Relations,Functions and Graphs Exercise 2.2 | Basic Mathematics Solution [NEB UPDATED]

1. Let X= {a,b,c} and Y ={p,r,s}. Determine which of the following relations from X to Y are functions. Give reason for your answer:

Solution:

a. The relation R₁ is not a function from X to Y because the element a of x has two images p and r in y but p ≠ r.

b. The relation R₂ is not a function from X to Y because the element c of x has no image in y.

c. The relation R₃ is a function from X to Y because each element of x has unique image in y.

d. The relation R₄ is a function from X to Y because each element of x has a unique image in y.

e. The relation R₅ is a function from X to Y because each element of x has unique image in y.

 2. If f: A →B where A and B ⊂ R, is defined by f(x) = 1-x, find the images of 1, $\left( {\frac{3}{2}} \right)$, -1 2 ∈A.

Solution:

Here, f(x) = 1 – x and x = 1, 3/2, –1, 2 ϵ A.

So, image, f(1) = 1 – 1 = 0

f$\left( {\frac{3}{2}} \right)$ = 1 – $\frac{3}{2}$ = $ - \frac{1}{2}$

f(–1) = 1 – (–1) = 2

And f(2) = 1 – 2 = –1 

3. Let i) f(x) = x+2 ii) f(x) = 2|x|+3x in the interval of -1≤x≤2. Find

 Solution:

Here,

(i) f(x) = x+2

f(x) = x + 2, where –1 ≤ x ≤ 2.

Now,

1. f(–1) = –1 + 2 = 1
2. f(0) = 0 + 2 = 2
3. f(1) = 1 + 2 = 3
4. f(2) = 2 + 2 = 4
5. f(–2) and f(3) are not defined because –2 and 3 does not lie in –1 ≤ x ≤ 2. 

(ii) f(x) = 2|x|+3x

f(x) = 2|x| + 3x, where –1 ≤ x ≤ 2.

Now,

a. f(–1) = 2|–1| + 3(–1) = 2 – 3 = –1

b. f(0) = 2|0| + 3.0 = 0 + 0 = 0

c. f(1) = 2|1| + 3.1 = 2 + 3 = 5

d. f(2) = 2|2| + 3.2 = 4 + 6 = 10

e. f(–2) and f(3) are not defined because –2 and 3 does not lie in –1 ≤ x ≤ 2 .

 

4 (i) Let the function f: R → R be defined by

$(x) = \left\{ {\begin{array}{*{20}{c}}{3 + 2x{\rm{ for  - 1/2}} \le x < 0}\\{3 - 2x{\rm{ for 0}} \le x < 1/2}\\{ - 3 - 2x{\rm{ for x}} \ge {\rm{1/2}}}\end{array}} \right\}$

Find: a) f(-1/2) b)f(0)    c) f(1/2)                d)$\frac{{f(h) - f(0)}}{h}{\rm{ for 0}} \le \~h < 1/2$

Solution:

a. f$\left( { - \frac{1}{2}} \right)$ = 3 + 2 $\left( { - \frac{1}{2}} \right)$ = 3 – 1 = 2 

b. f(0) = 3 – 2.0 = 3 

c. f$\left( {\frac{1}{2}} \right)$ = –3 – 2 $\left( { - \frac{1}{2}} \right)$ = – 3 – 1 = –4. 

(d) $\frac{{{\rm{f}}\left( {\rm{h}} \right) - {\rm{f}}\left( 0 \right)}}{{\rm{h}}}$ = $\frac{{\left( {3 - 2{\rm{h}}} \right) - \left( {3 - 2.0} \right)}}{{\rm{h}}}$              [0≤h<$\frac{1}{2}$]

 = – 2

 

4. (ii)

Solution:

a. f(2) = 4.2 – 2 = 6

 

b. f(1) = 4.1 – 2 = 2

 

c. f(0) = 2.0 = 0

 

d. f(–1) = 2(–1) = –2

 

e. $\frac{{{\rm{f}}\left( {\rm{h}} \right) - {\rm{f}}\left( 1 \right)}}{{\rm{h}}}$ = $\frac{{\left( {4{\rm{h}} - 2} \right) - \left( {4.1 - 2} \right)}}{{\rm{h}}}$ [since, 1< h ]

= 4(h-1) / h 

 

5.

Solution:

Here, A = {–1, 0, 2, 4, 6}

(i) y = f(x) = $\frac{{\rm{x}}}{{{\rm{x}} + 2}}$

y = f(x) = $\frac{{\rm{x}}}{{{\rm{x}} + 2}}$

So, range of f = {f(–1), f(0), f(2), f(4), f(6)}

= $\left\{ { - \frac{1}{{ - 1 + 2}},\frac{0}{{0 + 2}},{\rm{\: }}\frac{2}{{2 + 2}},{\rm{\: }}\frac{4}{{4 + 2}},{\rm{\: }}\frac{6}{{6 + 2}}} \right\}$

= $\left\{ { - 1,{\rm{\: }}0,{\rm{\: }}\frac{1}{2},{\rm{\: }}\frac{2}{3},{\rm{\: }}\frac{3}{4}} \right\}$

 

(ii) y = f(x) = $\frac{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}}{{{\rm{x}} + 2}}$

Solution:

y = f(x) = $\frac{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}}{{{\rm{x}} + 2}}$

So, Range of f = {f(–1), f(0), f(2), f(4), f(6)}

= $\left\{ {\frac{{ - 1\left( { - 1 + 1} \right)}}{{ - 1 + 2}},{\rm{\: }}\frac{{0\left( {0 + 1} \right)}}{{0 + 2}},{\rm{\: }}\frac{{2\left( {2 + 1} \right)}}{{2 + 2}},{\rm{\: }}\frac{{4\left( {4 + 1} \right)}}{{4 + 2}},{\rm{\: }}\frac{{6\left( {6 + 1} \right)}}{{6 + 2}}} \right\}{\rm{\: \: }}$

= $\left\{ {0,{\rm{\: }}0,{\rm{\: }}\frac{3}{2},{\rm{\: }}\frac{{10}}{3},{\rm{\: }}\frac{{21}}{4}} \right\}$ = $\left\{ {0,{\rm{\: }}\frac{3}{2},{\rm{\: }}\frac{{10}}{3},{\rm{\: }}\frac{{21}}{4}} \right\}$

 

6. Determine whether the function f and g defined below are equal or not.

a)f(x) = x² where A={x:x=-1,-2,-3}

b) Venn diagram

Solution:

Here, f(x) = x² where A = {–1, –2, –3}

So, f(1) = (–1)² = 1

f(–2) = (–2)² = 4

f(–3) = (–3)² = 9

So, domain f = {–1, –2, –3} and Range f = {1, 4, 9}

And. From Venn diagram we have,

g(–1) = 1

g(–2) = 4

g(–3) = 9

So, Domain g = {–1, –2, –3} and Range g = {1, 4, 9}

Thus, f and g have the same domain {–1, –2, –3} and f(x) = g(x) for all x ϵ A so the functions f and g are equal.

 

7. Determine whether each of the following functions is one to one.

a.

Solution:

Here, A = {–2, –1, 0, 1, 2}, B = {4, 0, 1} and f:A →B is f(x) = x², so f is not one – one because f(–2) = 4 = f(2) but  –2 ≠ 2.

 

b.

Solution:

Here, A = set of positive integers.

So, A = {1, 2, 3, ….}

And B = Set of squares of positive integers.

So, B = {1, 4, 9, …} and the function f:A→B is defined by f(x) = x².

To show: f is one – one.

Let x ≠ y where x, y ϵ A then to show f(x) ≠ f(y).

Since, x ≠ y.

= x² ≠ y² [x, y ϵ z⁺]

= f(x) ≠ f(y) [f(x) = x²]

Hence, f is one–one. 

c.

Solution:

Here, f:[–2, –2] → R is f(x) = x².

To show: f is one – one.

Let x, y ϵ [–2, –2] such that x ≠ y.

Then, to show: f(x) ≠ f(y)

Since, x ≠ y.

But x² = y² [–2, 2 ϵ [–2, –2], –2≠2 But (–2)² = 2² = 4]

So, f(x) = f(y)

Hence, x ≠ y → f(x) = f(y). So is not one – one.

 

d.

Solution:

Here, f:[0, 3] → R is f(x) = x² is one – one because x, y ϵ [0, 3], x ≠ y implies f(x) ≠ f(y).

 

8. Examine whether the following function are one to one, onto, both or neither.

(i)

Solution:

Here, A = {1, –3, 3}, B = {1, 9}

And f:A→B is f(x) = x² is not one – one because –3 ≠ 3 but f(–3) = 9 = f(3). Moreover, the function f in onto because every element of B has pre–image in A. Hence f is not one–one but is onto.

 

(ii)

Solution:

Here, f: N→ N is f(x) = 2x.

To show: f is one – one and onto.

a.

f is one – one.

Let x, y ϵ N such that x ≠ y. Then to show: f(x) ≠ f(y).

Since, x ≠ y

= 2x ≠ 2y

= f(x) ≠ f(y)  [ since, f(x) = 2x]

 Therefore, f is one to one 

 

b.

f is onto.

Let y ϵ N such that y = f(x) = 2x

So, y = 2x.

= x = $\frac{{\rm{y}}}{2}$ ϵ N, if y is odd.

So, f is not onto.

Hence, the function is one–one but not onto..

 

(iii)

Solution:

Here, f:(–2, 2) → R is f(x) = x²

a.

The function is not one–one because.

–1 ≠ 1 but f(–1) = 1 = f(1)

 

b.

The function is not one–one because

f(–2, 2) = (0, 4) ≠ R.

Hence, f is neither one–one or onto.

 

(iv)

Solution:

Here, f:Q→Q is f(x) = 6x + 5.

To show: f is one–one and onto

 

a.

f is one – one and onto.

Let x,y ϵ Q (i.e. domain Q)

Such that x ≠ y. Then to show f(x) ≠ f(y)

Since, x ≠ y.

= 6x ≠ 6y

= 6x + 5 ≠ 6y + 5

= f(x) ≠ f(y)

So, f is one – one.

OR

 

b.

f is one–one: Let x, y ϵ Q such that f(x) = f(y). Then show: x = y.

Since f(x) = f(y)

= 6x + 5 = 6y + 5 →x = y.         [f is one–one]

 

b.

f is onto

Let y ϵ Q (Codomain Q)

Such that y = f(x) = 6x + 5

Then to show: x ϵ Q.

Since, y = 6x + 5

= 6x = y – 5

So, x = $\frac{{{\rm{y}} - 5}}{6}$                ϵ Q         [y ϵ Q]

And f$\left( {\frac{{{\rm{y}} - 5}}{6}} \right)$ = $6\left( {\frac{{{\rm{y}} - 5}}{6}} \right)$ + 5 = y – 5 + 5 = y

 

So, y is the image of $\frac{{\left( {{\rm{y}} - 5} \right)}}{6}.$

So, f is onto.

Hence, f is one–one and onto.

 

(v)

Solution:

Here, f:R→ R is f(x) = x³.

To show: f is one–one and onto.

a.

f is one–one.

Let x, y ϵ R(i.e. domain R) such that x ≠ y. Then to show: f(x) ≠ f(y).

Since x ≠ y.

= x³ ≠ y³ [cubes of two different real numbers is different]

= f(x) ≠ f(y)

So, f is one–one.

 

b.

f is onto

Let y ϵ R (i.e. Codomain R)

Such that y = f(x) = x³

To show: x ϵ R

Since y = x³

So, x = $\sqrt[3]{{\rm{y}}}$ ϵ R [Cube root of different real number are different]

So, x ϵ R.

And f($\sqrt[3]{{\rm{y}}}$) = ${\left( {{{\rm{y}}^3}} \right)^{\frac{1}{3}}}$ = y.So, f is onto.

Hence, f is one–one and onto.

 

9.

a.

Solution:

Here A = {–2, –1, 0, 1, 2}, B ={1, $\frac{1}{6}$, $\frac{2}{3}$} and f(x) = $\frac{{{{\rm{x}}^2}}}{6}$.

So, range of f = {f(–2), f(–1), f(0), f(1), f(2)} = $\left\{ {\frac{4}{6}.\frac{1}{6}.0.\frac{1}{6},{\rm{\: }}\frac{4}{6}} \right\}$

So, range of f = $\left\{ {0,{\rm{\: }}\frac{1}{6},{\rm{\: }}\frac{2}{3}} \right\}{\rm{\: }}$

The function is not one–one because f(–2) = $\frac{2}{3}$ = f(2) but –2 ≠ 2.

However, the function is onto because f(A) = B.

 

b.

Solution:

Here, A = {–1, 0, 1, 2, 3, 4}, B = $\left\{ { - 1,{\rm{\: }}0,{\rm{\: }}\frac{4}{5},{\rm{\: }}\frac{5}{7},{\rm{\: }}1,{\rm{\: }}2,{\rm{\: }}3} \right\}$ and f:A→B is f(x) = $\frac{{\left( {{\rm{x}} + 1} \right)}}{{2{\rm{x}} - 1}}$.

Now,

(i)

Range of f = {f(–1), f(0), f(1), f(2), f(3), f(4)}

= $\{ \frac{{ - 1 + 1}}{{2\left( { - 1} \right) - 1}},{\rm{\: }}\frac{{0 + 1}}{{2.0 - 1}},{\rm{\: }}\frac{{1 + 1}}{{2.1 - 1}},{\rm{\: }}\frac{{2 + 1}}{{2.2 - 1}},{\rm{\: }}\frac{{3 + 1}}{{2.3 - 1}},{\rm{\: }}\frac{{4 + 1}}{{2.4 - 1}}$

= $\left\{ {0,{\rm{\: }} - 1,{\rm{\: }}2,{\rm{\: }}1,{\rm{\: }}\frac{4}{5},{\rm{\: }}\frac{5}{7}} \right\}$

So, Range of f = $\left\{ { - 1,{\rm{\: }}0,{\rm{\: }}\frac{4}{5},{\rm{\: }}\frac{5}{7},{\rm{\: }}\frac{1}{2}} \right\}$

 

(ii)

The function f is one–one because each element if A has distinct image in B, But the function is not onto because the element 3 ϵ B does not have a pre-image in A.

 

(iii)

The function can be made one–one and onto both is 3 ϵ B. i.e. if

A = {–1, 0, 1, 2, 3, 4}, B = $\left\{ { - 1,{\rm{\: }}0,{\rm{\: }}\frac{4}{5},{\rm{\: }}\frac{5}{7},{\rm{\: }}1,{\rm{\: }}2} \right\}$ and

f:A→ B is f(x) = $\frac{{{\rm{x}} + 1}}{{2{\rm{x}} - 1}}$

Then the function f is both one–one and onto.