Mathematics Chapter 1 – Functions | Nepal eNotes

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MATHEMATICS • CHAPTER 1

Chapter 1: Functions

Cartesian product, relations, functions, composite functions, inverse functions and types of functions — reconstructed from the supplied 58-page handwritten notes.

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PDF Page 1

Basic Concepts

Cartesian Product

If A = {a, b} and B = {1, 2}, then:

A × B = {(a,1), (a,2), (b,1), (b,2)}

Relation

Let A and B be two non-empty sets. A relation from A to B is a subset of the Cartesian product A × B.

Function

A function is a relation in which every element of the first set is associated with a unique element of the second set. It is denoted by:

f : A → B

Composite Function

The combination of two functions is called a composite function. If f : A → B and g : B → C, then the composite function from A to C is written as:

(g ∘ f)(x) = g(f(x))

PDF Pages 2–21

Exercise 1.1 – Composite Functions

PDF Pages 2–3

1. Composite Relations from Mapping Diagrams

(a)

Given:

f = {(1,2), (3,5), (2,1)}
g = {(2,3), (5,2), (1,3)}

Following each element first through f and then through g:

1 → 2 → 3

3 → 5 → 2

2 → 1 → 3

∴ g ∘ f = {(1,3), (3,2), (2,3)}

(b)

Given:

g = {(1,2), (2,3), (3,4)}
h = {(2,3), (3,4), (4,5)}

1 → 2 → 3

2 → 3 → 4

3 → 4 → 5

∴ h ∘ g = {(1,3), (2,4), (3,5)}

(c)

The source gives:

f = {(b,q), (a,p), (c,r)}
g ∘ f = {(b,1), (a,3), (c,3)}

Therefore, the second mapping is:

g = {(q,1), (p,3), (r,3)}

PDF Pages 4–6

2. Write the Formula for (f ∘ g)(m)

(a) f(m) = (m+2)/4, g(m) = 2m

(f ∘ g)(m) = f(g(m))

= f(2m)

= (2m+2)/4

∴ (f ∘ g)(m) = (m+1)/2

(b) f(m) = 3m+2, g(m) = (m−3)/2

(f ∘ g)(m) = 3[(m−3)/2] + 2

= (3m−9+4)/2

∴ (f ∘ g)(m) = (3m−5)/2

(c)

The source substitutes g(m) into a linear fractional expression for f and simplifies to:

(f ∘ g)(m) = (m+7)/3

(d)

The handwritten calculation gives:

(f ∘ g)(m) = (8m−13)/20

The exact coefficients in parts (c) and (d) are faint in the scan, but the final simplified source expressions above are clearly readable.

PDF Pages 6–8

3. Evaluate Composite Functions

(a) g(m)=3m+2 and h(m)=−4m

(g ∘ h)(m) = g(−4m)

= −12m + 2

For the value used in the source, m = −4:

= 48 + 2

∴ 50

(b) g(m)=(3m−1)/2 and h(m)=2m/5

(g ∘ h)(m) = [3(2m/5)−1]/2

= (6m−5)/10

For m = −4:

= (−24−5)/10

∴ −2.9

(c)

The source evaluates another composite expression at m = −4 and obtains:

PDF Pages 8–9

4. Find (f ∘ g)(m) and (g ∘ f)(m)

(a) f(m)=2m, g(m)=m+4

(f ∘ g)(m)=2(m+4)=2m+8

(g ∘ f)(m)=2m+4=2m+4

(b) f(m)=4m+6, g(m)=2m−1

(f ∘ g)(m)=4(2m−1)+6=8m+2

(g ∘ f)(m)=2(4m+6)−1=8m+11

(c) f(m)=−m+4, g(m)=m+2

(f ∘ g)(m)=−(m+2)+4=−m+2

(g ∘ f)(m)=−m+4+2=−m+6

(d) f(m)=2m+3, g(m)=3m−1

(f ∘ g)(m)=2(3m−1)+3=6m+1

(g ∘ f)(m)=3(2m+3)−1=6m+8

PDF Pages 9–12

5–7. Further Short Questions

5. If f(m)=3m+b and f(f(2))=12, find b.

f(2)=6+b

f(f(2))=3(6+b)+b

=18+4b

18+4b=12

4b=−6

∴ b=−3/2

6. Functional-relation question

Page 11 gives a function f : R → R with a relation involving f(3m), f(3) and f(m). The handwritten source concludes f(1)=0 during its working.

The exact final quantity requested in Question 6 is not sufficiently clear in the scan, so only the readable source step has been preserved.

7(a). Decompose h(m)=(8−9m)² as h=f∘g

Take g(m)=8−9m and f(m)=m².

Then f(g(m))=(8−9m)².

7(b). Decompose h(m)=1/√(3m+7) as h=f∘g

Take g(m)=√(3m+7) and f(m)=1/m.

Then f(g(m))=1/√(3m+7).

PDF Pages 12–21

Long Type Questions – Composite Functions

1. Given f(m)=3m−9, h(m)=m+3 and g(m)=2m+1

The source evaluates several combinations such as f∘h, h∘h and nested compositions by direct substitution.

(f ∘ h)(m)=f(m+3)=3(m+3)−9=3m

(h ∘ h)(m)=h(m+3)=m+6=m+6

2. Evaluation of composite functions at specified values

Pages 13–15 evaluate combinations such as h∘f, f∘f and nested composites by first simplifying the function and then substituting the required value.

3(a) f(m)=m²+1 and g(m)=4m−3

(f ∘ g)(m)=(4m−3)²+1

=16m²−24m+10

∴ (f ∘ g)(m)=16m²−24m+10

(g ∘ f)(m)=4(m²+1)−3

∴ (g ∘ f)(m)=4m²+1

4(a) f(m)=1/(1−m)

The source forms f∘f and evaluates it at a specified value.

f(f(m)) = f[1/(1−m)]

= 1 / [1 − 1/(1−m)]

= (m−1)/m

At m=1/2:

∴ f(f(1/2))=−1

4(b) f(m)=(m−1)/(m+1)

The source simplifies f(f(m)) to:

f(f(m))=−1/m

Hence at m=4:

f(f(4))=−1/4

5(b) Self-inverse linear fractional function

The handwritten source takes:

y=f(m)=(am−b)/(cm−a)

and algebraically proves that applying the same function again returns the original variable:

f(y)=m

6. If f(m)=2m²−5m+4 and 2f(m)=f(2m)

2(2m²−5m+4)=2(2m)²−5(2m)+4

4m²−10m+8=8m²−10m+4

4m²=4

m²=1

∴ m=±1

7. If g(h(m))=m² and g(m)=m−1

g(h(m))=h(m)−1

h(m)−1=m²

∴ h(m)=m²+1

PDF Pages 22–50

Exercise 1.2 – Inverse Functions

PDF Page 22

1. Find the Inverse of the Given Relation / Function

(a)

f = {(1,2), (2,3), (3,4), (4,5)}

Interchanging the coordinates of each ordered pair:

f⁻¹ = {(2,1), (3,2), (4,3), (5,4)}

(b)

The source applies the same rule to the second relation: interchange the first and second entries in every ordered pair.

PDF Pages 22–26

2. Find the Inverse Formula

(a) f(m)=9m

y=9m

m=y/9

∴ f⁻¹(m)=m/9

(b) f(m)=2m+5

y=2m+5

m=(y−5)/2

∴ f⁻¹(m)=(m−5)/2

(c) f(m)=(m+4)/2

2y=m+4

m=2y−4

∴ f⁻¹(m)=2m−4

(d) g(m)=(3m−2)/4

4y=3m−2

3m=4y+2

∴ g⁻¹(m)=(4m+2)/3

(e) g(m)=(8m+7)/5

5y=8m+7

8m=5y−7

∴ g⁻¹(m)=(5m−7)/8

(f) f(m)=(m−1)/(m+1)

y=(m−1)/(m+1)

ym+y=m−1

m(y−1)=−(y+1)

m=(y+1)/(1−y)

∴ f⁻¹(m)=(m+1)/(1−m)

(g) h={(m,2m+1): m∈R}

y=2m+1

m=(y−1)/2

∴ h⁻¹(m)=(m−1)/2

PDF Pages 26–28

3–4. Verify Inverse Functions and Evaluate

3. f(m)=m−3 and g(m)=m+3

f⁻¹(m): y=m−3 ⇒ m=y+3

∴ f⁻¹(m)=m+3=g(m)

Similarly, g⁻¹(m)=m−3=f(m).

Hence f and g are inverse functions.

4. Trigonometric function evaluation

Pages 27–28 use a function of the form f(m)=3−4 sin m and evaluate it at standard angles using sin 30°, sin 45° and related values.

The exact sub-question labels are faint, but the source clearly uses standard trigonometric values for direct substitution.

PDF Pages 29–50

Long Type Questions – Inverse Functions

1(a) f(m)=m−4

f⁻¹(m)=m+4

f⁻¹(2)=6

f⁻¹(0)=4

f⁻¹(−5)=−1

1(b) f(m)=2m+1

f⁻¹(m)=(m−1)/2

f⁻¹(5)=2

f⁻¹(0)=−1/2

f⁻¹(−11)=−6

1(c) Linear fractional form

The source rearranges a function of the form (4m−2)/5 and obtains its inverse by interchanging x and y, then evaluates the inverse at specified values.

1(d) f(m)=2m+7

f⁻¹(m)=(m−7)/2

Therefore:

f⁻¹(m+2)=(m−5)/2

3. One-one and onto functions

Pages 35–37 calculate inverses of several one-one and onto functions, then evaluate compositions involving the inverses.

4(a) f(m)=3m+4 and g(m)=2(m+1)

(f ∘ g)(m)=3(2m+2)+4=6m+10

(g ∘ f)(m)=2[(3m+4)+1]=6m+10

Hence f ∘ g = g ∘ f.

f⁻¹(m)=(m−4)/3

f⁻¹(2)=−2/3

4(b) f(m)=2m−3

f⁻¹(m)=(m+3)/2

The source verifies:

f(f⁻¹(m))=m

and

f⁻¹(f(m))=m.

6(a) f(m)=4m+7 and inverse-composition evaluation

The source uses g⁻¹(m)=(m+5)/3 and solves:

f(g⁻¹(m))=15

4[(m+5)/3]+7=15

∴ m=1

6(b) f(m)=4m−17 and g(m)=(2m+8)/5

The source equates f(f(m)) with g⁻¹(m) and obtains:

m=6

6(c) Linear fractional inverse equation

The handwritten solution produces two admissible solutions:

m=5 or m=2

6(d) f(m)=m/(2m−3)

The source first finds:

f⁻¹(m)=3m/(2m−1)

Then solves f(m)=f⁻¹(m).

Source result: m=2

6(e) f(m)=(m−2)/(2m+1), g(m)=1/m

Solving f⁻¹(m)=g(f(m)), the source obtains:

m=±√5

9. f(m)=cos m and g(m)=m²

(g ∘ f)(m)=cos²m

(f ∘ g)(m)=cos(m²)

Therefore:

g ∘ f ≠ f ∘ g

10. f(m)=(1−m)/(1+m)

The source shows that the function is self-inverse and hence:

f(f(tan θ)) = tan θ

PDF Pages 51–53

Types of Functions

1. Constant Function

A function written in the form f(x)=c, where c is a constant, is called a constant function. Its graph is a horizontal straight line.

2. Linear Function

A function written in the form f(x)=ax+b, where a and b are constants, is called a linear function.

3. Identity Function

The function f(x)=x is called the identity function. Its graph is a straight line through the origin.

4. Quadratic Function / Parabola

A function of the form f(x)=ax²+bx+c, where a, b and c are constants and a≠0, is called a quadratic function.

5. Cubic Function

A function of the form f(x)=ax³+bx²+cx+d, where a, b, c and d are constants and a≠0, is called a cubic function.

PDF Pages 54–58

Exercise 1.3

1. If f(m)=c and f(2)=5, find c.

Since f(m)=c is a constant function:

f(2)=c=5

∴ c=5

2. Prove that a composition is an identity function

The source takes functions of the form:

g(m)=(m−2)/3, h(m)=3m+2

(g ∘ h)(m)=g(3m+2)

=[(3m+2)−2]/3

=3m/3

∴ (g ∘ h)(m)=m

Hence g∘h is an identity function.

3. Long question involving composite/inverse functions

Pages 55–56 use inverse-function substitution to reduce a nested composite equation to a linear equation in the unknown.

4. Identity-function verification

Pages 57–58 take:

f(m)=(3m+1)/9

find the inverse function, and verify:

f⁻¹(f(m))=m

The final page also evaluates the inverse at a specified value and records:

f⁻¹(3)=26/3

Some symbols and coefficients in the last few exercise questions are faint. The clearly readable function forms, proof targets and source results are retained without inventing missing text.