(\frac{1}{x}\) = 4
⇒ x² – 4x + 1 = 0
⇒ x = 2 ± √3.

Case – (ii):
When y = 6
⇒ x + \(\frac{1}{x}\) = 4
⇒ x² – 6x + 1 = 0
⇒ x = 3 ± 2√2
∴ The roots are 2 ± √3, 3 ± 2√2.

ii) Given equation is
2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0 ………………. (1)
Let f(x) = 2x⁵ + x⁴ – 12x³ – 12x² + x + 2
(1) is an odd degree reciprocal equation of class one.
∴ x = – 1 is a root of (1)
x + 1 is a factor of f(x).
We divide f(x) with x + 1.
By synthetic division,

∴ f(x) = (x + 1) (2x⁴ – x³ – 11x² – x + 2)
Let g(x) = 2x⁴ – x³ – 11x² – x + 2
g(x) = 0 is an even degree reciprocal equation 0f class one.
Dividing g(x) = 0 by x²
We get 2x² – x – 11 – \(\frac{1}{x}+\frac{2}{x^2}\) = 0
⇒ \(2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)\) – 11 = 0
⇒ \(2\left(x+\frac{1}{x}\right)^2-\left(x+\frac{1}{x}\right)\) – 15 = 0
(Put x + \(\frac{1}{x}\) = y)
⇒ 2y² – y – 15 = 0
⇒ (2y + 5) (y – 3) = 0
∴ y = 3 or y = – 1.

Case – (i) :
When y = 3
x + \(\frac{1}{x}\) = 3
⇒ x² – 3x + 1 = 0
⇒ x = \(\frac{3 \pm \sqrt{5}}{2}\)

Case – (ii):
When y = \(\frac{-5}{2}\)
x + \(\frac{1}{x}\) = \(\frac{-5}{2}\)
⇒ 2x² + 2 = – 5x
⇒ 2x² + 5x + 2 = 0
⇒ (2x + 1) (x + 2) = 0
∴ x = – \(\frac{-1}{2}\) or x = – 2
∴ The roots are – 1, 2, \(\frac{3 \pm \sqrt{5}}{2}\).

Students must practice this TS Intermediate Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 4 Theory of Equations Ex 4(b)

I.
Question 1.
Solve x³ – 3x² – 16x + 48 = 0, given that the sum of two roots is zero.
Solution:
Let α, β, γ be the roots of
x³ – 3x² – 16x + 48 = 0 ………….(1)
Given that the sum of two roots is zero.
Let α + β = 0 …………(2)
But from (1) we have α + β + γ = 3
⇒ γ = 3
Hence αβγ = – 48
⇒ αβ = – 16
We know that,
(α + β)² – (α – β)² = 4αβ
⇒ (α – β)² = 64
⇒ α – β = ± 8.

i) When α – β = 8
⇒ 2α = 8 (∵ from (2))
⇒ α = 4
∴ β = – 4.

ii) When α – β = – 8
⇒ 2α = – 8 (∵ from (2))
⇒ α = – 4
∴ β = 4.
The roots of given equation are 4, – 4 and 3.

Question 2.
Find the condition that x³ – px² + qx – r = 0 may have the sum of two of its roots zero.
Solution:
Given equation is x³ – px² + qx – r = 0
Let α, β, γ be the roots.
∴ α + β + γ = p …………..(1)
Given sum of two of its roots is zero.
∴ (1) ⇒ α + 0 = p
i. e„ α = p
Substituting in given equation, we get p³ – p³ + pq – r = 0
⇒ pq – r = 0
⇒ Pq = r.

Question 3.
Given that the roots of x³ + 3px² + 3qx + r = 0 are in
i) A.P., show that 2p³ – 3qp + r = 0
ii) G.P., show that p³r = q³
iii) H.P., show that 2q³ = r (3pq – r).
Solution:
Given cubic equation is
x³ + 3px² + 3qx + r = 0 ……………..(1)
Let α, β, γ be its roots.

i) When the roots are in A.P. :
i.e., 2β = α + γ
from (1) α + β + γ = – 3p
⇒ (α + γ) + P = – 3p
⇒ 2β + β = – 3p
⇒ 3β = – 3p
⇒ β = – P
Substituting in given equation, we get
– p³ + 3p³ – 3pq + r = 0
⇒ 2p³ – 3pq + r = 0.

ii) When the roots are in G.P. :
∴ β² = αγ
from (1) αβγ = – r
⇒ β³ = – r
⇒ β = – r^1/3
Substituting in (1), we get
(- r^1/3)³ + 3pr^2/3 – 3qr^1/3 + r = 0
⇒ 3pr^2/3 = 3qr^1/3
⇒ P³r = q³.

iii) When the roots are in H.P. :
β = \(\frac{2 \alpha \gamma}{\alpha+\gamma}\)
⇒ β² = \(\frac{2 \alpha \beta \gamma}{(\alpha+\beta+\gamma)-\beta}\)
⇒ β² = \(\frac{-2 r}{-3 p-\beta}\)
⇒ β² + 3pβ² = 2r (∵ β is a root of (1))
⇒ – 3qβ – r = 2r
β = – \(\frac{r}{q}\)
Substituting in (1) we get
\(\frac{-r^3}{q^3}+\frac{3 p r^2}{q^2}-\frac{3 q r}{q}\) + r = 0
⇒ 2q³ = r(3pq – r).

Question 4.
Find the condition that x³ – px² + qx – r = 0 may have the roots in G.P.
Solution:
Let α, β, γ be the roots of
x³ – px² + qx – r = 0
If α, β, γ are in G.P., then
β² = αγ
(1) ⇒ αβγ = r
⇒ β³ = r
β = r^1/3
Substituting in (1), we get
(r^1/3)³ – p(r^1/3)² + q(r^1/3) – r = 0
⇒ pr^2/3 = qr^1/3
⇒ p³r² = q³r
⇒ q³ = p³r.

II.
Question 1.
Solve 9x³ – 15x² 7x – 1 = 0, given that two of its roots are equal.
Solution:
Given cubic equation is
9x³ – 15x² – 7x – 1 = 0 ……………. (1)
Suppose α, β, γ are the roots of (1)
∴ α + β + γ = \(\frac{15}{9}=\frac{5}{3}\)
αβ + βγ + γα = \(\frac{7}{9}\)
αβγ = \(\frac{1}{9}\)
According to the problem, α = β (∵ two of its roots are equal)
∴ 2α + γ = \(\frac{5}{3}\)
⇒ γ = \(\frac{5}{3}\) – 2α
Also, α² + 2αγ = \(\frac{7}{9}\)
⇒ α² + 2α (\(\frac{5}{3}\) – 2α) = \(\frac{7}{9}\)
⇒ 27α² – 30α + 7 = 0
⇒ (3α – 1) (9α – 7) = 0
∴ α = \(\frac{1}{3}\) or α = \(\frac{7}{9}\)

Case (i) :
when α = \(\frac{1}{3}\)
γ = \(\frac{5}{3}\) – 2α
= \(\frac{5}{3}\) – \(\frac{2}{3}\) = 1
∴ The roots are \(\frac{1}{3}\), \(\frac{1}{3}\), 1.

Case – (ii):
When α = \(\frac{7}{9}\)
γ = \(\frac{5}{3}\) – 2α
= \(\frac{5}{3}-\frac{14}{9}\) = \(\frac{1}{9}\)
Which is impossible as
αβγ = \(\frac{7}{9} \cdot \frac{7}{9} \cdot \frac{1}{9}\) ≠ \(\frac{1}{9}\)
∴ The roots are \(\frac{1}{3}\), \(\frac{1}{3}\), 1.

Question 2.
Given that one root of 2x³ + 3x² – 8x + 3 = 0 is double the other root, find the roots of equation.
Solution:
Given cubic equation is
2x³ + 3x² – 8x + 3= 0 ……………..(1)
Suppose α, β, γ are the roots of (1).
∴ α + β + γ = \(\frac{-3}{2}\)
αβ + βγ + γα = \(\frac{-8}{2}\) = – 4 …………….(2)
αβγ = \(\frac{-3}{2}\)
Given one root is double the other.
3α + γ = \(\frac{-3}{2}\)
⇒ γ = \(\frac{-3}{2}\) – 3α
Also from (2):
2α² – 3α (\(\frac{3}{2}\) + 3α) = – 4
14α² + 9α – 8 = 0
(2α – 1) (7α + 8) = 0
α = \(\frac{1}{2}\) or α = \(\frac{-8}{7}\).

Case (i):
When α = \(\frac{1}{2}\)
β = 2α = 2 (\(\frac{1}{2}\)) = 1
γ = \(\frac{-3}{2}\) – 3α
= \(\frac{-3}{2} \frac{-3}{2}\) = – 3.
∴ α = \(\frac{1}{2}\), β = 1 and γ = – 3
satisfies αβγ = \(\frac{-3}{2}\)
∴ The roots are \(\frac{1}{2}\), 1, – 3.

Case (ii):
When α = \(\frac{-8}{7}\)
β = 2α = \(\frac{-16}{7}\)
γ = \(\frac{-3}{2}\) – 3α
= \(\frac{-3}{2}+\frac{48}{7}=\frac{75}{14}\)
But α = \(\frac{-8}{7}\), β = \(\frac{-16}{7}\) and γ = \(\frac{75}{14}\) do not satisfy αβγ = \(\frac{-3}{2}\).
Hence the roots of given equation are \(\frac{1}{2}\), 1, – 3.

Question 3.
Solve x³ – 9x² + 14x + 24 = 0, given that two of the roots are in the ratio 3 : 2.
Solution:
Given cubic equation is
x³ – 9x² + 14x + 24 = 0 ……….(1)
Let α, β, γ be the roots of (1)
∴ α + β + γ = 9, αβ + βγ + γα = 14, αβγ = – 24 ……………..(2)
Given two roots are in the ratio 3 : 2,
let α : β = 3 : 2
⇒ β = \(\frac{2 \alpha}{3}\)
Now from (2) \(\frac{5 \alpha}{3}\) + γ = 9
⇒ γ = 9 – \(\frac{5 \alpha}{3}\)
Also, \(\frac{2}{3}\) α² + (9 – \(\frac{5 \alpha}{3}\)) \(\frac{5 \alpha}{3}\) = 14
⇒ 2α² + \(\frac{5 \alpha(27-5 \alpha)}{3}\) = 42
⇒ 19α² – 135α + 126 = 0
⇒ (19α – 21) (α – 6) = 0
⇒ α = \(\frac{21}{19}\) or α = 6.

Case (i):
When α = \(\frac{21}{19}\)
β = \(\frac{2}{3}(\alpha)=\frac{2}{3}\left(\frac{21}{19}\right)=\frac{14}{19}\)
γ = \(9-\frac{5 \alpha}{3}=9-\frac{5}{3}\left(\frac{21}{19}\right)=\frac{136}{19}\)
These values do not satisfy αβγ = – 24.

Case – (ii) :
When α = 6
β = \(\frac{2}{3}(\alpha)=\frac{2}{3}(6)\) = 4
γ = \(9-\frac{5 \alpha}{3}=9-\frac{5}{3}(6)\) = – 1
These values satisfy αβγ = – 24.
∴ The roots of given equation are 6, 4, – 1.

Question 4.
Solve the following equations, given that the roots of each are in A.P.
i) 8x³ – 36x² – 18x + 81 = 0
ii) x³ – 3x² – 6x + 8 = 0
Solution:
i) Given cubic equation is
8x³ – 36x² – 18x + 81 = 0 …………….(1)
Given the roots are in A.P.
∴ α – d, α, α + d be the roots.
∴ Sum of the roots 3α = \(\frac{36}{8}\)
⇒ α = \(\frac{3}{2}\)
∴ x – \(\frac{3}{2}\) is a factor of 8x³ – 36x² – 18x + 81
By synthetic division,

8x³ – 36x² – 18x + 81 = (x – \(\frac{3}{2}\)) (8x² – 24x – 54)
∴ Equation (1)
⇒ (x – \(\frac{3}{2}\)) (8x² – 24x – 54) = 0
⇒ (x – \(\frac{3}{2}\)) (2x + 3) (2x – 9) = 0
⇒ x = – \(\frac{3}{2}\) or x = \(\frac{3}{2}\) or x = \(\frac{9}{2}\)
∴ The roots are \(\frac{-3}{2}\), \(\frac{3}{2}\), \(\frac{3}{2}\).

ii) Given roots of cubic equation
x³ – 3x² – 6x + 8 = 0 ……………(1) are in G.P.
Let α – d, α, α + d be the roots.
∴ Sum of the roots 3α = 3
⇒ α = 1
∴ (x – 1) is a factor of x³ – 3x² – 6x + 8.
By synthetic division,

x³ – 3x² – 6x + 8 = (x – 1) (x² – 2x – 8)
∴ Equation (1)
⇒ (x – 1) (x² – 2x – 8) = 0
⇒ (x – 1) (x – 4) (x + 2) = 0
∴ x = 1 or x = 4 or x = – 2.
∴ The roots are – 2, 1, 4.

Question 5.
Solve the following equations, given that the roots of each are in GP.
i) 3x³ – 26x² + 52x – 24= 0
ii) 54x³ – 39x² – 26x + 16 = 0
Solution:
i) Given roots of cubic equation
3x³ – 26x² + 52x – 24 = 0 ……………. (1) are in G.P.
Let \(\frac{\alpha}{r}\), α, αr be the roots.
∴ Product of the roots α³ = \(\frac{24}{3}\) = 8
⇒ α = 2
∴ (x – 2) is a factor of 3x³ – 26x³ + 52x – 24
By synthetic division,

3x³ – 26x² + 52x – 24 = (x – 2) (3x² – 20x + 12)
∴ Equation (1) ⇒ (x – 2) (3x² – 20x + 12) = 0
⇒ (x – 2) (3x – 2) (x – 6) = 0
∴ x = 2 or x = \(\frac{2}{3}\) or x = 6
∴ The roots are \(\frac{2}{3}\), 2, 6.

ii) Given roots of cubic equation.
54x³ – 39x² – 26x + 16 = 0 (1) are in GP.
Let \(\frac{\alpha}{r}\), α, αr be the roots.
∴ Product of the roots α³ = \(\frac{-16}{54}\)
⇒ α³ = \(\frac{-2}{3}\)
∴ (x + \(\frac{2}{3}\)) is a factor 54x³ – 39x² – 26x + 16
By synthetic division,

54x³ – 39x² – 26x + 16 = (x + \(\frac{2}{3}\)) (54x² – 75x + 24)
∴ Equation (1),
(x + \(\frac{2}{3}\)) (54x² – 75x + 24) = 0
(x + \(\frac{2}{3}\)) (18x² – 25x + 8) = 0
(x + \(\frac{2}{3}\)) (9x – 8)(2x – 1) = 0
∴ x = – \(\frac{2}{3}\) or x = \(\frac{8}{9}\) or x = \(\frac{1}{2}\)
∴ The roots are \(\frac{8}{9}\), \(\frac{-2}{3}\), \(\frac{1}{2}\).

Question 6.
Solve the following equations, given that the roots of each are In H.P.
i) 6x³ – 11x² + 6x – 1 = 0
ii) 15x³ – 23x² – 9x – 1 = 0
Solution:
i) Given cubic equation is
6x³ – 11x² + 6x – 1 = 0 …………..(1)
Put y = \(\frac{1}{x}\)
∴ (1) ⇒ \(\frac{6}{y^3}-\frac{11}{y^2}+\frac{6}{y}\) – 1 = 0
⇒ y3 – 6y2 + 11y – 6 = 0 ………… (2)
Given roots of (1) are in H.P.
⇒ Roots of (2) are in AP.
Let a – d, a, a + d be the roots of (2),
∴ Sum of the roots, 3a = 6
⇒ α = 2
∴ (x – 2) is a factor of y³ – 6y² + 11y – 6
By synthetic division

∴ y³ – 6y² + 11y – 6 = (y – 2) (y² – 4y 3)
∴ Equation (2) = (y – 2) (y² – 4y + 3) = 0
⇒ (y – 2) (y – 3) (y – 1) = 0
∴ y = 1 or y = 2 ory = – 3
The roots of (2) are 1, 2, 3.
Hence the roots of (1) are 1, \(\frac{1}{2}\), \(\frac{1}{3}\).

ii) Given cubic equation is
15x³ – 23x² + 9x – 1 = 0 …………….(1)
put y = \(\frac{1}{x}\)
∴ (1) ⇒ y³ – 9y + 23y² – 15 = 0 ………..(2)
Given roots of (1) are in 1-LP.
⇒ Roots of (2) are in A.P.
Let a – d, a, a + d be the roots of (2),
∴ Sum of the roots, 3α = 9
⇒ α = 3
∴ (y – 3) is a factor of y³ – 9y + 23y² – 15.
By synthetic division,

∴ y³ – 9y + 23y² – 15 = (y – 3) (y² – 6y + 5)
∴ Equation (2) = (y – 3) (y2 – 6y + 5) = 0
⇒ (y – 3) (y – 1) (y – 5) = 0
∴ y = 1 or y = 3 or y = 5
∴ The roots of (2) are 1, 3, 5.
Hence the roots of (2) are 1, \(\frac{1}{3}\), \(\frac{1}{5}\).

Question 7.
Solve the following equations, given that they have multiple roots.
i) x⁴ – 6x³ + 13x² – 24x + 36 = 0
ii) 3x⁴ + 16x³ + 24x² – 16 = 0
Solution:
i) Given equation,
x⁴ – 6x³ + 13x² – 24x + 36 = 0 …………..(1)
Let f(x) = x⁴ – 6x³ + 13x² – 24x + 36
f’(x) = 4x³ – 18x² + 26x – 24
= 2 (2x³ – 9x² + 13x – 12)
f’(3) = 2(54 – 81 + 39 – 12)
⇒ f'(3) = o
Now
f(3) = 81 – 162 + 117 – 72 + 36
= f(3) = 0
∴ (x – 3) is a factor of f(x) and f’(x).
∴ 3 is the repeated root of f(x) = 0.
Now we divide f(x) by (x – 3) by using synthetic division.

∴ f(x) = (x – 3) (x – 3) (x2 + 4)
∴ Equation (1)
⇒ f(x) = 0
⇒ (x – 3) (x – 3) (x² + 4) = 0
∴ x = 3 or x² + 4 = 0
⇒ x = ±2i
∴ The roots of given equation are 3, 3, ± 2i.

ii) Given equation is
3x⁴ + 16x³ + 24x² – 16 = 0 …………..(1)
Let f(x) = 3x⁴ + 16x³+ 24x² – 16
⇒ f'(x) = 12x³ + 48x² + 48x
= 12 (x³ + 4x² + 4x)
= 12x (x + 2)²
⇒ f’ (- 2) = 0
Also f(- 2) = 3(16) + 16(- 8) + 24(4) – 16 = 0
∴ (x + 2) is a factor of f(x) and f'(x).
∴ – 2 is a repeated root of f(x) = 0.
Now we divide f(x) by (x + 2) using synthetic division.

∴ f(x) = (x + 2) (x + 2) (3x² + 4x – 4)
Equation (1)
⇒ f(x) = 0
⇒ (x + 2) (x + 2) (3x² + 4x – 4) = 0
⇒ (x + 2) (x + 2) (3x – 2) (x + 2) = 0
⇒ x = – 2 or x = \(\frac{2}{3}\)
∴ The roots of given equation are – 2, – 2, – 2, \(\frac{2}{3}\).

III.
Question 1.
Solve x⁴ + x³ – 16x² – 4x + 48 = 0, given that the product of two of the roots is 6.
Solution:
Given equation is
x⁴ + x³ – 16x² – 4x + 48 = 0 ………….(1)
Let α, β, γ, δ be the roots
∴ x⁴ + x³ – 16x² – 4x + 48 = (x + α) (x – β) (x – γ) (x – δ) ……….(2)
∴ Sum of the roots α + β + γ + δ = – 1
and product of roots ⇒ αβγδ = 48 …………..(3)
Given product of two roots = 6
Let αβ = 6
∴ γδ = \(\frac{48}{\alpha \beta}\)
γδ = 8
Let α + β = a and γ + δ = b
Now (2)
⇒ x⁴ + x³ – 16x² – 4x + 48 = (x² – (α + β) x + αβ) (x² – (γ + δ) x + γδ)
⇒ x⁴ + x³ – 16x² – 4x + 48 = (x² – ax + 6) (x² – bx + 8)
Comparing like terms.
we get, a + b = – 1 and
8a – 6b = 5 ……………..(4)
⇒ 4a + 3b = 2 (5)
(5) ⇒ 4a + 3 (- 1 – a) = 2 (∵ from (4))
⇒ a = 5
∴ b = – 6
∴ x⁴ + x³ – 16x² – 4x + 48 = (x² – 5x + 6) (x² + 6x + 8)
= (x – 2) (x – 3) (x + 2) (x + 4)
∴ Equation (1),
⇒ (x – 2) (x – 3) (x + 2) (x + 4) = 0
∴ x = – 4; x = – 2 or x = 2 or x = 3
∴ The roots of the given equation are 2, 3, – 4, – 2.

Question 2.
Solve 8x⁴ – 2x³ – 27x² + 6x + 9 = 0 given that two roots have the same absolute value, but are opposite in sign.
Solution:
Given equation is
8x⁴ – 2x³ – 27x² + 6x + 9 = 0
⇒ x⁴ – \(\frac{1}{4} x^3-\frac{27}{8} x^2+\frac{3}{4} x+\frac{9}{8}\) = 0 ……………(1)
Let α, β, γ, δ be the roots of (1)
∴ Sum of the roots α + β + γ + δ = \(\frac{1}{4}\)
and product of roots αβγδ = \(\frac{9}{8}\)
But given two roots have same absolute value but are opposite sign.
Let α = – β
⇒ α + β = 0
∴ γ + δ = \(\frac{-1}{4}\)
Let αβ = a and γδ = b
Now(x – α) (x – β) = x² – (α + β)x + αβ
⇒ (x – α) (x – β) = x² + a …………(2)
Also (x – γ) (x – δ) = x² – (γ + δ)x + γδ
= (x – γ) (x – δ) = x² – \(\frac{1}{4}\) x + b ………….(3)
From (1), (2) and (3)
x⁴ – \(\frac{1}{4} x^3-\frac{27}{8} x^2+\frac{3}{4} x+\frac{9}{8}\) = (x² + a) (x² – \(\frac{1}{4}\) x + b)
Comparing like terms,
\(\frac{3}{4}=\frac{-a}{4}\) and ab = \(\frac{9}{8}\)
a = – 3
∴ b = \(\frac{9}{8(-3)}\)
b = \(\frac{-3}{8}\)
∴ (2) ⇒ (x – α) (x – β) = x² – 3
& (3) ⇒ (x – γ) (x – δ) = (x² – \(\frac{1}{4}\) x + \(\frac{3}{8}\))
⇒ \(\frac{1}{8}\) (8x² – 2x – 3)
⇒ (x – γ) (x – δ) = \(\frac{1}{8}\) (2x + 1) (4x – 3)
(x – γ) (x – δ) = (x + \(\frac{1}{2}\)) (x – \(\frac{3}{4}\))
∴ Equation (1)
(x² – 3) (x + \(\frac{1}{2}\)) (x – \(\frac{3}{4}\)) = 0
⇒ x = ± √3 or x = – \(\frac{1}{2}\) or x = \(\frac{3}{4}\)
∴ The roots of given equation are – √3, √3, – \(\frac{1}{2}\), \(\frac{3}{4}\).

Question 3.
Solve 18x³ + 81x² + 121x + 60 = 0 given that one root is equal to half the sum of the remaining roots.
Solution:
Given equation is
18x³ + 81x² + 121x + 60 = 0 ……………(1)
Let α, β, γ, δ be the roots
∴ Sum of roots, α + β + γ = \(\frac{-81}{18}=\frac{-9}{2}\)
αβ + βγ + γδ = \(\frac{121}{18}\)
and product of roots αβγ = \(\)
given one root is equal to halt of the sum of the remaining roots.
∴ Let α = \(\frac{\beta+\gamma}{2}\)
∴ α + 2α = \(\frac{-9}{2}\)
⇒ 3α = \(\frac{-9}{2}\)
⇒ α = \(\frac{-9}{2}\)
∴ x + \(\frac{3}{2}\) is a factor of 18x³ + 81x² + 121x + 60.
By synthetic division,

∴ 18x³ + 81x² + 121x + 60 = (x + \(\frac{3}{2}\)) (18x² + 54x + 40)
= (x + \(\frac{3}{2}\)) (9x² + 27x + 60)
∴ 18x³ + 81x² + 121x + 60 = 2 (x + \(\frac{3}{2}\)) (3x + 4) (3x + 5)
∴ Equation (1),
⇒ 2 (x + \(\frac{3}{2}\)) (3x + 4) (3x + 5) = 0
∴ x = \(\frac{-3}{2}\) or x = \(\frac{-4}{3}\) or x = \(\frac{-5}{3}\).
∴ The roots of given equation are \(\frac{-3}{2}\), \(\frac{-4}{3}\), \(\frac{-5}{3}\).

Question 4.
Find the condition In order that the equation ax⁴ + 4bx³ + 6cx² + 4dx + e = 0 may have two pairs of equal roots.
Solution:
Given equation is
ax⁴ + 4bx³ + 6cx² + 4dx + e = 0 ………………..(1)
Given (1) has two pairs of equal roots.
∴ Let α, α, β, β be the root of (1).
(1) ⇒ x⁴ + \(\frac{4 b}{a} x^3+\frac{6 c}{a} x^2+\frac{4 d}{a} x+\frac{e}{a}\) = 0

3abc = 2b³ + a²d and ad² = eb².
∴ The required conditions are 2b³ + a²d = 3abc and ad² = eb².

Question 5.
i) Show that x⁵ – 5x³ + 5x² – 1 = 0 has three equal roots and and this root.
ii) Find the repeated roots of x⁵ – 3x⁴ – 5x³ + 27x² – 32x + 12 = 0.
Solution:
i) Given equation is x⁵ – 5x³ + 5x² – 1 = 0
Let f(x) = x⁵ – 5x³ + 5x² – 1
f’(x) = 5x⁴ – 15x² + 10x
f”(x) = 20x³ – 30x + 10
f”(1) = 20 – 30 + 10 = 0
Similarly, f’(1) = 0 and f(1) = 0
∴ (x – 1) is a factor of f”(x), f’(x) & f(x).
Thus f(x) = 0 has three equal roots and it is ‘1’.

ii) Given equation is
x⁵ – 3x⁴ – 5x³ + 27x² – 32x + 12 = 0 …………(1)
Let f(x) = x⁵ – 3x⁴ – 5x³ + 27x² – 32x + 12
f’(x) = 5x⁴ – 12x³ – 15x² + 54x – 32
f’(1) = 5 – 12 – 15 + 54 – 32 = 0
Similarly f'(1) = 0 and f(1) = 0
∴ (x – 1) is a factor of f”(x), f'(x) & f(x).
Thus f(x) = 0 has three equal roots and it is ‘1’.
By synthetic division,

∴ f(x) = (x – 1)² (x³ – x² – 8x + 12)
Let g(x) = x³ – x² – 8x + 12
g’(x) = 3x² – 2x – 8
g’(2) = 3(4) – 2(2) – 8 = 0
and g(2) = 2³ – 2² – 8(2) + 12 = 0.
∴ (x – 2) is a factor of g(x) and g’(x).
∴ 2 is a multiple root of g(x) = 0.
By synthetic division,

∴ g(x) = (x – 2)² (x + 3)
∴ f(x) = (x – 1)² (x – 2)² (x + 3)
The roots of given equation are 1, 1, 2, 2, 3.
Hence repeated roots are 1 and 2.

Question 6.
Solve the equation 8x³ – 20x² + 6x + 9 = 0 given that the equation has multiple roots.
Solution:
Given equation is 8x³ – 20x² + 6x + 9 = 0 …………..(1)
Let f(x) = 8x³ – 20x² + 6x + 9
f’(x) = 24x² – 40x + 6
= 2 (12x² – 20x + 3)
= 2 (2x – 3) (6x – 1)
\(f\left(\frac{3}{2}\right)=8\left(\frac{27}{8}\right)-20\left(\frac{9}{4}\right)+6\left(\frac{3}{2}\right)+9\)
= 27 – 45 + 9 + 9 = 0
∴ f(\(\frac{3}{2}\)) = 0
∴ f(x) and f'(x) has a common factor ‘2x – 3’.
∴ \(\frac{3}{2}\) is a multiple root of f(x) = 0.
By synthetic division,

∴ 8x³ – 20x² + 6x + 9 = 0
(x – \(\frac{3}{2}\))² (8x + 4) = 0
⇒ x = \(\frac{3}{2}\) or x = \(\frac{-1}{2}\)
∴ The roots of given equation are \(\frac{-1}{2}\), \(\frac{3}{2}\), \(\frac{3}{2}\).

Students must practice this TS Intermediate Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(d)

I.
Question 1.
Find the number of ways of arranging the letters of the words.
i) INDEPENDENCE
ii) MATHEMATICS
iii) SINGING
iv) PERMUTATION
v) COMBINATION vQ INTERMEDIATE
Solution:
i) Given word is INDEPENDENCE
The given 12 letters word INDEPENDENCE contains 3N’s, 2D’s and 4I’s.
Hence they can be arranged in \(\frac{12 !}{2 ! \times 3 ! \times 4 !}\) ways.

ii) Given word is MATHEMATICS
The given 11 letters word MATHEMATICS contains 2 M’s, 2 A’s and 2 T’s.
Hence they can be arranged in \(\frac{11 !}{2 ! \cdot 2 ! \cdot 2 !}\) ways.

iii) Given word is SINGING.
The given 7 letters word SINGING contains 2 I’s, 2 N’s and 2 G’s.
Hence they can be arranged in \(\frac{7 !}{2 ! \cdot 2 ! \cdot 2 !}\) ways.

iv) Given Word is PERMUTATION
The given 11 letters word PERMUTATION contains 2T’s.
Hence they can be arranged in \(\frac{11 !}{2 !}\) ways.

v) Given word is COMBINATION.
Given 12 letters word COMBINATION contains 2 O’s, 2 I’s, 3 N’s.
Hence they can be arranged in \(\frac{11 !}{2 ! \cdot 2 ! \cdot 2 !}\) ways.

vi) Given word is INTERMEDIATE.
Given 12 letters word INTERMEDIATE’ contains 2 I s, 2 T’s, 3 E’s.
Hence they can be arranged in \(\frac{12 !}{3 ! \times 2 ! \times 2 !}\) ways.

Question 2.
Find the number of 7- digit numbers that can be formed using 2, 2, 2, 3, 3, 4, 4.
Solution:
The 7 digit numbers that can be formed using three 2’s, two 3’s and two 4’s are \(\frac{7 !}{3 ! \cdot 2 ! \cdot 2 !}\).

II.
Question 1.
Find the number of 4 – letter words that can be formed using the letters of the word RAMANA.
Solution:
Given word is RAMANA.
The given word RAMANA has 6 letters which contains 3 A’s and rest are different.
In forming 4 letter words, these cases arises.

Case – (i) :
When all are different i.e., R, M, N, A.
∴ Number of 4 letter words formed = 4! = 24.

Case – (ii) :
When two are alike (i.e., A, A) and two are different, i.e., selected from R, M, N.
Two different letters are selected in \({ }^3 \mathrm{C}_2\) ways.
∴ Number of 4 letter words formed are \({ }^3 C_2 \times \frac{4 !}{2 !}\) = 36.

Case (iii):
When three are alike (i.e., A, A, A) and one different letter, selected out of R, M, N.
Now one different letters is selected in \({ }^3 \mathrm{C}_1\) ways.
∴ Number of 4 letter words formed = \({ }^3 \mathrm{C}_1 \times \frac{4 !}{3 !}\) = 12
∴ Total number of 4 letter words formed are 24 + 36 + 12 = 72.

Question 2.
How many numbers can be formed using all digits 1, 2, 3, 4, 3, 2, 1 such that even digits always occupy even places?
Solution:
Given digits 1, 2, 3, 4, 3, 2, 1 contains 3 even digits i.e., 2, 4, 2 and 3 even places.
Number of ways of arranging 3 even digits 2, 4, 2 in three even places is \(\frac{3 !}{2 !}\) 3.
The remaining 4 digits 1, 3, 3, 1 in remaining 4 places can be arranged in \(\frac{4 !}{2 ! \cdot 2 !}\) ways.
∴ Number of numbers formed using all the digits 1, 2, 3, 4, 3, 2, 1 such that even digits always occupy even places = 3 × \(\frac{4 !}{2 ! \cdot 2 !}\) = 18.

Question 3.
In a library, there are 6 copies of one book, 4 copies each of two different books, 5 copies each of three different books and 3 copies each of two different books. Find the number of ways of arranging all these books in a shelf in a single row.
Solution:
There are 6 copies of one book, 4 copies each of two different books, 5 copies each of three different books and 3 copies of two different books.
∴ Total number of books = 6 + 4(2) + 5(3) + 3(2) = 35
∴ Number of ways of arranging these 35 books = \(\frac{35 !}{6 ! \times 4 ! \times 4 ! \times 5 ! \times 5 ! \times 5 ! \times 3 ! \times 3 !}\)
= \(\frac{35 !}{6 ! \cdot(4 !)^2 \cdot(5 !)^3 \cdot(3 !)^2}\).

Question 4.
A book store has ‘m’ copies each of ‘n’ different books. Find the number of ways of arranging these books in a shelf in a single row.
Solution:
Book store has m’ copies each of n’ different books.
∴ Total number of books = mn
∴ Number of ways of arranging these mn books = \(\frac{(m n) !}{m ! \times m ! \times \underbrace{\ldots \ldots m !}_{n \text { times }}}=\frac{(m n) !}{(m !)^n}\)

Question 5.
Find the number of 5-digit numbers that can be formed using the digits 0, 1, 1, 2, 3.
Solution:
Given digits are 0, 1, 1, 2, 3.
The ten thousand’s place of a ‘5’ digit number can be filled by any of non-zero digit in 4 ways.
The remaining 4 places can be filled with remaining 4 digits in 4! ways.
∴ Number of ways = 4 × 4!
Since there are two 1’s in every arrangement, the number of 5 digited numbers formed are \(\frac{4 \times 4 !}{2 !}\) = 48.

Question 6.
In how many ways can the letters of the word CHEESE be arranged so that no two E’s come together ?
Solution:
Given word is ‘CHEESE’.
As no two E’s come together, first arrange letters C, H, S.
Number of ways of arranging C, H, S is 3!.
∴ Number of gaps formed are 4.
Number of ways of arranging 3 E’s in 4 gaps = \(\frac{{ }^4 P_3}{3 !}\)
∴ Required number of ways = 3! × \(\frac{{ }^4 P_3}{3 !}\) = 24.

III.
Question 1.
Find the number of ways of arranging the letters of the word ASSOCIATIONS. In how many of them
i) all the three S’s together
ii) the two A’s do not come together.
Solution:
Given word is ‘ASSOCIATIONS’.
The given 12 letters word ‘ASSOCIATIONS’ contains 2 A’s, 3 S’s, 2 O’s and 2 I s.
∴ Number of ways of arranging them = \(\frac{12 !}{2 ! \cdot 2 ! \cdot 2 ! \cdot 3 !}\).

i) All the three S’s come together :
Treat 3 S’s as one unit. This unit with remain¬ing 9 letters becomes 10 entities which contain 2 A’s, 2 O’s and 2 I’s.
Number of ways of arranging so that all the three S’s come together = \(\frac{10 !}{2 ! \cdot 2 ! \cdot 2 !}\).

ii) The two A’s do not come together :
As 2 A’s do not come together, arrange remaining 10 letters.
Remaining 10 letters contains 3 S’s, 2 O’s and 2 I’s.
∴ Number of ways of arranging = \(\frac{10 !}{3 ! \cdot 2 ! \cdot 2 !}\)
∴ Number of gaps formed are ’11’.
2 A’s in these 11 gaps can be arranged in up \(\frac{{ }^{11} \mathrm{P}_2}{2 !}\) ways.
∴ Required number of ways of arranging = \(\frac{10 !}{3 ! \times 2 ! \times 2 !} \times \frac{{ }^{11} P_2}{2 !}\).

Question 2.
Find the number of ways of arranging the letters of the word MISSING so that the two S’s are together and the two l’s are together.
Solution:
Given word is MISSING.
Treat 2 S’s as 1 unit and 2 l’s as another unit.
These 2 unit with remaining 3 letters be comes 5 entities.
Number of ways of arranging them = 5!
2 S’s and 21’s can arrange themselves in only 1 way.
∴ Required nunther of ways of arranging = 5! = 120.

Question 3.
If the letters of the word AJANTA are per muted in all possible ways and the words thus formed are arranged In dictionary order, find the ranks of the words
(i) AJANTA
(ii) JANATA.
Solution:
Given word MANTA in Alphabetic order is AAAJNT.

i) Rank of ‘AJANTA’:
In the dictionary order first comes that words which begin with letter A. Also the second place is to be filled by A.
The remaining 4 places can be filled in 4! ways.
On proceeding like this, we get
The number of word begin with AA is 4! = 24
The number of word begin with AJAA is 2! = 2
The number of word begin with AJANA = 1
Next word is AJANTA
∴ Rank of word AJANTA = 24 + 2 + 1 + 1 = 28.

ii) Rank of JANATA :
In the dictionary order first comes that words which begin with the letter A.
The remaining 5 places to be filled in \(\frac{5 !}{2 !}\) ways.
On proceeding like this
The number of words begin with A = \(\frac{5 !}{2 !}\) = 60.
The number of words begin with JAA = 3! = 6
The number of words begin with JANAA = 1! = 1
Next word is JANATA
∴ Rank of word JANATA = 60 + 6 + 1 + 1 = 68.

Students must practice this TS Intermediate Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(a)

I.
Question 1.
If \({ }^{\mathrm{n}} \mathrm{P}_3\) = 1320, find n.
Solution:
Given \({ }^{\mathrm{n}} \mathrm{P}_3\) = 1320
⇒ n(n – 1) (n – 2) = 12 × 11 × 10
On comparing, we have n = 12.

Question 2.
If \({ }^n P_7\) = 42 P5, find n.
Solution:
Given \({ }^n P_7\) = 42 . \({ }^n P_5\)
⇒ \(\frac{n !}{(n-7) !}=42 \cdot \frac{n !}{(n-5) !}\)
⇒ \(\frac{n !}{(n-7) !}=42 \cdot \frac{n !}{(n-5)(n-6)(n-7) !}\)
⇒ (n – 5) (n – 6) – 42
⇒ (n – 5) (n – 6) = 7 × 6
On comparing largest integers, we have n – 5 = 7
⇒ n = 12.

Question 3.
If \({ }^{(n+1)} P_5:{ }^n P_6\) = 2 : 7, find n.
Solution:
Given \({ }^{(n+1)} P_5:{ }^n P_6\) = 2 : 7
⇒ \(\frac{(n+1) !}{(n-4) !}: \frac{n !}{(n-6) !}\) = 2 : 7
⇒ \(7 \cdot \frac{(n+1) !}{(n-4) !}=2 \cdot \frac{n !}{(n-6) !}\)
⇒ \(\frac{7 \cdot(n+1) \cdot n !}{(n-4)(n-5)(n-6) !}=2 \cdot \frac{n !}{(n-6) !}\)
⇒ 7(n + 1) = 2 (n – 4) (n – 5)
⇒ 2n² – 25n + 33 = 0
⇒ (2n – 3) (n – 11) = 0
(∵ 2n – 3 ≠ 0 as an integer)
⇒ n – 11 = 0
⇒ n = 11.

Question 4.
If \({ }^{12} P_5+5 \cdot{ }^{12} P_4={ }^{13} P_t\) find r.
Solution:
Given, \({ }^{12} P_5+5 \cdot{ }^{12} P_4={ }^{13} P_t\)

Alternate method:
We know that
\({ }^n P_r={ }^{(n-1)} P_r+r \cdot{ }^{(n-1)} P_{r-1}\)
∴ \({ }^{12} \mathrm{P}_5+5 \cdot{ }^{12} \mathrm{P}_4={ }^{13} \mathrm{P}_{\mathrm{r}}\)
\({ }^{13} P_5={ }^{13} P_r\)
∴ r = 5.

Question 5.
If \({ }^{18} P_{(r-1)}{ }^{17} P_{(r-1)}\) = 9 : 7, find r.
Solution:
Given \({ }^{18} P_{(r-1)}{ }^{17} P_{(r-1)}\) = 9 : 7

19 – r = 14
r = 5.

Question 6.
A man has 4 sons and there are 5 schools within his reach. In how many ways can he admit his sons In the schools so that no two of them will he in the same school.
Solution:
The first son can be admitted to any one of the 5 schools ¡n 5 ways.
The second son can be admitted to any one of the remaining 4 schools in 4 ways.
Proceeding like this, the number of ways can he admit his sons in the schools so that no two of them will be in same school = 5 × 4 × 3 × 2 = \({ }^5 \mathrm{P}_4\) (or) 120.

II.
Question 1.
If there are 25 railway stations on a railway line, how many types of single second class tickets must be printed, so as to enable a passenger to travel from one station to another.
Solution:
Let the ticket printed to enable a passenger to travel be from station ‘x’ to station ‘y’.
As there are 25 railway stations on a railway line, ‘x’ can be filled in 25 ways and ‘y’ can be filled in remaining 24 ways.
∴ The required number of tickets = 25 × 24
= \({ }^{25} \mathrm{P}_2\) or 600.

Question 2.
In a class there are 30 students. On the New year day, every student posts a greeting card to all his/her classmates. Find the total number of greeting cards posted by them.
Solution:
Number of students in a class = 30.
As every student posts a greeting card to all his/her classmates,
the total number of greeting cards = \({ }^{30} \mathrm{P}_2\) = 870.

Question 3.
Find the number of ways of arranging the letters of the word TRIANGLE so that the relative positions of the vowels and conso-nants are not disturbed.
Solution:
The given word TRIANGLE has 3 vowels and 5 consonants.
Since the relative positions of vowels and consonants are not to be disturbed.
3 vowels can be arranged in their relative positions in 3! ways.
Similarly, 5 consonants can be arranged in their relative position in 5! ways.
∴ By fundamental principle of counting, the required number of ways = 3! × 5! = 720.

Question 4.
Find the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8, without repetition.
Solution:
Given digits are 0, 2, 4, 7, 8.
For a four digit number formed from given digits without repetition,
Thousand’s place can be filled by non-zero digit in 4 ways,
Hundred’s place can be filled by remaining 4 digits in 4 ways,
Ten’s place can be filled in 3 ways and units place can be filled in 2 ways.
∴ By fundamental principle of counting, the number of four digited numbers formed = 4 × 4 × 3 × 2 = 96
Out of these 96 numbers,
the numbers with ‘2’ in units place are 3 × 3 × 2 = 18
Similarly the numbers with ’2’ in ten s place are 18.
The numbers with ‘2’ in Hundred’s place are = 18.
The numberlrwith ‘2’ in thousands place are \({ }^4 \mathrm{P}_3\) = 24.
∴ The value obtained by adding ‘2’ in all the numbers = 18 (2) + 18 (20) + 18 (200) + 24 (2000)
= 18 × 2 (111) + 24 × 2 × 1000
Similarly the value obtained by adding ‘4’ is 18 × 4 (111) + 24 × 4 × 1000
The value obtained by adding ‘7’ is 18 (7) (111) + 24 × 7 × 1000
The value obtained by adding ‘8’ is 18 × 8 × 111 + 24 × 8 × 1000
∴ The sum of all numbers = 18 × 2 × 111 + 24 × 2 × 1000 + 18 × 4 × 111 + 24 × 4 × 1000 + 18 × 7 × 111 + 24 × 7 × 1000 + 18 × 8 × 111 + 24 × 8 × 1000
= 18 × 111 × (2 + 4 + 7 + 8) + 24 × 1000 × (2 + 4 + 7 + 8)
= 18 × 111 × 21 + 24 × 1000 × 21
= 3 × 6 × 111 × 21 +4 × 6 × 1000 × 21
= 21 × 6 (333 + 4000)
= 21 × 6 (4333) = 5,45,958.

Question 5.
Find the number of numbers that are greater than 4000 which can be formed using the digits 0, 2, 4, 6, 8 without repetition.
Solution:
Given digits are 0, 2, 4, 6, 8.
All the five digit numbers are greater than 4000.
In the case of four digit numbers, the numbers which start with 4 or 6 or 8 are greater than 4000.
The number of 4 digit numbers which starts with 4 or 6 or 8 using the given digits with out repetition are 3 × \({ }^4 P_3\) = 3 × 24 = 72.
The number of 5 digit numbers using the given digits with out repetition are 4 × 4! = 4 × 24 = 96
∴ The number of numbers which are greater than 4000 are 72 + 96 = 168.

Question 6.
Find the number of ways of arranging the letters of the word MONDAY so that no vowel occupies even place.
Solution:
Given word is MONDAY
Given word contains 2 vowels and 4 conso-nants.
Also given word contains 3 odd places and 3 even places.
. Since no vowel occupies even places, the two vowels in three odd places can be filled in \({ }^3 \mathrm{P}_2\) ways.
The four consonants can be filled in remain-ing four places in 4! ways.
∴ By fundamental principle of counting the required numbers of ways = \({ }^3 \mathrm{P}_2\) × 4! = 144.

Question 7.
Find the number of ways of arranging 5 different mathematics books, 4 different physics books and 3 different chemistry books such that the books of the same subject are together.
Solution:
Given number of Mathematics books = 5
Given number of Physics books = 4
Given number of Chemistry books = 3
Treat all Mathematics books as 1 unit, Phys¬ics books as 1 unit, Chemistry books as 1 unit.
The number of ways of arranging 3 units of books = 3!
5 different Mathematics books can be arranged in 5! ways.
4 different Physics books can be arranged in 4!
3 different Chemistry books can be arranged in 3! ways.
∴. By fundamental principle of counting, required number of ways of arranging books = 3! × 5! × 4! × 3!
= 6 × 120 × 24 × 6 = 1,03,680.

III.
Question 1.
Find the number of 5 letter words that can be formed using the letter of the word CONSIDER. How many of them begin with “C”, how many of them end with “R”, and how many of them begin with “C” and end with “R” ?
Solution:
Given word “CONSIDER” contains 8 different letters.
The number of 5 letter words that can be formed using the letters of word CONSIDER = \({ }^8 \mathrm{P}_5\)
Out of them,
i) If first place is filled by ‘C’, the remaining 4 places by remaining 7 letters can be filled in \({ }^7 \mathrm{P}_4\) ways.
Number of words begin which ‘C’ are P4.
ii) if last place is filled by ’R’, the remaining first four places by remaining 7 letters can be filled in \({ }^7 \mathrm{P}_4\) ways.
/. Number of words end with ’R’ are \({ }^7 \mathrm{P}_4\).
iii) If first place is filled with ‘C’ and last place is filled by ‘R’, the remaining 3 places between them by remaining 6 letters can be filled in \({ }^6 \mathrm{P}_3\) ways.
∴ Number of words begin with ‘C’ and end with ‘R’ are \({ }^6 \mathrm{P}_3\).

Question 2.
Find the number of ways of seating 10 students A₁, A₂, ………….., A₁₀ in a row such that
i) A₁, A₂, A₃ sit together
ii) A₁, A₂, A₃ sit in a specified order.
iii) A₁, A₂, A₃ sit together in a specified order.
Solution:
A₁, A₂, ……………., A₁₀ are the 10 students.
i) A₁, A₂, A₃ sit together :
Treat A₁, A₂, A₃ as 1 unit.
This unit with remaining 7 students can be arranged in 8! ways.
The students A₁, A₂, A₃ can be arranged in 3! ways.
∴ The number of ways of seating 10 students such that A₁, A₂, A₃ sit together = 8! × 3!.

ii) A₁, A₂, A₃ sit in a specified order :
In 10 positions remaining 7 students other than A₁, A₂, A₃ can be arranged in \({ }^{10} \mathrm{P}_7\)ways.
As A₁, A₂, A₃ sit in a specified order, they can be arranged in 3 gaps in only 1 way.
∴ The number of ways of A₁, A₂, A₂ sit in a 10 specified order = \({ }^{10} \mathrm{P}_7\).

iii) A₁, A₂, A₃ sit together in a specified order :
Treat A₁, A₂, A₃ as 1 unit and they are in a specified order.
This unit with remaining 7 students can be arranged in 8! ways.
∴ The number of ways of A₁, A₂, A₃ sit to-gether in specified order = 8!.

Question 3.
Find the number of ways in which 5 red balls, 4 black balls of different sizes can be arranged in a row so that
(i) no two balls of the same colour come together
(ii) the balls of the same colour come together.
Solution:
Given 5 red balls and 4 black balls are of different sizes.
i) No two balls of same colour come to-gether :
First arrange 4 black balls in row, which can be done in 4! ways × B × B × B × B ×
Then we find 5 gaps, to arrange 5 red balls. This arrangement can be done in 5! ways.
∴ By principle of counting total number of ways of arranging = 5! × 4!.

ii) The balls of same colour come together:
Treat all red balls as one unit and all black balls as another unit.
The number of ways of arranging these two units = 2!
The 5 red balls can be arranged in 5! ways while 4 black balls are arranged in 4! ways.
By fundamental principal of counting, the required number of ways = 2! × 4! × 5!.

Question 4.
Find the number of 4 – digit numbers that can be formed using the digits 1, 2, 5, 6, 7. How many of them are divisible by
i) 2
ii) 3
iii) 4
iv) 5
v) 25
Solution:
The number of 4 digit numbers that can be formed using the digits 1, 2, 5, 6, 7 with out repetition are \({ }^5 \mathrm{P}_4\) or 120.

i) Divisible by 2 :
A number with an even digit in its units place is divisible by ‘2’.
This can be done in 2 ways (with 2 or 6). The remaining 3 places can be filled with the remaining 4 digits in \({ }^4 P_3\) ways.
∴ The number of 4 digit numbers, divisible by 2 are 2 × \({ }^4 P_3\) = 48.

ii) Divisible by 3 :
A number is divisible by 3 only when the sum of digits in that number is a multiple of 3.
Sum of given 5 – digits is 21.
The only way to select 4 digits using digits such that their sum is a multiple of 3 is selecting 1, 2, 5, 7.
We can arrange then in 4! ways.
∴ The number of 4 digit numbers which are divisible by 3 are 4! = 24.

iii) Divisible by 4 :
A number is divisible by 4 only when the number in last two places (i.e., ten’s and units) is a multiple of 4.
∴ The last two places should be filled with one of the following.
12, 16, 52, 56, 72, 76.
i.e., this can be done in 6 ways.
The remaining 2 places can be filled with remaining 3 digits in \({ }^3 \mathrm{P}_2\) ways.
∴ The number of 4 digit numbers which are divisible by 4 are 6 × \({ }^3 \mathrm{P}_2\)= 36.

iv) Divisible by 5 :
A number is divisible by 5, if the units place is 0 or 5.
Hence the units place from given digits is filled with ‘5’ only.
The remaining 3 places by remaining 4 digits can be filled in \({ }^4 \mathrm{P}_3\) ways.
The number of 4 digit numbers which are divisible by 5 are \({ }^4 \mathrm{P}_3\) = 24.

v) Divisible by 25 :
A 4 digit number formed by using given digits is divisible by 25 if the number formed by ten’s and unit’s places is either 25 or 75. This can be done in 2 ways.
Now the remaining 2 places with remaining 3 digits can be filled in \({ }^3 \mathrm{P}_2\) ways.
∴ The number of 4 digit numbers which are divisible by 25 are 2 × \({ }^3 \mathrm{P}_2\) = 12.

Question 5.
If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the ranks of the words
i) REMAST
ii) MASTER.
Solution:
Given word is MASTER.
∴ The alphabetical order of the letters is A, E, M, R, S, T.
In the dictionary order, first we write all words beginning with ‘A’.
Filling first place with ‘A’, the remaining 5 places can be filled with the remaining 5 letters in 5! ways.
∴ The number of words begin with ‘A’ are 5! on proceeding like this, we get.

i) Rank of the word REMAST :
The number of words begin with A is 5! = 120
The number of words begin with E is 5! = 120
The number of words begin with M is 5! = 120
The number of words begin with RA is 4! = 24
The number of words begin with REA is 3! = 6
The next word is REMAST.
∴ Rank of word REMAST = (120) 3 + 24 + 6 + 1 = 391.

ii) Rank of the word MASTER :
The number of words begin with A is 5! = 120
The number of words begin with E is 5! = 120
The number of words begin with MAE is 3! = 6
The number of words begin with MAR is 3! = 6
The number of words begin with MASE is 2! = 2
The number of words begin with MASR is 2! = 2
The next word is MASTER.
∴ Rank of word MASTER = 2 (120) + 2 (6) + 2 (2) + 1 = 257.

Question 6.
If the letters of the word BRING are per-muted in all possible ways and the words thus formed are arranged in the dictionary order, then find the 59th word.
Solution:
Given word is BRING.
∴ The alphabetical order of the letters is B, G, I, N, R.
In the dictionary order, first we write all words beginning with B.
Clearly the number of words beginning with B are 4! = 24.
Similarly the number of words begin with G are 4! = 24.
Since the words begin with B and G sum to 48, the 59th word must start with I.
Number of words given with IB = 3! = 6
Hence the 59th word must start with IG.
Number of words begin with 1GB = 2! = 2
Number of words begin with IGN = 2! = 2
∴ Next word is 59th = IGRBN.

Question 7.
Find the sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition.
Solution:
Given digits are 1, 2, 4, 5, 6.
Hence the number of four digit numbers formed using given digits with out repetition are \({ }^5 P_4\) = 120.
Out of these 120 numbers the numbers with 2 in unit’s place = \({ }^4 P_3\)
the numbers with 2 in ten’s place = \({ }^4 P_3\)
the numbers with 2 in Hundred’s place = \({ }^4 P_3\)
The numbers with 2 in thousands place = \({ }^4 P_3\)
∴ The value obtained by adding ‘2’ in all-the numbers = \({ }^4 P_3\) × 2 + \({ }^4 P_3\) × 20 + \({ }^4 P_3\) × 200 + \({ }^4 P_3\) × 2000
= \({ }^4 P_3\) × 2 × 1111
Similarly the value obtained by adding T in all the numbers is \({ }^4 P_3\) × 1 × 1111
The value obtained by adding 4 in all the numbers is \({ }^4 P_3\) × 4 × 1111
The value obtained by adding 5 in all the numbers is \({ }^4 P_3\) × 5 × 1111
The value obtained by adding 6 in all the numbers is \({ }^4 P_3\) × 6 × 1111
.\ The sum of all the numbers = \({ }^4 P_3\) × 1 × 1111 + \({ }^4 P_3\) × 2 × 1111 + \({ }^4 P_3\) × 4 × 1111 + \({ }^4 P_3\) × 5 × 1111 + \({ }^4 P_3\) × 6 × 1111
= \({ }^4 P_3\) × (1111) (1 + 2 + 4 + 5 + 6)
= 24 × 1111 × 18 = 4,79,952
The sum of all 4 digit numbers that can be formed with given digits is 4,79, 952.

Question 8.
There are 9 objects and 9 boxes. Out of 9 objects, 5 cannot fit in three small boxes. How many arrangements can be made such that each object can be put in one box only.
Solution:
Given that there are 9 objects and 9 boxes.
As 5 objects out of 9, cannot fit in 3 small boxes, these 5 objects should be arranged in . remaining 6 boxes.
This can be done in \({ }^6 \mathrm{P}_5\) ways.
∴ The remaining 4 blanks are to be filled with remaining 4 objects.
This can be done in 4! ways.
∴ The required number of ways = \({ }^6 \mathrm{P}_5\) × 4! = 17,280.

Students must practice this TS Intermediate Maths 2A Solutions Chapter 1 Complex Numbers Ex 1(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Exercise 1(a)

I. Question 1.
If z₁ =(2, – 1); z₂ = (6, 3) find z₁ – z₂.
Solution:
z₁ = 2 – i; z₂ = 6 + 3i
z₁ – z₂ = (2 – i) – (6 + 3i)
= (2 – 6) + (- 1 – 3)i
= – 4 – 4i (or)
z₁ – z₂ = (- 4, – 4).

Question 2.
If z₁ = (3, 5) and z₂ = (2, 6) find z₁ . z₂.
Solution:
z₁ = 3 + 5i; z₂ = 2 + 6i
z₁ . z₂ = (3 + 5i) (2 + 6i)
= 6 + 18i + 10i + 30i²
= 6 – 30 + 28i
= – 24 + 28i = (- 24, 28).

Question 3.
Write the additive inverse of the following complex nwnbers:
i) (√3, 5)
ii) (- 6, 5) + (10, – 4)
iii) (2, 1) (- 4, 6)
Solution:
i) z₁ = √3 + 5
a + ib is additive inverse then
z₁ + a + ib = 0 + 0i
(√3 + a) + (5 + b)i = 0 + 0i
a = – √3, b = – 5
∴ Additive inverse of z₁ is – √3 – 5i (- √3, – 5).

ii) z₁ = – 6 + 5i; z₂ = 10 – 4i
z₁ + z₂ = (- 6 + 10) + i (5 – 4) = 4 + i.
Additive inverse be a + ib
(z₁ + z₂) + a + ib = 0 + 0i
(4 + a) + (1 + b)i = 0 + 0i
a = – 4
b = – 1
∴ – 4 – i = (- 4, – 1)

iii) z₁ = 2 + i; z₂ = – 4 + 6i
z₁z₂ = (2 + i) (- 4 + 6i)
= – 8 + 12i – 4i + 6i²
= – 8 – 6 + 8i
= – 14 + 8i
z₁z₂ + a + ib = 0 + 0i
(- 14 + a) + (8 + b) i = 0 + 0i
a = 14, b = – 8
∴ Additive inverse of z₁z₂ be 14 – 8i or (14, – 8).

II. Question 1.
If z₁ = (6, 3); z₂ =(2,- 1) find z₁/z₂.
Solution:
z₁ = 6 + 3i; z₂ = 2 – i

Question 2.
If z = (cos θ, sin θ) find z – \(\frac{1}{z}\).
Solution:
z – \(\frac{1}{z}\) = (cos θ + i sin θ) – \(\)
= cos θ + i sin θ – \(\frac{(\cos \theta-i \sin \theta)}{(\cos \theta+i \sin \theta)(\cos \theta-i \sin \theta)}\)
= cos θ + i sin θ – \(\frac{(\cos \theta-i \sin \theta)}{\left(\cos ^2 \theta+\sin ^2 \theta\right)}\)
= 2i sin θ
= θ + 2i sin θ
= (0, 2 sin θ).

Question 3.
Write the multiplicative inverse of the following complex numbers:
i) (3, 4)
ii) (sin θ, cos θ)
iii) (7, 24)
iv)(- 2, 1)
Solution:
i) z₁ = 3 + 4i
a + ib is multiplicative inverse of z₁
z (a + ib) = 1 + 0i
a + ib = \(\frac{1}{\mathrm{z}_1}\)
a + ib = \(\frac{1}{3+4 i}\)
a + ib = \(\frac{3-4 i}{(3+4 i)(3-4 i)}\)
= \(\frac{3-4 i}{9+16}=\frac{3}{25}-\frac{4}{25} i\)
= \(\left(\frac{3}{25}, \frac{-4}{25}\right)\)

ii) z = sin θ + i cos θ
a + ib is multiplicative inverse of z₁
z₁ . (a + ib) = 1 + 0i
a + ib = \(\frac{1}{\mathrm{z}_1}\)
= \(\frac{1}{\sin \theta+i \cos \theta}\)
= \(\frac{\sin \theta-i \cos \theta}{(\sin \theta+i \cos \theta)(\sin \theta-i \cos \theta)}\)
= \(\frac{\sin \theta-i \cos \theta}{\sin ^2 \theta+\cos ^2 \theta}\)
= (sin θ, – cos θ).

iii) z₁ = 7 + 24i
a + ib is multiplicative inverse of z₁
z₁ . a + ib = 1 + 0i
a + ib = \(\frac{1}{7+24 i}\)
= \(\frac{7-24 \mathrm{i}}{(7+24 \mathrm{i})(7-24 \mathrm{i})}\)
= \(\frac{7-24 i}{49+576}\)
= \(\frac{7}{625}-\frac{24}{625} 1=\left(\frac{7}{625}, \frac{-24}{625}\right)\)

iv) z₁ = – 2 + i
z₁ . (a + ib) = 1 + oi
z₁ = \(\frac{1}{a+i b}\)
a + ib = \(\frac{1}{z_1}\)
= \(\frac{1}{-2+1}\)
= \(\frac{-2-i}{(-2+i)(-2-i)}\)
= \(\frac{-2-i}{4+1}\)
= \(\left(\frac{-2}{5}, \frac{-1}{5}\right)\)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Exercise 3(a)

I.
Question 1.
Find the roots of the following equations.
i) x² – 7x + 12 = 0
ii) – x² + x + 2 = 0
iii) 2x² + 3x + 2 = 0
iv) √3x² + 10x – 8√3 = 0
v) 6√5x² – 9x – 3√5 = 0
Solution:
x² – 7x + 12 = 0
(x – 4) (x – 3) = 0
x = 4, 3.

ii) – x² + x + 2 = 0
x² – x – 2 = 0
(x – 2) (x + 1) = 0
x = 2, – 1

iii) 2x² + 3x + 2 = 0
x = \(\frac{-3 \pm \sqrt{9-16}}{4}\)
x = \(\frac{-3 \pm \sqrt{7} i}{4}\)

iv) √3x² + 10x – 8√3 = 0

v) 6√5x² – 9x – 3√5 = 0

Question 2.
Form quadratic equations whose roots are
i) 2, 5
ii) \(\frac{\mathbf{m}}{\mathbf{n}}, \frac{-\mathbf{n}}{\mathbf{m}}\) (m ≠ 0, n ≠ 0)
iii) \(\frac{\mathbf{p}-\mathbf{q}}{\mathbf{p}+\mathbf{q}}, \frac{-(\mathbf{p}+\mathbf{q})}{\mathbf{p}-\mathbf{q}}\) (p ≠ ± q)
iv) 7 ± 2√5
v) – 3 ± 5i
Solution:
i) α = 2, β = 5 → roots then quadratic equation be
(x – α) (x – β) = 0
(x – 2) (x – 5) = 0
x² – 7x + 10 = 0.

ii) α = \(\frac{\mathrm{m}}{\mathrm{n}}\), β = – \(\frac{\mathrm{n}}{\mathrm{m}}\)
(x – \(\frac{\mathrm{m}}{\mathrm{n}}\)) (x + \(\frac{\mathrm{n}}{\mathrm{m}}\)) = 0
x + x (\(\frac{\mathrm{n}}{\mathrm{m}}-\frac{\mathrm{m}}{\mathrm{n}}\)) – \(\frac{m}{n} \cdot \frac{n}{m}\) = 0
x² + x \(\frac{\left(n^2-m^2\right)}{n m}\) – 1 = 0
mnx² + x (n² – m²) – nnm = 0
x² – x(α + β) + αβ = 0.

iii) x² – \(\left(\frac{p-q}{p+q}-\frac{p+q}{p-q}\right)\) x – \(\left(\frac{p+q}{p-q}\right)\left(\frac{p-q}{p+q}\right)\) = 0
x – \(\left(\frac{(p-q)^2-(p+q)^2}{p^2-q^2}\right)\)x – 1 = 0
x + \(\frac{4 p q}{p^2-q^2}\)x – 1 = 0
(p² – q²)x² + 4px – (p² – q²) = 0.

iv) x² – x(7 + 2√5 + 7 – 2√5)+ (7 + 2√5) (7 – 2√5) = 0
x² – x(14) + (49 – 20) = 0
x² – 14x + 29 = 0.

v) x² – x (- 3 + 5i – 3 – 5i) + (- 3 + 5i) (- 3 – 5i) = 0
x² – x (- 6) + (9 + 25) = 0
x² + 6x + 34 = 0.

Question 3.
Find the nature of the roots of the following equations without finding the roots.
i) 2x² – 8x + 3 = 0
ii) 9x² – 30x + 25 = 0
iii) x² – 12x + 32 = 0
iv) 2x²2 – 7x + 10 = 0
Solution:
i) 2x² – 8x + 3 = 0
b² – 4ac = 64 – 24 > 0
roots are real and distinct.

ii) 9x² – 30x + 25
b² – 4ac = 900 – 4 × 9 × 25 = 0
Roots are real and equal.

iii) x² – 12x + 32 = 0
(12)² – 4 × 32 > 0
Roots are real and distinct.

iv) 2x² – 7x 10 = 0
(- 7)² – 4 × 2 × 10 < 0
Roots are imaginary.

Question 4.
If α, β are the roots of the equation ax² + bx + c = 0, find the value of following expressions in terms of a, b, c.
i) \(\frac{1}{\alpha}+\frac{1}{\beta}\)
ii) \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\)
iii) α⁴β⁷ + α⁷β⁴
iv) \(\left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2\), if c ≠ 0
v) \(\frac{\alpha^2+\beta^2}{\alpha^{-2}+\beta^{-2}}\), if c ≠ 0.
Solution:
i) α + β = \(\frac{-b}{a}\);
αβ = \(\frac{c}{a}\)
\(\frac{\alpha+\beta}{\alpha \beta}=\frac{\frac{-b}{a}}{\frac{c}{a}}=\frac{-b}{c}\)

ii) \(\frac{\alpha^2+\beta^2}{\alpha^2 \beta^2}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha^2 \beta^2}\)
= \(\frac{\frac{b^2}{a^2}-\frac{2 c}{a}}{\frac{c^2}{a^2}}\)
= \(\frac{b^2-2 c a}{c^2}\)

iii) α⁴β⁷ + α⁷β⁴
= α⁴β⁴ (β³ + α³)
= (αβ)⁴ [(α + β)³ – 3αβ (α + β)]
= \(\frac{c^4}{a^4}\left[\frac{-b^3}{a^3}+\frac{3 c}{a} \cdot \frac{b}{a}\right]\)
= \(\frac{c^4}{a^4}\left[\frac{3 a b c-b^3}{a^3}\right]\)

iv) \(\left(\frac{\alpha}{\beta}-\frac{\beta}{\alpha}\right)^2\)

v) \(\frac{\alpha^2+\beta^2}{\alpha^{-2}+\beta^{-2}}\)
= α²β²
= \(\frac{c^2}{a^2}\).

Question 5.
Find the values of m for which the following equations have equal roots.
i) x² – m(2x – 8) – 15 = 0
ii) (m + 1) x² + 2 (m + 3) x + (m + 8) = 0
iii) x² + (m + 3) x + (m + 6) = 0
iv) (3m + 1) x² + 2 (m + 1) x + m = 0
v) (2m + 1) x² + 2 (m + 3) x + m + 5 = 0
Solution:
i) x² – 2xm + 8m – 15 = 0
b² – 4ac = 0
(- 2m)² – 4 (8m – 15) = 0
4m² – 32m + 60 = 0
m² – 8m + 15=0
(m – 5) (m – 3) = 0
m = 3, 5.

ii) (m + 1)x² + 2(m + 3)x + (m + 8) = 0
b² – 4ac = 0
4(m + 3)² – 4(m + 8) (m + 1) = 0
(m + 3)² (m² + 9m + 8) = 0
6m + 9 – 9m – 8 = 0
– 3m + 1 = 0
m = \(\frac{1}{3}\).

iii) x² + (m + 3) x + (m + 6) = 0
b² – 4ac = 0
(m + 3)² – 4 (m + 6) = 0
m² e 6m +9-4m-24-0
m² + 2m – 15 = 0
(m + 5) (m – 3) = 0
m = 3, – 5.

iv) (3m + 1)x² + 2(m + 1)x + m = 0
b² – 4ac = 0
4(m + 1)² – 4m(3m + 1) = 0
m² + 2m + 1 – 3m² – m = 0
– 2m² + m + 1 = 0
2m² – m – 1 = 0
2m² – 2m + m – 1 = 0
2m (m – 1) + (m – 1) = 0
(m – 1) (2m + 1)= 0
m = 1, m = – \(-\frac{1}{2}\)

v) (2m + 1)x² + 2(m + 3)x + (m + 5) = 0.
b² – 4ac = 0
4(m + 3)² – 4(2m + 1) (m + 5) = 0
m² + 6m + 9 – 2m² – 11m – 5 = 0
– m² – 5m + 4 = 0
m² + 5m – 4 = 0
m = \(\frac{-5 \pm \sqrt{25+16}}{2}\)
m = \(\frac{-5}{2}\).

Question 6.
If α and β are the roots of equation x² + px + q = 0, form a quadratic equation where roots are (α – β)² and (α + β)².
Solution:
α + β = \(\frac{-b}{a}\); αβ = \(\frac{c}{a}\)
x² – [(α – β)² + (α + β)²]x + [(α – β) (α + β)]² = 0
x² – [2(α² + β²)]x + [α + β]² [(α + β)² – 4αβ] = o
x² – 2[(α – β)² – 2αβ]x + (α + β)² [(α + β)² – 4αβ] = o
x² – 2 \(\left[\frac{b^2-2 a c}{a^2}\right]\) x + \(\frac{b^2}{a^2}\left[\frac{b^2-4 a c}{a^2}\right]\) = 0
Here b = p, c = q, a = 1
x² – 2 (p² – 2q) x + p² (p² – 4q) = 0

Question 7.
If x² + bx + c = 0, x² + cx + b = 0 (b ≠ c) have a common root, then show that b + c + 1 = 0.
Solution:
x² + bx + c = 0
x² + cx + b = 0
α² is common root.
α² + bα + c = 0
α² + cα + b = 0
α (b – c) + c – b = 0
α (b – c) = b – c
α = 1
∴ 1 + b + c = 0.

Question 8.
Prove that roots of (x – a) (x – b) = h² are always real.
Solution:
(x – a) (x – b) = h²
x² – x (a + b) + ab – h² = 0
Discriminant = (a + b)² – 4 (ab – h²)
= (a + b)² – 4ab + 4h²
= (a – b)² + 4h² > 0
∴ Roots are real.

Question 9.
Find the condition that one root of the quadratic equation ax² + bx + c = 0 shall be n times the other, where n is positive integer.
Solution:
α + nα = – b/a
α . nα = \(\frac{c}{a}\)
α = \(\frac{-b}{a(n+1)}\)
\(\frac{n b^2}{a^2(n+1)^2}=\frac{c}{a}\)
nb² = ac (n + 1)²

Question 10.
Find two consecutive potive even Integers, the sum of whose squares in 340.
Solution:
2n, 2n + 2
(2n)² + (2n + 2)² = 340
4n² + 4n² + 8n + 4 = 340
8n² + 8n + 4 = 340
2n² + 2n + 1 = 85
2n² + 2n – 84 = 0
n² + n – 42 = 0
(n + 7) (n – 6) = 0
n = 6
12, 14 are two numbers.

II.
Question 1.
If x₁, x₂ are the roots of equation ax² + bx + c = 0 and c ≠ 0. Find the value of (ax₁ + b)^-2 + (ax₂ + b)^-2 in terms of a, b, c.
Solution:
ax₁² + bx₁ + c = 0
x₁ (ax₁ + b) + c = 0
(ax₁ + b) = \(\frac{-\mathrm{c}}{\mathrm{x}_1}\)
Similarly (ax₂ + b) = \(\frac{-\mathrm{c}}{\mathrm{x}_2}\)
∴ (ax₁ + b)^-2 + (ax₂ + b)^-2 = \(\)
= \(\frac{1}{c^2}\) [(x₁ + x₂)² – 2x₁x₂]
= \(\frac{1}{c^2}\left[\frac{b^2-2 a c}{a^2}\right]\)
= \(\frac{b^2-2 a c}{a^2 c^2}\)

Question 2.
If α, β are the roots of equation ax² + bx + c = 0, find a quadratic equation whose roots are α² + β² and α^-2 + β^-2.
Solution:

Solve the following equations:

Question 3.
2x⁴ + x³ – 11x² + x + 2 = 0
Solution:
2x⁴ + x³ – 10x² – x² + x + 2 = 0
x² (2x² + x – 10) – (x² – x – 2) = 0
x² [(2x + 5) (x – 2) – [(x – 2) (x + 1)] = 0
(x – 2) [x² (2x + 5) – x – 1] = 0
(x – 2) [2x³ + 5x² – x – 1] = 0
(x – 2) [(2x – 1) (x² + 3x + 1)] = 0
x = 2, \(\frac{1}{2}\); x² + 3x + 1 = 0
x = \(\frac{-3 \pm \sqrt{9-4}}{2}\)
x = \(\frac{-3 \pm \sqrt{5}}{2}\).

Question 4.
3^1+x + 3^1-x = 10
Solution:
Let 3^x = t
3. t + \(\frac{3}{t}\) = 10
3t² + 3 – 10t = 0
3t² – 10t + 3 = 0
3t² – 9t – t + 3 = 0
3t (t – 3) – 1 (t – 3) = 0
(3t – 1) (t – 3) = 0
t = \(\frac{1}{3}\), t = 3
3^x = 3^-1, 3^x = 3¹
x = – 1, 1.

Question 5.
4^{x – 1} – 3 . 2^{x – 1} + 2 = 0.
Solution:
\(\frac{4^x}{4}-\frac{3 \cdot 2^x}{2}\) + 2 = 0
4^x – 6 . 2^x + 8 = 0
2^x = t
t² – 6t + 8 = 0
(t – 4) (t – 2) = 0
2^x = t
t² – 6t + 8 = 0
(t – 4) (t – 2) = 0
2^x = 2²; 2^x = 2¹
x = 1, 2.

Question 6.
\(\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}\) when x ≠ 0, x ≠ 3.
Solution:
\(\sqrt{\frac{x}{x-3}}\) = t
t + \(\frac{1}{t}=\frac{5}{2}\)
2t² – 5t + 2 = 0
2t² – 4t – t + 2 = 0
2t (t – 2) – 1 (t – 2) = 0
(2t – 1) (t – 2) = 0
t = \(\frac{1}{2}\); t = 2
\(\frac{x}{x-3}\) = 4
x = 4x – 12
12 = 3x
x = 4
(or)
\(\frac{x}{x-3}=\frac{1}{4}\)
4x = x – 3
3x = – 3
x = – 1.

Question 7.
\(\sqrt{\frac{3 x}{x+1}}+\sqrt{\frac{x+1}{3 x}}\) = 2 when x ≠ 0, x ≠ 1.
Solution:
Let \(\sqrt{\frac{3 x}{x+1}}\) = t
t + \(\frac{1}{t}\) = 2
t² – 2t + 1 = 0
(t – 1)² = 0
t = 1
\(\frac{3 x}{x+1}\) = 1
3x = x + 1
x = \(\frac{1}{2}\).

Question 8.
2 \(\left(x+\frac{1}{x}\right)^2\) – 7 \(\left(x+\frac{1}{x}\right)\) + 5 = 0, when x ≠ 0.
Solution:
x + \(\frac{1}{x}\) = t
2t² – 7t + 5 = 0
2t² – 5t – 2t + 5 = 0
2t (t – 1) – 5 (t – 1) = 0
t = \(\frac{5}{2}\), t = 1 but x + \(\frac{1}{x}\) ≥ 2 ∀ x ∈ R⁺
x + \(\frac{1}{x}\) = \(\frac{5}{2}\) only possible
2x² – 5x + 2 = 0
(2x – 1) (x – 2) = 0
x = \(\frac{1}{2}\), 2
Now if x + \(\frac{1}{x}\) = 1
x² – x + 1 = o
x = \(\frac{1 \pm \sqrt{1-4}}{2}\)
x = \(\frac{1 \pm \sqrt{3} i}{2}\).

Question 9.
\(\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)\) + 6 = 0 when x ≠ o.
Solution:
\(\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)\) + 6 = 0
x + \(\frac{1}{x}\) = t
t² – 5t + 4 = 0
(t – 4) (t – 1) = 0
t = 4, 1
x + \(\frac{1}{x}\) = 4
x² – 4x + 1 = 0
x = \(\frac{4 \pm \sqrt{16-4}}{2}\)
x = 2 ± √3
x + \(\frac{1}{x}\) = 1
x² – x + 1 = 0
x = \(\frac{1 \pm \sqrt{1-4}}{2}\)
x = \(\frac{1 \pm \sqrt{3} i}{2}\).

Question 10.
Find a quadratic equation for which the sum of the roots is 7 and the sum of the squares of the roots is 25.
Solution:
α + β = 7, α² + β² = 25
(α + β)² – 2αβ= 25
49 – 2αβ = 25
24 = 2αβ
αβ = 12
x² – (7)x + 12 = 0.

Students must practice this TS Intermediate Maths 2A Solutions Chapter 1 Complex Numbers Ex 1(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Exercise 1(b)

I.
Question 1.
Write the following complex numbers in the form A + iB.
i) (2 – 3i) (2 + 3i)
ii) (1 + 2i)³
iii) \(\frac{a-i b}{a+i b}\)
iv) \(\frac{4+3 i}{(2+3 i)(4-3 i)}\)
v) (- √3 + √-2) (2√3 – i)
vi) (- 5i) \(\frac{i}{8}\)
vii) (- i) (2i)
viii) i⁹
ix) i^-19
x) 3 (7 + 7i) + i (7 + 7i)
xi) \(\frac{2+5 i}{3-2 i}+\frac{2-5 i}{3+2 i}\)
Solution:
i) z = (2 – 3i) (3 + 4i)
z = 6 + 8i – 9i + 12
z = 18 – i
= 18 + (- 1)i

ii) z = (1 + 2i)³
= 1³ + (2i)³ + 3 . 2i(1 + 2i)
= 1 – 8i + 6i (1 + 2i)
= (1 – 12) – 2i
= – 11 – 2i
= (- 11, – 2)

iii) z = \(\frac{a-i b}{a+i b}\)
z = \(\frac{(a-i b)(a-i b)}{(a+i b)(a-i b)}\)
= \(\frac{a^2-b^2-2 a b i}{a^2+b^2}\)
= \(\left(\frac{a^2-b^2}{a^2+b^2}, \frac{-2 a b}{a^2+b^2}\right)\)

iv) \(\frac{4+3 i}{(2+3 i)(4-3 i)}\)

v) (- √3 + √-2) (2√3 – i)
z = (- √3 + √2i) (2√3 – i)
= – 2 . 3 + √3i + 2√6i + √2
= (√2 – 6) + i (√3 + 2√6).

vi) z = (- 5i) (\(\frac{i}{8}\))
z = \(\) i²
z = \(\frac{5}{8}\) + 0i

vii) (- i) (2i)
z = (- i) (2i)
= 2 + 0i

viii) z = i⁹
z = (i²)⁴ . i
z = i = 0 + i . 1

ix) z = i^-19
z = \(\frac{1}{\left(\mathrm{i}^2\right)^9} \cdot \frac{1}{\mathrm{i}}\)
= \(\frac{-1}{i}=\frac{-1}{1^2}\)
z = i = 0 + 1 . i

x) z = 3 (7 + 7i) + i (7 + 7i)
= 21 + 21i + 7i + 7i²
= 21 – 7 + 28i
z = 14 + 28i

xi) \(\frac{2+5 i}{3-2 i}+\frac{2-5 i}{3+2 i}\)

Question 2.
Write the conjugate of the following complex numbers.
i) (3 + 4i)
ii) (15 + 3i) – (4 – 20i)
iii) (2 + 5i) (- 4 + 6i)
iv) \(\frac{5 \mathbf{i}}{7+1}\)
Solution:
i) z = 3 + 4i
\(\overline{\mathbf{z}}\) = 3 – 4i

ii) z = (15 + 31) – (4 – 201)
z = 11 + 23i
z = 11 – 23i

iii) z = (2 + 5i) (- 4 + 6i)
z = – 8 + 12i – 20i + 30i2
z = – 38 – 8i
z = – 38 + 8i

iv) \(\frac{5 \mathbf{i}}{7+1}\)

Question 3.
Simplify
i) i² + i⁴ + i⁶ + ………… + (2n + 1) terms
ii) i¹⁸ – 3 . i⁷ + i² (1 + i⁴) (- i)²⁶
Solution:
i) i² + i⁴ + i⁶ + …………… (2n + 1) terms
= i² + i⁴ + i⁶ + ……………. +
= -1 + 1+(-1) + 2n terms + i2
= 0 – 1 = – 1

ii) i¹⁸ – 3i⁷ + i² (1 + i⁴) (- i)²⁶
(i²)⁹ – 3i⁶i + i² (1 + i⁴) (i)²⁶ = – 1 + 3i + (- 1) (1 + 1) (- 1)¹³
= 3i – 1 + 2 = 3i + 1.

Question 4.
Find a square root for the following complex numbers.
i) 7 + 24i
ii) – 8 – 6i
iii) 3 + 4i
iv) – 47 + i 8√3
Solution:
i) z = 7 + 24i
Let square root of z be a + ib
a + ib = \(|\sqrt{7+24i}|\)
(a + ib)² = 7 + 24i
a² – b² + 2abi = 7 + 24i
a² – b² = 7, 2ab = 24 ……………..(1)
| a + ib | = \(|\sqrt{7+24i}|\)
Squaring on both sides,
| a + ib |2 = | 7 + 24i|
a² + b² = \(\sqrt{49+576}\)
a² + b² = \(\sqrt{625}\)

a² = 16
a = ± 4
2b² = 25 – 7
b² = 9
b = ± 3
a + ib = ± (4 + 3i)

ii) z = – 8 – 6i
Let square root of z be a + ib
a + ib = \(\sqrt{-8-6 \mathrm{i}}\)
Squaring on both sides,
(a + ib)² = – 8 – 61
a² – b² + 2abi = – 8 – 6i
a² – b² = – 8, 2ab = – 6 …………..(i)
| a + ib | = \(|\sqrt{-8-6 \mathbf{i}}|\)
Squaring on both sides,
| a + ib |² = \(|\sqrt{-8-6 \mathrm{i}}|^2\)
a² + b² = \(\sqrt{64+36}\)
a² + b² = 10 ………….(ii)
(i) + (ii)
2a² = 2
a² = 1 or a = ± 1
(ii) – (i)
2b² = 18
b² = 9
⇒ b = ± 3
a + ib = ± (1 – 3i).

iii) z = 3 + 4i
Let square root of z be a + ib
a + ib = \(\sqrt{3+4i}\)
Squaring on both sides,
(a + ib)² = 3 + 4i
a² – b² + 2abi = 3 + 4i
a² – b² = 3; 2ab = 4 …………..(i)
| a + ib | = \(|\sqrt{3+4 \mathrm{i}}|\)
Squaring on both sides,
| a + ib |² = | 3 + 4i |
a² + b² = \(\sqrt{9+16}\) …………..(ii)
a² – b² = 3

2a² = 8
a² = 4
⇒ a = ± 2
2b² = 2
⇒ b = ± 1
a + ib = ± (2 + i).

iv) z = – 47 + i 8√3
Let the square root of z be a + ib
(a + ib)² = – 47 + i 8√3
a² – b² = – 47, 2ab = 8√3
| a + ib | = \(|\sqrt{-47+i 8 \sqrt{3}}|\)
Squaring on bothsides,
a² + b² = \(\sqrt{(-47)^2+(8 \sqrt{3})^2}\)
= \(\sqrt{2209+192}=\sqrt{2401}\)

a² + b² = 49
a² – b² = – 47
2a² = 2
a² = 1
a = ± 1
2b² = 96
b² = 48
b = ± 4√3
a + ib = ± (1 + 4√3i).

Question 5.
Find the multiplicative inverse of the following complex numbers.
i) √5 + 3i
ii) – i
iii) i^-35
Solution:
i) z = √5 + 3i
Let a + ib be multiplicative inverse then (a + ib) . z = 1
z = \(\frac{1}{a+i b}\)
or a + ib = \(\frac{1}{\mathrm{z}}\)
a + ib = \(\frac{\overline{\mathbf{z}}}{(\mathrm{z} \overline{\mathrm{z}})}\)
a + ib = \(\frac{\bar{z}}{|z|^2}=\frac{\sqrt{5}-3 i}{5+9}\)
= \(\frac{1}{14}\) (√5 – 3i).

ii) z = – i
Let a + ib be multiplicative inverse then (a + ib) z = 1
a + ib = \(\frac{1}{\mathrm{z}}\)
= \(\frac{1}{-i}\)
= \(\frac{i}{-i \cdot i}\)
a + ib = i.

iii) z = i^-35
Let a + ib be multiplicative inverse then (a + ib) z = 1
a + ib = \(\frac{1}{\mathrm{z}}\)
= \(\frac{1}{i^{-35}}\) = i³⁵
(a + ib) = i³⁵
= (i²)¹⁷ i = – i.

II.
Question 1.
i) If (a + ib)² = x + iy, find x² + y².
ii) lf x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\) then, show that x² + y² = 4x – 3.
iii) If x + iy = \(\frac{1}{1+\cos \theta+i \sin \theta}\) then, show that 4x² – 1 = 0.
iv)If u + iv = \(\frac{2+i}{z+3}\) and z = x + iy find u, v.
Solution:
i) (a + ib) = x + iy
a² – b² + 2abi = x + iy ,
a² – b² = x
2ab = y
Now x2 = (a² – b²)²
y² = 4a²b²
x² + y² = (a² – b²)² . 4a²b² = (a² + b²)²

ii) x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\)

iii) x + iy = \(\frac{1}{1+\cos \theta+i \sin \theta}\)
x + iy = \(\frac{1+\cos \theta-i \sin \theta}{(1+\cos \theta)^2+\sin ^2 \theta}\)
x = \(\frac{1+\cos \theta}{2+2 \cos \theta}=\frac{(1+\cos \theta)}{2(1+\cos \theta)}\)
x = \(\frac{1}{2}\)
2x = 1
4x² = 1
4x² – 1 = 0.

iv) u + iv = \(\frac{2+i}{z+3}\)

Question 2.
i) If z = 3 + 5i then show that z³ – 10z² + 58z – 136 = 0
ii) If z = 2 – i√7 then show that 3z³ – 4z² + z + 88 = 0
iii) Show that \(\frac{2-i}{(1-2 i)^2}\) and \(\frac{-2-11 i}{25}\) are conjugate to each other.
Solution:
i) z = 3 + 5i
(z – 3)² = (5i)²
z² – 6z + 9 = 25i2
z² – 6z + 9 = – 25
z² – 6z + 34 = 0
z³ – 6z² + 34z = 0
(z³ – 10z² + 58z – 136) + 4z² – 24z + 136 = 0
(z³ – 10z² + 58z – 136) + 4 (z² – 6z + 34) = 0
(z³ – 10z² + 58z – 136) = 0

il) z = 2 – i√7
(z – 2)² = (- i – √7)²
z² – 4z + 4 = – 7
z² – 4z + 11 = 0
z³ – 4z² + 11z = 0
3z³ – 12z² + 33z = 0
(3z³ – 4z² + z + 88) + (- 8z² + 32z – 88) = 0
(3z³ – 4z² + z + 88) – 8 (z² – 4z + 11) = 0
3z³ – 4z² + z + 88 = 0.

iii) z = \(\frac{2-i}{(1-2 i)^2}\) (If z = a + ib, \(\overline{\mathbf{Z}}\) = a – ib)

Question 3.
i) If (x – iy)^1/3 = a – ib then show that \(\frac{x}{a}+\frac{y}{b}\) = 4(a² – b²)
ii) Write \(\left(\frac{a+i b}{a-i b}\right)^2-\left(\frac{a-i b}{a+i b}\right)^2\) in the form x + iy.
iii) If x and y are real numbers such that \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}\) determine the values of x and y.
Solution:
(x – iy)^1/3 = (a – ib)
x – iy = (a – ib)³
x – iy = a³ + ib³
x = a³ – 3ab²
– iy = ib³ – 3a²bi
\(\frac{x}{a}\) = a² – 3b²
y = b³ – 3a²b
\(\frac{x}{a}\) = a² – 3b²
\(\frac{y}{b}\) = b² – 3a²
\(\frac{x}{a}-\frac{y}{b}\) = 4(a² – b²).

ii) \(\left(\frac{a+i b}{a-i b}\right)^2-\left(\frac{a-i b}{a+i b}\right)^2\)

iii) \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}\)

4x + 9y – 3 = 0
2x – 7y – 3 = 10
Solving, we get
x = 3, y = – 1.

Question 4.
i) Find the least positive integer n, satisfying \(\left(\frac{1+1}{1-1}\right)^n\) = 1.
ii) If \(\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3\) = x + iy find x and y.
iii) Find the real values of θ in order that \(\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\) is a
a) real number
b) purely imaginary number
iv) Find the real values of x and y if \(\frac{x-1}{3+i}+\frac{y-1}{3-i}\) = i.
Solution:
i) \(\left(\frac{1+i}{1-i}\right)^n\) = 1
\(\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^n\) = 1
\(\left(\frac{2 \mathrm{i}}{2}\right)^{\mathrm{n}}\) = 1
iⁿ = 1
(∵ iⁿ = 1 = – 1 × – 1 = i² × i² = i⁴)
n = 4
i⁴ = 1
Least value of n = 4.

ii) \(\left(\frac{1+i}{1-i}\right)^3-\left(\frac{i-i}{1+i}\right)^3\) = x + iy
\(\left(\frac{(1+i)(1+i)}{(1+i)(1-i)}\right)^3-\left(\frac{(1-i)(1-i)}{(1+i)(1-i)}\right)^3\) = x + iy
\(\left(\frac{2 \mathrm{i}}{2}\right)^3-\left(\frac{-2 \mathrm{i}}{2}\right)^3\)
– i – i = x + iy
x = 0
y = – 2.

iii) z = \(\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\)
z = \(\frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1-2 i \sin \theta)(1+2 i \sin \theta)}\)
z = \(\frac{3-4 \sin ^2 \theta+8 i \sin \theta}{1+4 \sin ^2 \theta}\)
z is purely real ⇒ imaginary part = 0
\(\frac{8 \sin \theta}{1+4 \sin ^2 \theta}\) = 0
sin θ = 0
θ = nπ, n ∈ 1
z is purely imaginary ⇒ Real part zero
\(\frac{3-4 \sin ^2 \theta}{1+4 \sin ^2 \theta}\) = 0
sin θ = \(\frac{3}{4}\)
sin θ = \(\pm \frac{\sqrt{3}}{2}\)
sin θ = 2nπ ± \(\frac{\pi}{3}\), n ∈ 1.

iv) \(\frac{x-1}{3+i}+\frac{y-1}{3-1}\) = 1
\(\frac{(x-1)(3-i)}{9+1}+\frac{(y-1)(3+i)}{9+1}\) = 1
\(\frac{3(x-1)+3(y-1)}{10}+\frac{i(1-x+y-1)}{10}\) = 1
3x + 3y – 6 = 0
y – x = 10
Solving we get
x = – 4, y = 6.

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Students must practice this TS Intermediate Maths 2A Solutions Chapter 2 De Moivre’s Theorem Ex 2(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 2 De Moivre’s Theorem Exercise 2(a)

I.
Question 1.
If n is an integer then show that (1 + i)²ⁿ + (1 – i)²ⁿ = 2^{n + 1} cos \(\frac{n \pi}{2}\)
Solution:
L.H.S = (1 + i)²ⁿ + (1 – i)²ⁿ

Question 2.
Find the values of the following:
i) (1 + √3)³
ii) (1 – i)⁸
iii) (1 + i)¹⁶
iv) \(\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5-\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5\)
Solution:
i) z = (1 + i√3)³
= (\(\left(\frac{1}{2}+\mathrm{i} \frac{\sqrt{3}}{2}\right)^3\))³ . 2³
= [cos \(\frac{\pi}{3}\) + i sin \(\frac{\pi}{3}\)]³ . 2³
=[cos \(\frac{\pi}{3}\) . 3 + i sin \(\frac{\pi}{3}\) . 3] . 2³
= 8 [cos π + i sin π]
∴ z = – 8.

ii) z = (1 – i)⁸
z = \(\left[(\sqrt{2})\left[\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}}\right]\right]^8\)
z = (2)⁴ [cos \(\frac{\pi}{2}\) – i sin \(\frac{\pi}{2}\)]⁸
z = 16 [cos 2π – i sin 2π]
z = 16.

iii) z = (1 + i)¹⁶
z = \(\left[(\sqrt{2})\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\right]^{16}\)
z = 2⁸ [(cos \(\frac{\pi}{4}\) + i sin \(\frac{\pi}{4}\))¹⁶]
z = 2⁸ [cos 4π + i sin 4π]
z = 256.

iv) \(\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5-\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5\)

II.
Question 1.
If α, β are roots of the equation x² – 2x + 4 = 0 then for any n ∈ N show that αⁿ + βⁿ = 2ⁿ⁺¹ cos \(\frac{n \pi}{3}\).
Solution:
x² – 2x + 4 = 0
x = \(\frac{2 \pm \sqrt{4-16}}{2}\)
x = \(\frac{2 \pm 2 \sqrt{3} i}{2}\)
x = 1 ± √3i
α = 1 + √3i ; β = 1 – √3i

Question 2.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ then show that
i) cos 3α + cos 3β + cos 3γ = 3cos (α + β + γ)
ii) sin 3α + sin 3β + sin 3γ = 3sin(α + β + γ)
iii) cos (α + β) + cos (β + γ) + cos (γ + α) = 0
Solution:
1) cos α + cos β + cos γ = 0
sin α + sin β + sin γ = 0
(cos α + i sin α) + (cos β + i sin β) + (cos γ + i sin γ) = 0
A + B + C = 0
A³ + B³ + C³ = 3ABC
A = e^iα, B = e^iβ, C = e^iγ
A³ + B³ + C³ = e^3αi + e^3βi + e^3γi ……………(1)
= cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + isin 3γ …………..(1)
3ABC = 3e^{i(α + β + γ)}
= 3[cos (α + β + γ) + isin (α + β + γ)] …………..(2)
(1) = (2)
Comparing real and Imaginary parts
cos 3α + cos 3β . cos 3γ = 3 cos (α + β + γ)
sin3α. sin 3β + Sin 3γ = 3 sin (α + β + γ).

ii) cos (α + β) cos (β + γ) cos (γ + α) = 0
A + B + C = 0
\(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\) = 0
AB + BC + CA = 0
e^{i(α + β)} + e^{i(β + γ)} + e^{i(α + γ)} = o
cos (α + β) + i sin (α + β) + cos(β + γ) + isin(β + γ) + cos (α + γ) + isin (α + γ) = 0 + 0i
cos (α + β) cos (β + γ)) + cos (α + γ) = 0
sin (α + β) sin (β + γ)) + sin (α + γ) = 0.

Question 3.
If n is an integer and z = cis θ, (θ ≠ (2n + 1)\(\frac{\pi}{2}\)], then show that \(\frac{z^{2 n}-1}{z^{2 n}+1}\) = i tan nθ
Solution:
z = e^iθ
z²ⁿ = (e^iθ)²ⁿ
z²ⁿ = e^2nθi
z²ⁿ = cos 2nθ + isin2nθ – 1
z – 1 = cos 2nθ + isin 2nθ – 1
= – 2 sin²nθ + 2i sin nθ . cos nθ
= i 2 sin nθ [cos nθ + i sin nθ]
= 2i sin nθ [cos nθ + i sin nθ] …………..(1)
z²ⁿ + 1 = cos 2nθ + i sin 2nθ + 1
= 2 cos² nθ + 2 i sin nθ cos nθ
= 2 cos nθ [cos nθ + i sin bθ] ……………..(2)
\(\frac{z^{2 n}-1}{z^{2 n}+1}=\frac{2 i \sin n \theta}{2 \cos n \theta}\) = i tan nθ.

Question 4.
If (1 + x)ⁿ = a₀ + a₁x + a₂x² + ………….. + a_nxⁿ, then show that
i) a₀ – a₂ + a₄ – a₆ + ………….. = 2^n/2 cos \(\frac{n \pi}{4}\)
ii) a₁ – a₃ + a₅ ……………. = 2^n/2 sin \(\frac{n \pi}{4}\)
Solution:
(1 + x)ⁿ = a₀ + a₁x + a₂x² + ………….. + a_nxⁿ
(1 + i) = a₀ + a₁i + a₂i² + ………………… + a_niⁿ
(√2)ⁿ \(\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\)ⁿ = (a₀ – a₂ + a₄ – ………….) + i(a₁ – a₃ …………………)
Equating Real parts both sides
(√2)ⁿ cos \(\frac{n \pi}{4}\) = a₀ – a₂ + a₄ ……………
Equating Imaginary parts
(√2)ⁿ sin \(\frac{n \pi}{4}\) = a₁ – a₃ + a₅ ………………

Students must practice this TS Intermediate Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Exercise 3(b)

I.
Question 1.
If the quadratic equations ax² + 2bx + c = 0 and ax² + 2cx + b = 0, (b ≠ c) have a common root then show that a + 4b + 4c = 0.
Solution:
ax² + 2bx + c = 0
ax² + 2cx + c = 0
Let a be common root.
aα² + 2bα + c = 0
aα² + 2cα + b = 0
2α (b – c) + c – b = 0
(2α – 1) (b – c) = 0
α = \(\frac{1}{2}\)
\(\frac{a}{4}+\frac{2 b}{2}\) + c = 0
a + 4b + 4c = 0.

Question 2.
If x² – 6x + 5 = 0 and x² – 12x + p = 0 have a common root, then find p.
Solution:
x² – 6x + 5 = 0
(x – 5) (x – 1) = 0
x = 1, 5
Now x² – 12x + p = 0, can have 1 or 5 as root then 1 – 12 + p = 0
p = 11 (or)
25 – 60 + p = 0
p = 35.

Question 3.
If x² – 6x + 5 = 0 and x² – 3ax + 35 = 0 have a common root, then find a.
Solution:
(x² – 6x + 5) = 0
(x – 5) (x – 1) = 0
x = 5, 1
Now, 5, 1 satisfy
x² – 3ax + 35 = 0
25 – 15a + 35 = 0
60 = 15a
a = 4
1 – 3a + 35 = 0
3a = 36
a = 12.

Question 4.
If the equations x² + ax + b = 0 and x² + cx + d = 0 have a common root and the first equation have equal roots, then prove that 2 (b + d) = ac.
Solution:
x² + ax + b = 0
x² + cx + d = 0
x² + ax + b = 0 have equal roots.
⇒ a² – 4b = 0
a² = 4b
x² + ax + b = 0
x² + cx + d = 0
α (a – c) + b – d = 0
as 2α = – a
α = \({-a}{2}\)
\(\frac{a^2}{4}+c\left(\frac{-a}{2}\right)\) + d = 0
a² – 2ac + 4d = 0
4b + 4d = 2ac
or (b + d) = ac.

Question 5.
Discuss the signs of the following quadratic expressions when x is real.
i) x² – 5x + 4
ii) x² – x + 3
Solution:
i) f(x) = x² – 5x + 4
f(x) = (x – 4) (x – 1)
y = \(\left(x-\frac{5}{2}\right)^2+4-\frac{25}{4}\)
y = \(\left(x-\frac{5}{2}\right)^2-\frac{9}{4}\)
y + \(\frac{9}{4}\) = \(\left(x-\frac{5}{2}\right)^2\)

f(x) < 0
1 < x < 4 f(x) > 0
x > 4 and x < 1
(- ∞ < x < 1) ∪ (4 < x < ∞).

ii) x² – x + 3
f(x) = x² – x + 3
= \(\left(x-\frac{1}{2}\right)^2+3-\frac{1}{4}\)
y = \(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\)
y – \(\frac{11}{4}\) = \(\left(x-\frac{1}{2}\right)^2\)
Now discriminant = 1 – 12 < 0

f(x) > 0 ∀ x ∈ Real numbers.

Question 6.
For what values of x the following expressions are positive?
i) x² – 5x + 6
ii) 3x² + 4x + 4
iii) 4x – 5x² + 2
iv) x² – 5x + 14
Solution:
i) f(x) = x² – 5x + 6
f(x) > 0
x² – 5x + 6 > 0
(x – 3) (x – 1)> 0
x > 3 or x < 1
(- ∞ < x < 1) ∪ (3 < x < ∞)

ii) 3x² + 4x + 4
f(x) = 3x² + 4x + 4
f(x) > 0
3x² + 4x + 4> 0
x² + \(\frac{4}{3}\)x + \(\frac{4}{3}\) > 0
\(\left(x+\frac{2}{3}\right)^2+\frac{4}{3}-\frac{4}{9}\) > 0
\(\left(x+\frac{2}{3}\right)^2+\frac{8}{9}\) > 0 ∀ x ∈ Real numbers.

iii) f(x) = 4x – 5x² + 2
= [5x² – 4x – 2]

iv) f(x) = x² – 5x + 14
f(x) > 0
\(\left(x-\frac{5}{2}\right)^2+14-\frac{25}{4}\) > 0
\(\left(x-\frac{5}{2}\right)^2+\frac{31}{4}\) > 0.

Question 7.
For what values of x, the following expres¬sions are negative ?
i) x² – 7x + 10
ii) 15 + 4x – 3x²
iii) 2x² + 5x – 3
iv) x² – 5x – 6
Solution:
i) f(x) = x² – 7x + 10
f(x) < 0
(x – 5) (x – 2) < 0
2 < x < 5.

ii) f(x) = 15 + 4x – 3x²
f(x) < 0
15 + 4x – 3x² < 0
3x² – 9x + 5x – 15 > 0
3x (x – 3) + 5 (x – 3) > 0
(x – 3) (3x + 5) > 0
x > 3, x < \(\frac{-5}{3}\)
(- ∞ < x < \(\frac{-5}{3}\)) ∪ (3 < x < ∞).

iii) if(x) = 2x² + 5x – 3
f(x) < 0
2x² + 5x – 3 < 0
2x² + 6x – x – 3 < 0
2x (x + 3) – 1 (x + 3) < 0
(2x – 1) (x + 3) < 0
– 3 < x < \(\frac{1}{2}\).

iv) f(x) = x² – 5x – 6
f(x) < 0
x² – 5x – 6 < 0
(x – 6) (x + 1) < 0
– 1 < x < 6.

Question 8.
Find the changes in the sign of the following expressions and find their extreme values.
i) x² – 5x + 6
ii) 15 + 4x – 3x²
Solution:
i) f(x) = x² – 5x + 6
= (x – 3) (x – 2)
f(x) < 0
2 < x < 3 and f(x) > 0
(- ∞ < x < 2) (3 < x < ∞)
y = \(\left(x-\frac{5}{2}\right)^2+\frac{6-25}{4}\)
y_min. = \(\frac{-1}{4}\)

ii) f(x) = 15 + 4x – 3x²
f(x) <0
15 + 4x – 3x² < 0
3x² – 4x – 15 < 0
3x² – 9x + 5x – 15 < 0
3x (x – 3) 5 (x – 3) < 0
(x – 3) (3x + 5) < 0
\(\frac{-3}{5}\) < x < 3 f(x) > 0
(- ∞ < x < \(\frac{-3}{5}\)) ∪ (3 < x < ∞)
y = 15 + 4x – 3x²

Question 9.
Find the maximum or minimum of the following expressions as x varies over R.
i) x² – x + 7
ii) 12x – x² – 32
iii) 2x+ 5 – 3x²
iv) ax² + bx + a (a, b ∈ R and a ≠ 0)
Solution:
i) y = x² – x + 7
y = (x – \(\frac{1}{2}\))² – \(\frac{1}{4}\) + 7
y = (x – \(\frac{1}{2}\))² + \(\frac{27}{4}\)
y_min = \(\frac{27}{4}\)

ii) y = 12x – x² – 32
y = – [x² – 12x + 32]
= – [(x – 6)² – 36 +32]
y = – [x – 6]² + 4
y_max = 4

iii) f(x) = 2x + 5 – 3x²
y = – 3 [x² – \(\frac{2}{3}\) x – \(\frac{5}{3}\)]
= – 3 \(\left[\left(x-\frac{1}{3}\right)^2-\frac{1}{9}-\frac{15}{9}\right]\)
= – 3 \(\left(x-\frac{1}{3}\right)^2+\frac{16}{3}\)
y_max = \(\frac{16}{3}\)

iv) f(x) = ax² + bx + a

II.
Determine the range of the following expressions.
Question 1.
i) \(\frac{x^2+x+1}{x^2-x+1}\)
ii) \(\frac{x+2}{2 x^2+3 x+6}\)
iii) \(\frac{(x-1)(x+2)}{x+3}\)
iv) \(\frac{2 x^2-6 x+5}{x^2-3 x+2}\)
Solution:
\(\frac{x^2+x+1}{x^2-x+1}\)
yx² – yx + y = x² + x + 1
x² (y² – 1) + x (- y – 1) + y – 1 = 0
Discriminant ≥ 0
(- y- 1)² – 4(y – 1) (y – 1) ≥ 0
(y² + 2y + 1) – 4 (y² – 2y + 1) ≥ 0
– 3y² +10y – 3 ≤ 0
3y² – 10y + 3 ≤ 0
3y² – 9y – y + 3 ≤ 0
3y (y – 3) – 1 (y – 3) ≤ 0
(3y – 1) (y – 3)≤ 0
\(\frac{1}{3}\) ≤ y ≤ 3.

ii) y = \([\frac{x+2}{2 x^2+3 x+6}/latex]
2x²y + 3xy + 6y – x – 2 = 0
x² (2y) + x (3y – 1) + 6y – 2 = 0
Discriminant ≥ 0
(3y – 1)² – 4 (2y) (6y – 2) ≥ 0
9y² – 6y + 1 – 48y² + 16y ≥ 0
– 39y² + 10y + 1 ≥ 0
39y² – 10y – 1 ≥ 0
39y² – 13y + 3y – 1 ≤ 0
13y (3y – 1) + (3y – 1) ≤ 0
(13y – 1) (3y – 1) ≤ 0
[latex]-\frac{1}{13}\) ≤ y ≤ \(\frac{1}{3}\).

iii) y = \(\frac{(x-1)(x+2)}{x+3}\)
y = \(\frac{x^2+x-2}{x+3}\)
yx + 3y = x² + x – 2
x² + x(1 – y) – 2 – 3y = 0
Discriminant ≥ 0
(1 – y)² + 4(2 + 3y) ≥ 0
y² – 2y + 1 + 8 + 12y ≥ 0
y² + 10y + 9 ≥ 0
(y + 1) (y + 9) ≥ 0
y ≥ – 1 or y ≤ – 9
(- ∞ < y ≤ – 9) ∪ (- 1 ≤ y < ∞)

iv) y = \(\frac{2 x^2-6 x+5}{x^2-3 x+2}\)
y (x² – 3x . 2) = 2x² – 6x + 5
x² (y – 2) + x(- 3y + 6) + 2y – 5 = 0
Discriminant ≥ 0
(6 – 3y)² – 4(y – 2) (2y – 5)≥0
9y² – 36y + 36 – 8y² + 36y – 40 ≥ 0
y² -4 ≥ 0
(y – 2) (y + 2) ≥ 0
y ≥ 2 or y ≤ – 2
(- ∞ < y ≤ – 2) ∪ (2 ≤ y < ∞)

Question 2.
Prove that \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\) does not lie between 1 and 4 if x ∈ Real number.
Solution:
y = \(\frac{1}{3 x+1}+\frac{1}{x+1}-\frac{1}{(3 x+1)(x+1)}\)
y = \(\frac{4 x+1}{3 x^2+4 x+1}\)
3yx² + x(4y – 4) + y – 1 = 0
Discriminant ≤ 0
[4 (y – 1)]² – 4 . 3y (y- 1) ≤ 0
16 (y- 1)² – 12y(y- 1) ≤ 0
4(y – 1) [4(y – 1) – 3y)]≤ 0
4(3 – 1) [y – 4] ≤ 0
(y – 1) (y – 4) ≤ 0
⇒ y ≤ 4 or y ≥ 1.

Question 3.
If x is real, prove that \(\frac{x}{x^2-5 x+9}\) lies between \(\frac{-1}{11}\) and 1.
Solution:
y = \(\frac{x}{x^2-5 x+9}\)
y (x² – 5x + 9) = x
x²y + x(- 5y – 1) + 9y = 0
Discriminant ≤ 0
(- 5y – 1)² – 4 . 9y² ≥ 0
25y² + 10y + 1 – 36y² ≥ 0
– 11y² + 10y + 1 ≥ 0
11y² – 10y – 1 ≤ 0
11y² – 11y + y – 1 ≤ 0
11y (y- 1) + (y- 1) ≤ 0
(y – 1) (11y + 1) ≤ 0
\(\frac{-1}{11}\) ≤ y ≤1.

Question 4.
If the expression \(\frac{x-p}{x^2-3 x+2}\) takes all values of x ∈ R, then find the bounds for p.
Solution:
y = \(\frac{x-p}{x^2-3 x+2}\)
y(x² – 3x + 2) = x – p
x²y + x(- 3y – 1) + p = 0
Discriminant ≥ 0
(- 3y- 1)² – 4y(2y + p) ≥ 0
9y² + 6y + 1 – 8y² – 4p ≥ 0
y² + y(6 – 4p)+ 1 ≥ 0
Discriminant< 0
(6 – 4p)² – 4 < 0
16p² – 48p + 36 – 4< 0
16p² – 48p + 32 < 0
p² – 48p + 32 < 0
p² – 3p + 2 < 0
(p – 2) (p – 10)
1 < p < 2.

Question 5.
If c² ≠ ab and the roots of (c² – ab) x² – 2 (a² – bc) x + (b² – ac) = 0 are equal, then show that a³ + b³ + c³ = 3abc or a = 0.
Solution:
Roots are equal = Discriminant = 0
4 (a² – bc)² – 4 (c² – ab) (b² – ac) = 0
a⁴ + b²c² – 2a²bc – b²c² + c³a + b³a – a²bc = 0
a⁴ – 2a²bc + c3a. b3a – a²bc = 0
a (a³ + b³ + c³ – 3abc = 0
a³ + b³ + c³ – 3abc = 0 or a = 0