Mahesh

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Theory of Equations Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Theory of Equations Important Questions

Question 1.
Form polynomial equation of the lowest degree, with roots 1, -1, 3 (May ’06)
Hint : Equation having roots, α, β, γ is [(x – α)(x – β)(x – γ) = 0
Solution:
Required equation is
(x – 1) (x + 1) (x – 3) = 0
⇒ (x² – 1) (x – 3) = 0
⇒ -3x² – x + 3 = 0

Question 2.
If 1, 1 α are the roots of
x³ – 6x² + 9x – 4 = 0, then find α. (May ’11)
Solution:
1, 1, α are roots of x³ – 6x² + 9x – 4 = 0
Sum = 1 + 1 + α = 6
α = 6 – 2 = 4

Question 3.
If -1, 2 and α are the roots of 2x³ + x² – 7x – 6 = 0, then find α (Mar 14, 13)
Solution:
-1, 2, α are roots of 2x³ + x² – 7x – 6 = 0
sum = -1 + 2 + α = –\(\frac{1}{2}\)
α = –\(\frac{1}{2}\) – 1 = –\(\frac{3}{2}\)

Question 4.
If 1, -2 and 3 are roots of x³ – 2x² + ax + 6 = 0, then find α. (Mar. ’04,)
Solution:
1, -2 and 3 are roots of x³ + x² + ax – 6 = 0
x³ – 2x + ax + 6 = 0
⇒ 1(-2) + (-2) 3 + 3. 1 = a
i.e., a = -2 – 6 + 3 = -5

Question 5.
If the product of the roots of 4x³ + 16x² – 9x – a = 0 is 9, then find a. (TS Mar. ’17, ’16)
Solution:
α, β, γ are the roots of
4x³ + 16x² – 9x – a = 0
αβγ = \(\frac{a}{4}\) = 9 ⇒ a = 36

Question 6.
Find the transformed equation whose roots are the negative of the roots of
x⁴ + 5x³ + 11x + 3 = 0
Solution:
Given f(x) = x⁴ + 5x³ + 11x + 3 = 0
We want an equation whose roots are
-α₁, α₂, α₃, α₄,
Required equation f(-x) = 0
⇒ (-x)⁴ + 5 (-x)³ + 11 (-x) + 3 = 0
⇒ x⁴ – 5x³ – 11x + 3 = 0

Question 7.
Form the polynomial equation of degree 3 whose roots are 2, 3 and, 6.
Solution:
The required polynomial equation is,
(x – 2) (x – 3) (x – 6) = 0
⇒ x³ – 11x² + 36x – 36 = 0

Question 8.
Let α, β, γ be the roots of Σα³
Solution:
Σα³ = α³ + β³ + γ³
= (α + β + γ)
= (α² + β² + γ² – αβ – βγ – γα) + 3αβγ
= (-p)(p² – 2q – q) – 3r
= -p(p² – 3q) – 3r
∴ Σα³ = -p³ + 3pq – 3r = 3pq – p³ – 3r

Question 9.
If α, β and 1 are the roots of x³ – 2x² – 5x + 6 = 0, then find α and β (AP Mår ’16, ’08)
Solution:
α, β and 1 are the roots of
x³ – 2x² – 5x + 6 = 0
Sum = α + β + 1 = 2 ⇒ α + β = 1
product = αβ = -6
(α – β)² = (α + β )² – 4αβ = 1 + 24 = 25
α – β = 5
α + β = 1
Adding 2α = 6 ⇒ α = 3
∴ α = 3 and β = -2

Question 10.
If α, β and γ are the roots of x³ – 2x² + 3x – 4 = 0, then find
i) Σα²β²
ii) Σαβ (α + β)
Solution:
Since α, β, γ are the roots of
x³ – 2x² + 3x – 4 = 0
then α + β + γ = 2
αβ + βγ + γα = 3
αβγ = 4

i) Σα²β² = α²β² + β²γ² + γ²α²
= (αβ + βγ + γα)² – 2αβγ(α + β + γ)
= 9 – 2. 2.4 = 9 – 16 = -7

ii) Σαβ (α + β) = α²β + β²γ + γ²α + αβ² + βγ² + γα²
= (αβ + βγ + γα)(α + β + γ) – 3αβγ
= 2.3 – 3.4 = 6 – 12 = -6.

Question 11.
Solve the x³ – 3x² – 6x + 8 = 0 equation, given that the roots of each are in A.P. (‘Mar. ’07)
Solution:
The roots of x³ – 3x² – 6x + 8 = 0 are in A.P
Suppose a – d, a, a + d be the roots
Sum = a – d + a + a + d = 3
3a = 3
⇒ a = 1
∴ (x – 1) is a factor of x³ – 3x² – 6x + 8 = 0

⇒ x² – 2x – 8 = 0
⇒ x² – 4x + 2x – 8 = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ x = 4, -2
∴ The roots are -2, 1, 4.

Question 12.
Solve x⁴ – 4x² + 8x + 35 = 0, given that 2 + i \(\sqrt{3}\) is a root (AP Mar. ’15)
Solution:
Let 2 + i \(\sqrt{3}\) is one root
⇒ 2 – i \(\sqrt{3}\) is another root.
The equation having roots
2 ± i \(\sqrt{3}\) is x² – 4x + 7 = 0
∴ x² – 4x + 7 is a factor of

∴ The roots of the given equation are 2 ± i\(\sqrt{3}\), -2 ± i

Question 13.
Find the polynomial equation whose roots are the reciprocals of the roots of x⁴ – 3x³ + 7x² + 5x – 2 = 0 (TS Mar. ’15, ’11)
Solution:
Given equation is
f(x) = x⁴ – 3x³ + 7x² + 5x – 2 = 0
Required equation is f(\(\frac{1}{x}\)) = 0
i.e. \(\frac{1}{x^{4}}\) – \(\frac{3}{x^{3}}\) + \(\frac{7}{x^{2}}\) + \(\frac{5}{x}\) – 2 = 0
Multiplying with x⁴
⇒ 1 – 3x + 7x² + 5x³ – 2x⁴ = 0
i.e., 2x⁴ – 5x³ – 7x² + 3x – 1 = 0

Question 14.
Find the polynomial equation whose roots are the translates of those of x⁵ – 4x⁴ + 3x² – 4x + 6 = 0 by -3. (TS Mar. 16)
Solution:
Given equation is
f(x) = x⁵ – 4x⁴ + 3x² – 4x + 6 = 0
Required equation is f(x + 3) = 0
(x + 3)⁵ – 4(x + 3)⁴ + 3(x + 3)²

Required equation is
x⁵ + 11x⁴ + 42x³ + 57x² – 13x – 60 = 0

Question 15.
Show that x⁵ – 5x³ + 5x² – 1 = 0 has three equal roots and find this root. (TS Mar. ’17)
Solution:
Let f(x) = x⁵ – 5x³ + 5x² – 1
f'(x) = 5x⁴ – 15x² + 10x
= 5x (x³ – 3x + 2)
f'(1) = 5(1) (1 – 3 + 2) = 0
f(1) = 1 – 5 + 5 – 1 = 0
x – 1 is a factor of f'(x) and f(x)
∴ 1 is a repeated’ root of f(x).

⇒ 1 is a root of above equation
(∵ sum of the coefficients is zero)
∴ 1 is the required root.

Question 16.
Solve the 8x³ – 36x² – 18x + 81 = 0 equation, given that the roots of each are in AP. (Mar. 04’)
Soluution:
Given the roots of 8x³ – 36x² – 18x + 81 = 0 are in AP.
Let the roots be a – d, a, a + d
Sum of the roots = a – d + a + a + d
= \(\frac{36}{8}\) = \(\frac{9}{2}\)
i.e., 3a = \(\frac{9}{2}\) ⇒ a = \(\frac{3}{2}\)
∴ (x – \(\frac{3}{2}\)) is a factor of

⇒ 8x² – 24x – 54 = 0
⇒ 4x² – 12x – 27 = 0
⇒ 4x² – 18x + 6x – 27 = 0
⇒ 2x(2x – 9) + 3 (2x – 9) = 0
⇒ (2x + 3) (2x – 9) = 0
⇒ x = –\(\frac{3}{2}\), \(\frac{9}{2}\)
The roots are –\(\frac{3}{2}\), \(\frac{3}{2}\), \(\frac{9}{2}\)

Question 17.
Solve the 3x³ – 26x² + 52x – 24 = 0 equations, given that the roots of each are in GP.
(TS Mar. ’15)
Solution:
Given equation is 3x³ – 26x² + 52x – 24 = 0
The roots are in G.P.
Suppose \(\frac{a}{r}\), a, ar are the roots.
Product = \(\frac{a}{r}\).a.ar = –\(\left(-\frac{24}{3}\right)\)
a³ = 8 ⇒ a = 2
∴ (x – 2) is a factor of 3x³ – 26x² + 52x – 24

⇒ 3x³ – 20x + 12 = 0
⇒ 3x² – 18x – 2x + 12 = 0
⇒ 3x (x – 6) -2 (x – 6) = 0
⇒ (3x – 2)(x – 6) = 0
⇒ x = \(\frac{2}{3}\), 6
∴ The roots are \(\frac{2}{3}\), 2, 6.

Question 18.
Solve 18x³ + 81x² + 121x + 60 = 0 given that one root is equal to half the sum of the
remaining roots. (May ’11; Mar. ’05)
Solution:
Suppose α, β, γ are the roots of 18x³ + 81x² + 121x + 60 = 0.

Question 19.
Solve the equation
2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0 (AP Mar.17, 16; Mar. ‘08, 07)
Solution:
Given f(x) = 2x⁵ + x⁴ – 12x³ – 12x² + x + 2 = 0
This is an odd degree reciprocal equation of first type.
∴ -1 is a root.
Dividing f(x) with x + 1

Dividing f(x) by (x + 1) we get
2x⁴ – x³ – 11x² – x + 2 = 0
Dividing by x²

Substituting in (1), required equation is
2(a² – 2) – a – 11 = 0 .
2a² – 4 – a – 11 = 0
2a² – a – 15 = 0
(a – 3)(2a + 5) = 0
a = 3 or \(-\frac{5}{2}\)
Case (i) a = 3

Question 20.
Find the roots of
x⁴ – 16x³ + 86x² – 176x + 105 = 0
Solution:
Let f(x) = x⁴ – 16x³ + 86x² – 176x + 105
Now, f(1) = 1 – 16 + 86 – 176 + 105 = 0
∴ 1 is a root of f(x) = 0
⇒ x – 1 is a factor of f(x)

∴ f(x) = (x – 1) (x³ – 15x² + 71x – 105)
= (x – 1) g(x) where
g(x) = x³ – 15x² + 71x – 105
g(1) = 1 – 15 + 71 – 105 = -48 ≠ 0
g(2) = -15 ≠ 0
g(3) = 27 – 135 + 213 – 105 = 0
∴ 3 is a root of g(x) =0
⇒ x – 3 is a factor of g(x)

∴ g(x) = (x – 3) (x² – 12x + 35)
= (x – 3) (x – 5) (x – 7)
∴ f(x) = (x – 1) (x – 3) (x – 5) (x – 7)
∴ 1, 3, 5, 7 are the roots of f(x) = 0.

Question 21.
Solve 4x³ – 24x² + 23x + 18 = 0, given that the roots of this equation are in arithmetic progression (Mar. ’14; May ’06)
Solution:
Let a – d, a, a + d are the roots of the given equation
Now, sum of the roots
a – d + a + a + d = \(\frac{24}{4}\)
3a = 6
a = 2
Product of the roots (a – d) a (a + d) = \(-\frac{18}{4}\)
a(a² – d) = \(-\frac{9}{2}\)
2(4 – d²) = \(-\frac{9}{2}\)
4(4 – d²) = -9
16 – 4d² = -9
4d² = 25
d = ±\(\frac{5}{2}\)
∴ roots are –\(\frac{1}{2}\), 2 and \(\frac{9}{2}\)

Question 22.
Find the polynomial equation whose roots are the squares of the roots of x⁵ + 4x³ – x² + 11 = 0 by -3. (Mar. ’06)
Solution:
Let f (x) ≡ x⁵ + 4x³ – x² + 11
The required equation is f(x + 3) = 0

The required equation is
x⁵ + 15x⁴ + 94x³ + 305x² + 507x + 353 = 0

Question 23.
Solve the equation 6x⁴ – 35x³ + 62x² – 35x + 6 = 0. (May. 13)
Solution:
We observe that the given equation is an even degree reciprocal equation of class one. On dividing both sides of the given equation by x², we get

Then the above equation reduces to
6(y² – 2) – 35y + 62 = 0
i.e., 6y² – 35y + 50 = 0
i.e., (2y – 5)(3y – 10) = 0.
Hence the roots of 6y² – 35y + 50 = 0 are \(\frac{5}{2}\) and \(\frac{10}{3}\).

Hence the roots of the given equation are \(\frac{1}{2}\), \(\frac{1}{3}\), 2 and 3.

Question 24.
Solve x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0. (Mar. ’13)
Solution:
Given equation is
x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0 is a reciprocal equation of odd degree and of class two.
∴ 1 is a root of the given equation.
⇒ (x – 1) is a factor of
x⁵ + 4x³ + 5x² – 4x² + 1 = 0

Question 25.
Form the polynomial equation of degree 3 whose roots are 2, 3 and 6. (Mar. ’02)
Solution:
The required polynomial equation is,
(x – 2) (x – 3)(x – 6) = 0
⇒ x³ – 11x³ + 36x – 36 = 0

Question 26.
Find the relation between the roots and the coefficients of the cubic equation
3x³ – 10x² + 7x + 10 = 0.
Solution:
3x³ – 10x² + 7x + 1o = 0 ———- (1)
On.comparing (1) with
ax³ + bx² + cx + d = 0,
we have

Question 27.
Write down the relations between the roots and the coefficients of the bi-quadratic equation.
x⁴ – 2x³ + 4x² + 6x – 21 = 0
Solution:
Given equation is
x⁴ – 2x³ + 4x² + 6x – 21 = 0 —— (1)
On comparing (1) with
ax⁴ + bx³ + cx² + dx + c = 0,
we have

Question 28.
If 1, 2, 3 and 4 are the roots of x⁴ + ax³ + bx² + cx + d = 0, then find the
values of a, b, c and d.
Solution:
Given that the roots of the given equation are 1, 2, 3 and 4. Then
x⁴ + ax³ + bx² + cx + d
≡ (x – 1) (x – 2) (x – 3) (x – 4) = 0
≡ x⁴ – 10x³ + 35x² – 50x + 24 = 0
On equating the coefficients of like powers of x, we obtain
a = -10, b = 35, c = -50, d = 24

Question 29.
if a, b, c are the roots of
x³ – px² + qx- r = 0 and r ≠ 0, then find \(\frac{1}{\mathbf{a}^{2}}+\frac{1}{\mathbf{b}^{2}}+\frac{1}{\mathrm{c}^{2}}\) interms of p, q, r.
Solution:
Given that a, b, c are the roots of
x³ – px² + qx – r = 0, then
a + b + c = p, ab + bc + ca = q, abc = r

Question 30.
Find the sum of the squares and the sum of the cubes of the roots of the equation x³ – px² + qx – r = 0 in terms of p, q, r.
Solution:
Let α, β, γ be the roots of the given equation then α + β + γ = p, αβ + βγ + γα = q, αβγ = r
Sum of the squares of the roots is α² + β² + γ²
= (α + β + γ)² – 2(αβ + βγ + γα) = p² – 2q
Sum of the cubes of the roots is α³ + β³ + γ³
= (α + β + γ) (α² + β² + γ² – αβ – βγ – γα) + 3αβγ
= p(p² – 2q – q) + 3r
= p(p² – 3q) + 3r

Question 31.
Obtain the cubic equation, whose roots are the sqüares of the roots of the equation, x³ + p₁x² + P₂x + p₃ = 0
Solution:
The required equation is, f(\(\sqrt{x}\)) =0

Question 32.
Let α, β, γ be the roots of
x³ + px² + qx + r = 0. Then find the
i) Σα²
ii) Σ\(\frac{1}{\alpha}\)
iii) Σα³
iv) Σβ²γ²
v) Σ(α + β) (β + γ) (γ + α)
Solution:
since α, β, γ are the roots of the equation, we have α + β + γ = – p,
αβ + βγ + γα = q, αβγ = -r.

i) Σα²
Solution:
Σα² = α² + β² + γ²
= (α + β + γ) – 2(αβ + βγ + γα)
= p² – 2q

ii) Σ\(\frac{1}{\alpha}\)
Solution:

iii) Σα³
Solution:

iv) Σβ²γ²
Solution:

v) (α + β) (β + γ) (γ + α)
Solution:
We know, α + β + γ = -p
⇒ α + β = -p – r and β + γ = -p – α
= γ + α = -p – β
∴ (α + β) (β + γ)(γ + α)
= (-p – γ) (-p – α) (-p – β)
= -p³ – p²(α + β + γ) – p(αβ + βγ + γα) – αβγ
= -p³ + p³ – pq + r = r – pq .

Question 33.
Let α, β, γ be the roots of
x³ + ax² + bx + c = 0 then find Σα² + Σβ²
Solution:
Since α, β, γ are roots of the given equation,

Question 34.
If α, β, γ are the roots of x³ + px² + qx + r = 0, then form the cubic equation whose roots are .
α(β + γ), β(γ + α), γ(α + β)
Solution:
Let α, β, γ be the roots of the given equation.
we have, α + β + γ = – p, αβ + βγ + γα = q, αβγ = -r
Let y = α(β + γ)
= αβ + αγ + γβ – βγ

Question 35.
Solve x³ – 3x² – 16x + 48 = 0
Solution:
Let f(x) = x³ – 3x² – 16x + 48
by inspection, f(3) = 0
Hence 3 is a root of 1(x) = 0
Now we divide f(x) by (x – 3)

Question 36.
Find the roots of
x⁴ – 16x³ + 86x² – 176x + 105 = 0 (Mar. ‘02)
Solution:
Let f(x) = x⁴ – 16x³ + 86x² – 176x + 105
Now if (1) = 1 – 16 + 86 – 176 + 105 = 0
∴ 1 is a root of f(x) = 0
⇒ x – 1 is a factor of f(x)

∴ g(x) = (x – 3)(x² – 12x + 35)
= (x – 3) (x – 5) (x – 7)
∴ f(x) = (x – 1) (x – 3) (x – 5) (x – 7)
∴ 1, 3, 5, 7 are the roots of f(x) = 0.

Question 37.
Solve x³ – 7x² + 36 = 0, given one root being twice the other.
Solution:
Let α, β, γ be the root of the equation
x³ – 7x² + 36 = 0 and
let β = 2α
Now, we have, α + β + γ = 7
⇒ 3α + γ = 7 —— (1)
αβ + βγ + γα = 0
⇒ 2α² + 3αγ = 0 —— (2)
αβγ = -36 ⇒ 2α²γ = -36 —— (3)
From (1) and (2), we have
2α² + 3α(7 – 3α) = 0
i.e., α² – 3α = 0 (or) α(α – 3) = 0
∴ α = 0 or α = 3
Since α = 0 does not satisfy the given equation.
∴ α = 3, so β = 6 and γ = -2
∴ The roots are 3, 6, -2.

Question 38.
Given that 2 is a root of
x³ – 6x² + 3x + 1o = 0, find the other roots.
Solution:
Let f(x) = x³ – 6x² + 3x + 10
Since 2 is a root of f(x) = 0, we divide f(x) by (x – 2)

∴ -1, 2, and 5 are the roots of the given equation.

Question 39.
Given that two roots of
4x³ + 20x² – 23x + 6 = 0 are equal, find all the roots of the given equation.
Solution:
Let α, β, γ are the roots of
4x³ + 20x² – 23x + 6 = 0
Given two roots are equal, let α = β

⇒ 12α² + 4α – 23 = 0
⇒ (2α – 1) (6α + 23) = 0
α = \(\frac{1}{2}\), α = \(\frac{-23}{6}\)
On verfication, we get that is a root of (1)
α = \(\frac{1}{2}\) is roots of (1)
(2) ⇒ roots are \(\frac{1}{2}\), \(\frac{1}{2}\), -6

Question 40.
Given that the sum of two roots of
x⁴ – 2x³ + 4x² + 6x – 21 = 0 is zero find the roots of the equation.
Solution:
Let α, β, γ, δ are the roots of given equation, since sum of two is zero.
α + β = 0
Now α + β + γ + δ = 2 ⇒ γ + δ = 2
Let αβ = p, γδ = q
The equation having the roots α, β is

∴ Roots are –\(\sqrt{3}\), \(\sqrt{3}\), 1 – i \(\sqrt{6}\) and 1 + i \(\sqrt{6}\)

Question 41.
Solve 4x³ – 24x² + 23x + 18 = 0, given that the roots of this equation are in arithmetic progression. (Mar. 14, May ‘06)
Solution:
Let a – d, a, a + d are the roots of the given equation
Now, sum of the roots
a – d + a + a + d = \(\frac{24}{4}\)
3a = 6
a = 2
Product of the roots (a – d) a (a + d) = \(\frac{-18}{4}\)
a(a² – d²) = \(\frac{-9}{2}\)
2(4 – d²) = \(\frac{-9}{2}\)
4(4 – d²) = -9
16 – d² = -9
4d² = 25
d = ± \(\frac{5}{2}\)
∴ roots are –\(\frac{1}{2}\), 2 and \(\frac{9}{2}\)

Question 42.
Solve x³ – 7x² + 14x – 8 = 0, given that the roots are in geometric progression.
Solution:
Let \(\frac{a}{r}\), a, ar be the roots of the given equation. Then .

Hence a = 2. On substituting a = 2 in (1), we obtain
\(\frac{2}{r}\) + 2 + 2r = 7
i.e., 2r² – 5r + 2 = 0
i.e., (r – 2) (2r – 1) = 0
Therefore r = 2 or r = \(\frac{1}{2}\)
Hence the roots of the given equation are 1, 2 and 4.

Question 43.
Solve x⁴ – 5x³ + 5x² + 5x – 6 = 0 given that the product of two of its roots is 3.
Solution:
Let α, β, γ, δ be the roots of the given equation.
Product of the roots αβγδ = -6
Given αβ = 3 (∵ Product of two roots is 3)
∴ α, β, γ, δ = -6
γδ = -2
Let α + β, γ + δ = q
The equation having the roots α,β is
x² – (α + β) x + αβ = 0
x² + px + 3 = 0
The equation having the roots γ, δ is
x² – (γ + δ)x + γδ = 0
x² – qx – 2 = 0
∴ x⁴ – 5x³ + 5x² + 5x – 6
= (x² – px + 3)(x² – qx – 2)
= x⁴ – (p + q)x³ + (1 + pq)x² + (2p – 3q)x – 6
Comparing the like terms,
p + q = 5, 2p – 3q = 5
∴ 2p – 3q = 5
3p + 3q = 15
5p = 20 ⇒ p = 4
∴ q = 1
Now x² – 4x + 3 = 0 ⇒ (x – 3)(x – 1) = 0
⇒ x = 1, 3
x² – x – 2 = 0 ⇒ (x – 2)(x + 1) = 0
⇒ x = -1, 2
∴ The roots are -1, 2, 1, 3

Question 44.
Solve x⁶ + 4x³ – 2x² – 12x + 9 = 0, Given that it has two pairš of equal roots.
Solution:
Given equation is
x⁴ + 4x³ – 2x² – 12x + 9 = 0
Let the roots be α, α, β, β
Sum of the roots, 2(α + β) = -4
⇒ α + β = -2
Let αβ = p
The equation having roots α, β is
x² – (α + β)x + αβ = 0
i.e. x² + 2x + p = 0
∴ x⁴ + 4x³ – 2x² – 12x + 9
= [x² – (α + β)x + αβ]²
= (x² + 2x + p)²
= x⁴ + 4x³ + (2p + 4)x² + 4px + p²
Comparing coefficients of x on both sides
4p = -12 ⇒ p = – 3
x² + 2x + p = 0 ⇒ x² + 2x – 3 = 0
⇒ (x + 3)(x – 1) = 0
⇒ x = -3, 1
∴ The roots of the given equation are -3, -3, 1, 1

Question 45.
Prove that the sum of any two of the roots of the equation x⁴ + px³ + qx² + rx + s = 0 is equal to the sum of the remaining two roots of the equation iff p³ – 4pq + 8r = 0.
Solution:
Suppose that the sum of two of the roots of the given equation is equal to the sum of the remaining two roots.
Let α, β, γ, δ be roots of the given equation such that α + β = γ + δ

From these equations, we have

Then equations (1), (2) and (4) are satisfied. In view of (5), equation (3) is also satisfied. Hence (x² + bx + c)(x² + bx + d) = x⁴ + 2bx³ + (b² + c + d)x² + b(c + d)x + cd = x⁴ + px³ + qx² + rx + s
Hence the roots of the given equation are α₁, β₁, γ₁ and δ₁. where α₁ and β₁ are the roots of the equations x² + bx + c = 0 and γ₁ and δ₁ are those of the equation x² + bx + d = 0.
We have α₁ + β₁ = -b = γ₁ + δ₁.

Question 46.
Form the polynomial equation of degree 4 whose roots are
4 + \(\sqrt{3}\), 4 – \(\sqrt{3}\), 2 + i and 2 – i
Solution:
The equation having roots 4 + \(\sqrt{3}\), 4 – \(\sqrt{3}\) is
x² – 8x + 13 = 0
The equation having roots 2 + i, 2 – i is
x² – 4x + 5 = 0.
The required equation is
(x² – 8x + 13) (x² – 4x + 5) = 0
∴ x⁴ – 12x³ + 50x² – 92x + 65 = 0

Question 47.
Solve 6x⁴ – 13x³ – 35x² – x + 3 = 0 given that one of its root is 2 + \(\sqrt{3}\).
Solution:
2 + \(\sqrt{3}\) is a root 2 – \(\sqrt{3}\) is also a root.
The equation having roots

Question 48.
Find the polynomial equation of degree 4 whose roots are the negatives of the roots of x⁴ – 6x³ + 7x² – 2x + 1 = 0
Solution:
Let f(x) ≡ x⁴ – 6x³ + 7x² – 2x + 1
The required equation is f(-x) = 0
i.e., (-x)⁴ – 6(-x)³ + 7(-x)² + 2(-x) + 1 = 0
∴ x⁴ + 6x³ + 7x² + 2x + 1 = 0

Question 49.
Find the algebraic equation of the degree 4 whose roots are 3 times the roots of the equation
6x⁴ – 7x³ + 8x² – 7x + 2 = 0
Solution:
Let f(x) ≡ 6x⁴ – 7x³ + 8x² – 7x +2
The required equation is f\(\left(\frac{x}{3}\right)\) = o

Question 50.
Form the equation whose roots are m times the roots of the equation x³ + \(\frac{x^{2}}{4}\) – \(\frac{x}{16}\) + \(\frac{1}{72}\) = 0 and deduce the case when m = 12.
Solution:

Question 51.
Find the algebraic equation of degree 5 whose roots are the translates of the roots of x⁵ + 4x³ – x² + 11 = 0 by -3. (Mar. ’06)
Solution:
Let f (x) ≡ x⁵ + 4x³ – x² + 11
The required equation is f(x + 3) = 0

The required equation is
x⁵ + 15x⁴ + 94x³ + 305x² + 507x + 353 = 0

Question 52.
Find the algebraic equation of degree 4 whose roots are the translates of the roots
4x⁴ + 32x³ + 83x² + 76x + 21 = 0 by 2.
Solution:
Let f(x) ≡ 4x⁴ + 32x³ + 83x² + 76x + 21
The required equation is f(x – 2) = 0

The required equation is
4x⁴ – 13x² + 9 = 0

Question 53.
Find the polynomial equation whose roots are the reciprocals of the roots of the equation
x⁴ + 3x³ – 6x² + 2x -4 = 0
Solution:
Let f(x) ≡ x⁴ + 3x³ – 6x² + 2x – 4

Question 54.
Find the polynomial equation whose roots are the squares of the roots of x³ – x² + 8x – 6 = 0
Solution:
Let f(x) ≡ x³ – x² + 8x – 6 .
The required equation is f(\(\sqrt{x}\)) = o

Squaring on both sides
⇒ x(x² + 16x + 64) = x² + 12x + 36
= x³ + 6x² + 64x – x² – 12x – 36 = 0
∴ x³ + 15x² + 52x – 36 = 0

Question 55.
Show that 2x³ + 5x² + 5x + 2 = 0 is a reciprocal equation of class one.
Solution:
Given equation is 2x³ + 5x² + 5x + 2 = 0
P₀ = 2, p₁ = 5, P₂ = 5, p₃ = 2
Here P₀ = p₃, p₁ = p₂
∴ The equation 2x³ + 5x² + 5x + 2 = 0 is a reciprocal equation of class one.

Question 56.
Solve the equation
4x³ – 13x² – 13x + 4 = 0
Solution:
4x³ – 13x² – 13x + 4 = 0 is a reciprocal equation of first class and of odd degree.
Thus -1 is a root of the 9iven equation.

4x² – 17x + 4 = 0 ⇒ 4x² – 16x – x + 4 = 0
⇒ 4x(x – 4) – 1 (x – 4) = 0
⇒ (x – 4)(4x – 1) = 0
⇒ x = 4 or \(\frac{1}{4}\)
The roots are -1, 4, \(\frac{1}{4}\)

Question 57.
Solve the equation 6x⁴ – 35x³ + 62x² – 35x + 6 = 0. (May ‘13)
Solution:
We observe that the given equation is an even degree reciprocal equation of class one.
On dividing both sides 6f the given equation by x², we get

Then the above equation reduces to
6(y² – 2) – 35y + 62 = 0
i.e., 6y² – 35y + 50 = 0
i.e., (2y – 5) (3y – 10) = 0.
Hence the roots of 6y² – 35y + 50 = 0 are \(\frac{5}{2}\) and \(\frac{10}{3}\).
Therefore x + \(\frac{1}{x}\) = \(\frac{5}{2}\) and x + \(\frac{1}{x}\) = \(\frac{10}{3}\)
i.e., 2x² – 5x + 2 = 0 and 3x² – 10x + 3 = 0.
The roots of these equations are respectively

Hence the roots of the given equation are \(\frac{1}{2}\), \(\frac{1}{3}\), 2 and 3.

Question 58.
Solve x⁵ – 5x⁴ – 9x³ – 9x² + 5x – 1 = 0. (Mar ’13)
Solution:
Given equation is
x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0 is a reciprocal equation’of odd degree and of class two.
∴ 1 is a root of the given equation.
⇒ (x – 1) is a factor of
x⁵ – 5x⁴ + 9x³ – 9x² + 5x – 1 = 0

Question 59.
Solve the equation
6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0
Solution:
Given equation is
6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0 is a reciprocal equation of second class and of even degree.
∴ x² – 1 is a factor of
6x⁶ – 25x⁵ + 31x⁴ – 31x² + 25x – 6 = 0

∴ (1) becomes 6(y² – 2) – 25(y) + 37 = 0
⇒ 6y² – 12 – 25y + 37 = 0
⇒ 6y² – 25y + 25 = 0
⇒ 6y² – 15y – 10y + 25 = 0
⇒ 3y(2y – 5) – 5(2y – 5) = 0
⇒ (2y – 5)(3y – 5) = 0

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Quadratic Expressions Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Quadratic Expressions Important Questions

Question 1.
Form quadratic equation whose root 7 ± 2\(\sqrt{5}\) (Mar. ’11, ’05)
Solution:
α + β = 7 + 2\(\sqrt{5}\) + 7 – 2\(\sqrt{5}\) = 14
αβ = (7 + 2\(\sqrt{5}\)) (7 – 2\(\sqrt{5}\)) = 49 – 20 = 29
The required equation is
x² – (α + β)x + αβ = 0
x² – 14x + 29 = 0

Question 2.
Form quadratic equation whose root -3 ± 5i. (Mar. ’07)
Solution:
α + β = -3 + 5i – 3 – 51 = -6
αβ = (-3 + 5i)(-3 – 5i)
= 9 + 25 = 34
The required equation is
x² – (α + β)x + αβ = 0
x² + 6x + 34 = 0

Question 3.
For what values of x, 15 + 4x – 3x² expressions are negative? (AP Mar. ’15)
Solution:
The roots of 15 + 4x – 3x² = 0 are
\(\frac{-4 \pm \sqrt{16+180}}{-6}\) i.e., \(\frac{-5}{3}\), 3
∴ The expression 15 + 4x – 3x² is negative if
x < \(\frac{-5}{3}\) or x > 31 ∵ a = -3 < 0

Question 4.
If α, β are the roots of the equation ax² + bx + c = 0, find the value \(\frac{1}{\alpha^{2}}\) + \(\frac{1}{\beta^{2}}\) expressions in terms of a, b, c. (AP & TS Mar. ‘16, 08)
Solution:

Question 5.
Form quadratic equation whose root
\(\frac{p-q}{p+q}\), \(\frac{-p+q}{p-q}\), (p ≠ ±q) (Mar. ’06)
Solution:
α + β = \(\frac{p-q}{p+q}\) – \(\frac{p+q}{p-q}\)

Question 6.
Find the values of m for which the following equations have equal roots?
i) x² – 15 – m(2x – 8) = 0. (AP Mar. ’17) (TS Mar. ’15 13)
Solution:
Given equation is x² – 15 – m(2x – 8) = 0
x² – 2mx + 8m – 15 = 0
a = 1, b = -2m, c = 8m – 15
b² – 4ac = (-2m)² – 4(1) (8m – 15)
= 4m² – 32m + 60
= 4(m² – 8m + 15)
= 4(m – 3)(m – 5)
Hint: If the equation ax² + bx + c = 0 has equal roots then its discriminant is zero.
∵ The roots are equal b² – 4ac = 0
⇒ 4(m – 3) (m – 5) = 0
⇒ m – 3 = 0 or m – 5 = 0
∴ m = 3 or 5

Question 7.
(m + 1)x² + 2(m + 3)x + (m + 8) = 0.
Solution:
Given equation is
(m + 1)x² + 2(m + 3)x + (m + 8) = 0
a = m + 1, b = 2(m + 3), c = m + 8
b² – 4ac = (2(m + 3)]² – 4(m + 1) (m + 8)]
= 4(m² + 6m + 9) – 4(m² + 8m + m + 8)
= 4m² + 24m + 36 – 4m² – 36 m – 32
= -12m + 4
= -4(3m – 1)
∵ The roots are equal ⇒ b² – 4ac = 0
⇒ -4(3m – 1) = 0
⇒ 3m – 1 = 0
⇒ 3m = 1
∴ m = \(\frac{1}{3}\)

Question 8.
If x is real, prove that \(\frac{x}{x^{2}-5 x+9}\) lies between 1 and \(\frac{-1}{11}\). (Mar. ‘14, 13, ‘08, ‘02; May 11, ‘07)
Solution:
Let y = \(\frac{x}{x^{2}-5 x+9}\) ⇒ yx² – 5yx + 9y = x
⇒ yx² + (-5y – 1)x + 9y = 0
x ∈ R ⇒ (-5y – 1)² – 4y(9y) ≥ 0
⇒ 25y² + 1 + 10y – 36y² ≥ 0
⇒ -11y² + 10y + 1 ≥ 0 —— (1)
⇒ -11y² + 10y + 1 = 0 ⇒ -11y² + 11y – y + 1 = 0
⇒ 11y(-y + 1) + 1(-y + 1) = 0
⇒ (-y + 1)(11y + 1) = 0 ⇒ y = 1, \(\frac{-1}{11}\)
-11y² + 10y + 1 ≥ 0
∴ y² coeff is be, but the exp is ≥ 0 from (1)
⇒ \(\frac{-1}{11}\) ≤ y ≤ 1 ⇒ y lies between 1 and \(\frac{-1}{11}\)

Question 9.
Theorem : The roots of ax² + bx + c = 0 are

(Mar. ’02)
Proof:
Given quadratic equation is ax² + bx + c = 0
⇒ 4a(ax² + bx + c) = 0
⇒ 4a²x² + 4abx + 4ac = 0
⇒ (2ax)² + 2(2ax) (b) + b² – b² + 4ac = 0
⇒ (2ax + b)² = b² – 4ac

Question 10.
Find the maximum value of the function \(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R.
Solution:
Let y = \(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\)
⇒ yx² + 2yx + 3y = x² + 14x + 9
⇒ (y – 1)x² + 2(y – 7)x + 3y – 9 = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 7)]² – 4(y – 1) (3y – 9) ≥ 0
⇒ 4[(y² – 14y + 49) [(3y² – 12y + 9)] ≥ 0
⇒ -2y² – 2y + 40 ≥ 0
⇒ y² + y – 20 ≤ 0
⇒ (y + 5) (y -4) ≤ 0
⇒ -5 ≤ y ≤ 4
⇒ y ∈ [-5, 4]
⇒ Maximum value of y = 4
∴ Maximum value of the function
\(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R is 4.

Question 11.
If x² – 6x + 5 = 0 and x² – 12x + p = 0 have a common root, then find p. (TS Mar. ’17)
Solution:
Given x² – 6x + 5 = 0, x² – 12x + p = 0 have a common root.
If α is the common root then
α² -6α + 5 = 0, α² – 12α + p = 0
α² – 6α + 5 = 0 ⇒ (α – 1) (α – 5) = 0
⇒ α = 1 or 5
If α = 1 then α² – 12α + p = 0
⇒ 1 – 12 + p = 0 ⇒ p = 11
If α = 5 then α² – 12α + p = 0
⇒ 25 – 60 + p = 0 ⇒ p = 35
∴ p = 11 or 35

Question 12.
If x₁, x₂ are the roots of the quadratic equation ax₂ + bx + c = 0 and c ≠ 0, find the value of (ax₁ + b)^-2 + (ax₂ + b)_-2 in terms of a, b, c. (TS Mar. ’17)
Solution:
x₁, x₂ are the roots of the equation
ax² + bx + c = 0

Question 13.
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 is real. (AP & TS Mar. ’16, AP Mar. 15, ’11) (AP Mar. ‘17)
Solution:

⇒ 3yx² + 4yx + y = 4x + 1
⇒ 3yx² + (4y – 4) x + (y – 1) = 0
x ∈ R ⇒ (4y – 4)² – 4(3y)(y – 1) ≥ 0
⇒ 16y² + 16 – 32y – 12y² + 12y ≥ 0
⇒ 4y² – 20y + 16 ≥ 0
4y² – 20y + 16 = 0
⇒ y² – 5y + 4 = 0
⇒ (y – 1)(y – 4) = 0
⇒ y = 1, 4
⇒ 4y² – 20y + 16 ≥ 0.
⇒ y ≤ 1 or y ≥ 4
⇒ y does not lie between 1 and 4
Since y² coeff is the and exp ≥ 0.

Question 14.
Solve the following equations: (T.S Mar. ‘15)
2x⁴ + x³ – 11x² + x + 2 = 0
Solution:
Dividing by x²

Substituting in (1)
2(a² – 2) + a – 11 = 0
⇒ 2a² – 4 + a – 11 = 0
⇒ 2a² + a – 15 = 0
⇒ (a + 3) (2a – 5) = 0
⇒ a + 3 = 0 or 2a – 5 = 0
a = -3 or 2a = 5 a = \(\frac{5}{2}\)

Case(i) : a = -3
x + \(\frac{1}{x}\) = -3
x² + 1 = -3
x² + 3x + 1 = 0
x = \(\frac{-3 \pm \sqrt{9-4}}{2}\) = \(\frac{-3 \pm \sqrt{5}}{2}\)

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

Question 15.
For what values of x, the following expressions are positive? (May ’11)
i) x² – 5x + 6
Solution:
x² – 5x + 6 = (x – 2) (x – 3)
Roots of x² – 5x + 6 = 0 are 2, 3 which are real.
The expression x² – 5x + 6 is positive if x < 2 or x > 3, ∴ a = 1 > 0.

ii) 3x² + 4x + 4
Solution:
Here a = 3, b = 4, c = 4,
Δ = b² – 4ac
= 16 – 48 = -32 < 0
∴ 3x² + 4x + 4 is positive ∀ x ∈ R,
∵ a = 3 >0 and Δ < 0
Hint: ax² + bx + c and ‘a’ have same sign ∀ x ∈ R, if Δ < 0

iii) 4x – 5x² + 2
Solution:
Roots of 4x – 5x² + 2 = 0 are

iv) x² – 5x + 14
Solution:
Here a = 1, b = -5, c = 14,
Δ = b² – 4ac
= 25 – 56 = -31 < 0
∴ Δ < 0 ∵ a = 1 > 0 and Δ < 0
⇒ x² – 5x + 14 is positive ∀ x ∈ R.

Question 16.
Determine the range of the \(\frac{x^{2}+x+1}{x^{2}-x+1}\) expressions. (Mar ’04)
Solution:
Let y = \(\frac{x^{2}+x+1}{x^{2}-x+1}\)
⇒ x²y – xy + y = x² + x + 1
⇒ x²y – xy + y – x² – x – 1 = 0
⇒ x²(y – 1) – x(y + 1) + (y – 1) = 0
x is real ⇒ b² – 4ac ≥ 0
⇒ (y + 1)² – 4(y – 1)² ≥ 0
⇒ (y + 1)² – (2y – 2)² ≥ 0
⇒ (y + 1 + 2y – 2)(y + 1 – 2y + 2) ≥ 0
⇒ (3y – 1)(-y + 3) ≥ 0
⇒ -(3y – 1) (y – 3) ≥ 0
a = coeff. of y² = -3 < 0., But
The expression ≥ 0
⇒ y lies between \(\frac{1}{3}\) and 3
∴ The range of \(\frac{x^{2}+x+1}{x^{2}-x+1}\) is \(\left[\frac{1}{3}, 3\right]\)

Question 17.
Theorem : Let α, β be the real roots of ax² + bx + c = 0 and α < β. Then
i) x ∈ R, α < x < β ⇒ ax² + bx + c and ’a’ have opposite signs.
ii) x ∈ R, x < α or x > β ⇒ ax² + bx + c and ‘a’ have the same sign. (Apr. ’96, ’93)
Proof:
α, β are the roots of ax² + bx + c = 0
⇒ ax² + bx + c = a(x – α) (x – β)
⇒ \(\frac{a x^{2}+b x+c}{a}\) = (x – α) (x – β)

i) Suppose x ∈ R, α < x < β
α < x < β ⇒ x – α > 0, x – β < 0
⇒ (x – α) (x – β) < 0
⇒ ax² + bx + c, a have opposite signs.

ii) Suppose x ∈ R, x < α
x < α < 13
⇒ x – α < 0, x – β < 0 ⇒ (x – α) (x – β) > 0
⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign.
Suppose x ∈ R, x > β
x > β > α
⇒ x – α > 0, x – β > 0
⇒ (x – α)(x – β) > 0
⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign,
∴ x ∈ R, x < α or x > β
⇒ ax² + bx + c and a have the same sign.

Question 18.
Theorem: Let f(x) = ax² + bx + c be a quadratic function. (Apr. ‘01)
i) If a > 0 then f(x) has minimum value at x = \(\frac{-b}{2 a}\) and the minimum value = \(\frac{4 a c-b^{2}}{4 a}\)
ii) If a < 0 then f(x) has maximum value at x = \(\frac{-b}{2 a}\) and the maximum value \(\frac{4 a c-b^{2}}{4 a}\)
Proof:
ax² + bx + c =

≤ \(\frac{4 a c-b^{2}}{4 a}\),
when a < 0
∴ If a < 0, then \(\frac{4 a c-b^{2}}{4 a}\) is the maximum fór f when x = \(\frac{-b}{2 a}\)
Second Proof : f(x) = ax² + bx + c
⇒ f'(x) = 2ax + b
⇒ f”(x) = 2a
If f'(x) = 0, then 2ax + b = 0 and hence x = \(-\frac{b}{2 a}\)
If a > 0, then f”(x) > 0 and hence ‘f’ has minimum value at x = \(-\frac{b}{2 a}\).
Minimum of

If a < 0,
then f”(x) < 0 and hence ‘f’ has maximum value at x = \(-\frac{b}{2 a}\) maximum of

Question 19.
Theorem : The roots of ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\).
Proof:
Given quadratic equation is ax² + bx + c = 0
⇒ 4a (ax² + bx + c) = 0
⇒ 4a²x² + 4abx + 4ac = 0
⇒ (2ax)² + 2(2ax) (b) + b² – b² + 4ac = 0
⇒ (2ax + b)² = b² – 4ac
⇒ 2ax + b = ±\(\sqrt{b^{2}-4 a c}\)
⇒ 2ax = -b ± \(\sqrt{b^{2}-4 a c}\)
⇒ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
The roots of ax² + bx + c = 0 are
\(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

Question 20.
Theorem : Let a, b, c ∈ R and a ≠ 0. Then the roots of ax² + bx + c = 0 are non-real complex numbers if and only if ax² + bx + c and a have the same sign for all x ∈ R.
Proof :
The condition for the equation
ax² + bx + c = o to have non-real complex roots is b² – 4ac < 0, i.e., 4ac – b² > 0.

Hence 4ac – b² > 0, so that b² – 4ac < 0. Thus b² – 4ac < 0 if and only if ax² + bx + c and a have the same sign for all real x.

Question 21.
Theorem: If the roots of ax² + bx + c = 0 are real and equal to α = \(\frac{-b}{2 a}\); then for α ≠ x ∈ R, ax² + bx + c and ‘a’ have the same sign.
Proof:
The roots are equal ⇒ b² – 4ac = 0
⇒ 4ac – b² = 0

∴ For α ≠ x ∈ R, ax² + bx + c and a have the same sign.

Question 22.
Theorem : Let α, β be the real roots of ax² + bx + c = 0 and α < β. Then
i) x ∈ R, α < x < 13 ⇒ ax² + bx + c and ’a’ have opposite signs.
ii) x ∈ R, x < α or x > β ⇒ ax² + bx + c and ‘a’ have the same sign.
Proof.
α, β are the roots of ax² + bx + c = 0
⇒ ax² + bx + c = a(x – α)(x – β)
⇒ \(\frac{a x^{2}+b x+c}{a}\) = (x – α)(x – β)

i) Suppose x ∈ R, α < x < β
α 0, x – β < 0
⇒ (x – α)(x – β) < 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) < 0
⇒ ax² + bx + c, a have opposite signs

ii) Suppose x ∈ R, x < α
x < α < β ⇒ x – α < 0, x – β < 0 ⇒ (x – α) (x – β) > 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign.
Suppose x ∈ R, x > 3
x > β > α ⇒ x – α > 0, x – β > 0
⇒ (x – α) (x – β) > 0
⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c and a have the same sign,
x ∈ R, x < α or x > β
⇒ ax² + bx + c and a have the same sign

Question 23.
Theorem : Let f(x) = ax² + bx+ c be a quadratic function.
i) If a > 0 then f(x) has minimum value at x = \(\frac{-\mathbf{b}}{2 a}\) and the minimum value = \(\frac{4 a c-b^{2}}{4 a}\)
ii) If a < 0 then f(x) has maximum value at x = \(\frac{-\mathbf{b}}{2 a}\) and the maximum value = \(\frac{4 a c-b^{2}}{4 a}\) (Apr. ’01)
Proof.
ax² + bx+ c =

Second Proof: f(x) = ax² + bx + c
⇒ f'(x) = 2ax + b ⇒ f”(x) = 2a
If f'(x) = 0, then 2ax + b = 0 and hence x = \(-\frac{b}{2 a}\)
If a > 0, then f”(x) > 0 and hence ‘f’ has minimum value at x = \(-\frac{b}{2 a}\)
Minimum of

If a < 0 then f”(x) < 0 and hence ‘f’ has maximum value at x = –\(\frac{b}{2 a}\)
maximum of

Question 24.
Theorem : A necessary and sufficient condition for the quadratic equations
a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 to have a common root is (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁) (b₁c₂ – b₂c₁).
Proof:
Necessity
Let α be a common root of the given equations.
Then a₁α² + b₁α + c₁ = 0 ——(1)
a₂α² + b₂α + c₂ = 0 —— (2)
On multiplying euqation (1) by a₂, equation (2) by a₁ and then subtracting the latter from the former, we get

On multiplying euqation (1) by b₂, equation (2) by b₁ and then subtracting the latter from the former, we get
α² (a₁b₂ – a₂b₁) = b₁c₂ – b₂c₁ —— (4)
On squaring both sides of equation (3) and using (4) we obtain


Therefore the given equations have the same roots.

Case (ii): a₁b₂ – a₂b₁ ≠ 0

Similarly we can prove that a₂α² + b₂α + c₂ = 0
Thus α is a common root of the given equations.

Question 25.
Find the roots of the equation 3x² + 2x – 5 = 0.
Solution:
The roots of the quadratic equation
ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Here a = 3, b = 2 and c = -5.
Therefore the roots of the given equation are

Hence 1 and –[/latex]\frac{5}{3}[/latex] are the roots of the given equation.
Another method
We can also obtain these roots in thè following way.
3x² + 2x – 5 = 3x² + 5x – 3x – 5
= x(3x + 5) -1 (3x + 5)
= (x – 1) (3x + 5)
= 3(x – 1) \(\left(x+\frac{5}{3}\right)\)
Since 1 and –[/latex]\frac{5}{3}[/latex] are the zeros of 3x² + 2x – 5, they are the roots of 3x² + 2x – 5 = 0.

Question 26.
Find the roots of the equation 4x² – 4x + 17 = 3x² – 10x – 17.
Solution:
Given equation can be rewritten as x² + 6x + 34 = 0. The roots of the quadratic equation
ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Here a = 1, b = 6 and c = 34
Therefore the roots of the given equation are

Hence the roots of the given equation are -3 + 5i and -3 – 5i

Question 27.
Find the roots of the equation
\(\sqrt{3}\)x² + 10x – 8\(\sqrt{3}\) = 0.
Solution:
Here a = \(\sqrt{3}\), b = 10, c = -8\(\sqrt{3}\)

Question 28.
Find the nature of the roots of 4x² – 20x + 25 = 0
Solution:
Here a =4, b = -20, c = 25
Hence Δ = b² – 4ac
= (-20)² – 4(4) (25) = 0
Since Δ is zero and a, b, c are real, the roots of the given equation are real and equal.

Question 29.
Find the nature of the roots of 3x² + 7x + 2 = 0
Solution:
Here a = 3, b = 7, c = 2
Hence Δ = b² – 4ac
= 49 – 4(3) (2)
= 49 – 24
= 25 = (5)² > 0
Since a, b, c are rational numbers and Δ = 5² is a perfect square and positive, the roots of the given equation are rational and unequal.

Question 30.
For what values of m, the equation x² – 2(1 + 3m)x + 7(3 + 2m) = 0 will have equal roots?
Solution:
The given equation will have equal roots iff its discriminant is 0.
Here Δ = [(-2(1 + 3m)]² – 4(1) [7 + (3 + 2m)]
= 4(1 + 9m² + 6m) – 28(3 + 28m)
= 9m² – 8m – 20
= (m – 2)(9m + 10)
Hence Δ = 0 ⇔ m = 2, \(\frac{-10}{9} .\)

Question 31.
If α and β are the roots of ax² + bx + c = 0, find the value of α² + β² and α³ + β³ in terms of a, b, c.
Solution:
α, β are the roots of ax² + bx + c = 0

Question 32.
Show that the roots of the equation x² – 2px + p² – q² + 2qr – r² = 0 are rational, given that p, q, r are rational.
Solution:
Here a = 1, b = -2p, c = p² – q² + 2qr – r²
∆ = b² – 4ac
= (-2p)² – 4(1) (p² – q² + 2qr – r²)
= 4p² – 4p² + 4q² + 8qr + 4r²
= 4(q – r)².
∵ p, q, r are rational, the coefficient of the given equation are rational and is a square of a rational number 2(q – r).
∴ The roots of the given equation are rational.

Question 33.
Form a quadratic equation whose roots are 2\(\sqrt{3}\) – 5 and -2\(\sqrt{3}\) – 5.
Solution:
Let α = 2\(\sqrt{3}\) – 5 and β = -2\(\sqrt{3}\) – 5
Then α + β = (2\(\sqrt{3}\) – 5) + (-2\(\sqrt{3}\) – 5) = -10
and αβ = (2\(\sqrt{3}\) – 5) + (-2\(\sqrt{5}\) – 5)
= (-5)² – (-2\(\sqrt{3}\))²
= 25 – 4 × 3
= 13
The required quadratic equation is
x² – (α + β)x + αβ = 0
⇒ x² – (-10)x + 13 = 0
⇒ x² + 10x + 13 = 0

Question 34.
Find the quadratic equation, the sum of whose roots is 1 and the sum of squares of the roots is 13.
Solution:
Let α, β be the roots of a required quadratic equation.
Then α + β = 1 and α² + β² = 13
Now αβ = \(\frac{1}{2}\)[(α + β)² – (α² + β²)]
= \(\frac{1}{2}\)[(1)² – (13)]
Therefore x² – (α + β)x + αβ = 0
⇒ x² – (1)x + (-6) = 0
⇒ x² – x – 6 = 0

Question 35.
Let α and β be the roots of the quadratic equation ax² + bx + c = 0, c ≠ 0, then form the quadratic equation whose roots are \(\frac{1-\alpha}{\alpha}\) and \(\frac{1-\beta}{\beta}\).
Solution:
From the given equation

Question 36.
Solve x^2/3 + x^1/3 – 2 = 0
Solution:
(x^1/3)² + (x^1/3) – 2 = 0
Let x^1/3 = a, a² + a – 2 = 0
⇒ (a +2)(a – 1) = 0 ⇒ a = 1, -2
Now, x^1/3 = 1 ⇒ x = (1)³ = 1
x^1/3 = -2 ⇒ x = (-2)³ = -8
∴ roots are 1, -8.

Question 37.
Solve 7^{1 + x} + 7^{1 – x} = 50 for real x.
Solution:
The given equation can be written as,
7.7^x + \(\frac{7}{7^{x}}\) – 50 = 0
Let 7^x = a
7a + \(\frac{7}{a}\) – 50 = 0
⇒ 7a² + 7 – 50a = 0
⇒ 7a² – 49a – a + 7 = 0
⇒ 7a(a – 7) – 1(a – 7) = 0
⇒ (a – 7)(7a – 1) = 0
∴ a = 7, \(\frac{1}{7}\)
Now, if a = 7 then 7^x = 7¹ ⇒ x = 1
a = \(\frac{1}{7}\) then 7^x = \(\frac{1}{7}\) = 7^-1
x = -1
∴ x = -1, 1

Question 38.
Solve

Solution:
On taking \(\sqrt{\frac{x}{1-x}}\) = a
a + \(\frac{1}{a}\) = \(\frac{13}{6}\)
⇒ \(\frac{a^{2}+1}{a}\) = \(\frac{13}{6}\)
⇒ \(\frac{a^{2}+1}{a}\) = \(\frac{13}{6}\)
⇒ 6a² + 6 = 13a
⇒ 6a² – 13a + 6 = 0
⇒ 6a² – 9a – 4a + 6 = 0
⇒ 3a(2a – 3) – 2(2a – 3) = 0
⇒ (2a – 3)(3a – 2) = 0
a = \(\frac{3}{2}\), a = \(\frac{2}{3}\)
If a = \(\frac{3}{2}\) then \(\sqrt{\frac{x}{1-x}}\) = \(\frac{3}{2}\)
⇒ \(\frac{x}{1-x}\) = \(\frac{9}{4}\)
⇒ 4x = 9 – 9x
⇒ 13x = 9
⇒ x = \(\frac{4}{13}\)
∴ x = \(\frac{9}{13}\), \(\frac{4}{13}\)

Question 39.
Find all number which exceed their square root by 12.
Solution:
Let the required number be ‘x’
given, x = \(\sqrt{x}\) + 12
⇒ x – 12= \(\sqrt{x}\)
Squaring on both sides
(x – 12)² = (\(\sqrt{x}\))²
⇒ x² – 24x + 144 = x
⇒ x² – 25x + 144 = 0
⇒ (x – 9)(x – 16) = 0
⇒ x = 9, 16
But x = 9 does not satisfy the given condition
x = 16
∴ The required numbér = 16.

Question 40.
If x² + 4ax + 3 = 0 and 2x² + 3ax – 9 = 0 have a common root, then find the values
of a and the common root.
Solution:
The quadratic equations

Substitute in the above equation
[(3) (2) – (-9) (1)]² = [(1) (3a) – (2) (4a)] [(4a) (-9) – (3a) (3)]
(15)² = (-5a) (-45a)
⇒ 225 = 225a² ⇒ a² = 1 ⇒ a = ±1

Case (i) : If a = 1, the given equations become x² + 4x + 3 = 0 and 2x² + 3x – 9 = 0
⇒ (x + 1)(x + 3) = 0 and (2x + 3)(x + 3) = 0
⇒ x = 3, -1 and -3, \(\frac{3}{2}\)
In this case, the common root of the given equations is -3

Case (ii) : If a = -1, the given equations become x² – 4x + 3 = 0 and 2x² – 3x – 9 = 0
⇒ (x – 1) (x – 3) = 0 and (2x + 3) (x – 3) = 0
⇒ x = 1, 3 and x = 3, –\(\frac{3}{2}\)
In this case, the common root of the given equation is 3.

Question 41.
Prove that there is unique pair of consecutive positive odd integers such that the sum of their squares is. 290 and find it.
Solution:
The difference of two consecutive odd integers is 2.
So, we have to prove that there is a unique positive odd integer ‘x’ such that
x² + (x + 2)² = 290 —– (1)
x² + (x + 2)² = 290
⇒ x² + x² + 4x + 4 = 290
⇒ 2x² + 4x – 286 = 0
⇒ x² + 2x – 143 = 0
⇒ (x + 13) (x – 11) = 0
⇒ x = -13, 11
Hence 11 is the only positive odd integer satisfying equation (1).
∴ 11, 13 is the unique pair of integers which satisfies the given condition.

Question 42.
The cost of a piece of cable wire is Rs. 35/-, If the length of the piece of wire is 4 meters more and each meter costs, Rs. 1/— less, the cost would remain unchanged. What is the length of the wire?
Solution:
Let the length of the piece of the wire be ‘l’ meters and the cost of each meter be Rs. x.
given lx. = 35 —— (1)
and (l + 4) (x – 1) = 35
⇒ lx – l + 4x – 4 = 35
⇒ 35 – l + 4x – 4 = 35
⇒ 4x = l + 4
⇒ x = \(\frac{l+4}{4}\)
Substitute x in (1), we get l[latex]\frac{l+4}{4}[/latex] = 35
⇒ l² + 4l = 140
⇒ l² + 4l – 140 = 0
⇒ l² + 14l – 10l – 140 = 0
⇒ l(l + 14) – 10(l + 14) = 0
⇒ (l – 10)(l + 14) = 0
⇒ l = -14, 10
Since length cannot be negative, l = 10
∴ The length of the piece of wire is 10 meters.

Question 43.
One fourth of a herd of goats was seen in the forest. Twice the square root of the number in the herd had gone up the hill and the remaining 15 goats were on the bank of the river. Find the total number of goats.
Solution:
Let the number of goats in the herd be ‘x’.
The number of goats seen in the forest = \(\frac{x}{4}\)
The number of goats gone upto the hill = \(2 \sqrt{x}\)
The number of goats on the bank of a river = 15
∴ \(\frac{x}{4}\) + 2\(\sqrt{x}\) + 15 = x
⇒ x + 8\(\sqrt{x}\) + 60 = 4x
⇒ 3x – 8\(\sqrt{x}\) – 60 = 0
Put \(\sqrt{x}\) = y
⇒ 3y² – 8y – 60 = 0
⇒ 3y² – 18y + 10y – 60 = 0
⇒ 3y(y – 6) + 10(y – 6) = 0
⇒ (3y + 10)(y – 6) = 0
⇒ y = 6, –\(\frac{10}{3}\)
Since y cannot be negative, y = 6
⇒ \(\sqrt{x}\) = 36
∴ x = 36
∴ Total number of goats = 36

Question 44.
In a cricket match Anil took one wicket less than twice the number of wickets taken by Ravi. If the product of the
number of wickets taken by them is 15, find the number of wickets taken by each of them.
Solution:
Let the number of wickets taken by Ravi be ‘x’ and the number of wickets taken by Anil is 2x – 1.
Given x(2x – 1) = 15
⇒ 2x² – x – 15 = 0
⇒ 2x² – 6x + 5x – 15 = 0
⇒ 2x(x – 3) + 5(x – 15) = 0
⇒ (x – 3)(2x + 5) = 0
⇒ x = 3, –\(\frac{5}{2}\)
Since the number of wickets be integer,
x = 3 and 2x – 1 = 2(3) – 1 = 5
∴ The number of wickets taken by Anil and Ravi are 5 and 3 respectively.

Question 45.
Some points on a plane are marked and they are connected pairwise by line segments. If the total number of line
segments formed is 10, find the number of marked points on the plane.
Solution:
Let the number of points marked on the plane be ’x’.
The total number of line segments actually formed is

Given \(\frac{x(x-1)}{2}\) = 10
⇒ x² – x = 20
⇒ x² – x – 20 = 0
⇒ (x – 5)(x + 4) = 0
Since x cannot be negatives x = 5
∴ The number of points marked on the plane is 5.

Question 46.
Suppose that the quadratic equations ax² + bx + c = 0 and bx² + cx + a = 0 have a common root. Then show that a³ + b³ + c³ = 3abc.
Solution:
Let α be the common root of the equations

Question 47.
For what values of x1 the expression x² – 5x – 14 is positive?
Solution:
Since x² – 5x – 14 = (x + 2) (x – 7), the roots of the equation -2 and 7
Here the coefficient of x² is positive.
Hence x² – 5x – 14 is positive when x < -2 or x > 7.

Question 48.
For what values ofx. the expression -6x² + 2x³ is negative?
Solution:
-6x² + 2x – 3 = 0 can be written as 6x² – 2x + 3 = 0

∴ The roots of -6x² + 2x – 3 = 0 are non-real complex numbers.
Hence coefficient of x² is -6, which is negative.
∴ -6x² + 2x – 3 < 0 for all x ∈ R.

Question 49.
Find the value of x at which the following expressions have maximum or minimum.

i) x² + 5x + 6
Solution:
Here a = 1 > 0, the expression has minimum value at x = \(\frac{-b}{2 a}\)
= \(\frac{-5}{2(1)}\)
= \(\frac{-5}{2}\)

ii) 2x – x² + 7
Solution:
Here a = -1 < 0, the expression has maximum value at x
= \(\frac{-b}{2 a}\)
= \(\frac{-2}{2 x-1}\) = 1

Question 50.
Find the maximum or minimum value of the quadratic expression
(i) 2x – 7 – 5x²
(ii) 3x² + 2x + 17 (Mar. ’14)
Solution:
i) Compare the given equation with ax² + bx + c, we get
a = -5, b = 2, c = -7
Here a = -5 < 0, the given expression has maximum value at x = \(\frac{-b}{2 a}\) = \(\frac{-2}{2(-5)}\)
= \(\frac{1}{5}\)
and maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(-5) \cdot(-7)-(?)^{2}}{4(-5)}\)
= \(\frac{140-4}{-20}\) = \(\frac{-34}{5}\)

(ii) Compare 3x² + 2x + 11 with
ax² + bx + c, we get a = 3, b = 2, c = 11
∵ a = 3 > 0, the given expression has minimum value at

Question 51.
Find the changes in the sign of 4x – 5x² + 2 for x ∈ R and find the extreme value.
Solution:
The roots of 4x – 5x² + 2 = 0
⇒ 5x² – 4x – 2 = 0
roots are = \(\frac{2 \pm \sqrt{14}}{5}\)
∴ \(\frac{2-\sqrt{14}}{5}\) < x < \(\frac{2+\sqrt{14}}{5}\) the sign of 4x – 5x² + 2 is positive.
x < \(\frac{2-\sqrt{14}}{5}\) or x > \(\frac{2+\sqrt{14}}{5}\), the sign of 4x – 5x² + 2 is negative.
Since a = -5 < 0, then maximum value = \(\frac{4 a c-b^{2}}{4 a}\)
= \(\frac{4(-5)(2)-(4)^{2}}{4(-5)}\)
= \(\frac{-56}{-20}\) = \(\frac{14}{5}\)
Hence extreme value = \(\frac{14}{5}\)

Question 52.
Find the solution set of x² + x – 12 ≤ 0 by both algebraic and graphical methods.
Solution:
Algebraic Method : We have x² + x – 12 = (x + 4) (x – 3).
Hence -4 and 3 are the roots of the equation x² + x – 12 = 0.
Since the coefficient of x² in the quadratic expression x² + x – 12 = 0 is positive, x² + x – 12 is negative if -4 < x < 3 and positive if either x < -4 or x > 3.
Hence x² + x – 12 ≤ 0 ⇔ -4 ≤ x ≤ 3.
Therefore the solution set is
{x ∈ R : -4 ≤ x ≤ 3}.
Graphical Method: Let y = f(x) = x² + x – 12.
The values of y at some selected values of s are given in the following table:

The graph of the function y = f(x) is drawn using the above tabulated values. This is shown in Fig.

Therefore the graph of y = f(x) we observe that
y = x² – x – 12 < 0 if f -4 ≤ x ≤ 3.
Hence the solution set is {x ∈ R : -4 ≤ x ≤ 3}.

Question 53.
Find the set of values of x for which the inequalities x² – 3x – 10 < 0, 10x – x² – 16
> 0 hold simultaneously.
Solution:
∵ x² – 3x – 10 < 0
⇒ x² – 5x + 2x – 10 < 0
⇒ x(x – 5) + 2(x – 5) < 0
⇒ (x – 5)(x + 2) < 0
∴ x² coeff. is the and expression is <0
⇒ x ∈ (-2, 5) ——- (1)
Now 10x – x² – 16 > 0
⇒ x² – 10x + 16 < 0
⇒ (x – 2) (x – 8) < 0
∴ x² coeff is the ana expression is < 0
⇒ x ∈ (2, 8) ——- (2)
Hence x² – 3x – 10 < 0 and 10x – x² – 16 > 0
⇔ x ∈ (-2, 5) ∩ (2, 8)
∴ The solution set is {x/x ∈ R : 2 < x < 5}

Question 54.
Solve the inequation \(\sqrt{x+2}\) > \(\sqrt{8-x^{2}}\).
Solution:
When a, b ∈ R and a ≥ 0, b ≥ 0 then \(\sqrt{a}\) > \(\sqrt{b}\) ⇔ a > b ≥ 0
∴ \(\sqrt{x+2}\) = \(\sqrt{8-x^{2}}\)
⇔ x + 2 > 8 – x² ≥ 0 and x > -2, |x| < 2\(\sqrt{2}\)
We have (x + 2) > 8 – x²
⇔ x² + x – 6 > 0
⇔ (x + 3)(x – 2) > 0
⇔ x ∈ (-∞, -3)(2, ∞) .
We have 8 – x² ≥ 0
⇔ x² ≤ 8 ⇔ |x| < 2 \(\sqrt{2}\) ⇔ x ∈ [-2 \(\sqrt{2}\), 2 \(\sqrt{2}\)] Also x + 2 > 0 ⇔ x > -2
Hence x + 2 > 8 – x² ≥ 0
⇔ x ∈ ((-∞, -3) ∪ (2, ∞)) ∩ [-2 \(\sqrt{2}\), 2 \(\sqrt{2}\)] and x > -2
⇔ x ∈ (2, 2 \(\sqrt{2}\))
∴ The solution set is [x ∈ R : -2 < x ≤ 2 \(\sqrt{2}\)]

Question 55.
Solve the equation
\(\sqrt{(x-3)(2-x)}\) < \(\sqrt{4 x^{2}+12 x+11}\).
Solution:
The given inequation is equivalent to the following two inequalities.
(x – 3)(2 – x) ≥ 0 and
(x – 3)(2 – x) < 4x² + 12x + 11
(x – 3)(2 – x) ≥ 0 ⇔ (x – 2)(x – 3) ≤ 0
⇔ 2 ≤ x ≤ 3
-x² + 5x – 6 < 4x² + 12x + 11
⇔ 5x² + 7x + 17 > 0
The discriminant = b² – 4ac = 49 – 340 < 0 ⇒ x ∈ R : a = 5 > 0 and expressive is the
Hence the solution set of the given inequation is {x ∈ R : 2 ≤ x ≤ 3}

Question 56.
Solve the inequation

Solution:

⇔ either 6 + x – x² = 0, 2x + 5 ≠ 0 and x + 4 ≠ 0 or 6 + x – x² > 0
and \(\frac{1}{2 x+5}\) ≥ \(\frac{1}{x+4}\)
We have 6 + x – x² = -(x² – x – 6)
= -(x + 2) (x – 3)
Hence 6 + x + x² = 0 ⇔ x = -2 or x = 3
2x + 5 = 0 ⇔ x = –\(\frac{5}{2}\)
x + 4 = 0 ⇔ x = -4
∴ 6 + x – 2x² > 0 ⇔ -2 < x < 3 x ∈ (-2, 3), 2x + 5 > -4 + 5 = 1 > 0 and x + 4 > -2 + 4 = 2 > 0
x + 4 > -2 + 4 = 2 > 0
x ∈ (-2, 3), \(\frac{1}{2 x+5}\) ≥ \(\frac{1}{x+4}\)
⇔ 2x + 5 ≤ x + 4,
2x + 5 ≤ x + 4 ⇔ x ≤ 1
Hence 6 + x – x² > 0 and \(\frac{1}{2 x+5}\) ≥ \(\frac{1}{x+4}\)
⇔ -2 < x ≤ -1
∴ The solution set is {-2, 3} ∪ (-2, -1)
= [-2, -1] ∪ {3}

Question 57.
Find the maximum value of the function
\(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R. (May ’05)
Solution:
Let y = \(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\)
⇒ yx² + 2yx + 3y = x² + 14x + 9
⇒ (y – 1)x² + 2(y – 7)x + 3y – 9 = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 7)²] – 4(y – 1) (3y – 9) ≥ 0
⇒ 4[(y² – 14y + 49) [(3y² – 12y + 9)] ≥ 0
⇒ -2y² – 2y + 40 ≥ 0
⇒ y² + y – 20 ≤ 0
⇒ (y + 5) (y – 4) ≤ 0
⇒ -5 ≤ y ≤ 4
⇒ y ∈ [-5, 4]
∴ Maximum value of y = 4
∴ Maximum value of the function
\(\frac{x^{2}+14 x+9}{x^{2}+2 x+3}\) over R is 4.

Question 58.
Show that none of the values of the function over \(\frac{x^{2}+34 x-71}{x^{2}+2 x-7}\) over R lies between 5 and 9. (Mar. 2005)
Solution:
Let y = \(\frac{x^{2}+34 x-7}{x^{2}+2 x-7}\)
⇒ x² + 2x – 7 = x² + 34x – 71
⇒ (y – 1)x² + 2(y – 17)x + (-7y + 71) = 0
Since x ∈ R, discriminant ≥ 0
⇒ [2(y – 17)]² – 4(y – 1)(-7y + 71)] ≥ 0
⇒ 4[(y² – 34y + 289) – (-7y² + 78y – 71)] ≥ 0
⇒ 8y² – 112y + 360 ≥ 0
⇒ y² – 14y + 45 ≥ 0
⇒ (y – 5) (y – 9) ≥ 0
⇒ y ≤ 5 or y ≥ 9
⇒ y does not lies between 5 and 9, since expression is ≥ 0 and y² coeff is the
∴ None of the values of the function
\(\frac{x^{2}+34 x-7}{x^{2}+2 x-7}\) over R lies between 5 and 9.

Question 59.
Solve the inequation
\(\sqrt{x^{2}-3 x-10}\) > (8 – x)
Solution:
(i) \(\sqrt{x^{2}-3 x-10}\) > (8 – x)
⇒ x² – 3x – 10 ≥ 0 and (i) (8 – x < 0)
or (ii) x² – 3x – 10 > (8 – x)² and (8 – x ≥ 0)
Now x² – 3x – 10 = (x – 5) (x + 2)
Hence x² – 3x – 10 ≥ 0
⇒ x ∈ (-∞, -2] ∪ [5, ∞)
8 – x < 0 ⇒ x ∈ (8, ∞)
∴ x² – 3x – 10 ≥ 0 and 8 – x < 0
⇔ x ∈ (-∞, -2] ∪ [5, ∞) and x ∈ (8, ∞)
⇔ x ∈ (8, ∞)

(ii) x² – 3x – 10 > (8 – x)²
⇔ x² – 3x – 10 > 64 + x² – 16x
⇔ 13x > 74

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A De Moivre’s Theorem Important Questions

Question 1.
Find the value of (1 – i)⁸  (Mar. ’07)
Solution:

Question 2.
If x = cis θ, then find the value of [x⁶ + \(\frac{1}{x^{6}}\)]
Solution:
∵ x = cos θ + i sin θ
⇒ x⁶ = (cos θ + i sin θ)⁶
= cos 6θ + i sin 6θ
⇒ \(\frac{1}{x^{6}}\) = cos 6θ – i sin 6θ
∴ x⁶ + \(\frac{1}{x^{6}}\) = 2 cos 6θ

Question 3.
If A, B, C are angles of a triangle such that x = cis A, y = cis B, z = cis C, then find the value of XYZ. (AP Mar. ‘16, ’15)
Solution:
∴ A, B, C are angles of a triangle
⇒ A + B + C = 180° ——- (1)
x = cis A, y = cis B, Z = cis C
⇒ xyz = cis(A + B + C)
= cos(A + B + C) + i sin(A + B + C)
= cos(180°) + i sin (180°) .
= -1 + i(0) = -1
∴ xyz = -1

Question 4.
If 1, ω, ω² are the cube roots of unity, then prove that \(\frac{1}{2+\omega}\) – \(\frac{1}{1+2 \omega}\) = \(\frac{1}{1+\omega}\)
Solution:
ω is a cube root of unity
1 + ω + ω² = 0 and ω³ = 1

Question 5.
(2 – ω) (2 – ω²) (2 – ω¹⁰) (2 – w¹¹) = 49. (TS Mar. ’17)
Solution:
∵ 1, ω, ω² are the cube roots of unity,
ω³ = 1 and 1 + ω + ω² = 0
2 – ω¹⁰ = 2 – ω⁹ . ω
= 2 – (ω³)³ . ω²
= 2 – (1)³ ω² = 2 – ω²
(2 – ω)(2 – ω²) = 4 – 2ω – 2ω² + ω³
= 4 – 2(ω + ω²) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1 = 7
∴ (2 – ω)(2 – ω²)(2 – ω¹⁰)(2 – ω¹¹)
= (2 – ω) (2 – ω²) (2 – ω) (2 – ω²)
= ((2 – ω) (2 – ω²))²
= 7² = 49

Question 6.
If α, β are the roots of the equation x² – 2x + 4 = 0 then for any n ∈ N show that αⁿ + βⁿ = 2^{n + 1} cos \(\left(\frac{n \pi}{3}\right)\) (Mar. ’14)
Solution:

Question 7.
Show that one value of

is -1.
Solution:

Question 8.
If n is a positive integer, show that (1 + i)ⁿ + (1 – i)ⁿ = 2^{\(\frac{n+2}{2}\)} cos \(\left(\frac{\mathrm{n} \pi}{4}\right)\). (A.P) (Mar. ’15)
Solution:

Question 9.
If n is an integer then show that (1 + cos θ + i sin θ)ⁿ + (1 + cos θ – i sin θ)ⁿ = 2^{n + 1} cosⁿ (θ/2) cos \(\left(\frac{n \theta}{2}\right)\) (May. ’11) (TS & AP Mar. ’17)
Solution:
L.H.S.

Question 10.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ. Prove that cos² α + cos² β + cos² γ = \(\frac{3}{2}\) = sin² α + sin² β + ² γ. (AP. Mar. ’16; TS Mar. ’15, ‘13)
Solution:
Let x = cos α + i sin α
y = cos β + i sin β
z = cos γ + i sin γ
∴ x + y + z = (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 + i. 0 = 0
(x + y + z)² = 0
⇒ x² + y² + z² + 2(xy + yz + zx) = 0
x² + y² + z² = -2(xy + yz + zx)


-i(sin α + sin β + sin γ) = 0 – i. 0 = 0
Substituting in (1)
x² + y² + z² = 0.
(cos α + i sin α)² + (cos β + i sin β)² + (cos γ + i sinγ)² = 0
(cos 2α + i sin 2α) + (cos β + i sin 2β) + (cos 2γ + i sin 2γ) = 0
(cos 2α + cos 2β + cos 2γ) + i (sin 2α + sin 2β + sin 2γ) = 0
Equating real parts
cos 2α + cos 2β + cos 2γ = 0
2cos²α – 1 + 2 cos²β – 1 + 2 cos²γ – 1 = 0
2 (cos²α + cos²β + cos²γ) = 3
cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
∵ ²α + cos²β + cos²γ = \(\frac{3}{2}\)
⇒ (1 – sin²α) + (1 – sin²β) + (1 – sin²γ) = \(\frac{3}{2}\)
⇒ sin²α + sin²β + sin²γ = \(\frac{3}{2}\)
∴ cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
= sin²α + sin²β + sin²β

Question 11.
If 1, ω, ω² are the cube roots of unity prove that
i) (1 – ω + ω²)⁶ + (1 – ω² + ω)⁶ = 128
= (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
ii) (a + b) (aω + bω²) (aω² + bω) = a³ + b³
iii) x² + 4x + 7 = 0 where x = ω – ω² – 2.
Solution:
∵ 1, ω, ω² are the cube roots of unity
⇒ 1 + ω + ω² = 0 and ω³ = 1

i) (1 – ω + ω²)⁶ + (1 – ω² + ω)⁶
= (-ω – ω)⁶ + (-ω² – ω²)⁶
= (-2ω)⁶ + (-2ω²)⁶
= 2⁶(ω⁶ + ω¹²)
= 2⁶(1 + 1) = 2⁶ × 2 = 2⁷ = 128

Again (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
= (1 + ω² – ω)⁷ + (1 + ω – ω²)⁷
= ( -ω – ω)⁷ + (-ω² – ω²)⁷
= (-2ω)⁷ + (-2ω²)⁷
= (-2)⁷ (ω⁷ + ω¹⁴)
= (-2)⁷ (ω + ω²)
= -(2)⁷ (-1)
= 2⁷ = 128

ii) (a+b) (a]ω + bω²)(aω² + bω) (AP) (Mar. ’17)
Solution:
= (a + b) [a²ω³ + abω⁴)4 + abω² + b²ω³]
= (a + b) [a² + ab(ω² + ω⁴) + b²]
= (a + b) [a² + ab(ω² + ω) + b²]
= (a + b) (a² – ab + b²) = a³ + b³

iii) x = ω – ω² – 2
⇒ x + 2 = ω – ω²
⇒ (x + 2)² = ω² + ω⁴ – 2ω³
⇒ x² + 4x + 4 = ω² + ω – 2
⇒ x² + 4x + 4 = (-1) – 2 = -3
⇒ x² + 4x + 7 = 0

Question 12.
Simplify \(\frac{(\cos \alpha+i \sin \alpha)^{4}}{(\sin \beta+i \cos \beta)^{8}}\)
Solution:


= (cos α + i sin α)⁴ (cos β – i sin β)^-8 [i⁸ = (i²)⁴ = (-1)⁴ = 1]
= (cos 4α + i sin 4α) (cos 8β + i sin 8β)
= cos (4α + 8β) + i sin (4α + 8β)

Question 13.
If m, n are integers and x = cos α + i sin α, y = cos β + i sin β, then prove that
x^m yⁿ + \(\frac{1}{x^{m} y^{n}}\) = 2 cos (mα + nβ) and
x^m yⁿ – \(\frac{1}{x^{m} y^{n}}\) = 2i sin (mα + nβ).
Solution:
∵ x = cos α + i sin α, y = cos β + i sin β
⇒ x^m = (cos α + i sin α)^m = cos mα + i sin mα
yⁿ = (cos β + i sin β)ⁿ = cos nβ + i sin nβ
∴ x^m yⁿ = (cos mα + i sin mα)(cos nβ + i sin nβ)
= cos (mα + nβ) + i sin (mα + nβ) ——— (1)
\(\frac{1}{x^{m} \cdot y^{n}}\) = cos (mα + nβ) – i sin (mα + nβ) —— (2)
By adding (1) and (2).
x^myⁿ + \(\frac{1}{x^{m} y^{n}}\) = 2 cos (mα + nβ)
By subtracting (2) from (1)
x^myⁿ – \(\frac{1}{x^{m} y^{n}}\) = 2 sin (mα + nβ)

Question 14.
If n is a positive integer, show that (1 + i)ⁿ + (1 – i)ⁿ = 2^{\(\frac{n+2}{2}\)} cos \(\left(\frac{n \pi}{4}\right)\) (A.P.) (Mar. ‘15)
Solution:

Question 15.
If n is an integer then show that (1 + cos θ + i sin θ)ⁿ + (1 + cos θ – i sin θ)ⁿ = 2^{n + 1} + cosⁿ (θ/2) cos \(\left(\frac{\mathrm{n} \theta}{2}\right)\). (May ’11)
Solution:
L.H.S.

Question 16.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ, Prove that cos² α + cos² β + cos² γ = \(\frac{3}{2}\) = sin² α +
sin² β + sin² γ. (A.P. Mar. ‘16, T.S. Mar. ‘15, ’13)
Solution:
Let x = cos α + i sin α
y = cos β + i sin β
z = cos γ + i sin γ
∴ x + y + z = (cos α + cos β + cos γ) + i(sin α + sin β + sin γ)
= 0 + i.0 = 0
(x + y + z)² = 0
⇒ x² + y² + z² + 2(xy + yz + zx) = 0
x² + y² + z² = -2(xy + yz + zx)

Similarly \(\frac{1}{y}\) = cos β – i sin β
\(\frac{1}{z}\) = cos γ – i sin γ
∴ \(\frac{1}{x}\) + \(\frac{1}{y}\) + \(\frac{1}{z}\) = (cos α + cos β + cos γ) – i(sin α + sin β + sin γ) = 0 – i. 0 = 0
Substituting in (1)
x² + y² + z² = 0
(cos α + i sin β)² + (cos β + i sin β)² + (cos γ + i sin γ)² = o
(cos 2α + i sin 2α) + (cos 2β + i sin 2β) + (cos 2γ + i sin 2γ) = 0
(cos 2α + cos 2β + cos 2γ) + i (sin 2α + sin 2β + sin 2γ) = 0
Equating real parts
cos 2α + cos 2β + cos 2γ = 0
2 cos²α – 1 + 2 cos²β – 1 + 2 cos²γ – 1 = 0
2 (cos²α + cos²β + cos²γ) = 3
cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
∵ cos²α + cos²β + cos²γ = \(\frac{3}{2}\)
⇒ (1 – sin²α) + (1 – sin²β) + (1 – sin²γ) = \(\frac{3}{2}\)
⇒ sin²α + sin²β + sin²γ = 3 – \(\frac{3}{2}\) = \(\frac{3}{2}\)
∴ cos² α + cos²β + cos²γ = \(\frac{3}{2}\)
= sin² α + sin²β + sin²β

Question 17.
Find all the values of (\(\sqrt{3}\) + i)^1/4.
Solution:

Question 18.
Find all the roots of the equation
x¹¹ – x⁷ + x⁴ – 1 = 0.
Solution:
x¹¹ – x⁷ + x⁴ – 1 = 0
⇒ x⁷(x⁴ – 1) + 1 (x⁴ – 1) = 0
⇒ (x⁴ – 1) (x⁷ + 1) = 0
Case(i) : x⁴ – 1 = 0
x⁴ = 1 = (cos o + i sin 0)
⇒ x⁴ = (cos 2kπ + i sin 2kπ)
∴ x = (cos 2kπ + i sin 2kπ)^1/4
⇒ x = cis \(\left(\frac{2 k \pi}{4}\right)\) = cis \(\frac{\mathrm{k} \pi}{2}\), k = 0, 1, 2, 3.

Case (ii): x⁷ + 1 = 0
⇒ x⁷ = -1 = cos π + i sin π
⇒ x⁷ = cos (2kπ + n) + i sin (2kπ + π)
∴ x = [cos (2k + 1)π + i sin (2k + 1)π]^1/7
⇒ x = cis(2k + 1)\(\frac{\pi}{7}\), k = 0, 1, 2, 3, 4, 5, 6
The values of x are

Question 19.
If 1, ω, ω² are the cube roots of unity prove that (TS. Mar. ‘16)
i) (1 – ω + (ω²)⁶ + (1 – ω² + ω)⁶ = 128
= (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
ii) (a + b)(aω + bω²)(aω² + bω) = a³ + b³
iii) x² + 4x + 7 = 0 where x = ω – ω² – 2.
Solution:
∵ 1, ω, ω² are the cube roots of unity
⇒ 1 + ω + ω² = 0 and ω³ = 1
i) (1 – ω + ω²)⁶ + (1 – ω² + ω)⁶
= (-ω – ω)⁶ +(-ω² – ω²)⁶
= (-2ω)⁶ + (-2w²)⁶
= 2⁶ (ω⁶ + ω¹²)
= 2⁶(1 + 1) = 2⁶ × 2 = 2⁷ = 128 .
Again (1 – ω + ω²)⁷ + (1 + ω – ω²)⁷
= (1 + ω² – ω)⁷ + (1 + ω – ω²)⁷
= (- ω – ω) + (-ω² – ω²)⁷
= (-2ω)⁷ + (-2ω²)⁷
= (-2)⁷ (ω⁷ + ω¹⁴)
= (-2)⁷(ω + ω²)
= -(2)⁷ (-1)
= 2⁷ = 128

ii) (a + b) (aω + bω²) (aω² + bω)
= (a + b) [a²ω³ + abω⁴ + abω² + b²ω³]
= (a + b) [a² + ab(ω² + ω⁴) + b²]
= (a + b) [a² + ab(ω² + ω) + b²]
= (a + b) (a² – ab + b²) = a³ + b³

iii) x = ω – ω² – 2
= x + 2 = ω – ω²
(x + 2)² = ω² – 2ω³
⇒ x² +4x + 4 = ω² + ω – 2
⇒ x² + 4x + 4 = (-1) – 2 = -3.
⇒ x² + 4x + 7 = 0

Question 20.
If α, β are the roots of the equation x² + x + 1 = 0 then prove that α⁴ + β⁴ + α^-1β^-1 = 0
Solution:
Since α, β are the complex cube roots of unity.
We take α = ω, β = ω²
∴ α⁴ + β⁴ + α^-1 + β^-1
= ω⁴ + (ω²)⁴ + \(\frac{1}{\omega} \cdot \frac{1}{\omega^{2}}\)
= ω + ω² + \(\frac{1}{\omega^{3}}\)
= (-1) + \(\frac{1}{1}\)
= -1 + 1 = 0
∴ α⁴ + β⁴ + α^-1 + β^-1 = 0

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Complex Numbers Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2A Complex Numbers Important Questions

Question 1.
If z₁ = -1, z₂ = i then find Arg \(\left(\frac{z_{1}}{z_{2}}\right)\) (AP Mar. 17) (TS Mar.’ 16; May ‘11)
Solution:
Z₁ = -1 = cos π + i sin π
⇒ Arg z₁ = π
z₂ = i = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
⇒ Arg z₂ = \(\frac{\pi}{2}\)
⇒ Arg \(\left(\frac{z_{1}}{z_{2}}\right)\) = Arg z₁ – Arg z₂ = π – \(\frac{\pi}{2}\).
= \(\frac{\pi}{2}\)

Question 2.
If z = 2 – 3i, show that z² – 4z + 13 = 0. (Mar. ‘08)
Solution:
∴ z = 2 – 3i
⇒ z – 2 = -3i
⇒ (z – 2)² = (-3i)²
⇒ z² – 4z + 4 = 9i²
⇒ z² – 4z + 4 = 9(-1)
⇒ z² – 4z + 13 = 0

Question 3.
Find the multiplicative inverse of 7 + 24i. (TS Mar. 16)
Solution:
Since (x + iy)\(\left[\frac{x-i y}{x^{2}+y^{2}}\right]\) = 1, it follows that the multiplicative inverse of (x + iy) is \(\frac{x-i y}{x^{2}+y^{2}}\)
Hence the multiplicative inverse of 7 + 24i is

Question 4.
Write the following complex numbers in the form A + iB. (2 – 3i) (3 + 4i)  (AP Mar. ’17)
Solution:
(2 – 3i) (3 + 4i) = 6 + 8i – 9i – 12i²
= 6 – i + 12
= 18 – i = 18 + i(-1)

Question 5.
Write the following complex numbers in the form A + iB. (1 + 2i)³ (TS Mar. ’17)
Solution:
(1 + 2i)³ = 1 + 3.i².2i + 3.1. 4i² + 8i³
= 1 + 6i – 12 – 8i
= -11 – 2i = (-11) + i(-2)

Question 6.
Write the conjugate of the following complex number \(\frac{5 i}{7+i}\) (AP Mar. ’15)
Solution:

Question 7.
Find a square root for the complex number 7 + 24i. (Mar. ‘14)
Solution:
7 + 24i

Question 8.
Find a square root for the complex number 3 + 4i  (Mar. ’13)
Solution:

Question 9.
Express the following complex numbers in modulus amplitude form. 1 – i (AP Mar. 15)
Solution:
1 – i
Let 1 – i = r (cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = 1
r sin θ = -1
⇒ θ lies in IV quadrant .
Squaring and adding
r² (cos² θ + sin² θ) = 1 + 1 = 2
r² = 2 ⇒ r = \(\sqrt{2}\)
tan θ = -1
⇒ θ = -π/4

Question 10.
Express the complex numbers in modulus — amplitude form 1 + i\(\sqrt{3}\) (TS Mar. ’17)
Solution:
1 + i\(\sqrt{3}\) = r (cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = 1 —– (1)
r sin θ = \(\sqrt{3}\) —– (2)
θ lies in I quadrant
Squaring and adding (1) and (2)
r² (cos² θ – sin² θ) = 1 + 3
r² = 4 ⇒ r = 2
Dividing (2) by (1)

Question 11.
If the Arg \(\overline{\mathbf{z}}_{1}\) and Arg \(z_{2}\) are \(\frac{\pi}{5}\) and \(\frac{\pi}{3}\) respectively, find (Arg z₁ + Arg z₂) (AP Mar. ’16)
Solution:
Arg \(\overline{\mathbf{z}}_{1}\) = \(\frac{\pi}{5}\) ⇒ Arg z₁ = – Arg z₁ = – \(\frac{\pi}{5}\)
Arg z₂ = \(\frac{\pi}{3}\)
∴ Arg z₁ + Arg z₂ = – \(\frac{\pi}{5}\) + \(\frac{\pi}{3}\)
= \(\frac{-3 \pi+5 \pi}{15}\) = \(\frac{2 \pi}{15}\)

Question 12.
If |z – 3 + i| = 4 determine the locus of z. (May. ’14)
Solution:
Let z = x + iy
Given |z – 3 + i| = 4
|x + iy – 3 + i| = 4
⇒ (x – 3) + i(y + 1) = 4
⇒ \(\sqrt{(x-3)^{2}+(y+1)^{2}}\) = 4
⇒ (x – 3)² + (y + 1)² = 16
⇒ x² – 6x + 9 + y² + 2y + 1 = 16
⇒ x² + y² – 6x + 2y – 6 = 0
∴ The locus õf z is x² + y² – 6x + 2y – 6 = 0

Question 13.
The points P, Q denote the complex numbers z₁, z₂ in the Argand diagram. O is the origin. If z₁z₂ + z₂z₁ = 0, show that POQ = 90°. (Mar. ‘07)
Solution:
Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂

Question 14.
Find the real and imaginary parts of the complex number \(\frac{a+i b}{a-i b}\). (TS Mar. 15)
Solution:

Question 15.
Write z = –\(\sqrt{7}\) + i\(\sqrt{21}\) in the polar form. (Mar. ’11)
Solution:

Since the given point lies in the second quadrant we look for a solution of tan θ = –\(\sqrt{3}\) that lies in \(\left[\frac{\pi}{2}, \pi\right]\), we find that θ = \(\frac{2 \pi}{3}\) is such a solution.

Question 16.
z = x + iy and the point P represents z in the Argand plane and \(\left|\frac{z-a}{z+\bar{a}}\right|\) = 1, Re (a) ≠ 0, then find the locus of P. (TS Mar. ’17)
Solution:
Let z = x + iy and a = α + iβ

Locus of P is x = 0 i.e., Y – axis

Question 17.
If x + iy = \(\frac{1}{1+\cos \theta+i \sin \theta}\), show that 4x² – 1 = 0 (AP Mar. ’16, TS Mar. ’17, ’15, ’06 )
Solution:

Equating real parts on both sides, we have
x = \(\frac{1}{2}\)
2x = 1
⇒ 4x² = 1
4x² – 1 = 0

Question 18.
If (\(\sqrt{3}\) + 1)¹⁰⁰ = 2⁹⁹ (a + ib), then show that a² + b² = 4. (AP Mar. ‘16)
Solution:

Question 19.
Show that the points in the Argand diagram represented by the complex numbers 2 + 2i, -2 – 2i, 2\(\sqrt{3}\) + 2\(\sqrt{3}\)i are the vertices of an equilateral triangle. (Mar ‘07)
Solution:
A (2, 2), B (-2, -2), C (-2\(\sqrt{3}\), 2\(\sqrt{3}\)) represents the given complex number in the Argand diagram.

Question 20.
Show that the 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. (June 04) (TS Mar. ’16; AP Mar.’15 ’05; May ’05)
Solution:
A(-2, 7), B(-\(\frac{3}{2}\), \(\frac{1}{2}\)), C(4, -3), D(\(\frac{7}{2}\), \(\frac{7}{2}\)) represents the given complex numbers in the Argand diagram.


∴ AB² = BC² = CD² = DA²
⇒ AB = BC = CD = DA
AC² = (-2 – 4)² + (7 + 3)²
= 36 +100 = 136
(BD)² = (-\(\frac{3}{2}\) – \(\frac{7}{2}\))² + (\(\frac{1}{2}\) – \(\frac{7}{2}\))²
= 25 + 9 = 34
AC ≠ BD
A, B, C, D are the vertices of a Rhombus.

Question 21.
Show that the points in the Argand diagram represented by the complex numbers z₁, z₂, z₃ are collinear, if and only if there exists three real numbers p, q, r not all zero, satisfying pz₁ + qz₂ + rz₃ = 0 and p + q + r = 0. (Mar. ‘07)
Solution:
pz₁ + qz₂ + rz₃ = 0
⇔ rz₃ = -pz₁ – qz₂
⇔ z₃ = \(\frac{-p z_{1}-q z_{2}}{r}\) ∵ r ≠ 0
∵ p + q + r = 0
⇔ r = -p – q
⇔ z₃ = \(-\frac{\left(p z_{1}+q z_{2}\right)}{-(p+q)}\)
⇔ z₃ = \(\frac{p z_{1}+q z_{2}}{p+q}\)
⇔ z₃ = \(\frac{p z_{1}+q z_{2}}{p+q}\)
⇔ z₃ divides line segment joining z₁, z₂ in the ratio q : p
⇔ z₁, z₂, z₃ are collinear

Question 22.
If the amplitude of \(\left(\frac{z-2}{z-6 i}\right) \frac{\pi}{2}\), find its locus. (Mar. ’06)
Solution:
Let z = (x + iy)

Hence a = 0 and b > 0
∴ x(x – 2) + y(y – 6) = 0
or x² + y² – 2x – 6y = 0.

Question 23.
If x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\) then show that x² + y² = 4x – 3 (TS Mar ’17)
Solution:

Equating real and imaginary parts on both sides, we have

Question 24.
Express \(\frac{4+2 i}{1-2 i}\) + \(\frac{3+4 i}{2+3 i}\) in the form a + ib, a ∈ R, b ∈ R.
Solution:

Question 25.
Find the real and imaginary parts of the complex number \(\frac{a+i b}{a-i b}\) (TS Mar ’15)
Solution:

Question 26.
Express (1 – 3)³ (1 + i) in the form of a + ib.
Solution:
(1 – i)³ (1 + j) = (1 – j)² (1 – i) (1 + j)
= (1 + i² – 2i) (1² – i²)
= (1 – 1 – 2i) (1 + 1)
= 0 – 4i = 0 + i (-4)

Question 27.
Find the multiplicative inverse of 7 + 24i.  (TS. Mar. ’16 )
Solution:
Since (x + iy)\(\left[\frac{x-i y}{x^{2}+y^{2}}\right]\) = 1, it follows that the multiplicative inverse of

Question 28.
Determine the locus of z, z ≠ 2i, such that Re\(\left(\frac{z-4}{z-2 i}\right)\) = 0
Solution:
Let z = x + iy

Hence the locus of the given point representing the complex number is the circle with (2, 1) as centre and \(\sqrt{5}\) units as radius, excluding the point (0, 2).

Question 29.
If 4x + i (3x – y) = 3 -6i where x and y are real numbers, then find the values of x and y.
Solution:
∵ 4x + i(3x – y) = 3 – 6i
Equating real and imaginary parts, we get 4x = 3 and 3x – y = -6
4x = 3 and 3x – y = -6
⇒ x = 3/4 and 3\(\left(\frac{3}{4}\right)\) – y = -6
\(\frac{9}{4}\) + 6 = y
⇒ y = \(\frac{33}{4}\)
∴ x = \(\frac{3}{4}\) and y = \(\frac{33}{4}\)

Question 30.
If z = 2 – 3i, show that z² – 4z + 13 = 0. (Mar. ’08)
Solution:
∴ z = 2 – 3i
⇒ z – 2 = – 3i
⇒ (z – 2)² = (-3i)²
⇒ z² – 4z + 4 = 9i²
⇒ z² – 4z + 13 = 0
⇒ z² – 4z + 4 = 9

Question 31.
Find the complex conjugate of (3 + 4i) (2 – 3i).
Solution:
The given complex number is
(3 + 4i) (2 – 3i) = 6 + 8i – 9i – 12i²
= 6 – i – 12(-1) = 18 + i
Its complex conjugate is 18 + i

Question 32.
Show that z₁ = \(\frac{2+11 i}{25}\), z₂ = \(\frac{-2+i}{(1-2 i)^{2}}\), are conjugate to each other.
Solution:

Since, this complex number is the conjugate of \(\frac{2+11 i}{25}\), the two given complex numbers
are conjugate to each other.

Question 33.
Find the square root of (-5 + 12i).
Solution:
We have \(\sqrt{a+i b}\) =

In this example a = -5, b = 12

Question 34.
Write z = –\(\sqrt{7}\) + i\(\sqrt{21}\) in the polar form. (Mar ’11)
Solution:

Since the given point lies in the second quadrant we look for a solution of
tan θ = – \(\sqrt{3}\) that lies in [\(\frac{\pi}{2}\), π] we find that θ = \(\frac{2 \pi}{3}\) is such a solution.
∴ –\(\sqrt{7}\) + i\(\sqrt{21}\) = 2\(\sqrt{7}\) cis \(\frac{2 \pi}{3}\)
(or) 2\(\sqrt{7}\)(cos \(\frac{2 \pi}{3}\) + i sin \(\frac{2 \pi}{3}\))

Question 35.
Express -1 – i in polar form with principle value of the amplitude.
Solution:
Let -1 – i = r (cos θ + i sin θ), then
-1 = r cos θ, -1 = r sin θ, tan θ = 1 ——— (1)
∴ r² = 2
⇒ r = ±\(\sqrt{2}\)
Since θ is positive, -π < θ < π, the value θ satisfying the equation (1) is
θ = -135° = \(\frac{-3 \pi}{4}\)

Question 36.
If the amplitude of \(\left(\frac{z-2}{z-6 i}\right) \frac{\pi}{2}\), find its locus.  (Mar. ’06)
Solution:

By hypothesis, amplitude of a + ib = \(\frac{\pi}{2}\)
So \(\frac{\pi}{2}\) = tan^-1 \(\frac{b}{a}\)
Hence a = 0 and b > 0
∴ x(x – 2) + y(y – 6) = 0
or x² + y² – 2x – 6y = 0.

Question 37.
Show that the equation of any circle in the complex plane is of the form z\(\overline{\mathbf{z}}\) + b\(\overline{\mathbf{z}}\) + b\(\overline{\mathbf{z}}\) + c = 0, 1(b ∈ C, c ∈ R).
Solution:
Assume the general form of the equation of a circle in cartesian co-ordinates as
x² + y² + 2gx + 2fy + c = 0, (g, f ∈ R) —— (1)
To write this equation in the complex variable form, let (x, y) = z.
Then \(\frac{z+\bar{z}}{2}\) = x, \(\frac{z-\bar{z}}{2 i}\)
= y = \(\frac{-i(z-\bar{z})}{2}\)
∴ x² + y² = |z|² = z\(\overline{\mathbf{z}}\)
Substituting these results in equation (1), we obtain
z\(\overline{\mathbf{z}}\) + g(z + \(\overline{\mathbf{z}}\)) + f(z – \(\overline{\mathbf{z}}\))(-i) + c = 0
i.e., z\(\overline{\mathbf{z}}\) + (g – if)z + (g + if)\(\overline{\mathbf{z}}\) + c = 0 ——-(2)
If (g + if) = b, then equation (2) can be written as z\(\overline{\mathbf{z}}\) + \(\overline{\mathbf{b}}\)z + b\(\overline{\mathbf{z}}\) + c = 0

Question 38.
Show that the complex numbers z satisfying z² + \((\overline{\mathbf{z}})^{2}\) = 2 constitute a hyperbola.
Solution:
Substituting z = x + iy in the given equation
z² + (\(\overline{\mathbf{z}}\))² = 2, we obtain the cartesian form of the given equation.
∴ (x + iy)² + (x – iy)² = 2
i.e., x² – y² + 2ixy + x² – y² – 2ixy = 2
i.e., x² – y² = 1.
Since, this equation denotes a hyperbola, all the complex numbers satisfying

lie on the hyperbola x² – y² = 1.

Question 39.
Show that the points in the Argand diagram represented by the complex numbers 1 + 3i, 4 – 3i, 5 – 5i are collinear.
Solution:
Let the three complex numbers be represented in the Argand plane by the points P, Q, R respectively. Then P = (1, 3), Q = (4, -3), R = (5, -5). The slope of the line segment joining P,Q is \(\frac{3+3}{1-4}\) = \(\frac{6}{-3}\) = -2.
Similarly the slope of the line segment joining Q, R is \(\frac{-3+5}{4-5}\) = \(\frac{2}{-1}\) = -2.
Since the slope of PQ is the slope of QR, the points P, Q and R are collinear.

Question 40.
Find the equation of the straight line joining the points represented by (- 4 + 3i), (2 – 3i) in the Argand plane.
Solution:
Take the given points as
A = -4 + 3i = (-4, 3)
B = 2 – 3i = (2, -3)
Then equation of the straight line \(\overleftrightarrow{\mathrm{AB}}\) is
y – 3 = \(\frac{3+3}{-4-2}\)(x + 4)
i.e., x + y + 1 = 0.

Question 41.
z = x + iy represents a point in the Argand plane, find the locus of z. Such that |z| = 2.
Solution:
|z| = 2, z = x + iy
if \(\sqrt{x^{2}+y^{2}}\) = 2
if \(\sqrt{x^{2}+y^{2}}\) = 2
if and only if x² + y² = 4
The equation x² + y² = 4 represents the circle with centre at the origin (0, 0) and radius 2 units.
∴ The locus of |z| = 2 is the circle
x² + y² = 4

Question 42.
The point P represents a complex number z in the Argand plane. If the amplitude of z is \(\frac{\pi}{4}\), determine the locus of P.
Solution:
Let z = x + iy.
By hypothesis, amplitude of z = \(\frac{\pi}{4}\)
Hence tan^-1 \(\left(\frac{y}{x}\right)\) = \(\frac{\pi}{4}\) and \(\frac{y}{x}\) = tan \(\frac{\pi}{4}\)
Hence x = y
∴ The locus of P is x = y.

Question 43.
If the point P denotes the complex number z = x + iy in the Argand plane and if \(\frac{z-i}{z-1}\) is a purely imaginary number, find the locus of P.
Solution:
We note that the quotient \(\frac{z-i}{z-1}\) is not defined if z = 1.

∴ The locus of P is the circle
x² + y² – x – y = 0
excluding the point (1, 0).

Question 44.
Describe geometrically the following subsets of C.
i) {z ∈ C| |z – 1 + i| = 1}
ii) {z ∈ C| |z + i| ≤ 3|
Solution:
i) Let S = {z ∈ C| z – 1 + i| = 1}
If we write z = (x, y), then
S = {(x, y) ∈ R²||x + iy – 1 + i| = 1}
= {x, y) ∈ R² || x + i(y – 1)| ≤ 3}
= {(x, y) ∈ R² || (x – 1)² + (y + 1)² = i}
Hence S is a circle with centre (1, -1) and radius 1 unit.

ii) Let S’ = {z ∈ C || z + i| ≤ 3}
Then S = {(x, y ∈ R² || x + iy + i| ≤ 3}
= {(x, y) ∈ R² || x² + i(y + 1) ≤ 3}
= {(x, y) ∈ R² || x² + (y + 1)² ≤ 9}
Hence S’ is the closed circular disc with centre at (0, -1) and radius 3 units.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 10 Random Variables and Probability Distributions to solve questions creatively.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Formulas

Random variable:
→ Suppose S is the sample space of a random experiment then any function X : S → R is called a random variable.

→ Let S be a sample space and X : S → R.be a random variable. The function F : R → R defined by F(x) = P(X ≤ x), is called probability distribution function of the random variable X.

→ A set ‘A’ is said to be countable if there exists a bijection from A onto a subset of N.

→ Let S be a sample space. A random variable X : S → R is said to be discrete or discontinuous if the range of X is countable.

→ If X : S → R is a discrete random variable with range {x₁, x₂, x₃, ……………. } then
\(\sum_{r=1}^{\infty}\) P(x = x_r) = 1

→ Let X : S → R be a discrete random variable with range {x₁, x₂, x₃, …………….} If Σx_r P(X = x_r) exists, then Σx_r. P(X = x) is called the mean of the random variable X. It is denoted by µ or x . If Σ (x_r – µ)² P(X = X_r) exists, then Σ(x_r – µ)² P(X = X_r) is called variance of the random variable X. It is denoted by σ².

→ The positive square root of the variance is called the standard deviation of the Fandom variable x. It is denoted by σ.

Binomial distribution:
→ Let n be a positive integer and p be a real number-such that 0 ≤ p ≤ 1. A random variable x with range {0, 1, 2, 3, ……….. n} is said to follows (or have) binomial distribution or Bernoulli distribution with parameters n and p if P (X = r) = ⁿC_r p^r q^{n – r} for r = 0, 1, 2, ………. n where q = 1 – p. Its Mean µ = np and variance σ² = npq. n and p are called. parameters of the Binomial distribution.

Poisson distribution:
→ Let λ > 0 be a real, number. A, random variable x with range {0, 1, 2, ……… n} is said to follows (have) poisson distribution with parameter λ if P(X = r) = \(\frac{e^{-\lambda} \lambda^{r}}{r !}\) for r = 0, 1, 2, ……………. . Its Mean = λ and variance = λ. Its parameter is λ.

→ If X : S → R is a discrete random variable with range {x₁ x₂, x₃, …. } then \(\sum_{r=1}^{\infty}\) P (X = x_r) = 1

→ Let X : S → R be a discrete random variable with range {x₁ x₂, x₃, …..} .If Σ x_r P(X = x_r) exists, then Σ x_r P(X = x_r) is called the mean of the random variable X. It is denoted by or x.

→ If Σ(x_r – μ)² P(X = x_r) exists, then Σ (x_r – μ)² P(X = x_r) is called variance of the random variable X. It is denoted by σ². The positive square root of the variance is called the standard deviation of the random variable X. It is denoted by σ

→ If the range of discrete random variable X is {x₁ x₂, x₃, …. x_n, ..} and P(X = x_n) = P_n for every Integer n is given then σ² + μ² = Σx_n²P_n

Binomial Distribution:
A random variable X which takes values 0, 1, 2, ., n is said to follow binomial distribution if its probability distribution function is given by
P(X = r) = ⁿc_rp^rq^n-r, r = 0,1,2, ……………. , n where p, q > 0 such that p + q = 1.

→ If the probability of happening of an event in one trial be p, then the probability of successive happening of that event in r trials is p^r.

→ Mean and variance of the binomial distribution

• The mean of this distribution is \(\sum_{i=1}^{n}\) X_ip_i = latex]\sum_{X=1}^{n}[/latex] X. ⁿC_xq^n-xp^X = np,
• The variance of the Binomial distribution is σ² = npq and the standard deviation is σ = \(\sqrt{(n p q)}\).

→ The Poisson Distribution : Let X be a discrete random variable which can take on the values 0, 1, 2,… such that the probability function of X is given by
f(x) = P(X = x) = \(\frac{\lambda^{x} e^{-\lambda}}{x !}\), x = 0, 1, 2, ………….
where λ is a given positive constant. This distribution is called the Poisson distribution and a random variable having this distribution is said to be Poisson distributed.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 9 Probability to solve questions creatively.

Intermediate 2nd Year Maths 2A Probability Formulas

→ An experiment which can be repeated any number of times under essentially identical conditions and which is associated with a set of known results, is called a random experiment or trail if the result of any single repetition of the experiment ism certain and is any one of the associated set.

→ A combination of elementary events in a trial is called an event.

→ The list of all elementary events in a trail is called list of exhaustive events.

→ Elementary events are said to be equally likely if they have the same chance of happening.

→ If there are n exhaustive equally likely elementary events in a trail and m of them are
favourable to an event A, then \(\frac{m}{n}\) is called the probability of A. It is denoted by P(A).

→ P(A) = \(\frac{\text { Number of favourable cases (outcomes) with respect } A}{\text { Number of all cases (outcomes) of the experiment }}\)

→ The set of all possible outcomes (results) in a trail is called sample space for the trail. It is denoted by ‘S’. The elements of S are called sample points.

→ Let S be a sample space of a random experiment. Every subset of S is called an event.

→ Let S be a sample space. The event Φ is called impossible event and the event S is called certain event in S.

→ Two events A, B in a sample space S are said to be disjoint or mutually exclusive if A ∩ B = Φ

→ Two events A, B in a sample space S are said to be exhaustive if A ∪ B = S.

→ Two events A, B in a sample space S are said to be complementary if A ∪ B = S, A ∩ B = Φ. The complement B of A is denoted by Ā (or) A^c.

→ Let S be a finite sample space. A real valued function P: p(s) → R is said to be a probability function on S if (i) P(A) ≥ 0 ∀ A ∈ p(s) (ii) p(s) = 1

→ A, B, ∈ p(s), A ∩ B = Φ ⇒ P (A ∪ B) = P(A) + P(B). Then P is called probability function and for each A ∈ p(s), P(A) is called the probability of A.

→ If A is an event in a sample space S, then 0 ≤ P(A) ≤ 1.

→ If A is an event in a sample space S, then the ratio P(A) : P̄(Ā) is called the odds favour to A and P̄(A): P(A) is called the odds against to A.

→ If A, B are two events in a sample space S, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

→ If A, B are two events in a sample space, then P(B – A) = P(B) – P(A ∩ B) and P(A – B) = P(A) – P(A ∩ B) .

→ If A, B, C are three events in a sample space S, then
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C).

→ If A, B are two events in a sample space then the event of happening B after the event A happening is called conditional event It is denoted by B/A.

→ If A, B are two events in a sample space S and P(A) ≠ 0, then the probability of B after the event A has occured is called conditional probability of B given A. It is denoted by P\(\left(\frac{B}{A}\right)\)

→ If A, B are two events in a sample space S such that P(A) ≠ o then \(P\left(\frac{B}{A}\right)=\frac{n(A \cap B)}{n(A)}\)

→ Let A, B be two events in a sample space S such that P(A) ≠ 0, P(B) ≠ 0, then

• \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)
• \(P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}\)

→ Multiplication theorem on Probability: If A and B are two events of a sample space 5 and P(A) > 0, P(B) > 0 then P(A n B) = P(A), P(B/A) = P(B). P(A/B).

→ Two events A and B are said to be independent if P(A ∩ B) = P(A). P(B). Otherwise A, B are said to be dependent.

→ Bayes’ theorem : Suppose E₁, E₂ ……… E_n are mutually exclusive and exhaustive events of a Random experiment with P(E_i) > 0 for ī = 7, 2, ……… n in a random experiment then we have

\(p\left(\frac{E_{k}}{A}\right)\) = \(\frac{P\left(E_{k}\right) P\left(\frac{A}{E_{k}}\right)}{\sum_{i=1}^{n} P\left(E_{i}\right) P\left(\frac{A}{E_{i}}\right)}\) for k = 1, 2 …… n.

Random Experiment:
If the result of an experiment is not certain and is any one of the several possible outcomes, then the experiment is called Random experiment.

Sample space:
The set of all possible outcomes of an experiment is called the sample space whenever the experiment is conducted and is denoted by S.

Event:
Any subset of the sample space ‘S’ is called an Event.

Equally likely Events:
A set of events is said to be equally likely if there is no reason to expect one of them in preference to the others.

Exhaustive Events:
A set of events is said to be exhaustive of the performance of the experiment always results in the occurrence of at least one of them.

Mutually Exclusive Events:
A set of events is said to the mutually exclusive if happening of one of them prevents the happening of any of the remaining events.

Classical Definition of Probability:
If there are n mutually exclusive equally likely elementary events of an experiment and m of them are favourable to an event A then the probability of A denoted by P(A) is defined as min.

Axiomatic Approach to Probability:
Let S be finite sample space. A real valued function P from power set of S into R is called probability function if
P(A) ≥ 0 ∀ A ⊆ S
P(S) = 1, P(Φ) = 0;
(3) P(A ∪ B) = P(A) + P(B) if A ∩ B = Φ. Here the image of A w.r.t. P denoted by P(A) is called probability of A.

Note:

• P(A) + P(A̅) = 1
• If A₁ ⊆ A₂, then P(A₁) < P(A₂) where A₁, A₂ are any two events.

Odds in favour and odds against an Event:
Suppose A is any Event of an experiment. The odds in favour of Event A is P(A̅) : P(A). The odds against of A is P(A̅) : P(A).

Addition theorem on Probability:
If A, B are any two events in a sample space S, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B).
If A and B are exclusive events

• P(A ∪ B) = P(A) + P(B)
• P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C).

Conditional Probability:
If A and B are two events in sample space and P(A) ≠ 0. The probability of B after the event A has occurred is called the conditional probability of B given A and is denoted by P(B/A).
P(B/A) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{A})}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}\)
Similarly
n(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{B})}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)

Independent Events:
The events A and B of an experiment are said to be independent if occurrence of A cannot influence the happening of the event B.
i.e. A, B are independent if P(A/B) = P(A) or P(B/A) = P(B).
i.e. P(A ∩ B) = P(A) . P(B).

Multiplication Theorem:
If A and B are any two events in S then
P(A ∩ B) = P(A) P(B/A) if P(A)≠ 0.
P (B) P(A/B) if P(B) ≠ 0.
The events A and B are independent if
P(A ∩ B) = P(A) P(B).
A set of events A₁, A₂, A₃ … A_n are said to be pair wise independent if
P(A_i n A_j) = P(A_i) P(A_j) for all i ≠ J.

Theorem:
Addition Theorem on Probability. If A, B are two events in a sample space S
Then P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Proof:
FRom the figure (Venn diagram) it can be observed that
(B – A) ∪ (A ∩ B) = B, (B – A) ∩ (A ∩ B) = Φ

∴ P(B) = P[(B – A) ∪ (A ∩ B)]
= P(B – A) + P(A ∩ B)
⇒ P(B – A) = P(B) – P(A ∩ B) ………(1)

Again from the figure, it can be observed that
A ∪ (B – A) = A ∪ B, A ∩ (B – A) = Φ
∴ P(A ∪ B) = P[A ∪ (B – A)]
= P(A) + P(B – A)
= P(A) + P(B) – P(A ∩ B) since from (1)
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Theorem:
Multiplication Theorem on Probability.
Let A, B be two events in a sample space S such that P(A) ≠ 0, P(B) ≠ 0, then
i) P(A ∩ B) = P(A)P\(\left(\frac{B}{A}\right)\)
ii) P(A ∩ B) = P(B)P\(\left(\frac{A}{B}\right)\)
Proof:
Let S be the sample space associated with the random experiment. Let A, B be two events of S show that P(A) ≠ 0 and P(B) ≠ 0. Then by def. of confidential probability.
P\(\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}\)
∴ P(B ∩ A) = P(A)P\(\left(\frac{B}{A}\right)\)
Again, ∵P(B) ≠ 0
\(\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)
∴ P(A ∩ B) = P(B) . P\(\left(\frac{\mathrm{A}}{\mathrm{B}}\right)\)
∴ P(A ∩ B) = P(A) . P\(\left(\frac{B}{A}\right)\) = P(B).P\(\left(\frac{\mathrm{A}}{\mathrm{B}}\right)\)

Baye’s Theorem or Inverse probability Theorem
Statement:
If A₁, A₂, … and A_n are ‘n’ mutually exclusive and exhaustive events of a random experiment associated with sample space S such that P(A_i) > 0 and E is any event which takes place in conjuction with any one of A_i then
P(A_k/E) = \(\frac{P\left(A_{k}\right) P\left(E / A_{k}\right)}{\sum_{i=1}^{n} P\left(A_{i}\right) P\left(E / A_{i}\right)}\), for any k = 1, 2, ………. n;
Proof:
Since A₁, A₂, … and A_n are mutually exclusive and exhaustive in sample space S, we have A_i ∩ A_j = for i ≠ j, 1 ≤ i, j ≤ n and A₁ ∪ A₂ ∪…. ∪ A_n = S.

Since E is any event which takes place in conjunction with any one of A_i, we have
E = (A₁ ∩ E) ∪ (A₂ ∩ E) ∪ …………….. ∪(A_n ∩ E).

We know that A₁, A₂, ……… A_n are mutually exclusive, their subsets (A₁ ∩ E), (A₂ ∩ E) , … are also
mutually exclusive.

Now P(E) = P(E ∩ A₁) + P(E ∩ A₂) + ………. + P(E ∩ A_n) (By axiom of additively)
= P(A₁)P(E/A₁) + P(A₂)P(E/A₂) + ………… + P(A_n)P(E/A_n)
(By multiplication theorem of probability)
= \(\sum_{i=1}^{n}\)P(A_i)P(E/A_i) …………..(1)

By definition of conditional probability
P(A_k/E) = \(\frac{P\left(A_{k} \cap E\right)}{P(E)}\) for
= \(\frac{P\left(A_{k}\right) P\left(E / A_{k}\right)}{P(E)}\)
(By multiplication theorem)
= \(\frac{P\left(A_{k}\right) P\left(E / A_{k}\right)}{\sum_{i=1}^{n} P\left(A_{I}\right) p\left(E / A_{i}\right)}\) from (1)
Hence the theorem

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 8 Measures of Dispersion to solve questions creatively.

Intermediate 2nd Year Maths 2A Measures of Dispersion Formulas

→ The arithmetic mean of ungrouped data = \(\frac{\Sigma x_{i}}{n}\) where n is the number of observations.

→ The median of ungrouped data
First expressing the n data points in the ascending order of magnitude.

→ If n is odd then \(\frac{n+1}{2}\)the data points in the median of the given ungrouped data.

→ If n is even then the average of \(\frac{n}{2}, \frac{n+2}{2}\) the data points in the median of the given ungrouped data.

• Range,
• Mean deviation,
• Variance and,
• Standard deviation are some measures of dispusion for ungrouped and grouped data.

→ Range is defined as the difference between the maximum value and the minimum value of the series of observations.

Mean Deviation for ungrouped distribution:
→ Mean deviation about the mean = \(\frac{\text { Sum of the absolutevalues of deviationsfrom } \bar{x}}{\text { Number of Observations }} .\)
= \(\frac{\sum\left|x_{i}-\bar{x}\right|}{n}\) where x̅ is the mean

→ Mean Deviation about the median = \(\frac{\sum\left|x_{i}-{median}\right|}{n}\)

Mean Deviation for grouped data:
→ Mean Deviation about mean = \(\frac{1}{N}\) Σf_i|x – x̅_i|
Where N = Σf_i and x̅ is the mean

→ Mean Deviation about median = \(\frac{1}{N}\) Σf_i|x_i – median|
When N = Σf_i

→ Ungrouped data variance σ² = \(\frac{1}{n}\) = Σ(x_i – x̅)²
Standard deviation σ = \(\sqrt{\frac{1}{n} \sum\left(x_{i}-\bar{x}\right)^{2}}\)

→ Discrete frequency distribution variance σ² = \(\frac{1}{n}\) = Σf_i(x_i – x̅)² where x̅ is the mean]
Standard deviation σ = \(\frac{1}{N}\) \(\sqrt{\sum f_{i}\left(x_{i}-\bar{x}\right)^{2}}\)

→ Continuous frequency distribution standard deviation σ = \(\frac{1}{N} \sqrt{N \sum f_{i} x_{i}{ }^{2}-\left(\sum f_{i} x_{i}\right)^{2}}\)
(or) σ = \(\frac{h}{N}\) \(\sqrt{N \Sigma f_{i} y_{i}^{2}-\left(\sum f_{i} y_{i}\right)^{2}}\)

→ Co-efficient of variation = \(\frac{\sigma}{x}\) × 100 (x̅ ≠ 0)

→ If each observation in a data is. multiplied by a constant K, then the variance of the resulting observations is K² times that or the variance of original observations.

→ If each of the observations x₁, x₂, ………, x_n is increased by K, where K is a positive or negative number then the variance remains unchanged.

Measure of Central Tendency:
1. Mathematical Average:
a) Arithmetic mean (A.M.)
b) Geometric mean (G.M.)
c) Harmonic mean (H.M.)

2. Averages of Position:
a) Median
b) Mode

Arithmetic Mean:
(1) Simple arithmetic mean in individual series
(i) Direct method: If the series in this case be x₁, x₂, …………. x_n then the arithmetic mean x̄ is given by
x̄ = \(\frac{\text { Sum of the series }}{\text { Number of terms }}\)
i.e x̄ = \(\frac{x_{1}+x_{2}+x_{3}+\ldots+x_{n}}{n}=\frac{1}{n} \sum_{i=1}^{n} x_{i}\)

(2) Simple arithmetic mean in continuous series If the terms of the given series be x₁, x₂, …………. x_n and the corresponding frequencies be f₁, f₂, …………. f_n then the arithmetic mean x̄ is given by,
x̄ = \(\frac{f_{1} x_{1}+f_{2} x_{2}+\ldots+f_{n} x_{n}}{f_{1}+f_{2}+\ldots+f_{n}}=\frac{\sum_{i=1}^{n} f_{i} x_{i}}{\sum_{i=1}^{n} f_{i}}\)

Continuous Series:
If the series is continuous then x_ii’s are to be replaced by m_i’s where m_i’s are the mid values of the class intervals.

Mean of the Composite Series:
If x̄_i,(i = 1, 2, …………… k) are the means of k-component series of sizes n_i(i = 1,2,….,k) respectively, then the mean x̄ of the composite series obtained on combining the component series is given by the formula x̄ = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}+\ldots+n_{k} \bar{x}_{k}}{n_{1}+n_{2}+\ldots+n_{k}}=\frac{\sum_{i=1}^{n} n_{i} \bar{x}_{i}}{\sum_{i=1}^{n} n_{i}}\)

Geometric Mean:
If x₁, x₂, …………. x_n are n values of a variate x, none of them being zero, thengeometric mean (G.M.) is given by G.M. (x₁, x₂, …………. x_n)^1/n

In case of frequency distribution, G.M. of n values x₁, x₂, …………. x_n of a variate x occurring with frequency f₁, f₂, …………. f_n is given by G.M = (x₁^f₁, x₂^f₂, …………. x_n^f_n)^1/N, where N = f₁ + f₂ + ……….. + f_n

Continuous Series:
If the series is continuous then xjj’s are to be replaced by m_ii’s where m_ii’s are the mid values of the class intervals.

Harmonic Mean:
The harmonic mean of n items x₁, x₂, …………. x_n is defined as H.M. = \(\frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots .+\frac{1}{x_{n}}}\)
If the frequency distribution is f₁, f₂, …………. f_n respectively, then H.M = \(\frac{f_{1}+f_{2}+f_{3}+\ldots . .+f_{n}}{\left(\frac{f_{1}}{x_{1}}+\frac{f_{2}}{x_{2}}+\ldots .+\frac{f_{n}}{x_{n}}\right)}\)

Median:
The median is the central value of the set of observations provided all the observations are arranged in the ascending or descending orders. It is generally used, when effect of extreme items is to be kept out.

(1) Calculation of median
(i) Individual series: If the data is raw, arrange in ascending or descending order. Let n be the number of observations.
If n is odd, Median = value of \(\left(\frac{n+1}{2}\right)^{t h}\) item.
If n is even, Median = \(\frac{1}{2}\)[value of \(\left(\frac{n}{2}\right)^{\text {th }}\) item + value of \(\left(\frac{n}{2}+1\right)^{\text {th }}\) item]

(ii) Discrete series: In this case, we first find the cumulative frequencies of the variables arranged in ascending or descending order and the median is given by
Median = \(\left(\frac{n+1}{2}\right)^{t h}\) observations, where n is the cumulative frequency.

(iii) For grouped or continuous distributions: In this case, following formula can be used.
(a) For series in ascending order, Median = l + \(\frac{\left(\frac{N}{2}-C\right)}{f}\) × i
Where l = Lower limit of the median class
f = Frequency of the median class
N = The sum of all frequencies
I = The width of the median class
C = The cumulative frequency of the class preceding to median class.

(b) For series in descending order
Median = u – \(\left(\frac{\frac{N}{2}-C}{f}\right)\) × i, where u = upper limit of the median class, N = \(\sum_{i=1}^{n}\)f_i
As median divides a distribution into two equal parts, similarly the quartiles, quintiles, deciles and percentiles divide the distribution respectively into 4, 5, 10 and 100 equal parts. The th quartile is given by Q = l + \(\left(\frac{j \frac{N}{4}-C}{f}\right)\)i:j= 1,2,3. Q₁ is the lower quartile, Q₂ is the median and Q₃ is called the upper quartile.

(2) Lower qualities

• Discrete series: Q₁ = size of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) item
• Continuous series : Q₁ = l + \(\frac{\left(\frac{N}{4}-C\right)}{f}\) × i

(3) Upper qualities

• Discrete series : Q₃ = size of \(\left[\frac{3(n+1)}{4}\right]^{\text {th }}\) item
• Continuous series : Q₃ = l + \(\frac{\left(\frac{3N}{4}-C\right)}{f}\) × i

Mode:
The mode or model value of a distribution is that value of the variable for which the frequency is maximum. For continuous series, mode is calculated as,
Mode = l₁ + \(\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right]\) × i
Where, l₁ = The lower limit of the model class
f₁ = The frequency of the model class
f₀ = The frequency of the class preceding the model class
f₂ = The frequency of the class succeeding the model class
i = The size of the model class.

Empirical relation :
Mean – Mode = 3(Mean – Median) ⇒ Mode = 3 Median – 2 Mean.

Measure of dispersion:
The degree to which numerical data tend to spread about an average value is called the dispersion of the data. The four measure of dispersion are

1. Range
2. Mean deviation
3. Standard deviation
4. Square deviation

(1) Range:
It is the difference between the values of extreme items in a series. Range = X_max – X_min

• The coefficient of range (scatter) = \(\frac{x_{\max }-x_{\min }}{x_{\max }+x_{\min }}\)
  Range is not the measure of central tendency. Range is widely used in statistical series relating to quality control in production.
• Range is commonly used measures of dispersion in case of changes in interest rates, exchange rate, share prices and like statistical information. It helps us to determine changes in the qualities of the goods produced in factories.

Quartile deviation or semi inter-quartile range:
It is one-half of the difference between the third quartile and first quartile i. e., Q.D. = \(\frac{Q_{3}-Q_{1}}{2}\) and coefficient of quartile deviation = \(\frac{Q_{3}-Q_{1}}{Q_{3}+Q_{1}}\), where Q₃ is the third or upper quartile and Q₁ is the lowest or first quartile.

(2) Mean Deviation:
The arithmetic average of the deviations (all taking positive) from the mean, median or mode is known as mean deviation.
Mean deviation is used for calculating dispersion of the series relating to economic and social inequalities. Dispersion in the distribution of income and wealth is measured in term of mean deviation.

• Mean deviation from ungrouped data (or individual series) Mean deviation = \(\frac{\sum|x-M|}{n}\) where |x – M| means the modulus of the deviation of the variate from the mean (mean, median or mode) and n is the number of terms.
• Mean deviation from continuous series: Here first of all we find the mean from which deviation is to be taken. Then we find the deviation dM =| x -M| of each variate from the mean M so obtained.
  • Next we multiply these deviations by the corresponding frequency and find the product f.dM and then the sum ΣfdM of these products.
  • Lastly we use the formula, mean deviation = \(\frac{\sum f|x-M|}{n}=\frac{\sum f d M}{n}\), where n = Σf.

(3) Standard Deviation:
Standard deviation (or S.D.) is the square root of the arithmetic mean of the square of deviations of various values from their arithmetic mean and is generally denoted by aread as sigma. It is used in statistical analysis.
(i) Coefficient of standard deviation: To compare the dispersion of two frequency distributions the relative measure of standard deviation is computed which is known as coefficient of standard deviation and is given by
Coefficient of S.D. = \(\frac{\sigma}{\bar{x}}\), where x̄ is the A.M.

(ii) Standard deviation from individual series σ = \(\sqrt{\frac{\sum(x-\bar{x})^{2}}{N}}\)
where, x̄ = The arithmetic mean of series
N = The total frequency.

(iii) Standard deviation from continuous series σ = \(\sqrt{\frac{\sum f_{i}\left(x_{i}-\bar{x}\right)^{2}}{N}}\)
where, x̄ = Arithmetic mean of series
x_i = Mid value of the class
f_i = Frequency of the corresponding
N = Σf = The total frequency

Short cut Method:
(i) σ = \(\sqrt{\frac{\sum f d^{2}}{N}-\left(\frac{\sum f d}{N}\right)^{2}}\)
(ii) σ = \(\sqrt{\frac{\sum d^{2}}{N}-\left(\frac{\sum d}{N}\right)^{2}}\)
where, d = x – A = Deviation from the assumed mean A
f = Frequency of the item
N = Σf = Sum of frequencies

(4) Square Deviation:
(i) Root mean square deviation
S = \(\sqrt{\frac{1}{N} \sum_{i=1}^{n} f_{i}\left(x_{i}-A\right)^{2}}\)
where A is any arbitrary number and S is called mean square deviation.

(ii) Relation between S.D. and root mean square deviation: If σ is the standard deviation and S is the root mean square deviation.
Then, S² = σ² + d².
Obviously, S² will be least when d = 0 i.e., x̄ = A
Hence, mean square deviation and consequently root mean square deviation is least if the deviations are taken from the mean.

Variance:
The square of standard deviation is called the variance. Coefficient of standard deviation and variance: The coefficient of standard deviation is the ratio of S.D. to A.M. i.e., \(\frac{\sigma}{x}\).
Coefficient of variance = coefficient of S.D.× 100 = \(\frac{\sigma}{x}\) × 100 .

Variance of the combined series :
If n₁,n₂ are the sizes, x̄₁, x̄₂ the means and σ₁, σ₂ the standard deviation of two series, then σ² = \(\frac{1}{n_{1}+n}\)[n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]
where d₁ = x̄₁ – x̄, d₂ = x̄₂ – x̄ and x̄ = \(\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 7 Partial Fractions to solve questions creatively.

Intermediate 2nd Year Maths 2A Partial Fractions Formulas

→ The quotient of two polynomials f(x) and Φ(x) where Φ(x) ≠ 0, is called a rational fraction.

→ If the degree of f(x) < the degree of Φ(x) in a rational fraction \(\frac{f(x)}{\phi(x)}\), then the rational fraction is called a proper fraction.

→ If the degree of f(x) ≥ the degree of Φ(x) in a rational fraction \(\frac{f(x)}{\phi(x)}\), then the rational fraction is called a proper fraction.

→ Let \(\frac{f(x)}{\phi(x)}\) be a proper fraction

→ When Φ(x) contains non-repeated linear factors only corresponding to every non-repeated linear factor (ax +b) of Φ(x) there exists a partial fraction of the form \(\frac{A}{a x+b}\) where A is a real number.

→ When Φ(x) contains repeated and non-repeated linear factors only corresponding to every repeated linear factor (ax + b)^p of Φ(x) there exist fractions of the form.
\(\frac{A_{1}}{a x+b}+\frac{A_{2}}{(a x+b)^{2}}+\ldots+\frac{A_{p}}{(a x+b)^{p}}\) Where A₁, A₂, A₃ ………… A_p are real numbers.

→ When Φ(x) contains non-repeated irreducible factors only. Corresponding to every non-repeated irreducible quadratic factor ax² + bx + c of Φ(x) of exists a partial fraction of the form \(\frac{A x+B}{a x^{2}+b x+C}\) where A and B are real numbers.

→ When Φ(x) contains repeated and non-repeated irreducible factors only. Corresponding to every repeated quadratic factor (ax² + bx + c0^p of Φ(x) there exists the partial fractions of the form
\(\frac{A_{1} x+B_{1}}{a x^{2}+b x+c}\) + \(\frac{A_{2} x+B_{2}}{\left(a x^{2}+b x+c\right)^{2}}\) + ……… + \(\frac{A_{p} x+\dot{B}_{p}}{\left(a x^{2}+b x+c\right)^{p}}\) where A₁, A₂, …….. A^p and B₁, B₂, ………. B_p are real numbers.

→ Let \(\frac{f(x)}{\phi(x)}\) be a improper fraction, then
\(\frac{f(x)}{\phi(x)}\) = Q(x) + \(\frac{R(x)}{\phi(x)}\) where Q(x) is quotient and R(x) is the remainder, and Degree R(x) < that of Φ(x).

→ Remainder obtained when f(x) is divided by x – a is f(a).
If degree of divisor is ‘n’, then the degree of remainder is (n – 1)
f(x), g(x) are two polynomials. If g(x) ≠ 0, then ∃ two polynomials q(x) , r(x) such that
\(\frac{f(x)}{g(x)}\) = q(x) + \(\frac{r(x)}{g(x)}\) if the degree of f(x) is > that of g(x)

II. Method of resolving proper fraction \(\frac{f(x)}{g(x)}\) into partial fractions.
Type 1 : When the denominator g(x) contains non-repeated factors i.e
g(x) = (x – a)(x – b)(x – c)
\(\frac{f(x)}{(x-a)(x-b)(x-c)}=\frac{\mathrm{A}}{\mathrm{x}-\mathrm{a}}+\frac{\mathrm{B}}{\mathrm{x}-\mathrm{b}}+\frac{\mathrm{C}}{\mathrm{x}-\mathrm{c}}\)

Type 2 : When the denominator g(x) contains repeated and non repeated linear factors
i.e g(x) = (x – a)²(x – b)
\(\frac{f(x)}{(x-a)^{2}(x-b)}=\frac{A}{x-a}+\frac{B}{(x-a)^{2}}+\frac{C}{(x-b)}\)

Type 3 : When the denominator g(x) contains non repeated irreducible quadratic factors
i.e g(x) = (ax² + bx + c)(x – d)

Type 4 : When the denominator g(x) contains repeated irreducible quadratic factors
i.e g(x) = (ax² + bx + c)²(x – d)
\(\frac{f(\mathrm{x})}{\left(\mathrm{ax}{ }^{2}+\mathrm{bx}+\mathrm{c}\right)^{2}(\mathrm{x}-\mathrm{d})}=\frac{\mathrm{Ax}+\mathrm{B}}{\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)^{2}}+\frac{\mathrm{E}}{\mathrm{x}-\mathrm{d}}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 6 Binomial Theorem to solve questions creatively.

Intermediate 2nd Year Maths 2A Binomial Theorem Formulas

→ Let n be a positive integer and x, a be real numbers then
(x + a)ⁿ = ⁿC₀. xⁿ. a⁰ + ⁿC₁. x^{n – 1}. a¹ + ⁿC₂. x^{n – 2}. a² + ……… + ⁿC_r. x^{n – r}. a^r + ……… + ⁿC_n. x⁰. aⁿ = \(\sum_{r=0}^{n}\) ⁿC_r.x^{n – r}. a^r. and (x – a)ⁿ – ⁿC₀. xⁿ – (x)ⁿ – ⁿC₁. x^{n – 1}a + ⁿC₂ x^{n – 2}. a² …….. + ( – 1)^{r n}C_r. x^{n – r}. a^r + ……… + (- 1)^{n n}C_n.aⁿ

→ The expansion of (x + a)ⁿ contains (n + 1) terms.

→ In the expansion, the coefficients ⁿC₀, ⁿC₁, ⁿC₂, ….. ⁿC_n are called binomial coefficients and these are simply denoted by C₀, C₁, C₂, …….. C_n,.

→ In the expansion, (r + 1)^th term is called the general term. It is denoted by T_{r + 1}.
∴ T_{r + 1} = ^cC_r. x^{n – r}. a^r, (0 ≤ r ≤ n)

→ The number of terms in the expansion of (a + b + c)ⁿ = \(\frac{(n+1)(n+2)}{2}\)

→ If n is even in the expansion of (x + a)ⁿ, the middle term = T_{\(\left(\frac{n}{2}+1\right)\)}

→ If n is odd in the expansion of (x + a)ⁿ, it has two middle terms which are T_{\(\left(\frac{n+1}{2}\right)\)}, T_{\(\left(\frac{n+3}{2}\right)\)}.

→ If \(\frac{(n+1)|x|}{|x|+1}\) = p, a positive integer then p^th and (p + 1)^th terms are the numerically greatest terms in the expansion of (1 + x)ⁿ.

→ If \(\frac{(n+1)|x|}{|x|+1}\) = P + F where p is a positive integer and 0 < F < 1 then (p + 1)^th the numerically greatest term in the expansion of (1 + x)ⁿ

→ C₀ + C₁ + C₂ + ………. + C_n = 2ⁿ

→ C₀ – C₁ + C₂ – C₃ + ……… + (- 1)ⁿC_n = 0

→ C₀ + C₂ + C ₄ + …………… = C₁ + C₃ + C₅ + …………….. = 2^{n – 1}

→ \(\sum_{r=0}^{n}\) ⁿC_r = 2ⁿ

→ \(\sum_{r=0}^{n}\) r. ⁿC_r = n. 2^{n – 1}

→ \(\sum_{r=2}^{n}\) r(r – 1). ⁿC_r = n(n – 1). 2^{n – 2}

→ \(\sum_{r=1}^{n}\) r² . ⁿC_r = n(n + 1). 2^{n – 2}

→ a. C₀ + (a + d). C₁ + (a + 2d). C₂ + ……… + (a + nd). C_n = (2a + nd) 2^{n – 1}

→ C₀C_r + C₁C_{r + 1} + C₂C_{r + 2} + ………… + C_{n – r}. C_n = ²ⁿC_{n + r}

→ If f(x) = (a₀ + a₁x + a₂x² + ……… a_mx^m)ⁿ then

• Sum of the coefficients = f(1)
• Sum of the coefficients of even powers of x is \(\frac{f(1)+f(-1)}{2}\)
• Sum of the coefficients of odd powers of x is \(\frac{f(1)-f(-1)}{2}\)

→ Let n be a positive integer and x is a ,real number such that |x| < 1 then

→ (1 – x)^-n = 1 + nx + \(\frac{n(n+1)}{2 !}\) x² + \(\frac{n(n+1)(n+2)}{3 !}\) x² + ……… + ……… + \(\frac{n(n+1)(n+2) \ldots \ldots(n+r-1)}{r !}\) x^r + ……. to ∞

→ (1 + x)^-n = 1,- nx + \(\frac{n(n+1)}{2 !}\) x² + …….. + \(\frac{(-1)^{r} n(n+1)(n+2) \ldots(n+r-1)}{r !}\) x^r + …….. ∞

→ If |x| < 1, then for p, q ∈ N

→ (1 – x)^-p/q = 1 + \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2}\) + ………. + \(\frac{p(p+q) \ldots \ldots(p+(r-1) q)}{r !}\) \(\left(\frac{x}{q}\right)^{r}\) + …….. ∞

→ (1 + x)^-p/q = 1 – \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2}\) + ………. + \(\frac{(-1)^{r} p(p+q) \ldots(p+(r-1) q)}{r !}\) \(\left(\frac{x}{q}\right)^{r}\) + …….. ∞

→ (1 + x)^-p/q = 1 + \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{(p)(p-q)}{1 .2}\left(\frac{x}{q}\right)^{2}\) + …………. + \(\frac{(p)(p-q)(p-2 q) \ldots \ldots .[p-(r-1) q]}{(r) !}\) \(\left(\frac{x}{\cdot q}\right)^{r}\) + ……… ∞

→ (1 – x)^-p/q = 1 – \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{(p)(p-q)}{1.2}\left(\frac{x}{q}\right)^{2}\) – …………. + (- 1)^r \(\) \(\left(\frac{x}{q}\right)^{r}\) + ……… ∞

Binomial Theorem for integral index:
If n is a positive integer then (x + a)ⁿ = ⁿC_o xⁿ + ⁿC₁ x^n-1 a + ⁿC₂ x^n-2 a² + . … + ⁿC_r x^n-ra^r + …… + ⁿC_naⁿ

→ The expansion of (x + a)ⁿ contains (n + 1) terms.

→ In the expansion, the sum of the powers of x and a in each term is equal to n.

→ In the expansion, the coefficients ⁿC₀, ⁿC₁. ⁿC₂………….. ⁿC_n are called binomial coefficients and these are simply denoted by C₀, C₁, C₂ …. C_N.
ⁿC₀ = 1, ⁿC_N = 1, ⁿC₁ = n, ⁿC_r = ⁿC_n-r

→ In the expansion, (r + 1)^th term is called the general term. It is denoted by
T_r+1. Thus T_r+1 = ⁿC_rx_n-ra_r

→ (x + a)ⁿ = \(\sum_{r=0}^{n}\)ⁿC_rx_n-ra_r

→ (x + a)ⁿ = \(\sum_{r=0}^{n}\)ⁿC_rx_n-r(-a)_r = \(\sum_{r=0}^{n}\)(-1)^{n n}C_rx_n-r(-a)_r = ⁿC₀xⁿ – ⁿC₁x^n-1a + ⁿC₁x^n-2a² – ……….. + (-1)^{n n}C_n aⁿ

→ (1 + x)ⁿ = \(\sum_{r=0}^{n}\)ⁿC_rx_r = ⁿC₀ + ⁿC₀x + ……….. + ⁿC_n xⁿ = C₀ + C₁x + C₂x² + ………. + C_nxⁿ

→ Middle term(s) in the expansion of (x + a)ⁿ.

• If n is even, then (\(\frac{n}{2}\) + 1)th term is the middle term
• If n is odd, then \(\frac{n+1}{2}\) th and \(\frac{n+3}{2}\) th terms are the middle terms.

→ Numerically greatest term in the expansion of (1 + x)ⁿ :

• If \(\frac{(n+1)|x|}{|x|+1}\) = p, a integer then plu1 and (p + 1) th terms are the numerically greatest terms in the expansion of (1 + x).
• If \(\frac{(n+1)|x|}{|x|+1}\) = p + F where pis a positive integer and 0< F < 1 then (p+1) th term is the numerically greatest term in the expansion of (1 + x).

→ Binomial Theorem for rational index: If n is a rational number and
|x| < 1, then 1 + nx + \(\frac{n(n-1)}{2 !}\) x² + \(\frac{n(n-1)(n-2)}{3 !}\)x³ + ………… = (1 + x)ⁿ

→ If |x| < 1 then

• (1 + x)^-1= 1 – x + x² – x³ + … + (-1)^rx^r + …….
• (1 – x)^-1 = 1 + x + x² + x³ + … + x^r + …….
• (1 + x)^-2 = 1 – 2x + 3x² – 4x³ + … + (-1)^r (r + 1)x^r + ………..
• (1 – x)^-2 = 1 + 2x + 3x² + 4x³ + … +(r + 1)x^r + ….
• (1 – x)^-n = 1 – nx + \(\frac{n(n-1)}{2 !}\) x² – \(\frac{n(n-1)(n-2)}{3 !}\)x³ + …………..
• (1 – x)^-n = 1 + nx + \(\frac{n(n-1)}{2 !}\) x² + \(\frac{n(n-1)(n-2)}{3 !}\)x³ + ………

→ If |x| < 1 an dn is a positive integer, then

• (1 – x)^-n = 1 + ⁿC₁x + ⁽ⁿ⁺¹⁾C₂x² + ⁽ⁿ⁺²⁾C₃x³ + ………….
• (1 + x)^-n = 1 – ⁿC₁x + ⁽ⁿ⁺¹⁾C₂x² – ⁽ⁿ⁺²⁾C₃x³ + ………….

→ When |x| < 1
(1 – x)^-p/q = 1 + \(\frac{p}{1 !}\left(\frac{x}{q}\right)+\frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2}+\frac{p(p+q)(p+2 q)}{3 !}\left(\frac{x}{q}\right)^{3}\) + …………………∞

→ When |x| < 1
(1 + x)^-p/q = 1 – \(\frac{p}{1 !}\left(\frac{x}{q}\right)_{+} \frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2} \quad \frac{p(p+q)(p+2 q)}{3 !}\left(\frac{x}{q}\right)^{3}\) + …………………∞

Binomial Theorem:
Let n be a positive integer and x, a be real numbers, then (x + a)ⁿ = ⁿC₀.xⁿa° + ⁿC₁.xⁿa¹ + ⁿC₂.x^n-1a² + …………… + ⁿC_r.x^n-ra^r + ……….. + ⁿC_n.x⁰aⁿ
Proof :
We prove this theorem by u sing the principle of mathematical induction (on n).
When n = 1, (x + a)¹ =(x + a)¹ = x + a = ¹C₀x¹a°+ ¹C₁x°a¹
Thus the theorem is true for n = 1
Assume that the theorem is true for n = k ≥ 1 (where k is a positive integer). That is
(x+a)^k = ^kC₀x^k.a⁰ + ^kC₁x^k-2a² + ^kC₂.x^k-2a² + …+ ^kC_r.x^k-r.a^r + ………….. + ^kCx⁰a^k

Now we prove that the theorem is true when n = k + 1 also
(x + a)^k+1 = (x + a)(x + a)^k

Therefore the theorem is true for n = k + 1
Hence, by mathematical induction, it follows that the theorem is true of all positive integer n

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 5 Permutations and Combinations to solve questions creatively.

Intermediate 2nd Year Maths 2A Permutations and Combinations Formulas

→ Fundamental principle: If a work can be performed in m different ways and another work can be performed in n different ways, then the two works simultaneously can be performed in mn different ways.

→ If n is a non-negative integer then

• 0 ! = 1
• n ! = n (n – 1) ! if n > 0

→ The number of permutations of n dissimilar things taken ‘r at a time is denoted by ⁿP_r and ⁿP_r = \(\frac{n !}{(n-r) !}\) for 0 ≤ r ≤ n.

→ If n, r positive integers and r ≤ n, then

• ⁿP_r = n. ^{n – 1}P_{r – 1} if r ≥ 1
• ⁿP_r = n.(n – 1) ^{n – 2}P_{r – 2} if r ≥ 2
• ⁿP_r = ^{n – 1}P_r + r. ^{(n – 1)}P_{(r – 1)}

→ The number of injections that can be defined from set A into set B is ^n(B)p_n(A) n(A) ≤ n(B)

→ The number of bijections that can be defined from set A into set B having same number of elements with A is n(A)!

→ The number of permutations of1n’ dissimilar things taken ‘r’ at a time.

• Containing a particular thing is (r) ^{n – 1}P_{r – 1}
• Not containing a particular thing is ^{n – 1}P_r
• Containing a particular thing in a particular place is ^{n – 1}P_{r – i}.

→ The number of functions that can be defined from set A into set B in [n(B)]^n(A)

→ The sum of the r – digited numbers that can be formed using the given ‘n’ distinct non-zero digits (r ≤ n ≤ 9) is ^{(n – 1)}P_{(r – 1)} × sum of all n digits × 111 …… 1 (r times)

→ In the above, if ‘0’ is one among the given n digits, then the sum is ^{(n – 1)}P_{(r – 1)} × sum of the digits × 111 … 1 (r times) ^{(n – 2)}P_{(r – 2)} × sum of the digits × 111 … 1 [(r – 1) times]

→ The number of permutations of n dissimilar things taken r at a time when repetitions are allowed any number of times is n^r.

→ The number of circular permutations of n dissimilar things is (n – 1) !

→ In case of hanging type circular permutations like garlands of flowers, chains of beads etc., the number of circular permutations of n things is \(\frac{1}{2}\) [(n – 1) !]

→ If in the given n things, p alike things are of one kind, q alike things are of the second kind, r alike things are of the third kind and the rest are dissimilar, then the number of permutations (of these n things) is \(\frac{n !}{p ! q ! r !}\)

→ The number of combinations of n things taken r at a time is denoted by ⁿC_r and ⁿC_r = \(\frac{n !}{(n-r) ! r !}\) for 0 ≤ r ≤ n and \(\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}\) and ⁿP_r = r! ⁿC_r

→ If n, r are integers and 0 ≤ r ≤ n then ⁿc_r = ⁿC_{r – 1}

→ Let n, r, s are integers and 0 ≤ r ≤ n, 0 ≤ s ≤ n. If ⁿC_r = ⁿC_s then r = s or r + s = n.

→ The number of ways of dividing ‘m + n’ things (m ≠ n) into two groups containing m, n things is ^{(m + n)}C_m = ^{(m + n)}C_m = \(\frac{(m+n) !}{m ! n !}\)

→ The number of ways of dividing (m + n + p) things (m, n, p are distinct) into 3 groups of m, n, p things is \(\frac{(m+n+p) !}{m ! n ! p !}\).

→ The number of ways of dividing mn things into m equal groups is \(\frac{(m n) !}{(n !)^{m} m !}\)

→ The number of ways if distributing mn things equally to m persons is \(\frac{(m n) !}{(n !)^{m}}\)

→ If p alike things are of one kind, q alike things are of the second kind and r alike things are of the third kind, then the number of ways of selecting one or more things out of them is (p + 1) (q + 1) (r +1) – 1.

→ If m is a positive integer and m = p₁^α₁, p₁^α₂ …… p_k^α_k where p₁, p₂ …… p_k are distinct primes and α₁, α₂, …. α_k are non-negative integers, then the number of divisors of m is (α₁ + 1) (α₂ + 1) ……… (α_k + 1). [This includes 1 and m].

→ The total number of combinations of n different things taken any number of times is 2ⁿ.

→ The total number of combinations of n different things taken one or more at a timers 2ⁿ – 1.

→ The number of diagonals in a regular polygon of n sides is \(\frac{n(n-3)}{2}\)

→ Permutations are arrangements of things taken some or all at a time.

→ In a permutation, order of the things is taken into consideration.

→ ⁿp_r represents the number of permutations (without repetitions) of n dissimilar things taken r at a time.

→ Fundamental Principle: If an event is done in ‘m’ ways and another event is done in ‘n’ ways, then the two events can be together done in mn ways provided the events are independent.

→ ⁿp_r = \(\frac{n !}{(n-r) !}\) = n(n – 1)(n – 2) …………. (n – r + 1)

→ ⁿp₁ = n, ⁿp₂ = n(n-1), ⁿp₃ = n(n-1)(n-2).

→ ⁿp_r = n!

→ \(\frac{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}-1}}\)ⁿp_n = n!

→ ⁿ⁺¹p_r = r. ⁿp_r-1 + ⁿp_r

→ ⁿp_r = r. ^n-1p_r-1 + ⁿp_r

→ ⁿp_r = r.^n-1p_r-1 + ^n-1p_r

→ The number of permutations of n things taken r at a time containing a particular thing is r × ^n-1p_r-1

→ The number of permutations of n things taken r at a time not containing a particular thing is ^n-1p_r

• The number of permutations of n things taken r at a time allowing repetitions is n^r.
• The number of permutations of n things taken not more than r at a time allowing repetitions is \(\frac{n\left(n^{\mathrm{r}}-1\right)}{(n-1)}\)

→ The number of permutations of n things of which p things are of one kind and q things are of another kind etc., is \(\frac{n !}{p ! q ! \ldots \ldots \cdots}\)

→ The sum of all possible numbers formed out of all the ‘n’ digits without zero is (n-1)! (sum of all the digits) (111 ……….. n times).

→ The sum of all possible numbers formed out of all the ‘n’ digits which includes zero is [(n -1)! (sum of all the digits) (111 …………… n times)] – [( n – 2)! (sum of all the digits)(111 ………….. (n -1) times]

→ The sum of all possible numbers formed by taking r digits from the given n digits which do not include zero is ^n-1p_r-1 (sum of all the digits)(111 …………… r times).

→ The sum of all possible numbers formed by taking r digits from the given n digits which include zero is ^n-1p_r-1(sum of all the digits)(111 ……………….. r times) – ^n-2p_r-2(sum of all the digits)(111 ……………….. (r-1) times).

• The number of permutations of n things when arranged round a circle is (n -1)!
• In case of necklace or garland number of circular permutations is \(\frac{(n-1) !}{2}\)

→ Number of permutations of n things taken r at a time in which there is at least one repetition is n^r – ⁿp_r.

→ The number of circular permutations of ‘n’ different things taken ‘r’ at a time is \(\frac{{ }^{n} \mathrm{P}_{\mathrm{r}}}{\mathrm{r}}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 4 Theory of Equations to solve questions creatively.

Intermediate 2nd Year Maths 2A Theory of Equations Formulas

→ If n is a non-negative integer and a₀, a₁, a₂, ……….. a_n are real or complex numbers and a₀ ≠ 0, then the expression f(x) = a₀xⁿ + a₁x^{n – 1} + a₂x^{n – 2} + ……. + a_n is called a polynomial in x of degree n.

→ f(x) = a₀xⁿ + a₁x^{n – 1} + a₂x^{n – 2} + ……. + a_n = 0 is called a polynomial equation in x of degree n (a₀ ≠ 0). Every non-constant polynomial equation has atleast one root.

→ If f(α) = 0 then α is called a root of the equation f(x) = 0.

→ If f(α) = 0 then (x – α) is a factor of f(x).

Relation between roots and coefficients of an equation:
→ If α β γ are the roots of x³ + p₁x² + p₂x + p₃ = 0 then sum of the roots s₁ = α + β + γ = – p₁.
Sum of the products of two roots taken at a time s₂ = αβ + βγ + γα = p₂.
Product of all the roots, s₃ = αβγ = – p₃.

→ If α, β, γ, δ are the roots of x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0 then sum of the roots s₁ = α + β + γ + δ = – p₁.
Sum of the products of roots taken two at a time
s₂ = αβ + αγ + αδ + βγ + βδ + γδ = p₂.
Sum of the products of roots taken three at a time .
s₃ = αβγ + βγδ + γδα + δαβ = – p₃.
Product of the roots, s₄ = αβγδ = p₄.

→ For a cubic equation, when the roots are

• In A.P., then they are taken as a – d, a, a + d.
• In G.P., then they are taken as \(\frac{a}{r}\), a, ar.
• In H.P., then they are taken as \(\frac{1}{a-d}, \frac{1}{a^{\prime}}, \frac{1}{a+d}\).

→ For a bi quadratic equation, if the roots are

• In A.P., then they are taken as a – 3d, a – d, a + d, a + 3d.
• In C.P., then they are taken as \(\frac{1}{a-3 d}, \frac{1}{a-d}, \frac{1}{a+d}, \frac{1}{a+3 d}\).

→ In an equation with real coefficients, imaginary roots occur in conjugate pairs.

→ In an equation with rational coefficients, irrational roots occur in pairs of conjugate surds.

→ The equation whose roots are those of the equation f(x) = 0 with contrary signs is f(- x) = 0.

→ The equation whose roots are multiplied by kfa 0) of those of the proposed equation f(x) = 0 is f \(\left(\frac{x}{k}\right)\) = 0.

→ The equation whose roots are reciprocals of the roots of f(x) = 0 is f \(\left(\frac{1}{x}\right)\) = 0.

→ The equation whose roots are exceed by h than those of f(x) = 0 is f(x – h) = 0.

→ The equation whose roots are diminished by h than those of f(x) 0 is f(x + h) = 0.

→ The equation whose roots are the square of the roots of f(x) = 0 is obtained by eliminating square root from f(√x) = 0.

→ If f(x) = p₀xⁿ + p₁x^{n – 1} + p₂x^{n – 2} + ……. + p_n = 0 then to eliminate the second term,
f(x) = 0 can be transformed to f(x + h) = 0 where h = \(\frac{-p_{1}}{n \cdot p_{0}}\).

→ If an equation is unaltered by changing x into \(\frac{1}{x}\) then it is a reciprocal equation.

→ A reciprocal equation f(x) = p₀xⁿ + p₁x^{n – 1} + …… + pⁿ = 0 is said to be a reciprocal equation of first class p_i = p_{n – i} for all i.

→ A reciprocal equation f(x) = p₀xⁿ + p₁x^{n – 1} + …… + pⁿ = 0 is said to be a reciprocal equation of second class if p_i = – p_{n – i} for all i.

→ For an odd degree reciprocal equation of class one, – 1 is a root and for an odd degree reciprocal equation of class two, 1 is a root.

→ For an even degree reciprocal equation of class two, 1 and – 1 are roots.

→ If f(x) = 0 is an equation of degree ‘n’ then to eliminate r^th term, f(x) = 0 can be transformed to f(x + h) = 0 where h is a constant such that f^{(n – r + 1)}(h) = 0 i.e.,(n – r + 1)^th derivative of f(h) is zero.

→ Every n^th degree equation has exactly n roots real or imaginary.

→ Relation between, roots and coefficients of an equation.

(i) If α, β, γ are the roots of x³ + p₁x² + p₂x + p₃ = 0 the sum of the roots s₁ = α + β + γ = -p₁.
Sum of the products of two roots taken at a time s₂ = αβ + βγ + γα = -p₂
Product of all the roots, s₃ = αβγ= – p₃.

(ii) If α, β, γ, δ are the roots of x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0 then

• Sum of the roots s₁ = a+P+y+S = -p₁.
  s₂ = αβ + αγ + αδ + βα + βδ + γδ = p₂.
• Sum of the products of roots taken three at a time
  s₃ = αβγ + βγδ + γδα + δαβ = – p₃.
• Product of the roots, s₄ = αβγδ = p₄

→ For the equation xⁿ + p₁x^n-1 + p₂x^n-2 + ……… + p_n = 0

• Σ α₂ = p₁² – 2p₂
• Σ α₃ = -p₁³ + 3p₁p₂ – 3p₃
• Σ α₄ =p₁⁴ – 4p₁²p² + 2p₂² + 4p₁p₃ – 4p₄
• Σ α₂β = 3p₃ – p₁p₂
• Σ α₂βγ = p₁p₃ – 4p₄

Note: For the equation x³ + p₁x² + p₂x + p₃ = 0 Σα₂β₂ — p₂ -2p₁p₃

→ To remove the second term from a nth degree equation, the roots must be diminished by \(\frac{-\mathrm{a}_{1}}{\mathrm{na}_{0}}\) and the resultant equation will not contain the term with x^n-1.

→ If α₁ , α₂ ………………. , α_n are the roots of f(x) = 0, the equation

• Whose roots are \(\) is f\(\left(\frac{1}{x}\right)\) = 0
• Whose roots are kα₁, kα₂ …,kα_n is f\(\left(\frac{x}{h}\right)\) = 0
• Whose roots are α₁ – h, α₂ – h, …. α_n – h is f(x + h) = 0.
• Whose roots are α₁ + h, α₂ + h, ………….. α_n + h is f(x – h) = 0
• Whose roots are α₁², α₂²…. α₁² is f (f√y) = 0

→ In any equation with rational coefficients, irrational roots occur in conjugate pairs.

→ In any equation with real coefficients, complex roots occur in conjugate pairs.

→ If α is r – multiple root of f(x) = 0, then a is a (r – 1) – multiple root of f¹(x) = 0 and (r-2) – Multiple root of f ¹¹(x) = 0 and non multiple root of f^r-1(x) =0.

→ If f(x) = xⁿ + p₁x^n-1 + …………. + p_n-1x + p_n and f(a) and f(b) are of opposite sign, then at least
one real root of f(x) =0 lies between a and b.

(a) For a cubic equation, when the roots are

• In A.P., then they are taken as a – d, a, a + d
• In G.P., then are taken as \(\frac{a}{r}\), a, ar
• In H.P., then they are taken as \(\frac{1}{a-d}, \frac{1}{a}, \frac{1}{a+d}\)

(b) For a bi quadratic equation, if the roots are

• In A.P., then they are taken as a – 3d, a + d, a + 3d
• In G.P., then they are \(\frac{a}{d^{3}}, \frac{a}{d}\), ad, ad³
• In H.P., then they are taken as \(\frac{1}{a-3 d}, \frac{1}{a-d}, \frac{1}{a+d}, \frac{1}{a+3 d}\)

→ It an equation is unaltered by changing x into \(\frac{1}{x}\), then it is a reciprocal equation.

• A reciprocal equation f (x) = p₀xⁿ + p₁x^n-1 + ……………….. + p_n = 0 is said to be a reciprocal equation of first class p_i = p_n-i for all i.
• A reciprocal equation f (x) = p₀xⁿ + p₁x^n-1 + ……………….. + p_n = 0 = 0 is said to be a reciprocal equation of second class p_i = p_n-i for all i.
• For an odd degree reciprocal equation of class one, -1 is a root and for an odd degree reciprocal equation of class two, 1 is a root.
• For an even degree reciprocal equation of class two, 1 and -1 are roots.

→ If f(x) = 0 is an equation of degree ‘n’ then to eliminate rth term, .f(x) = 0 can be transformed to f(x+h) = 0 where h is a constant such that f^(n-r+1)(h) =0 i.e., (n – r + 1)^th derivative of f(h) is zero.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 3 Quadratic Expressions to solve questions creatively.

Intermediate 2nd Year Maths 2A Quadratic Expressions Formulas

→ If a, b, c are real or complex numbers and a ≠ 0, then the expression ax² + bx + c is called a quadratic expression in the variable x.
Eg: 4x² – 2x + 3

→ If a, b, c are real or complex numbers and a ≠ 0, then ax² + bx + c = 0 is called a quadratic equation in x.
Eg: 2x² – 5x + 6 = 0

→ A complex number α is said to be a root or solution of the quadratic equation ax² + bx + c = 0 if aα² + bα + c = 0

→ The roots of ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

→ If α, β are roots of ax² + bx + c = 0, then α + β = \(\frac{-b}{a}\) and αβ = \(\frac{C}{a}\)

→ The equation of whose roots are α, β is x² – (α + β)x + αβ = 0

→ Nature of the roots: ∆ = b² – 4ac is called the discriminant of the quadratic equation ax² + bx + c = 0. Let α, β be the roots of the quadratic equation ax² + bx + c = 0
Case 1: If a, b, c are real numbers, then

• ∆ = 0 ⇔ α = β = \(\frac{-b}{2 a}\) (a repeated root or double root)
• ∆ > 0 ⇔ α and β are real and distinct.
• ∆ < 0, ⇔ α and β are non- real complex numbers conjugate to each other.

Case 2: If a, b, c are rational numbers, then

• ∆ = 0 ⇔ α and β are rational and equal (ei) α = \(\frac{-b}{2 a}\), a double root or a repeated root.
• ∆ > 0 and is a square of a rational number ⇔ α and β are rational and distinct.
• ∆ > 0 but not a square of a rational number ⇔ α and β are conjugate surds.
• ∆ < 0, ⇔ α and β are non- real ⇔ α and β are non-real con conjugate complex numbers.

→ Let a, b and c are rational numbers, α and β be the roots of the equations ax² + bx + c = 0. Then

• α, β are equal rational numbers if ∆ = 0.
• α, β are distinct rational numbers if ∆ is the square of a non zero rational numbers.
• α, β are conjugate surds if ∆ > 0 and ∆ is not the square of a nonzero square of a rational number.

→ If a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 have two same roots, then \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

→ If α, β are roots of ax² + bx + c = 0,

• the equation whose roots are \(\frac{1}{\alpha}, \frac{1}{\beta}\) is f \(\left(\frac{1}{x}\right)\) = 0. If c ≠ 0 (ie) αβ ≠ 0
• the equation whose roots are α + k, β + k is f(x – k) = 0
• the equation whose roots are kα and kβ is f\(\left(\frac{x}{k}\right)\) = 0
• the equation whose roots are equal but opposite in sign is f(-x) = 0
  (ie) the equation whose roots are – α, – β is f(-x) = 0.

→ If the roots of ax² + bx + c = 0 are complex roots then for x ∈ R, ax² + bx + c and ‘a’ have the same sign.

→ If α and β (α < β) are the roots of ax² + bx + c = 0 then

• ax² + bx + c and ‘a’ are of opposite sign when α < x < β
• ax² + bx + c and ‘a’ are of the same sign if x < α or x > β.

→ Let f(x) = ax² + bx + c be a quadratic function

• If a > 0 then f(x) has minimum value at x = \(\frac{-b}{2 a}\) and the minimum value is given by \(\frac{4 a c-b^{2}}{4 a}\)
• If a < 0 then f(x) has maximum value at x = \(\frac{-b}{2 a}\) and the maximum value is given by \(\frac{4 a c-b^{2}}{4 a}\)

→ A necessary and sufficient condition for the quadratic equation a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 to have a common root is (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁) (b₁c₂ – b₂c₁).

→ If a₁b₂ – a₂b₁ = 0 then common root of a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 is \(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\).

→ The standard form of a quadratic ax² + bx + c = 0 where a, b, c ∈ R and a ≠ 0

→ The roots of ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

→ For the equation ax² + bx + c = 0, sum of the roots = \(-\frac{b}{a}\), product of the roots = \(\frac{c}{a}\).

→ If the roots of a quadratic are known, the equation is x² – (sum of the roots)x +(product of the roots)= 0

→ “Irrational roots” of a quadratic equation with “rational coefficients” occur in conjugate pairs. If p + √q is a root of ax² + bx + c = 0, then p – √q is also a root of the equation.

→ “Imaginary” or “Complex Roots” of a quadratic equation with “real coefficients” occur in conjugate pairs. If p + iq is a root of ax² + bx + c = 0. Then p – iq is also a root of the equation.

→ Nature of the roots of ax² + bx + c = 0

→ Two equations a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 have exactly the same roots if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

→ The equations a₁x² + b₁x + c₁ = 0, a2x² + b₂x + c₂ = 0 have a common root, if (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁)(b₁c₂ – b₂c₁) and the common root is \(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\) if a₁b₂ ≠ a₂b₁

→ If f(x) = 0 is a quadratic equation, then the equation whose roots are

• The reciprocals of the roots of f(x) = 0 is f\(\left(\frac{1}{x}\right)\) = 0
• The roots of f(x) = 0, each ‘increased’ by k is f(x – k) = 0
• The roots of f(x) = 0, each ‘diminished’ by k is f(x + k) = 0
• The roots of f(x) = 0 with sign changed is f(-x) = 0
• The roots of f(x) = 0 each multiplied by k(≠0) is f\(\left(\frac{x}{k}\right)\) = 0

→ Sign of the expression ax² + bx + c = 0

• The sign of the expression ax² + bx + c is same as that of ‘a’ for all values of x if b² – 4ac ≤ 0 i.e. if the roots of ax² + bx + c = 0 are imaginary or equal.
• If the roots of the equation ax² + bx + c = 0 are real and different i.e b² – 4ac > 0, the sign of the expression is same as that of ‘a’ if x does not lie between the two roots of the equation and opposite to that of ‘a’ if x lies between the roots of the equation.

→ The expression ax² + bx + c is positive for all real values of x if b² – 4ac < 0 and a > 0.

→ The expression ax² + bx + c has a maximum value when ‘a’ is negative and x = –\(\frac{\mathrm{b}}{2 \mathrm{a}}\). Maximum value of the expression = \(\frac{4 a c-b^{2}}{4 a}\)

→ The expression ax² + bx + c has a maximum value when ‘a’ is positive and x = –\(\frac{\mathrm{b}}{2 \mathrm{a}}\). Minimum value of the expression = \(\frac{4 a c-b^{2}}{4 a}\)

Theorem 1:
If the roots of ax² + bx + c = 0 are imaginary, then for x ∈ R , ax² + bx + c and a have the same sign.
Proof:
The root are imaginary
b² – 4ac < 0 4ac – b² > 0
\(\frac{a x^{2}+b x+c}{a}=x^{2}+\frac{b}{a} x+\frac{c}{a}=\left(x+\frac{b}{2 a}\right)^{2}+\frac{c}{a}-\frac{b^{2}}{4 a^{2}}=\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a^{2}}\)
∴ For x ∈ R, ax² + bx + c = 0 and a have the same sign.

Theorem 2.
If the roots of ax² + bx + c = 0 are real and equal to α = \(\frac{-b}{2 a}\), then α ≠ x ∈ R ax² + bx + c and a will have same sign.
Proof:
The roots of ax² + bx + c = 0 are real and equal
⇒ b² = 4ac ⇒ 4ac – b² = 0
\(\frac{a x^{2}+b x+c}{a}\) = x + \(\frac{b}{a}\)x + \(\frac{c}{a}\)
= \(\left(x+\frac{b}{2 a}\right)^{2}+\frac{c}{a}-\frac{b^{2}}{4 a^{2}}\)
= \(\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a^{2}}\)
= \(\left(x+\frac{b}{2 a}\right)^{2}\) > 0 for x ≠ \(\frac{-b}{2 a}\) = α
For α ≠ x ∈ R, ax² + bx + c and a have the same sign.

Theorem 3.
Let be the real roots of ax² + bx + c = 0 and α < β. Then
(i) x ∈ R, α < x< β ax² + bx + c and a have the opposite signs
(ii) x ∈ R, x < α or x> β ax² + bx + c and a have the same sign.
Proof:
α, β are the roots of ax² + bx + c = 0
Therefore, ax² + bx + c = a(x – α)(x – β)
\(\frac{a x^{2}+b x+c}{a}\) = (x – α)(x – β)

(i) Suppose x ∈ R, α < x < β
⇒ x < α < β then x – α < 0, x – β < 0 ⇒ (x – α)(x – β) > 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c, a have a same sign

(ii) Suppose x ∈ R, x > β, x > β > α then x – α > 0, x – β > 0
⇒ (x – α)(x – β) > 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c, a have same sign
∴ x ∈ R, x < α or x > β ⇒ ax² + bx + c and a have the same sign.