Category: Inter 1st Year

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(g) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(g)

Examine whether the following systems of equations are consistent or inconsistent and if consistent find the complete solutions.

Question 1.
x + y + z = 4
2x + 5y – 2z = 3
x + 7y – 7z = 5
Solution:

ρ(A) = 2, ρ(AB) = 3
ρ(A) ≠ ρ(AB)
∴ The given system of equations are in consistent.

Question 2.
x + y + z = 6
x – y + z = 2
2x – y + 3z = 9
Solution:

Question 3.
x + y + z = 1
2x + y + z = 2
x + 2y + 2z = 1
Solution:

ρ(A) = 2 = ρ(AB) < 3
The given system of equations are consistent and have infinitely many solutions.
The solutions are given by [(x, y, z) 1x = 1, y + z = 0].

Question 4.
x + y + z = 9
2x + 5y + 7z = 52
2x + y – z = 0
Solution:

∴ ρ(A) = ρ(AB) = 3
The given system of equations are consistent have a unique solution.
∴ Solution is given by x = 1, y = 3, z = 5.

Question 5.
x + y + z = 6
x + 2y + 3z = 10
x + 2y + 4z = 1
Solution:
Augmented matrix A = \(\left[\begin{array}{cccc}
1 & 1 & 1 & 6 \\
1 & 2 & 3 & 10 \\
1 & 2 & 4 & 1
\end{array}\right]\)
By R₂ → R₂ – R₁, R₃ → R₃ – R₂, we obtain

∴ ρ(A) = ρ(AB) = 3
The given system of equations are consistent.
They have a unique solution.
∴ Solution is given by x = -7, y = 22, z = -9.

Question 6.
x – 3y – 8z = -10
3x + y – 4z = 0
2x + 5y + 6z = 13
Solution:
The Augmented matrix

ρ(A) = ρ(AB) = 2 < 3
∴ The given system of equations are consistent have infinitely many solutions.
x + y = 2 and y + 2z = 3
Taking z = k, y = 3 – 2z = 3-2k
x = 2 – y
= 2 – (3 – 2k)
= 2 – 3 + 2k
= 2k – 1
∴ The solutions are given by x = -1 + 2k, y = 3 – 2k, z = k where ‘k’ is any scalar.

Question 7.
2x + 3y + z = 9
x + 2y + 3z = 6
3x + y + 2z = 8
Solution:

∴ ρ(A) = ρ(AB) = 3
The given system of equations are consistent have a unique solution.
∴ Solution is given by x = \(\frac{35}{18}\), y = \(\frac{29}{18}\), z = \(\frac{5}{18}\)

Question 8.
x + y + 4z = 6
3x + 2y – 2z = 9
5x + y + 2z = 13
Solution:

∴ ρ(A) = ρ(AB) = 3
∴ The given system of equations are consistent have a unique solution.
∴ Solution is given by x = 2, y = 2, z = \(\frac{1}{2}\)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(f) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(f)

I. Find the rank of each of the following matrices.

Question 1.
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|\) = 0 – 0 = 0
and |1| = 1 ≠ 0
∴ ρ(A) = 1

Question 2.
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1 – 0 = 1 ≠ 0
∴ ρ(A) = 2

Question 3.
\(\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right|\) = 0 – 0 = 0
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 4.
\(\left[\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right|\) = 0 – 1 = -1 ≠ 0
∴ ρ(A) = 2

Question 5.
\(\left[\begin{array}{ccc}
1 & 0 & -4 \\
2 & -1 & 3
\end{array}\right]\)
Solution:
\(\left|\begin{array}{cc}
1 & -4 \\
2 & 3
\end{array}\right|\) = 3 + 8 = -11 ≠ 0
∴ ρ(A) = 2

Question 6.
\(\left[\begin{array}{lll}
1 & 2 & 6 \\
2 & 4 & 3
\end{array}\right]\)
Solution:
\(\left|\begin{array}{ll}
2 & 6 \\
4 & 3
\end{array}\right|\) = 6 – 24 = -18 ≠ 0
∴ ρ(A) = 2

II.

Question 1.
\(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
= 1(1 – 0) – 0(0 – 0) + 0(0 – 0)
= 1 – 0 + 0
= 1 ≠ 0
∴ ρ(A) = 3

Question 2.
\(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right|\)
= 1(6 – 0) – 2(8 + 1) + 0(0 + 3)
= 6 – 18
= -12 ≠ 0
∴ ρ(A) = 3

Question 3.
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\)
Solution:
Det A = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right|\)
= 1(6 – 4) – 2(4 – 3) + 0(8 – 9)
= 2 – 2 + 0
= 0
∴ ρ(A) ≠ 3, ρ(A) < 3
Take \(\left|\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right|\) = 3 – 4 = -1 ≠ 0
∴ ρ(A) = 2

Question 4.
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\), det A = 0, ρ(A) ≠ 3
All 2 × 2 sub-matrix det. is zero
∴ ρ(A) ≠ 2
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 5.
\(\left[\begin{array}{cccc}
1 & 2 & 0 & -1 \\
3 & 4 & 1 & 2 \\
-2 & 3 & 2 & 5
\end{array}\right]\)
Solution:
Take sub-matrix B = \(\left|\begin{array}{ccc}
1 & 2 & 0 \\
3 & 4 & 1 \\
-2 & 3 & 2
\end{array}\right|\)
= 1(8 – 3) – 2(6 + 2)
= 5 – 16
= -11 ≠ 0
Rank of the given matrix is 3.

Question 6.
\(\left[\begin{array}{cccc}
0 & 1 & 1 & -2 \\
4 & 0 & 2 & 5 \\
2 & 1 & 3 & 1
\end{array}\right]\)
Solution:
Take sub matrix A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
4 & 0 & 2 \\
2 & 1 & 3
\end{array}\right]\)
= -1(12 – 4) + 1(4 – 0)
= -8 + 4
= -4 ≠ 0
∴ ρ(A) = 3

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(e) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(e)

I.

Question 1.
Find the adjoint and inverse of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\)
Solution:

Question 2.
If A = \(\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]\), a² + b² + c² + d² = 1, then find the inverse of A.
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\), then find A^-1
Solution:

Question 4.
If A = \(\left|\begin{array}{ccc}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right|\), then show that the adjoint of A = 3A’ find A^-1.
Solution:

Question 5.
If abc ≠ 0, find the inverse of \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]\)
Solution:

II.

Question 1.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\) and B = \(\frac{1}{2}\left[\begin{array}{lll}
b+c & c-a & b-a \\
c-b & c+a & a-b \\
b-c & a-c & a+b
\end{array}\right]\) then show that ABA^-1 is a diagonal matrix.
Solution:

Question 2.
If 3A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
-2 & 2 & -1
\end{array}\right]\) then show that A^-1 = A’
Solution:

Question 3.
If A = \(\left[\begin{array}{rrr}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\), then show that A^-1 = A³
Solution:

∴ A⁴ = I
det A = 3(1) – 3(-2) + 4(-2) = 1
∵ A ≠ 0 ⇒ A^-1 exists
∵ A⁴ = I
Multiply with A^-1
A⁴ (A^-1) = I (A^-1)
⇒ A³ (AA^-1) = A^-1
⇒ A³ (I) = A^-1
∴ A^-1 = A³

Question 4.
If AB = I or BA = I, then prove that A is invertible and B = A^-1
Solution:
Given AB = I
⇒ AB| = |1|
⇒ |A| |B| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix and BA = I
⇒ |BA| = |I|
⇒ |B| |A| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix.
AB = I or BA = I, A is invertible.
∴ A^-1 exists.
AB = I
⇒ A^-1AB = A^-1I
⇒ IB = A^-1
⇒ B = A^-1
∴ B = A^-1

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(d)

I.

Question 1.
Find the determinants of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & 1 \\
1 & -5
\end{array}\right]\)
Solution:
det A = ad – bc
= 2(-5) – 1(1)
= -10 – 1
= -11

(ii) \(\left[\begin{array}{cc}
4 & 5 \\
-6 & 2
\end{array}\right]\)
Solution:
det A = 4(2) – (-6)(5)
= 8 + 30
= 38

(iii) \(\left[\begin{array}{cc}
\mathbf{i} & 0 \\
0 & -\mathbf{i}
\end{array}\right]\)
Solution:
det A = -i² – 0
= 1 – 0
= 1

(iv) \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)
Solution:
det A = 0(0 – 1) – 1(0 – 1) + 1(1 – 0)
= 1 + 1
= 2

(v) \(\left[\begin{array}{ccc}
1 & 4 & 2 \\
2 & -1 & 4 \\
-3 & 7 & 6
\end{array}\right]\)
Solution:
det A = 1(-6 – 28) – 4(12 + 12) + 2(14 – 3)
= -34 – 96 + 22
= -108

(vi) \(\left[\begin{array}{ccc}
2 & -1 & 4 \\
4 & -3 & 1 \\
1 & 2 & 1
\end{array}\right]\)
Solution:
det A = 2(-3 – 2) + 1(4 – 1) + 4(8 + 3)
= -10 + 3 + 44
= 37

(vii) \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
4 & -1 & 7 \\
2 & 4 & -6
\end{array}\right]\)
Solution:
det A = 0 since R₁ and R₃ are proportional.

(viii) \(\left[\begin{array}{lll}
a & h & g \\
\text { h } & b & f \\
g & f & c
\end{array}\right]\)
Solution:
det A = a(bc – f²) – h(ch – fg) + g(hf – bg)
= abc – af² – ch² + fgh + fgh – bg²
= abc + 2fgh – af² – bg² – ch²

(ix) \(\left[\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{b} & \mathbf{c} & \mathbf{a} \\
\mathbf{c} & \mathbf{a} & \mathbf{b}
\end{array}\right]\)
Solution:
det A = a(bc – a²) – b(b² – ac) + c(ab – c²)
= abc – a³ – b³ + abc + abc – c³
= 3abc – a³ – b³ – c³

(x) \(\left[\begin{array}{ccc}
1^{2} & 2^{2} & 3^{2} \\
2^{2} & 3^{2} & 4^{2} \\
3^{2} & 4^{2} & 5^{2}
\end{array}\right]\)
Solution:
det A = \(\left|\begin{array}{ccc}
1 & 4 & 9 \\
4 & 9 & 16 \\
9 & 16 & 25
\end{array}\right|\)
= 1(225 – 256) – 4(100 – 144) – 9(64 – 81)
= -31 + 176 – 153
= -184 + 176
= -8

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right]\) and det A = 45 then find x.
Solution:
det A = 45
\(\left|\begin{array}{ccc}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right|\) = 45
⇒ 3x + 24 = 45
⇒ 3x – 45 + 24 = 0
⇒ 3x – 21 = 0
⇒ x = 7

II.

Question 1.
Show that \(\left|\begin{array}{lll}
b c & b+c & 1 \\
c a & c+a & 1 \\
a b & a+b & 1
\end{array}\right|\) = (a – b)(b – c)(c – a)
Solution:

Question 2.
Show that \(\left|\begin{array}{ccc}
\mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} \\
\mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} \\
\mathbf{a} & \mathbf{b} & \mathbf{c}
\end{array}\right|\) = a³ + b³ + c³ – 3abc
Solution:

= (a + b + c) [(-ac + b²) – (-c² + ab) + (-bc + a²)]
= (a + b + c) (-ac + b² + c² – ab – bc + a²)
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= a³ + b³ + c³ – 3abc

Question 3.
Show that \(\left|\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & \mathbf{z} & \mathbf{x}+\mathbf{y}
\end{array}\right|\) = 4xyz
Solution:
L.H.S = \(\left|\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & \mathbf{z} & \mathbf{x}+\mathbf{y}
\end{array}\right|\)
= (y + z) [(z + x) (x + y) – yz] – x[y(x + y) – yz] + x[yz – z(z + x)]
= (y + z) (zx + yz + x² + xy – yz) – x(xy + y² – yz) + x(yz – z² – zx)
= (y + z) (zx + x² + xy) – x(xy + y² – yz) + x(yz – z² – zx)
= xyz + x²y + xy² + xz² + x²z + xyz – x²y – xy² + xyz + xyz – xz² – x²z
= 4xyz
= R.H.S

Question 4.
If \(\left|\begin{array}{ccc}
a & a^{2} & 1+a^{3} \\
b & b^{2} & 1+b^{3} \\
c & c^{2} & 1+c^{3}
\end{array}\right|\) = 0 and \(\left|\begin{array}{ccc}
a & a^{2} & 1 \\
b & b^{2} & 1 \\
c & c^{2} & 1
\end{array}\right|\) ≠ 0 then show that abc = -1
Hint: If each element in a row (column) of a square matrix is the sum of two numbers, then its discriminant can be expressed as the sum of discriminants of two square matrices.
Solution:

Question 5.
Without expanding the determinant, prove that
(i) \(\left|\begin{array}{ccc}
a & a^{2} & b c \\
b & b^{2} & c a \\
c & c^{2} & a b
\end{array}\right|=\left|\begin{array}{ccc}
1 & a^{2} & a^{3} \\
1 & b^{2} & b^{3} \\
1 & c^{2} & c^{3}
\end{array}\right|\)
Solution:

(ii) \(\left|\begin{array}{ccc}
a x & b y & c z \\
x^{2} & y^{2} & z^{2} \\
1 & 1 & 1
\end{array}\right|=\left|\begin{array}{ccc}
a & b & c \\
x & y & z \\
y z & z x & x y
\end{array}\right|\)
Solution:

(iii) \(\left|\begin{array}{lll}
1 & b c & b+c \\
1 & c a & c+a \\
1 & a b & a+b
\end{array}\right|=\left|\begin{array}{ccc}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{array}\right|\)
Solution:
L.H.S = \(\left|\begin{array}{ccc}
1 & b c & b+c \\
1 & c a & c+a \\
1 & a b & a+b
\end{array}\right|\)

= (b – a) (c – a) (c + a – b – a)
= (a – b) (b – c) (c – a)
∴ LHS = RHS

Question 6.
If ∆₁ = \(\left|\begin{array}{ccc}
a_{1}^{2}+b_{1}+c_{1} & a_{1} a_{2}+b_{2}+c_{2} & a_{1} a_{3}+b_{3}+c_{3} \\
b_{1} b_{2}+c_{1} & b_{2}^{2}+c_{2} & b_{2} b_{3}+c_{3} \\
c_{3} c_{1} & c_{3} c_{2} & c_{3}^{2}
\end{array}\right|\) and ∆₂ = \(\left|\begin{array}{lll}
a_{1} & b_{2} & c_{2} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right|\), then find the value of \(\frac{\Delta_{1}}{\Delta_{2}}\)
Solution:

Question 7.
If ∆₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \alpha & 1
\end{array}\right|\), ∆₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\) and ∆₁ = ∆₂, then show that cos²α + cos²β + cos²γ = 1
Solution:
∆₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \alpha & 1
\end{array}\right|\)
= 1(1 – cos²γ) – cos α (cos α – cos β cos γ) + cos β (cos α cos γ – cos β)
= 1 – cos²γ – cos²α + cos α cos β cos γ + cos α cos β cos γ – cos²β
= 1 – cos²γ – cos²α – cos²β + 2 cos α cos β cos γ
∆₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\)
= 0(0 – cos²γ) – cos α (0 – cos γ cos β) + cos β (cos α cos γ – 0)
= cos α cos β cos γ + cos α cos β cos γ
= 2 cos α cos β cos γ
Given ∆₁ = ∆₂
1 – cos²α – cos²β – cos²γ + 2 cos α cos β cos γ = 2 cos α cos β cos γ
1 – cos²α – cos²β – cos²γ = 0
1 = cos²α + cos²β + cos²γ

III.

Question 1.
Show that \(\left|\begin{array}{ccc}
\mathbf{a}+\mathbf{b}+2 \mathbf{c} & \mathbf{a} & \mathbf{b} \\
\mathbf{c} & \mathbf{b}+\mathbf{c}+\mathbf{2} \mathbf{a} & \mathbf{b} \\
\mathbf{c} & \mathbf{a} & \mathbf{c}+\mathbf{a}+\mathbf{2} \mathbf{b}
\end{array}\right|\) = 2(a + b + c)³
Solution:

Question 2.
Show that \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|^{2}\) = \(\left|\begin{array}{ccc}
2 b c-a^{2} & c^{2} & b^{2} \\
c^{2} & 2 a c-b^{2} & a^{2} \\
b^{2} & a^{2} & 2 a b-c^{2}
\end{array}\right|\) = (a³ + b³ + c³ – 3abc)²
Solution:

Question 3.
Show that \(\left|\begin{array}{ccc}
a^{2}+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|\) = (a – 1)³
Solution:

Question 4.
Show that \(\left|\begin{array}{ccc}
a & b & c \\
a^{2} & b^{2} & c^{2} \\
a^{3} & b^{3} & c^{3}
\end{array}\right|\) = abc(a – b)(b – c)(c – a)
Solution:

Question 5.
Show that \(\left|\begin{array}{ccc}
-2 \mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{c}+\mathbf{a} \\
\mathbf{a}+\mathbf{b} & -\mathbf{2} \mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{c}+\mathbf{a} & \mathbf{c}+\mathbf{b} & -2 \mathbf{c}
\end{array}\right|\) = 4(a + b) (b + c) (c + a)
Solution:

∴ (c + a) is a factor for ∆
Similarly a + b, b + c are also factors ∆.
∵ ∆ is a third-degree expression in a, b, c.
∆ = k(a + b) (b + c) (c + a),
where k is a non-zero scalar.
Put a = 1, b = 1, c = 1, then
\(\left|\begin{array}{ccc}
-2 & 2 & 2 \\
2 & -2 & 2 \\
2 & 2 & -2
\end{array}\right|\) = k(1 + 1) (1 + 1) (1 + 1)
⇒ -2(4 – 4) – 2(-4 – 4) + 2(4 + 4) = 8k
⇒ 16 + 16 = 8k
⇒ k = 4
∴ ∆ = 4(a + b) (b + c) (c + a)
Hence \(\left|\begin{array}{ccc}
-2 \mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{c}+\mathbf{a} \\
\mathbf{a}+\mathbf{b} & -\mathbf{2} \mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{c}+\mathbf{a} & \mathbf{c}+\mathbf{b} & -2 \mathbf{c}
\end{array}\right|\) = 4(a + b) (b + c) (c + a)

Question 6.
Show that \(\left|\begin{array}{lll}
\mathbf{a}-\mathbf{b} & \mathbf{b}-\mathbf{c} & \mathbf{c}-\mathbf{a} \\
\mathbf{b}-\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{c}-\mathbf{a} & \mathbf{a}-\mathbf{b} & \mathbf{b}-\mathbf{c}
\end{array}\right|\)
Solution:
L.H.S = \(\left|\begin{array}{ccc}
0 & 0 & 0 \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\) = 0
By R₁ → R₁ + (R₂ + R₃)

Question 7.
Show that \(\left|\begin{array}{ccc}
1 & a & a^{2}-b c \\
1 & b & b^{2}-c a \\
1 & c & c^{2}-a b
\end{array}\right|\) = 0
Solution:

Question 8.
Show that \(\left|\begin{array}{lll}
\mathbf{x} & \mathbf{a} & \mathbf{a} \\
\mathbf{a} & \mathbf{x} & \mathbf{a} \\
\mathbf{a} & \mathbf{a} & \mathbf{x}
\end{array}\right|\) = (x + 2a) (x – a)²
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(c)

I.

Question 1.
If A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
-1 & 1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & 1 & -2
\end{array}\right]\), then find (AB’)’.
Solution:

Question 2.
If A = \(\left[\begin{array}{cc}
-2 & 1 \\
5 & 0 \\
-1 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-2 & 3 & 1 \\
4 & 0 & 2
\end{array}\right]\) then find 2A + B’ and 3B’ – A.
Solution:

Question 3.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
-5 & 3
\end{array}\right]\), then find A + A’ and A.A’
Solution:

Question 4.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
2 & 5 & 6 \\
3 & x & 7
\end{array}\right]\) is a symmetric matrix, then find x.
Hint: ‘A’ is a symmetric matrix ⇒ A^T = A
Solution:
A is a symmetric matrix
⇒ A’ = A

Equating 2nd row, 3rd column elements we get x = 6.

Question 5.
If A = \(\left[\begin{array}{ccc}
0 & 2 & 1 \\
-2 & 0 & -2 \\
-1 & x & 0
\end{array}\right]\) is a skew-symmetric matrix, find x.
Solution:
∵ A is a skew-symmetric matrix
⇒ A^T = -A

Equating second-row third column elements we get x = 2.

Question 6.
Is \(\left[\begin{array}{ccc}
0 & 1 & 4 \\
-1 & 0 & 7 \\
-4 & -7 & 0
\end{array}\right]\) symmetric or skewsymmetric?
Solution:

∴ A is a skew-symmetric matrix.

II.

Question 1.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that A.A’ = A’. A = I₂
Solution:

From (1), (2) we get A.A’ = A’. A = I₂

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & 5 & 3 \\
2 & 4 & 0 \\
3 & -1 & -5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
2 & -1 & 0 \\
0 & -2 & 5 \\
1 & 2 & 0
\end{array}\right]\) then find 3A – 4B’.
Solution:

Question 3.
If A = \(\left[\begin{array}{cc}
7 & -2 \\
-1 & 2 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-2 & -1 \\
4 & 2 \\
-1 & 0
\end{array}\right]\) then find AB’ and BA’.
Solution:

Question 4.
For any square matrix A, Show that AA’ is symmetric.
Solution:
A is a square matrix
(AA’)’ = (A’)’A’ = A.A’
∵ (AA’)’ = AA’
⇒ AA’ is a symmetric matrix.

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(b)

I.

Question 1.
Find the following products wherever possible.
Hint: (1 × 3) by (3 × 1) = 1 × 1
(i) \(\left[\begin{array}{lll}
-1 & 4 & 2
\end{array}\right]\left[\begin{array}{l}
5 \\
1 \\
3
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
2 & 1 & 4 \\
6 & -2 & 3
\end{array}\right]\left[\begin{array}{l}
1 \\
2 \\
1
\end{array}\right]\)
(iii) \(\left[\begin{array}{cc}
3 & -2 \\
1 & 6
\end{array}\right]\left[\begin{array}{cc}
4 & -1 \\
2 & 5
\end{array}\right]\)
(iv) \(\left[\begin{array}{lll}
2 & 2 & 1 \\
1 & 0 & 2 \\
2 & 1 & 2
\end{array}\right]\left[\begin{array}{ccc}
-2 & -3 & 4 \\
2 & 2 & -3 \\
1 & 2 & -2
\end{array}\right]\)
(v) \(\left[\begin{array}{ccc}
3 & 4 & 9 \\
0 & -1 & 5 \\
2 & 6 & 12
\end{array}\right]\left[\begin{array}{ccc}
13 & -2 & 0 \\
0 & 4 & 1
\end{array}\right]\)
(vi) \(\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]\left[\begin{array}{ccc}
2 & 1 & 4 \\
6 & -2 & 3
\end{array}\right]\)
(vii) \(\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
(viii) \(\left[\begin{array}{ccc}
0 & c & -b \\
-c & 0 & a \\
b & -a & 0
\end{array}\right]\left[\begin{array}{ccc}
a^{2} & a b & a c \\
a b & b^{2} & b c \\
a c & b c & c^{2}
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{ccc}
3 & 4 & 9 \\
0 & -1 & 5 \\
2 & 6 & 12
\end{array}\right]\left[\begin{array}{ccc}
13 & -2 & 0 \\
0 & 4 & 1
\end{array}\right]\)
First matrix is a 3 × 3 matrix and second matrix is 2 × 3 matrix.
No. of columns in the first matrix ≠ No. of rows in the second matrix.
∴ Matrix product is not possible.

(vi) \(\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]\left[\begin{array}{ccc}
2 & 1 & 4 \\
6 & -2 & 3
\end{array}\right]\)
No. of columns in first matrix = 1
No. of rows in second matrix = 2
No. of columns in the first matrix ≠ No. of rows in the second matrix
Multiplication of matrices is not possible.

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\), do AB and BA exist? If they exist, find them. Do A and B commute with respect to multiplication?
Solution:

AB ≠ BA
∴ A and B are not commutative with respect to the multiplication of matrices.

Question 3.
Find A² where A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\)
Solution:

Question 4.
If A = \(\left[\begin{array}{ll}
i & 0 \\
0 & i
\end{array}\right]\), find A².
Solution:

Question 5.
If A = \(\left[\begin{array}{cc}
i & 0 \\
0 & -i
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
0 & \mathbf{i} \\
\mathbf{i} & \mathbf{0}
\end{array}\right]\), and I is the unit matrix of order 2, then show that
(i) A² = B² = C² = -I
(ii) AB = -BA = -C
Solution:

Question 6.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\), find AB. Find BA if it exists.
Solution:
Given A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\)

The order of AB is 2 × 3
BA does not exist since no. of columns in B ≠ No. of rows in A.

Question 7.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & k
\end{array}\right]\) and A² = 0, then find the value of k.
Solution:

II.

Question 1.
If A = \(\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]\) then find A⁴.
Solution:

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right]\) then find A³.
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & -1 \\
3 & -1 & 1
\end{array}\right]\), then find A³ – 3A² – A – 3I, where I is unit matrix of order 3 × 3.
Solution:

Question 4.
If I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) and E = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\), show that (aI + bE)³ = a³I + 3a²bE, Where I is unit matrix of order 2.
Solution:

III.

Question 1.
If A = [a₁, a₂, a₃,], then for any integer n ≥ 1 show that Aⁿ = \(\left[\begin{array}{lll}
a_{1}, & a_{2}^{n}, & a_{3}^{n}
\end{array}\right]\)
Solution:
Given A = diag [a₁, a₂, a₃,] = \(\left[\begin{array}{ccc}
a_{1} & 0 & 0 \\
0 & a_{2} & 0 \\
0 & 0 & a_{3}
\end{array}\right]\)
Aⁿ = diag \(\left[\begin{array}{lll}
a_{1}^{n} & a_{2}^{n} & a_{3}^{n}
\end{array}\right]=\left[\begin{array}{ccc}
a_{1}^{n} & 0 & 0 \\
0 & a_{2}^{n} & 0 \\
0 & 0 & a_{3}^{n}
\end{array}\right]\)
This problem can be should by using Mathematical Induction
put n = 1
A¹ = \(\left[\begin{array}{ccc}
a_{1} & 0 & 0 \\
0 & a_{2} & 0 \\
0 & 0 & a_{3}
\end{array}\right]\)
∴ The result is true for n = 1
Assume the result is true for n = k
A^k = \(\left[\begin{array}{ccc}
a_{1}^{k} & 0 & 0 \\
0 & a_{2}^{k} & 0 \\
0 & 0 & a_{3}^{k}
\end{array}\right]\)
Consider

∴ The result is true for n = k + 1
Hence by the Principle of Mathematical Induction, the statement is true ∀ n ∈ N

Question 2.
If θ – φ = \(\frac{\pi}{2}\), then show that \(\left[\begin{array}{cc}
\cos ^{2} \theta & \cos \theta \sin \theta \\
\cos \theta \sin \theta & \sin ^{2} \dot{\theta}
\end{array}\right]\) \(\left[\begin{array}{cc}
\cos ^{2} \phi & \cos \phi \sin \phi \\
\cos \phi \sin \phi & \sin ^{2} \phi
\end{array}\right]\) = 0
Solution:

Question 3.
If A = \(\left[\begin{array}{rr}
3 & -4 \\
1 & -1
\end{array}\right]\) then show that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), for any integer n ≥ 1 by using Mathematical Induction.
Solution:
We shall prove the result by Mathematical Induction.

∴ The given result is true for n = k + 1
By Mathematical Induction, the given result is true for all positive integral values of n.

Question 4.
Give examples of two square matrices A and B of the same order for which AB = 0 but BA ≠ 0.
Solution:

Question 5.
A Trust fund has to invest ₹ 30,000 in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds if the trust fund must obtain an annual total interest of (a) ₹ 1800 (b) ₹ 2000
Solution:
Let the first bond be ‘x’ and the second bond be 30,000 – x respectively
The rate of interest is 0.05 and 0.07 respectively.
(a) \([x, 30,000-x]\left[\begin{array}{l}
0.05 \\
0.07
\end{array}\right] \quad=[1800]\)
[0.05x + 0.07(30,000 – x)] = 1800
\(\frac{5}{100} x+\frac{7}{100}(30,000-x)=1800\)
5x + 21,0000 – 7x = 1,80,000
-2x = 1,80,000 – 2,10,000 = -30,000
x = 15,000
∴ First bond = 15,000
Second bond = 30,000 – 15,000 = 15,000

(b) \(\left[\begin{array}{ll}
x & 30,000-x
\end{array}\right]\left[\begin{array}{l}
0.05 \\
0.07
\end{array}\right]=[2000]\)
[0.05x + 0.07(30,000 – x)] = [2000]
\(\frac{5 x}{100} \times \frac{7}{100}(30,000-x)=2000\)
5x + 2,10,000 – 7x = 2,00,000
-2x = 2,00,000 – 2,10,000
-2x = -10,000
x = 5,000
∴ First bond = 5000
Second bond = 30,000 – 5000 = 25,000

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(a)

I.

Question 1.
Write the following as a single matrix.
(i) \(\left[\begin{array}{lll}
2 & 1 & 3
\end{array}\right]+\left[\begin{array}{lll}
1 & 0 & 0
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
3 & 9 & 0 \\
1 & 8 & -2
\end{array}\right]+\left[\begin{array}{ccc}
4 & 0 & 2 \\
7 & 1 & 4
\end{array}\right]\)
(iii) \(\left[\begin{array}{c}
0 \\
1 \\
-1
\end{array}\right]+\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\)
(iv) \(\left[\begin{array}{cc}
-1 & 2 \\
2 & -2 \\
3 & 1
\end{array}\right]-\left[\begin{array}{cc}
0 & 1 \\
-1 & 0 \\
-2 & 1
\end{array}\right]\)
Solution:

Question 2.
If A = \(\left[\begin{array}{cc}
-1 & 3 \\
4 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
3 & -5
\end{array}\right]\), X = \(\left[\begin{array}{ll}
x_{1} & x_{2} \\
x_{3} & x_{4}
\end{array}\right]\) and A + B = X, then find the values of x₁, x₂, x₃ and x₄.
Solution:
A + B = X

∴ x₁ = 1, x₂ = 4, x₃ = 7, x₄ = -3

Question 3.
If A = \(\left[\begin{array}{ccc}
-1 & -2 & 3 \\
1 & 2 & 4 \\
2 & -1 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -2 & 5 \\
0 & -2 & 2 \\
1 & 2 & -3
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
-2 & 1 & 2 \\
1 & 1 & 2 \\
2 & 0 & 1
\end{array}\right]\) then find A + B + C.
Solution:

Question 4.
If A = \(\left[\begin{array}{ccc}
3 & 2 & -1 \\
2 & -2 & 0 \\
1 & 3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & -1 & 0 \\
2 & 1 & 3 \\
4 & -1 & 2
\end{array}\right]\) and X = A + B then find X.
Solution:

Question 5.
If \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]=\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\) then find the values of x, y, z and a.
Solution:
Given \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]=\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\)
∴ x – 3 = 5 ⇒ x = 3 + 5 = 8
2y – 8 = 2 ⇒ 2y = 8 + 2 = 10 ⇒ y = 5
z + 2 = -2 ⇒ z = -2 – 2 = -4
a – 4 = 6 ⇒ a = 4 + 6 = 10

II.

Question 1.
If \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\) then find the values of x, y, z and a.
Solution:
Given \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\)
∴ x – 1 = 1 ⇒ x = 1 + 1 = 2
5 – y = 3 ⇒ y = 5 – 3 = 2
z – 1 = 4 ⇒ z = 4 + 1 = 5
a – 5 = 0 ⇒ a = 5

Question 2.
Find the trace of \(\left[\begin{array}{ccc}
1 & 3 & -5 \\
2 & -1 & 5 \\
1 & 0 & 1
\end{array}\right]\)
Solution:
Trace of \(\left[\begin{array}{ccc}
1 & 3 & -5 \\
2 & -1 & 5 \\
1 & 0 & 1
\end{array}\right]\) = Sum of the diagonal elements
= 1 – 1 + 1
= 1

Question 3.
If A = \(\left[\begin{array}{rrr}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & -6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) find B – A and 4A – 5B.
Solution:

Question 4.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\) find 3B – 2A.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Mathematical Induction Solutions Exercise 2(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Mathematical Induction Solutions Exercise 2(a)

Using mathematical induction, prove each of the following statements for all n ∈ N.

Question 1.
1² + 2² + 3² + …… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let p(n) be the given statement:
1² + 2² + 3² + ….. + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Since 1² = \(\frac{(1)(1+1)(2 \times 1+1)}{6}\)
⇒ 1 = 1 the formula is true for n = 1
i.e., p(1) is true.
Assume the statement p(n) is true for n = k
i.e., 1² + 2² + 3² + …… + 1k² = \(\frac{k(k+1)(2 k+1)}{6}\)
We show that the formula is true for n = k + 1
i.e., We show that p(k + 1) = \(\frac{(k+1)(k+2)(2 k+3)}{6}\)
(Where p(k) = 1² + 2² + 3² + … + k²)
We observe that
p(k + 1) = 1² + 2² + 3² + …… + (k)² + (k + 1)² = p(k) + (k + 1)²
Since p(k) = \(\frac{k(k+1)(2 k+1)}{6}\)
We have p(k + 1) = p(k) + (k + 1)²

∴ The formula holds for n = k + 1
∴ By the principle of mathematical induction, p(n) is true for all n ∈ N
i.e., the formula 1² + 2² + 3² + ……. + n² = \(\frac{n(n+1)(2 n+1)}{6}\) for all n ∈ N

Question 2.
2.3 + 3.4 + 4.5 + …… up to n terms = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)
Solution:
The nth term in the given series is (n + 1) (n + 2)
Let p(n) be the statement:
2.3 + 3.4 + 4.5 + …… + (n + 1) (n + 2) = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)
and let S(n) be the sum on the left-hand side.
Since S(1) = 2.3 = \(\frac{(1)(1+6+11)}{3}\) = 6
∴ The statement is true for n = 1
Assume that the statement p(n) is true for n = k
i.e., S(k) = 2.3 + 3.4 + …… + (k + 1) (k + 2) = \(\frac{k\left(k^{2}+6 k+11\right)}{3}\)
We show that the statement is true for n = k + 1
i.e., We show that S(k + 1) = \((k+1)\left[\frac{(k+1)^{2}+6(k+1)+11}{3}\right]\)
We observe that
S(k + 1) = 2.3 + 3.4 + 4.5 + + (k + 1) (k + 2) + (k + 2) (k + 3)
= S(k) + (k + 2) (k + 3)

∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
i.e., 2.3 + 3.4 + 4.5 + ……. + (n + 1) (n + 2) = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)

Question 3.
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)
Solution:
Let p(n) be the statement:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)
and let S(n) be the sum on the L.H.S.
Since S(1) = \(\frac{1}{1.3}=\frac{1}{1(2+1)}=\frac{1}{1.3}\)
∴ P(1) is true.
Assume that the statement p(n) is true for n = k
i.e., S(k) = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 k-1)(2 k+1)}\) = \(\frac{k}{2 k+1}\)
We show that the statement p(n) is true for n = k + 1
i.e., we show that s(k + 1) = \(\frac{k+1}{2(k+1)+1}\)
We observe that

∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
i.e., \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)

Question 4.
4³ + 8³ + 12³ + …… up to n terms = 16n²(n + 1)².
Solution:
4, 8, 12,….. are in A.P., whose nth term is (4n)
Let p(n) be the statement:
4³ + 8³ + 12³ + ………. + (4n)³ = 16n²(n + 1)²
and S(n) be the sum on the L.H.S.
S(1) = 4³ = 16(1²) (1 + 1)² = 16(4) = 64 = 4³
∴ p(1) is true
Assume that the statement p(n) is true for n = k
i.e., S(k) = 4³ + 8³ + (12)³ + …… + (4k)³ = 16k²(k + 1)²
We show that the statement is true for n = k + 1
i.e., We show that S(k + 1) = 16(k + 1)² (k + 2)²
We observe that
S(k + 1) = 4³ + 8³ + 12³ + …… + (4k)³ + [4(k + 1)]³
= S(k) + [4(k + 1)]³
= 16k² (k + 1)² + 4³ (k + 1)³
= 16(k + 1)² [k² + 4(k + 1)]
= 16(k + 1)² [k² + 4k + 4]
= 16(k + 1)² (k + 2)²
= 16(k + 1)² \((\overline{k+1}+1)^{2}\)
∴ The formula holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e.,) 4³ + 8³ + 12³ + …… + (4n)³ = 16n²(n + 1)²

Question 5.
a + (a + d) + (a + 2d) + ……. up to n terms = \(\frac{n}{2}\) [2a + (n – 1)d]
Solution:
Let p(n) be the statement:
a + (a + d) + (a + 2d) + …… + [a + (n – 1)d] = \(\frac{n}{2}\) [2a + (n – 1)d]
and let the sum on the L.H.S. is denoted by S(n)
Since S(1) = a = \(\frac{1}{2}\) [2a + (1 – 1)d] = a
∴ p(1) is true.
Assume that the statement is true for n = k
(i.e.,) S(k) = a + (a + d) + (a + 2d) + ……. + [a + (k – 1)d] = \(\frac{k}{2}\) [2a + (k – 1 )d]
We show that the statement is true for n = k + 1
(i.e.,) we show that S(k + 1) = \(\left(\frac{k+1}{2}\right)[2 a+k d]\)
We observe that
S(k + 1) = a + (a + d) + (a + 2d) + …… + [a + (k – 1)d] + (a + kd)
= S(k) + (a + kd)
= \(\frac{k}{2}\) [2a + (k – 1)d] + (a + kd)
= \(\frac{k[2 a+(k-1) d]+2(a+k d)}{2}\)
= \(\frac{1}{2}\) [2ak + k(k – 1)d + 2a + 2kd]
= \(\frac{1}{2}\) [2a(k + 1) + k(k – 1 + 2)d]
= \(\frac{1}{2}\) (k + 1)(2a + kd)
∴ The statement holds for n = k + 1
∴ By the principle of mathematical inductions,
p(n) is true for all n ∈ N
(i.e.,) a + (a + d) + (a + 2d) + …… + [a + (n – 1)d] = \(\frac{n}{2}\) [2a + (n – 1)d]

Question 6.
a + ar + ar² + ……… up to n terms = \(\frac{a\left(r^{n}-1\right)}{r-1}\); r ≠ 1
Solution:
Let p(n) be the statement:
a + ar + a.r² + …… + a. r^n-1 = \(\frac{a\left(r^{n}-1\right)}{r-1}\), r ≠ 1
and let S(n) be the sum on the L.H.S
Since S(1) = a = \(\frac{a\left(r^{1}-1\right)}{r-1}\) = a
∴ p(1) is true
Assume that the statement is true for n = k
(i.e) S(k) = a + ar + ar² + ……… + a . r^k-1 = \(\frac{a\left(r^{k}-1\right)}{r-1}\)
We show that the statement is true for n = k + 1
(i.e) S(k + 1) = \(\frac{a\left(r^{k+1}-1\right)}{r-1}\)
Now S(k + 1) = a + ar + ar² + ……. + a r^k-1 + ar^k
= S(k) + a . r^k

∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e) a + ar + ar² + ……. + a.r^n-1 = \(\frac{a\left(r^{n}-1\right)}{r-1}\), r ≠ 1

Question 7.
2 + 7 + 12 + ……. + (5n – 3) = \(\frac{n(5 n-1)}{2}\)
Solution:
Let p(n) be the statement:
2 + 7 + 12 + ……. + (5n – 3) = \(\frac{n(5 n-1)}{2}\)
and let S(n) be the sum on the L.H.S
Since S(1) = 2 = \(\frac{1(5 \times 1-1)}{2}=\frac{4}{2}\) = 2
∴ p(1) is true
Assume that the statement is true for n = k
(i.e) S(k) = 2 + 7 + 12 + …….. + (5k – 3) = \(\frac{k(5 k-1)}{2}\)
We have to show that S(k + 1) = \(\frac{(k+1)(5 k+4)}{2}\)
We observe that S(k + 1) = 2 + 7 + 12 + ……. + (5k – 3) + (5k + 2)
= S(k) + (5k + 2)
= \(\frac{k(5 k-1)}{2}\) + (5k + 2)
= \(\frac{5 k^{2}-k+2(5 k+2)}{2}\)
= \(\frac{1}{2}\) [5k² + 9k + 4]
= \(\frac{1}{2}\) (k + 1) (5k + 4)
= \(\frac{1}{2}\) (k + 1) [5(k + 1) – 1]
∴ p(k + 1) is true
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N.
(i.e.,) 2 + 7 + 12 + …… + (5n – 3) = \(\frac{n(5 n-1)}{2}\)

Question 8.
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\frac{2 n+1}{n^{2}}\right)\) = (n + 1)²
Solution:
Let p(n) be the statement:
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\frac{2 n+1}{n^{2}}\right)\) = (n + 1)²
and let S(n) be the product on the LHS
since S(1) = 1 + 3 = 4 = (1 + 1)² = 4
∴ P(a) 4 time for n = 1
Assume that p(n) is true for n = k

= (k + 1)² + 2k + 3
= k² + 2k + 1 + 2k + 3
= k² + 4k + 4
= (k + 2)²
= (k + 1 + 1)²
∴ P(n) is true for n = k + 1
By the principle of Mathematical Induction,
p(n) is true & n ∈ N

Question 9.
(2n + 7) < (n + 3)²
Solution:
Let p (n) be the statement
When n = 1, 9 < 16
∴ p(n) is true for n = 1
Assume p (n) is true for n = k
(2k + 7) < (k + 3)²
We show that p(n) is true for n = k + 1
2(k + 1) + 7 = 2k + 7 + 2
< (k + 3)² + 2
< k² + 6k + 9 + 2 + 2k + 5 – 2k – 5
< (k + 4)² – (2k + 5)
< (k + 4)²
< (k + 1 + 3)²
∴ p(n) is true for n = k + 1
By the principle of Mathematical Induction
p(n) is true ∀ n ∈ N

Question 10.
1² + 2² + …… + n² > \(\frac{n^{3}}{3}\)
Solution:
Let P(n) by the statement
when n = 1, 1 > \(\frac{1}{3}\)
∴ p(n) is true for n = 1
Assume p (n) is true for n = k
1² + 2² + …… + k² > \(\frac{k^{3}}{3}\)
We show that p(n) is true for n = k + 1

∴ p(n) is true for n = k + 1
By the principle of Mathematical Induction,
p(n) is true ∀ n ∈ N

Question 11.
4ⁿ – 3n – 1 is divisible by 9.
Solution:
Let p(n) be the statement:
4ⁿ – 3n – 1 is divisible by 9
Since 4¹ – 3(1) – 1 = 0 is divisible by 9.
The statement is true for n = 1
Assume that p(n) is true for n = k
(i.e) 4^k – 3k – 1 is divisible by 9
Then 4^k – 3k – 1 = 9t, for some t ∈ N ……..(1)
Show that the statement p(n) is true for n = k + 1
(i.e.,) we show that S(k + 1) = 4^k+1 – 3(k+1) – 1 is divisible by 9
From (1), we have
4^k = 9t + 3k + 1
∴ S(k + 1) = 4 . 4^k – 3(k + 1) – 1
= 4(9t + 3k + 1) – 3k – 3 – 1
= 4(9t) + 9k
= 9[4t + k]
Hence s(k + 1) is divisible by 9
Since 4t + k is an integer
∴ 4^k+1 – 3(k+1) – 1 is divisible by 9
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ k
(i.e.,) 4ⁿ – 3n – 1 is divisible by 9

Question 12.
3 . 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17.
Solution:
Let p(n) be the statement:
3. 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17
Since 3 . 5²⁽¹⁾⁺¹ + 2³⁽¹⁾⁺¹
= 3 . 5³ + 2⁴
= 3(125) + 16
= 375 + 16
= 391
= 17(23) is divisible by 17
∴ The statement is true for n = 1
Assume that the statement is true for n = k
(i.e) 3 . 5^2k+1 + 2^3k+1 is divisible by 17
Then 3 . 5^2k+1 + 2^3k+1 = 17t, for some t ∈ N ……..(1)
Show that the statement p(n) is true for n = k + 1
(i.e.,) We have to show that
\(\text { 3. } 5^{2(k+1)+1}+2^{3(k+1)+1}\) is divisible by 17
From (1) we have

Here 25t + 2^3k+1 is an integer
∴ \(\text { 3. } 5^{2(k+1)+1}+2^{3(k+1)+1}\) is divisible by 17
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e.,) 3 . 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17.

Question 13.
1.2.3 + 2.3.4 + 3.4.5 + ……. upto n terms = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
Solution:
The nth term of the given series is (n) (n + 1) (n + 2)
Let p(n) be the statement:
1.2.3 + 2.3.4 + 3.4.5 +……. + (n) (n+1) (n+2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
and S(n) be the sum on the L.H.S.
∵ S(1) = 1.2.3 = \(\frac{(1)(1+1)(1+2)(1+3)}{4}\) = 1.2.3
∴ p(1) is true
Assume that the statement p(n) is true for n = k
(i.e) S(k) = 1.2.3 + 2.3.4 + 3.4.5 + ……. + k(k + 1) (k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\)
We show that the statement is true for n = k + 1
(i.e) We show that S(k + 1) = \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
We observe that
S(k + 1) = 1.2.3 + 2.3.4 + …… + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)
= S(k) + (k + 1) (k + 2) (k + 3)
= \(\frac{k(k+1)(k+2)(k+3)}{4}\) + (k + 1)(k + 2)(k + 3)
= (k + 1)(k + 2)(k + 3) \(\left(\frac{k}{4}+1\right)\)
= \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e.,) 1.2.3 + 2.3.4 + 3.4.5 + ……. + (n)(n + 1)(n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)

Question 14.
\(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}\) + …. up to n terms = \(\frac{n}{24}\) [2n² + 9n + 13]
Solution:
The nth term of the given series is \(\frac{1^{3}+2^{3}+3^{3}+\ldots .+n^{3}}{1+3+5+\ldots+(2 n-1)}\)
Let p(n) be the statement :

and let S(n) be the sum on the L.H.S.
∵ S(1) = \(\frac{1^{3}}{1}=\frac{1}{24}(2+9+13)=1=\frac{1^{3}}{1}\)
∴ p(1) is true
Assume that p(k) is true
(i.e.,) S(k) = \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\ldots+\frac{1^{3}+2^{3}+\ldots \pm k^{3}}{1+3+\ldots+(2 k-1)}\) = \(\frac{k}{24}\) [2k² + 9k + 13]
We show that p(k + 1) is true
(i.e,) we show that


∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n
(i.e.,) \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\ldots+\frac{1^{3}+2^{3}+\ldots \ldots+n^{3}}{1+3+\ldots+(2 n-1)}\) = \(\frac{n}{24}\) [2n² + 9n + 13]

Question 15.
1² + (1² + 2²) + (1² + 2² + 3²) + ……. up to n terms = \(\frac{n(n+1)^{2}(n+2)}{12}\)
Solution:
The nth term of the given series is (1² + 2² + 3² + …… + n²)
Let p(n) be the statement:
1² + (1² + 2²) + (1² + 2² + 3²) + ……. + (1² + 2² + …… + n²) = \(\frac{\mathrm{n}(\mathrm{n}+1)^{2}(\mathrm{n}+2)}{12}\)
and the sum on the LH.S. is denoted by S(n).
Since S(1) = 1² = \(\frac{1(1+1)^{2}(1+2)}{12}\) = 1 = 1²
∴ p(1) is true.
Assume that the statement is true for n = k
(i.e.,) S(k) = 1² + (1² + 2²) + ……. + (1² + 2² + ……. + k²)
= \(\frac{k(k+1)^{2}(k+2)}{12}\)
We show that S(k + 1) = \(\frac{(k+1)(k+2)^{2}(k+3)}{12}\)
We observe that


∴ The statement holds for n = k + 1.
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N.
(i.e.,) 1² + (1² + 2²) + (1² + 2² + 3²) + …….. (1² + 2² + ………. + n²) = \(\frac{n(n+1)^{2}(n+2)}{12}\)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(c)

I.

Question 1.
Find the domains of the following real-valued functions.
(i) f(x) = \(\frac{1}{\left(x^{2}-1\right)(x+3)}\)
Solution:
f(x) = \(\frac{1}{\left(x^{2}-1\right)(x+3)}\) ∈ R
⇔ (x² – 1) (x + 3) ≠ 0
⇔ (x + 1) (x – 1) (x + 3) ≠ 0
⇔ x ≠ -1, 1, -3
∴ Domain of f is R – {-1, 1, -3}

(ii) f(x) = \(\frac{2 x^{2}-5 x+7}{(x-1)(x-2)(x-3)}\)
⇔ (x – 1) (x – 2) (x – 3) ≠ 0
⇔ x ≠ 1, x ≠ 2, x ≠ 3
∴ Domain of f is R – {1, 2, 3}

(iii) f(x) = \(\frac{1}{\log (2-x)}\)
Solution:
f(x) = \(\frac{1}{\log (2-x)}\)
⇔ log (2 – x) ≠ 0 and 2 – x > 0
⇔ (2 – x) ≠ 1 and 2 > x
⇔ x ≠ 1 and x < 2
x ∈ (-∞, 1) ∪ (1, 2) (or) x ∈ (-∞, 2) – {1}
∴ Domain of f is {(-∞, 2) – {1}}

(iv) f(x) = |x – 3|
Solution:
f(x) = |x – 3| ∈ R
⇔ x ∈ R
∴ The domain of f is R

(v) f(x) = \(\sqrt{4 x-x^{2}}\)
Solution:
f(x) = \(\sqrt{4 x-x^{2}}\) ∈ R
⇔ 4x – x² ≥ 0
⇔ x(4 – x) ≥ 0
⇔ x ∈ [0, 4]
∴ Domain of f is [0, 4]

(vi) f(x) = \(\frac{1}{\sqrt{1-x^{2}}}\)
Solution:
f(x) = \(\frac{1}{\sqrt{1-x^{2}}}\) ∈ R
⇔ 1 – x² > 0
⇔ (1 + x) (1 – x) > 0
⇔ x ∈ (-1, 1)
∴ Domain of f is {x/x ∈ (-1, 1)}

(vii) f(x) = \(\frac{3^{x}}{x+1}\)
Solution:
f(x) = \(\frac{3^{x}}{x+1}\) ∈ R
⇔ 3^x ∈ R, ∀ x ∈ R and x + 1 ≠ 0
⇔ x ≠ -1
∴ Domain of f is R – {-1}

(viii) f(x) = \(\sqrt{x^{2}-25}\)
Solution:
f(x) = \(\sqrt{x^{2}-25}\) ∈ R
⇔ x² – 25 ≥ 0
⇔ (x + 5) (x – 5) ≥ 0
⇔ x ∈ (-∞, -5] ∪ [5, ∞)
⇔ x ∈ R – (-5, 5)
∴ Domain of f is R – (- 5, 5)

(ix) f(x) = \(\sqrt{x-[x]}\)
Solution:
f(x) = \(\sqrt{x-[x]}\) ∈ R
⇔ x – [x] ≥ 0
⇔ x ≥ [x]
⇔ x ∈ R
∴ Domain of f is R.

(x) f(x) = \(\sqrt{[x]-x}\)
Solution:
f(x) = \(\sqrt{[x]-x}\) ∈ R
⇔ [x] – x ≥ 0
⇔ [x] ≥ x
⇔ x ≤ [x]
⇔ x ∈ Z
∴ The domain of f is z (Where z denotes a set of integers)

Question 2.
Find the ranges of the following real-valued functions.
(i) log|4 – x²|
Solution:
Let y = f(x) = log|4 – x²|
f(x) ∈ R
⇔ 4 – x² ≠ 0
⇔ x ≠ ±2
∵ y = log|4 – x²|
⇒ |4 – x²| = e^y
∵ e^y > 0 ∀ y ∈ R
∴ The range of f is R.

(ii) \(\sqrt{[x]-x}\)
Solution:
Let y = f(x) = \(\sqrt{[x]-x}\)
f(x) ∈ R
⇔ [x] – x ≥ 0
⇔ x ≤ [x]
⇔ x ∈ z
∴ Domain of f is z. Then range of f is {0}

(iii) \(\frac{\sin \pi[x]}{1+[x]^{2}}\)
Solution:
Let f(x) = \(\frac{\sin \pi[x]}{1+[x]^{2}}\) ∈ R
⇔ x ∈ R
∴ The domain of f is R
For x ∈ R, [x] is an integer,
sin π[x] = 0, ∀ x ∈ R [∵ sin nπ = 0, ∀ n ∈ z]
∴ Range of f is {0}

(iv) \(\frac{x^{2}-4}{x-2}\)
Solution:
Let y = f(x) = \(\frac{x^{2}-4}{x-2}\) ∈ R
⇔ y = \(\frac{(x+2)(x-2)}{x-2}\)
⇔ x ≠ 2
∴ The domain of f is R – {2}
Then y = x + 2, [∵ x ≠ 2 ⇒ y ≠ 4]
Then its range R – {4}

(v) \(\sqrt{9+x^{2}}\)
Solution:
Let y = f(x) = \(\sqrt{9+x^{2}}\) ∈ R
The domain of f is R
When x = 0, f(0) = √9 = 3
For all values of x ∈ R – {0}, f(x) > 3
∴ The range of f is [3, ∞)

Question 3.
If f and g are real-valued functions defined by f(x) = 2x – 1 and g(x) = x² then find
(i) (3f – 2g)(x)
(ii) (fg) (x)
(iii) \(\left(\frac{\sqrt{f}}{g}\right)(x)\)
(iv) (f + g + 2) (x)
Solution:
(i) (3f – 2g)(x)
f(x) = 2x – 1, g(x) = x²
(3f – 2g) (x) = 3f(x) – 2g(x)
= 3(2x – 1) – 2x²
= -2x² + 6x – 3

(ii) (fg) (x)
= f(x) . g(x)
= (2x – 1) (x²)
= 2x³ – x²

(iii) \(\left(\frac{\sqrt{f}}{g}\right)(x)\)
\(\frac{\sqrt{f(x)}}{g(x)}=\frac{\sqrt{2 x-1}}{x^{2}}\)

(iv) (f + g + 2) (x)
= f(x) + g(x) + 2
= (2x – 1) + x² + 2
= x² + 2x + 1
= (x + 1)²

Question 4.
If f = {(1, 2), (2, -3), (3, -1)} then find
(i) 2f
(ii) 2 + f
(iii) f²
(iv) √f
Solution:
Given f = {(1, 2), (2, -3), (3, -1)}
(i) 2f = {(1, 2 × 2), (2, 2(-3), (3, 2(-1))}
= {(1, 4), (2, -6), (3, -2)}

(ii) 2 + f = {(1, 2 + 2), (2, 2 + (-3)), (3, 2 + (-1)}
= {(1, 4), (2, -1), (3, 1)}

(iii) f² = {(1, 22), (2, (-3)2), (3, (-1)2)}
= {(1, 4), (2, 9), (3, 1)}

(iv) √f = {(1, √2)}
∵ √-3 and √-1 are not real

II.

Question 1.
Find the domains of the following real-valued functions.
(i) f(x) = \(\sqrt{x^{2}-3 x+2}\)
Solution:
f(x) = \(\sqrt{x^{2}-3 x+2}\) ∈ R
⇔ x² – 3x + 2 ≥ 0
⇔ (x- 1) (x – 2) ≥ 0
⇔ x ∈ (-∞, 1 ] ∪ [2, ∞]
∴ The domain of f is R – (1, 2)

(ii) f(x) = log(x² – 4x + 3)
Solution:
f(x) = log(x² – 4x + 3) ∈ R
⇔ x² – 4x + 3 > 0
⇔ (x – 1) (x – 3) > 0
⇔ x ∈ (-∞, 1) ∪ (3, ∞)
∴ Domain of f is R – [1, 3]

(iii) f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\)
Solution:
f(x) = \(\frac{\sqrt{2+x}+\sqrt{2-x}}{x}\) ∈ R
⇔ 2 + x ≥ 0, 2 – x ≥ 0, x ≠ 0
⇔ x ≥ -2, x ≤ 2, x ≠ 0
⇔ -2 ≤ x ≤ 2, x ≠ 0
⇔ x ∈ [-2, 2] – {0}
Domain of f is [-2, 2] – {0}

(iv) f(x) = \(\frac{1}{\sqrt[3]{(x-2)} \log _{(4-x)} 10}\)
Solution:
f(x) = \(\frac{1}{\sqrt[3]{(x-2)} \log _{(4-x)} 10}\) ∈ R
⇔ 4 – x > 0, 4 – x ≠ 1 and x – 2 ≠ 0
⇔ x < 4, x ≠ 3, x ≠ 2
∴ Domain of f is (-∞, 4) – {2, 3}

(v) f(x) = \(\sqrt{\frac{4-x^{2}}{[x]+2}}\)
Solution:
f(x) = \(\sqrt{\frac{4-x^{2}}{[x]+2}}\) ∈ R
Case (i) 4 – x² ≥ 0 and [x] + 2 > 0 (or) Case (ii) 4 – x² ≤ 0 and [x] + 2 < 0
Case (i): 4 – x² ≥ 0 and [x] + 2 > 0
⇔ (2 – x) (2 + x) ≥ 0 and [x] > -2
⇔ x ∈ [-2, 2] and x ∈ [-1, ∞]
⇔ x ∈ [-1, 2] ……..(1)
Case (ii): 4 – x² ≤ 0 and [x] + 2 < 0
⇔ (2 + x) (2 – x) ≤ 0 and [x] < – 2
⇔ x ∈ (-∞, -2] ∪ [2, ∞] and x ∈ (-∞, -2)
⇔ x ∈ (-∞, -2) ……(2)
from (1) and (2),
Domain of f is (-∞, -2) ∪ [-1, 2]

(vi) f(x) = \(\sqrt{\log _{0.3}\left(x-x^{2}\right)}\)
Solution:
f(x) = \(\sqrt{\log _{0.3}\left(x-x^{2}\right)}\) ∈ R
Then log_0.3(x – x²) ≥ 0
⇒ x – x² ≤ (0.3)⁰
⇒ x – x² ≤ 1
⇒ -x² + x – 1 ≤ 0
⇒ x² – x + 1 ≥ 0
This is true for all x ∈ R ……..(1)
and x – x² ≥ 0
⇒ x² – x ≤ 0
⇒ x(x – 1) ≤ 0
⇒ x ∈ (0, 1) …….(2)
From (1) and (2)
Domain of f is R ^ (0, 1) = (0, 1)
∴ The domain of f is (0, 1)

(vii) f(x) = \(\frac{1}{x+|x|}\)
Solution:
f(x) = \(\frac{1}{x+|x|}\) ∈ R
⇔ x + |x| ≠ 0
⇔ x ∈ (0, ∞)
∵ |x| = x, if x ≥ 0
|x| = -x, if x < 0
∴ The domain of f is (0, ∞)

Question 2.
Prove that the real valued function f(x) = \(\frac{x}{e^{x}-1}+\frac{x}{2}+1\) is an even function on R \ {0}.
Solution:
f(x) ∈ R, e^x – 1 ≠ 0
⇒ e^x ≠ 1
⇒ x ≠ 0

⇒ f(x) is an even function on R – {0}

Question 3.
Find the domain and range of the following functions.
(i) f(x) = \(\frac{\tan \pi[x]}{1+\sin \pi[x]+\left[x^{2}\right]}\)
Solution:
f(x) = \(\frac{\tan \pi[x]}{1+\sin \pi[x]+\left[x^{2}\right]}\) ∈ R
⇔ x ∈ R, since [x] is an integer tan π[x] and sin π[x] each is zero for ∀ x ∈ R and f(x) ∈ R
Domain of f is R
Its range = {0}

(ii) f(x) = \(\frac{x}{2-3 x}\)
Solution:

(iii) f(x) = |x| + |1 + x|
Solution:
f(x) = |x| + |1 + x| ∈ R
⇔ x ∈ R
∴ Domain of f is R
∵ |x| = x, if x ≥ 0
= -x, if x < 0
|1 + x| = 1 + x, if x ≥ -1
= -(1 + x) if x < -1
For x = 0, f(0) = |0| + |1 + 0| = 1
x = 1, f(1) = |1| + |1 + 1| = 1 + 2 = 3
x = 2, f(2) = |2| + |1 + 2| = 2 + 3 = 5
x = -2, f(-2) = |-2| + |1 + (-2)| = 2 + 1 = 3
x = -1, f(-1) = |-1| + |1 +(-1)| = 1 + 0 = 1
∴ The range of f is [1, ∞]

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(b)

I.

Question 1.
If f(x) = e^x and g(x) = log_ex, then show that f o g = g o f and find f^-1 and g^-1.
Solution:
Given f(x) = e^x and g(x) = log_ex
Now (f o g) (x) = f(g(x))
= f(log_ex) [∵ g(x) = \(\log _{e} x\)]
= \(e^{\left(\log _{e} x\right)}\)
= x
∴ (fog) (x) = x ………(1)
and (g o f) (x) = g(f(x))
= g(e^x) [∵ f(x) = e^x]
= log_e (e^x) [∵ g(x) = log_ex]
= x log_e (e)
= x(1)
= x
∴ (g o f) (x) = x …….(2)
From (1) and (2)
f o g = g o f
Given f(x) = e^x
Let y = f(x) = e^x ⇒ x = f^-1(y)
and y = e^x ⇒ x = log_e (y)
∴ f^-1(y) = log_e (y) ⇒ f^-1(x) = log_e (x)
Let y = g(x) = log_e (x)
∵ y = g(x) ⇒ x = g^-1(y)
∵ y = log_e (x) ⇒ x = e^y
∴ g^-1(y) = e^y ⇒ g^-1(x) = e^x
∴ f^-1(x) = log_e (x) and g^-1(x) = e^x

Question 2.
If f(y) = \(\frac{y}{\sqrt{1-y^{2}}}\), g(y) = \(\frac{y}{\sqrt{1+y^{2}}}\) then show that (fog) (y) = y
Solution:
f(y) = \(\frac{y}{\sqrt{1-y^{2}}}\) and g(y) = \(\frac{y}{\sqrt{1+y^{2}}}\)
Now, (fog) (y) = f(g(y))

∴ (fog) (y) = y

Question 3.
If f : R → R, g : R → R are defined by f(x) = 2x² + 3 and g(x) = 3x – 2, then find
(i) (fog)(x)
(ii) (gof) (x)
(iii) (fof) (0)
(iv) go(fof) (3)
Solution:
f : R → R, g : R → R and f(x) = 2x² + 3; g(x) = 3x – 2
(i) (f o g) (x) = f(g(x))
= f(3x – 2) [∵ g(x) = 3x – 2]
= 2(3x- 2)² + 3 [∵ f(x) = 2x² + 3]
= 2(9x² – 12x + 4) + 3
= 18x² – 24x + 8 + 3
= 18x² – 24x + 11

(ii) (gof) (x) = g(f(x))
= g(2x² + 3) [∵ f(x) = 2x² + 3]
= 3(2x² + 3) – 2 [∵ g(x) = 3x – 2]
= 6x² + 9 – 2
= 6x² + 7

(iii) (fof) (0) = f(f(0))
= f(2(0) + 3) [∵ f(x) = 2x² + 3]
= f(3)
= 2(3)² + 3
= 18 + 3
= 21

(iv) g o (f o f) (3)
= g o (f (f(3)))
= g o (f (2(3)² + 3)) [∵ f(x) = 2x² + 3]
= g o (f(21))
= g(f(21))
= g(2(21)² + 3)
= g(885)
= 3(885) – 2 [∵ g(x) = 3x – 2]
= 2653

Question 4.
If f : R → R, g : R → R are defined by f(x) = 3x – 1, g(x) = x² + 1, then find
(i) (f o f) (x² + 1)
(ii) f o g (2)
(iii) g o f (2a – 3)
Solution:
f : R → R, g : R → R and f(x) = 3x – 1 ; g(x) = x² + 1
(i) (f o f) (x² + 1)
= f(f(x² + 1))
= f[3(x² + 1) – 1] [∵ f(x) = 3x – 1]
= f(3x² + 2)
= 3(3x² + 2) – 1
= 9x² + 5

(ii) (f o g) (2)
= f(g(2))
= f(2² + 1) [∵ g(x) = x² + 1]
= f(5)
= 3(5) – 1
= 14 [∵ f(x) = 3x – 1]

(iii) (g o f) (2a – 3)
= g(f(2a – 3))
= g[3(2a – 3) – 1] [∵ f(x) = 3x – 1]
= g(6a – 10)
= (6a – 10)² + 1 [∵ g(x) = x² + 1]
= 36a² – 120a + 100 + 1
= 36a² – 120a + 101

Question 5.
If f(x) = \(\frac{1}{x}\), g(x) = √x for all x ∈ (0, ∞) then find (g o f) (x).
Solution:
f(x) = \(\frac{1}{x}\), g(x) = √x, ∀ x ∈ (0, ∞)
(g o f) (x) = g(f(x))
= g(\(\frac{1}{x}\)) [∵ f(x) = \(\frac{1}{x}\)]
= \(\sqrt{\frac{1}{x}}\)
= \(\frac{1}{\sqrt{x}}\) [∵ g(x) = √x]
∴ (gof) (x) = \(\frac{1}{\sqrt{x}}\)

Question 6.
f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) for all x ∈ R, find (g o f) (x).
Solution:
f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) ∀ x ∈ R
(g o f) (x) = g(f(x))
= g(2x – 1) [∵ f(x) = 2x – 1]
= \(\frac{(2 x-1)+1}{2}\)
= x [∵ g(x) = \(\frac{x+1}{2}\)]
∴ (g o f) (x) = x

Question 7.
If f(x) = 2, g(x) = x², h(x) = 2x for all x ∈ R, then find (f o (g o h)) (x).
Solution:
f(x) = 2, g(x) = x², h(x) = 2x, ∀ x ∈ R
[f o (g o h) (x)]
= [f o g (h(x))]
= f o g (2x) [∵ h(x) = 2x]
= f[g(2x)]
= f((2x)²) [∵ g(x) = x²]
= f(4x²) = 2 [∵ f(x) = 2]
∴ [f o (g o h) (x)] = 2

Question 8.
Find the inverse of the following functions.
(i) a, b ∈ R, f : R → R defined by f(x) = ax + b, (a ≠ 0).
Solution:
a, b ∈ R, f : R → R and f(x) = ax + b, a ≠ 0
Let y = f(x) = ax + b
⇒ y = f(x)
⇒ x = f^-1(y) ……..(i)
and y = ax + b
⇒ x = \(\frac{y-b}{a}\) ……..(ii)
From (i) and (ii)
f^-1(y) = \(\frac{y-b}{a}\)
⇒ f^-1(x) = \(\frac{x-b}{a}\)

(ii) f : R → (0, ∞) defined by f(x) = 5^x
Solution:
f : R → (0, ∞) and f(x) = 5^x
Let y = f (x) = 5^x
y = f(x) ⇒ x = f^-1(y) ……(i)
and y = 5^x ⇒ log₅ (y) = x ……..(ii)
From (i) and (ii)
f^-1(y) = log₅(y) ⇒ f^-1(x) = log5 (x)

(iii) f : (0, ∞) → R defined by f(x) = log₂ (x).
Solution:
f : (0, ∞) → R and f(x) = log₂ (x)
Let y = f(x) = log₂ (x)
∵ y = f(x) ⇒ x = f^-1(y) ……..(i)
and y = log₂(x) ⇒ x = 2y
From (i) and (ii)
f^-1(y) = 2y ⇒ f^-1(x) = 2x

Question 9.
If f(x) = 1 + x + x² + …… for |x| < 1 then show that f^-1(x) = \(\frac{x-1}{x}\)
Solution:
f(x) = 1 + x + x² + ……..

Question 10.
If f : [1, ∞) ⇒ [1, ∞) defined by f(x) = \(2^{x(x-1)}\) then find f^-1(x).
Solution:

II.

Question 1.
If f(x) = \(\frac{x-1}{x+1}\), x ≠ ±1, then verify (f o f^-1) (x) = x.
Solution:
Given f(x) = \(\frac{x-1}{x+1}\), x ≠ ±1
Let y = f(x) = \(\frac{x-1}{x+1}\)
∵ y = f(x) ⇒ x = f^-1(y) ……(i)
and y = \(\frac{x-1}{x+1}\)

Question 2.
If A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f : A → B, g : B → C are defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}, then show that f and g are bijective functions and (gof)^-1 = f^-1 o g^-1.
Solution:
A = {1, 2, 3}, B = {α, β, γ},
f : A → B and f = {(1, α), (2, γ), (3, β)}
⇒ f(1) = α, f(2) = γ, f(3) = β
∵ Distinct elements of A have distinct f – images in B, f: A → B is an injective function.
Range of f = {α, γ, β} = B(co-domain)
∴ f : A → B is a surjective function.
Hence f : A → B is a bijective function.
B = {α, β, γ}, C = {p, q, r}, g : B → C and g : {(α, q), (β, r), (γ, p)}
⇒ g(α) = q, g(β) = r, g(γ) = p
∴ Distinct elements of B have distinct g – images in C, g : B → C is an injective function.
Range of g = {q, r, p} = C, (co-domain)
∴ g : B → C is a surjective function
Hence g : B → C is a bijective function
Now f = {(1, α), (2, γ), (3, β)}
g = {(α, q), (β, r), (γ, p)}
g o f = {(1, q), (2, p), (3, r)}
∴ (g o f)^-1 = {(q, 1), (r, 3), (p, 2)} ………(1)
g^-1 = {(q, α), (r, β), (p, γ)}
f^-1 = {(α, 1), (γ, 2),(β, 3)}
Now f^-1 o g^-1 = {(q, 1), (r, 3), (p, 2)} …….(2)
From eq’s (1) and (2)
(gof)^-1 = f^-1 o g^-1

Question 3.
If f : R → R, g : R → R defined by f(x) = 3x – 2, g(x) = x² + 1, then find
(i) (g o f^-1) (2)
(ii) (g o f)(x – 1)
Solution:
f : R → R, g : R → R and f(x) = 3x – 2
f is a bijective function ⇒ its inverse exists
Let y = f(x) = 3x – 2
∵ y = f(x) ⇒ x = f^-1(y) …….(i)
and y = 3x – 2
⇒ x = \(\frac{y+2}{3}\) ……..(ii)
From (i) and (ii)
f^-1(y) = \(\frac{y+2}{3}\)
⇒ f^-1(x) = \(\frac{x+2}{3}\)
Now (gof^-1) (2)
= g(f^-1(2))

∴ (g o f^-1) (2) = \(\frac{25}{9}\)

(ii) (g o f) (x -1)
= g(f(x – 1))
= g(3(x – 1) – 2) [∵ f(x) = 3x – 2]
= g(3x – 5)
= (3x – 5)² + 1 [∵ g(x) = x² + 1]
= 9x² – 30x + 26
∴ (g o f) (x – 1) = 9x² – 30x + 26

Question 4.
Let f = {(1, a), (2, c), (4, d), (3, b)} and g^-1 = {(2, a), (4, b), (1, c), (3, d)} then show that (gof)^-1 = f^-1 o g^-1
Solution:
f = {(1, a), (2, c), (4, d), (3, b)}
∴ f^-1 = {(a, 1), (c, 2), (d, 4), (b, 3)}
g^-1 = {(2, a), (4, b), (1, c), (3, d)}
∴ g = {(a, 2), (b, 4), (c, 1), (d, 3)}
(g o f) = {(1, 2), (2, 1), (4, 3), (3, 4)}
∴ (gof)^-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ……….(1)
f^-1 o g^-1 = {(2, 1), (4, 3), (1, 2), (3, 4)} ……..(2)
From eq’s (1) and (2), we observe (gof)^-1 = f^-1 o g^-1

Question 5.
Let f : R → R, g : R → R be defined by f(x) = 2x – 3, g(x) = x³ + 5 then find (f o g)^-1 (x).
Solution:
f : R → R, g : R → R and f(x) = 2x – 3 and g(x) = x³ + 5
Now (fog) (x) = f(g(x))
= f(x³ + 5) [∵ g(x) = x² + 5]
= 2(x³ + 5) – 3 [∵ f(x) = 2x – 3]
= 2x³ + 7
∴ (f o g) (x) = 2x³ + 7
Let y = (f o g) (x) = 2x³ + 7
∵ y = (fog)(x)
⇒ x = (fog)^-1 (y) …….(1)
and y = 2x³ + 7
⇒ x³ = \(\frac{y-7}{2}\)
⇒ x = \(\left(\frac{y-7}{2}\right)^{\frac{1}{3}}\) …..(2)
From eq’s (1) and (2),
(f o g)^-1 (y) = \(\left(\frac{y-7}{2}\right)^{\frac{1}{3}}\)
∴ (f o g)^-1 (x) = \(\left(\frac{x-7}{2}\right)^{\frac{1}{3}}\)

Question 6.
Let f(x) = x², g(x) = 2^x. Then solve the equation (f o g) (x) = (g o f) (x)
Solution:
Given f(x) = x² and g(x) = 2^x
Now (f o g) (x) = f(g(x))
= f(2^x) [∵ g(x) = 2^x]
= (2^x)²
= 2^2x [∵ f(x) = x²]
∴ (f o g) (x) = 2^2x ……(1)
and (g o f) (x) = g(f(x))
= g(x²) [∵ f(x) = x²]
= \((2)^{x^{2}}\) [∵ g(x) = 2^x]
∴ (g o f) (x) = \((2)^{x^{2}}\)
∵ (f o g) (x) = (g o f) (x)
⇒ 2^2x = \((2)^{x^{2}}\)
⇒ 2^x = x²
⇒ x² – 2x = 0
⇒ x(x – 2) = 0
⇒ x = 0, x = 2
∴ x = 0, 2

Question 7.
If f(x) = \(\frac{x+1}{x-1}\), (x ≠ ±1) then find (fofof) (x) and (fofofof) (x).
Solution:
f(x) = \(\frac{x+1}{x-1}\), (x ≠ ±1)
(i) (fofof) (x) = (fof) [f(x)]
= (fof) \(\left(\frac{x+1}{x-1}\right)\) [∵ f(x) = \(\left(\frac{x+1}{x-1}\right)\)]

(ii) (fofofof) (x) = f[(f o f o f) (x)]
= f [f(x)] {from (1)}

In the above problem if a number of f is even its answer is x and if a number of f is odd its answer is f(x).

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(a)

I.

Question 1.
If the function f is defined by

then find the values of
(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)
(v) f(-5)
Solution:
(i) f(3)
For x > 1, f(x) = x + 2
∴ f(3) = 3 + 2 = 5

(ii) f(0)
For -1 ≤ x ≤ 1, f(x) = 2
∴ f(0) = 2

(iii) f(-1.5)
For -3 < x < -1, f(x) = x – 1
∴ f(-1.5) = -1.5 – 1 = -2.5

(iv) f(2) + f(-2) For x > 1, f(x) = x + 2
∴ f(2) = 2 + 2 = 4
For -3 < x < -1, f(x) = x – 1
∴ f(-2)= -2 – 1 = -3
f(2) + f(-2) = 4 + (-3) = 1

(v) f(-5) is not defined, since domain of x is {X/X ∈ (-3, ∞)}

Question 2.
If f: R{0}R is defined by f(x) = \(x^{3}-\frac{1}{x^{3}}\); then show that f(x) + \(f\left(\frac{1}{x}\right)\) = 0.
Solution:
Given f(x) = \(x^{3}-\frac{1}{x^{3}}\) ……(i)
Now \(f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^{3}-\frac{1}{\left(\frac{1}{x}\right)^{3}}=\frac{1}{x^{3}}-x^{3}\) ……(2)
Add (1) and (2)
\(f(x)+f\left(\frac{1}{x}\right)=\left(x^{3}-\frac{1}{x^{3}}\right)+\left(\frac{1}{x^{3}}-x^{3}\right)\) = 0
∴ f(x) + \(f\left(\frac{1}{x}\right)\) = 0

Question 3.
If f : R → R is defined by f(x) = \(\frac{1-x^{2}}{1+x^{2}}\), then show that f(tan θ) = cos 2θ.
Solution:
Given f(x) = \(\frac{1-x^{2}}{1+x^{2}}\)

∴ f(tan θ) = cos 2θ

Question 4.
If f : R\{±1} → R is defined by f(x) = \(\log \left|\frac{1+x}{1-x}\right|\), then show that \(f\left(\frac{2 x}{1+x^{2}}\right)\) = 2f(x)
Solution:
f : R\{±1} → R and f(x) = \(\log \left|\frac{1+x}{1-x}\right|\)

∴ \(f\left(\frac{2 x}{1+x^{2}}\right)\) = 2f(x)

Question 5.
If A = {-2, -1, 0, 1, 2} and f : A → B is a surjection defined by f(x) = x² + x + 1, then find B.
Solution:
A = {-2, -1, 0, 1, 2} and f : A → B, f(x) = x² + x + 1
f : A → B is a surjection
f(-2) = (-2)² + (-2) + 1
= 4 – 2 + 1
= 3
f(-1) = (-1)² + (-1) + 1
= 1 – 1 + 1
= 1
f(0) = 0² + 0 + 1
= 0 + 0 + 1
= 1
f(1) = 1² + 1 + 1
= 1 + 1 + 1
= 3
f(2) = 2² + 2 + 1
= 4 + 2 + 1
= 7
∴ B = f(A) = {3, 1, 7}

Question 6.
If A = {1, 2, 3, 4} and f : A → R is a function defined by f(x) = \(\frac{x^{2}-x+1}{x+1}\), then find the range of f.
Solution:
A= {1, 2, 3, 4}

∴ Range of f = f(A) = \(\left\{\frac{1}{2}, 1, \frac{7}{4}, \frac{13}{5}\right\}\)

Question 7.
If f(x + y) = f(xy) ∀ x, y ∈ R then prove that f is a constant function.
Solution:
Given f(x + y) = f(x y), x, y ∈ R
take x = y = 0
⇒ f(0) = f(0) ………(1)
Let x = 1, y = 0
⇒ f(1) = f(0) ……..(2)
Let x = 1, y = 1
⇒ f(2) = f(1) ………(3)
from (1), (2), (3)
f(0) = f(1) = f(2)
⇒ f(0) = f(2)
Similarly f(3) = f(0)
f(4) = f(0)
and so on
f(n) = f(0)
∴ f is a constant function

II.

Question 1.
If A = {x | -1 ≤ x ≤ 1}, f(x) = x², g(x) = x³, which of the following are surjections?
(i) f : A → A
(ii) g : A → A
Solution:
(i) ∵ A = {x | -1 ≤ x ≤ 1} and f(x) = x²
This implies f(x) is a function from A to A
(i.e.,) f : A → A
Now let y ∈ A
If f(x) = y then x² = y
x = √y
So, if y = -1 then x = √-1 ∉ A
∴ f : A → A is not a surjection.

(ii) ∵ A = {x | -1 ≤ x ≤ 1} and g(x) = x³
⇒ g : A → A
Let y ∈ A. Then g(x) = y
⇒ x³ = y
⇒ x = \((y)^{1 / 3}\) ∈ A
So if y = -1 then x = -1 ∈ A
y = 0, then x = 0 ∈ A
y = 1, then x = 1 ∈ A
∴ g : A → A is a surjections.

Question 2.
Which of the following are injections or surjections or bijections? Justify your answers.
(i) f : R → R defined by f(x) = \(\frac{2 x+1}{3}\)
Solution:
f(x) = \(\frac{2 x+1}{3}\)
Let x₁, x₂ ∈ R
∵ f(x₁) = f(x₂)
⇒ \(\frac{2 x_{1}+1}{3}=\frac{2 x_{2}+1}{3}\)
⇒ 2x₁ + 1 = 2x₂ + 1
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂
∵ f(x₁) = f(x₂) ⇒ x₁ = x₂, ∀ x₁, x₂ ∈ R
So f(x) = \(\frac{2 x+1}{3}\), f : R → R is an injection
If y ∈ R (co-domain) then y = \(\frac{2 x+1}{3}\)
⇒ x = \(\frac{3 y-1}{2}\)
Then f(x) = \(\frac{2 x+1}{3}=\frac{2\left(\frac{3 y-1}{2}\right)+1}{3}=y\)
∴ f is a surjection
∴ f : R → R defined by f(x) = \(\frac{2 x+1}{3}\) is a bijection

(ii) f : R → (0, ∞) defined by f(x) = 2x
Solution:
Let x₁, x₂ ∈ R
∵ f(x₁) = f(x₂)
⇒ \(2^{x_{1}}=2^{x_{2}}\)
⇒ x₁ = x₂
∴ f(x₁) = f(x₂) ⇒ x₁ = x₂ ∀ x₁, x₂ ∈ R
∴ f(x) = 2x, f : R → (0, ∞) is injection
If y ∈ (0, ∞) and y = 2x ⇒ x = log₂ (y)
Then f(x) = 2x
= \(2^{\log _{2}(y)}\)
= y
∴ f is a surjection
Hence f is a bijection.

(iii) f : (0, ∞) → R defined by f(x) = log_ex
Solution:
Let x₁, x₂ e (0, ∞)
f(x₁) = f(x₂)
⇒ \(\log _{e}\left(x_{1}\right)=\log _{e}\left(x_{2}\right)\)
⇒ x₁ = x₂
∵ f(x₁) = f(x₂)
⇒ x₁ = x₂ ∀ x₁, x₂ ∈ (0, ∞)
∴ f(x) is injection.
Let y ∈ R.
y = log_ex ⇒ x = e^y
Then f(x) = log_ex
= log_e(e^y)
= y . log_ee
= y(1)
= y
∴ f is a surjection.
∴ f is a bijection.

(iv) f : [0, ∞) → [0, ∞) defined by f(x) = x².
Solution:
Let x₁, x₂ ∈ [0, ∞) (i.e.,) domain of f.
Now f(x₁) = f(x₂)
⇒ \(x_{1}^{2}=x_{2}^{2}\)
⇒ x₁ = x₂
∵ x₁, x₂ ≥ 0
∴ f(x) = x², f : {0, ∞) → {0, ∞) is injection
Let y ∈ (0, ∞), co-domain of f
Let y = x² ⇒ x = √y, ∵ y ≥ 0
Then f(x) = x²
= \((\sqrt{y})^{2}\)
= y
∴ f is surjection.
Hence f is a bijection.

(v) f : R → [0, ∞) defined by f(x) = x².
Solution:
Let x₁, x₂ ∈ R.
f(x₁) = f(x₂)
⇒ \(x_{1}^{2}=x_{2}^{2}\)
⇒ x₁ = ±x₂, ∵ x₁, x₂ ∈ R
Hence f is not injection
Let y ∈ [0, ∞)
y = x²
⇒ x = ±√y, where y ∈ [0, ∞)
Then f(x) = x²
= \((\sqrt{y})^{2}\)
= y
∴ f is surjection
Hence f is not a bijection

(vi) f : R → R defined by f(x) = x².
Solution:
Let x₁, x₂ ∈ R, (domain of f)
∴ f(x₁) = f(x₂)
⇒ \(x_{1}^{2}=x_{2}^{2}\)
⇒ x₁ = ±x₂, ∵ x₁, x₂ ∈ R
∴ f(x) is not injection
For elements that belong to (-∞, 0) codomain of f has no pre-image in f.
∴ f is not a surjection
Hence f is neither injection nor surjection.

Question 3.
Is g = {(1, 1) (2, 3) (3, 5) (4, 7)} is a function from A = {1, 2, 3, 4} to B = {1, 3, 5, 7}. If this is given by the formula g(x) = ax + b, then find a and b.
Solution:
A = {1, 2, 3, 4}; B = {1, 3, 5, 7}
g : {(1, 1), (2, 3), (3, 5), (4, 7)}
∵ g(1) = 1, g(2) = 3, g(3) = 5, g(4) = 7
So for each element a ∈ A, there exists a unique b ∈ B such (a, b) ∈ g
∴ g : A → B is a function
Given g(x) = ax + b, ∀ x ∈ A
g(1) = (a) + b = 1
⇒ a + b = 1 ……..(1)
g(2) = 2a + b = 3
⇒ 2a + b = 3 …….(2)
Solve (1) and (2)
a = 2, b = -1

Question 4.
If the function f : R → R defined by f(x) = \(\frac{3^{x}+3^{-x}}{2}\), then show that f(x + y) + f(x – y) = 2f(x) f(y).
Solution:
f : R → R and f(x) = \(\frac{3^{x}+3^{-x}}{2}\)

∴ f(x + y) + f(x – y) = 2 f(x).f(y)

Question 5.
If the function f : R → R defined by f(x) = \(\frac{4^{x}}{4^{x}+2}\), then show that f(1 – x) = 1 – f(x) and hence reduce the value of \(f\left(\frac{1}{4}\right)+2 f\left(\frac{1}{2}\right)+f\left(\frac{3}{4}\right)\)
Solution:


∴ f(1 – x) = 1 – f(x)

Question 6.
If the function f : {-1, 1} → {0, 2), defined by f(x) = ax + b is a surjection, then find a and b.
Solution:
f : {-1, 1} → {0, 2} and f(x) = ax + b is a surjection
Given f(-1) = 0 and f(1) = 2 (or) f(-1) = 2, f(1) = 0
Case (i):
f(-1) = 0 and f(1) = 2
a(-1) + b = 0 ⇒ -a + b = 0 ……..(1)
a(1) + b = 2 ⇒ a + b = 2 ……(2)
Solve eq’s (1) and (2), we get a = 1, b = 1
Case (ii):
f(-1) = 2 and f(1) = 0
a(-1) + b = 2 ⇒ -a + b = 2 ……(3)
a(1) + b = 0 ⇒ a + b = 0 ……….(4)
Solve eq’s (3) and (4), we get a = -1, b = 1
Hence a = ±1 and b = 1

Question 7.
If f(x) = cos (log x), then show that \(f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]=0\)
Solution:
Given f(x) = cos(log x)
\(f\left(\frac{1}{x}\right)=\cos \left(\log \left(\frac{1}{x}\right)\right)\)
= cos(log 1 – log x)
= cos(-log x)
= cos (log x) (∵ log 1 = 0)
Similarly
\(f\left(\frac{1}{y}\right)\) = cos (log y)
\(f\left(\frac{x}{y}\right)=\cos \log \left(\frac{x}{y}\right)\)
= cos (log x – log y)
and f(x y) = cos log (x y) = cos (log x + log y)
\(f\left(\frac{x}{y}\right)\) + f(x y) = cos (log x – log y) + cos (log x + log y)
= 2 cos (log x) cos (log y)
[∵ cos (A – B) + cos (A + B) = 2 cos A . cos B]
LHS = \(f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]\)
= cos (log x) cos (log y) – \(\frac{1}{2}\) [2 cos (log x) cos (log y)]
= 0

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