Relations,Functions and Graphs Exercise 2.3 | Basic Mathematics Solution [NEB UPDATED]

1

Solution:

Here,

a. f^–1(1) = {2}

b. f^–1(3) = {1, 3, 5}

c. f^–1(5) = {4}

d. f^–1(2) = ɸ

e. f^–1(1, 3, 5) = {1, 2, 3, 4, 5}

f. f^–1(2, 4) = ɸ

 

2.

a.

Solution:

Here, A = {0, 1, 2, 3}, B = {10, 13, 16, 19} and f:A→ B is f(0) = 10, f(1) = 13, f(2) = 16 and f(3) = 19.

Then f^–1:B → A as an ordered pair is {(10, 0), (13, 1), (16, 2), (19, 3)}

 

b.

Solution:

Here, x = {–1, 1, 2, 4}, Y = $\left\{ {\frac{1}{5},{\rm{\: }}\frac{2}{5},{\rm{\: }}\frac{1}{2},{\rm{\: }}1} \right\}{\rm{\: }}$

And f: X → Y is f(–1) = $\frac{1}{5}$, f(1) = 1, f(2) = $\frac{1}{2}$, f (4) = $\frac{2}{5}$.

Then, f^–1: Y → X as an ordered pairs is,

f^–1 = $\left\{ {\left( {\frac{1}{5},{\rm{\: }}1} \right),{\rm{\: }}\left( {1,{\rm{\: }}1} \right),{\rm{\: }}\left( {\frac{1}{2},{\rm{\: }}2} \right),{\rm{\: }}\left( {\frac{2}{5},{\rm{\: }}4} \right)} \right\}$

 

3.

a.

Solution:

Here, let y = f(x) = x + 1

So, y = f(x) → x = f^–1(y) …(i)

Also, y = x + 1.

So, x = y – 1.

Or, f^–1(y) = y – 1             [from (i)]

Hence, f^–1 (x) = x – 1

 

b.

Solution:

Here, let y = f(x) = 2x + 3

So, y = f(x) → x = f^–1(y) …(i)

Also, y = 2x + 3.

So, x = $\frac{{{\rm{y}} - 3}}{2}$

Or, f^–1 (y) = $\frac{{{\rm{y}} - 3}}{2}$  [from(i)]

Hence, f^–1(x) = $\frac{{{\rm{x}} - 3}}{2}$.

 

c.

Solution:

Here, let y = f(x) = 2x + 5

So, y = f(x) → x = f^–1(y) …(i)

Also, y = 2x + 5.

So, x = $\frac{{{\rm{y}} - 5}}{2}$

Or, f^–1 (y) = $\frac{{{\rm{y}} - 5}}{2}$  [from(i)]

Hence, f^–1(x) = $\frac{{{\rm{x}} - 5}}{2}$.

 

d.

Solution:

Here, let y = f(x) = x³ + 5

So, y = f(x) → x = f^–1(y) …(i)

Also, y = x³ + 5.

Or, x³ = y – 5

So, x = (y – 5)^1/3

Or, f^–1 (y) = (y – 5)^1/3  [from(i)]

Hence, f^–1(x) = (x – 5)^1/3.

 

e.

Solution:

Here, let y = f(x) = 3x – 2

So, y = f(x) = 3x – 2

So, y = f(x) → x = f^–1(y) …(i)

Also, y = 3x – 2

So, x = $\frac{{{\rm{y}} + 2}}{3}$

Or, f^–1 (y) = $\frac{{{\rm{y}} + 2}}{3}$  [from(i)]

Hence, f^–1(x) = $\frac{{{\rm{x}} + 2}}{3}$.

 

 

4.

Solution:

a.

Here, f(x) = 2x + 1 and g(x) = 3x – 1.

Now, (gof)(x) = g(f(x)) = g(2x + 1) = 3(2x + 1) –1 = 6x + 2.

And (fog)(x) = f(g(x)) = f(3x – 1) = 2(3x – 1) + 1 = 6x – 1.

 

b.

f(x) = 3x² – 4 and g(x) = 2x – 5

Now, (gof)(x) = g(f(x)) = g(3x² – 4) = 2(3x² – 4) – 5 = 6x² – 13.

And (fog)(x) = f(g(x)) = f(2x – 5) = 3(2x – 5)² – 4

 

c.

f(x) = x³ – 1 and g(x) = x²

Now, (gof)(x) = g(f(x)) = g(x³ – 1) = (x³ – 1)².

And (fog)(x) = f(x²) = (x²)³ – 1 = x⁶ – 1.

 

d.

f(x) = x² + 1 and g(x) = x⁵.

Now, (gof)(x) = g(f(x)) = g(x² + 1) = (x² + 1)⁵.

And (fog)(x) = f(g(x)) = f(x⁵) = (x⁵)² + 1 = x¹⁰ + 1.

 

5.

Solution:

a.

 For gof,

so, f(1) = 5, g(5) = 1.

f(2) = 6, g(6) = 2.

f(3) = 7, g(7) = 3.

f(4) = 6.

Now,

(gof)(1) = g(f(1)) = g(5) = 1.

(gof)(2) = g(f(2)) = g(6) = 2.

(gof)(3) = g(f(3)) = g(7) = 3.

(gof)(4) = g(f(4)) = g(6) = 2.

So, gof = {(1, 1), (2, 2), (3, 3), (4, 2)}

For, fog.

g(5) = 1, f(1) = 5.

g(6) = 2, f(2) = 6.

g(7) = 3, f(3) = 7.

Now, (fog)(5) = f(g(5)) = f(1) = 5.

(fog)(6) = f(g(6)) = f(2) = 6.

(fog)(7) = f(g(7)) = f(3) = 7.

So, fog = {(5, 5), (6, 6), (7, 7)}

 

b.

For gof,

so, f(1) = 2, f(3) = 5, f(4) = 1

And, g(2) = 3, g(5) = 1, g(1) = 3,

Now, (gof)(1) = g(f(1)) = g(2) = 3.

(gof)(3) = g(f(3)) = g(5) = 1.

(gof)(4) = g(f(4)) = g(1) = 3.

So, gof = {(1, 3), (3, 1), (4, 3)}

For, fog.

g(2) = 3, g(5) = 1, g(1) = 3.

f(3) = 5, f(1) = 2, f(4) = 1.

Now, (fog)(1) = f(g(1)) = f(3) = 5.

(fog)(2) = f(g(2)) = f(3) = 5.

(fog)(5) = f(g(5)) = f(1) = 2.

So, fog = {(1, 5), (2, 5), (5, 2)}

 

6.

a.

Solution:

Here, f : R → R is f(x) = $\frac{1}{{1 - {\rm{x}}}}$, (x≠1)

To show: (fof)(1/2) = –1.

Now, (fof)(x) = f(f(x)) = f $\left( {\frac{1}{{1 - {\rm{x}}}}} \right)$

= $\frac{1}{{1 - \left( {\frac{1}{{1 - {\rm{x}}}}} \right)}}$ = $\frac{{1 - {\rm{x}}}}{{1 - {\rm{x}} - 1}}$ = $\frac{{1 - {\rm{x}}}}{{ - {\rm{x}}}}$ = $\frac{{{\rm{x}} - 1}}{{\rm{x}}}$.

So, (fof)(x) = $\frac{{{\rm{x}} - 1}}{{\rm{x}}}$

So, (fof)(1/2) = $\frac{{\left( {\frac{1}{2} - 1} \right)}}{{\frac{1}{2}}}$ = $\frac{{1 - 2}}{{\frac{{\frac{2}{1}}}{2}}}$ = –1

Hence, (fof)(1/2) = –1.

 

b.

Solution:

Here, f:R→ R is f(x) = $\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}$, (x ≠ –1)

To show: (fof)(4) = –1/4.

Now, (fof)(x) = f(f(x)) = f $\left( {\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}}} \right)$

= $\frac{{\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}} - 1}}{{\frac{{{\rm{x}} - 1}}{{{\rm{x}} + 1}} + 1}}$ = $\frac{{{\rm{x}} - 1 - {\rm{x}} - 1}}{{{\rm{x}} - 1 + {\rm{x}} + 1}}$ = $\frac{{ - 2}}{{2{\rm{x}}}}$ = $ - \frac{1}{{\rm{x}}}$

So, (fof)(x) = $ - \frac{1}{{\rm{x}}}$

So, (fof)(4) = $ - \frac{1}{4}$.

 

7. Find the domain and range of the following function defined in the set of real numbers.

a. y = 3x + 1

Solution:  Here given function is, y = 3x + 1

Since y is defined for all real values of x, so the domain of the function is D(f) = R = (–∞, ∞).

Also, y can take all real values of the range of the function is;

R(f) = R = (–∞, ∞)

 

b. Y = x² – 1

Solution:

Given function is

Y = x² – 1

Since y is defined for all real values of x, so the domain of the function is D(f) = R = (–∞, ∞).

Also, y can take all real values of the range of the function is;

D(f) =  R(f) = R = (–∞, ∞).

since  Y  = x² – 1

therefore x²  =  y + 1 

since ,  x² ≥ 0 so, Y  + 1 ≥ 0

therefore y ≥ -1

hence, the range of the function  is R(f)  = [ -1 ,∞ 0)

 

c.  y = x³

 Solution:

Given function is:

y = x³

Since y is defined for all real values of x, so the domain of the function is D(f) = R = (–∞, ∞).
Since x is real x³ is real.

Also, y can take all real values of the range of the function is;

R(f) = R = (–∞, ∞).

 

d. y = –x² + 4x – 3

Solution:           

Given function is:

y = –x² + 4x – 3.

Since y is defined for all real values of x, so the domain of the function is D(f) = R = (–∞, ∞).
Again, y = –x² + 4x – 3 = –(x² – 4x) – 3 = –(x – 2)² + 4 – 1 = –(x – 2)² + 1.

Since, x – 2 ≥ 0 so y ≤ 1 for all real x, Hence, the range of the function is R(f) = (–∞, 1].

 

e. y = $\sqrt {{\rm{x}} - 2} {\rm{\: }}$

Solution:

Given function is:

y = $\sqrt {{\rm{x}} - 2} {\rm{\: }}$

Since, y is defined for the real values of x for which x – 2 .

So, x ≥2.

The domain, D(f) = [2, ∞)

Again, since xϵ [2, ∞) so the range is R(f) = [0, ∞).

 

f. y = $\frac{1}{{{\rm{x}} + 1}}{\rm{\: }}$

Solution:

Given function is:

y = $\frac{1}{{{\rm{x}} + 1}}{\rm{\: }}$

Since, y is defined for the real values of x except x + 1 = 0 .i.e x = –1.

Thus, the domain of the function is: D(f) = R – {–1}.

Since, Y = $\frac{1}{{{\rm{x}} + 1}}$

So, x + 1 = $\frac{1}{{\rm{y}}}$.

So, x = $\frac{1}{{\rm{y}}}$ – 1.

Since, x is not defined for y = 0. So the range of the function is,

R(f) = R – {0}.

 

j. y = $\frac{{{{\rm{x}}^2} - 16}}{{{\rm{x}} - 4}}{\rm{\: }}$

Solution:

Given function is:

y = $\frac{{{{\rm{x}}^2} - 16}}{{{\rm{x}} - 4}}{\rm{\: }}$

Since, y is not defined for x – 4 = 0 .i.e x = 4.

Thus, the domain of the function is: D(f) = R – {4}.

Again, if x ≠ 4. Then,

y = $\frac{{{{\rm{x}}^2} - 16}}{{{\rm{x}} - 4}}$ = x + 4.

Since, x = 4 is not defined for y = 8 will not be in the range,

Thus, the range is:

R(f) = R – {8}.

 

k. y =$\sqrt {{{\rm{x}}^2} - 2{\rm{x}} - 8} $

Solution:

Given function is:

y =$\sqrt {{{\rm{x}}^2} - 2{\rm{x}} - 8} $

Since, y is defined only when,,

X² – 2x – 8 ≥ 0

Or, x² – 4x + 2x – 8 ≥ 0

Or, x(x – 4) + 2(x – 4) ≥0.

So, (x – 4)(x + 2)≥0 …(i)

This is true only when

Either, x – 4 ≥ 0, x + 2 ≥ 0

= x ≥ 4                    x ≥ –2

OR,

x – 4 ≤ 0, x + 2 ≤ 0

= x ≤ 4                    x ≤ –2

= x ≥ 4 and x ≤ –2.         [x≥–2 and x ≤ 4 does not satisfy (i) if –2 < x < 4]

Thus, the fomain, D(f) = (–∞–2] U [4, ∞)

Again since,

y² = x² – 2x – 8.

= y² = (x – 1)² – 9

= y² + 9 = (x – 1)², are both sides positive,

Thus, the range is:

R(f) = {y:y ≥ 0} = [0, ∞)

 

l. y = $\sqrt {21 - 4{\rm{x}} - {{\rm{x}}^2}} $

Solution:

Given function is:

y = $\sqrt {21 - 4{\rm{x}} - {{\rm{x}}^2}} $

Since, y is defined only when,,

$21 - 4{\rm{x}} - {{\rm{x}}^2}$≥ 0

Or, 21 – 7x + 3x – x² ≥ 0

Or, 7(3 – x) + x(3 – x) ≥0.

So, (3 – x)(7 + x) ≥ 0 …(i)

This is true only when

Either, 3 – x ≥ 0, 7 + x ≥ 0

= 3 ≥ x                    7 ≥ –x

= x ≤ 3, x ≥ –7.

OR,

 3 – x ≤ 0, 7 + x ≤ 0

= 3 ≤ x          x ≤ –7

= x ≥ 3 and x ≤ –7.         [x≥3 and x ≤ –7 does not satisfy (i)]                       

= –7 ≤ x ≤ 3, i.e. [–7, 3]

Thus, the domain of the function is, D(f) = [–7, 3]

Again since,

y² = 21 – 4x – x². = 21 – (x² + 4x)

or, y² = 25 – (x+2)²      [(x+2)² = 25 – y²]

Since, (x+2)²≥ 0 for all real x.

SO, 25 – y² ≥ 0.                                [y² ≤ 25}

Since, y is positive square root so 0 ≤ y ≤ 5.

Hence, the range of f is:

R(f) = [0, 5]

 

m. y = $\frac{{\left| {{\rm{x}} - 1} \right|}}{{{\rm{x}} - 1}}{\rm{\: }}$

Solution:

Here, the given function is:

y = $\frac{{\left| {{\rm{x}} - 1} \right|}}{{{\rm{x}} - 1}}{\rm{\: }}$

Since, y is defined for all real values of x except at x = 1 so the domain D(f) = R – {–1}

Again, since y = $\frac{{\left| {{\rm{x}} - 1} \right|}}{{{\rm{x}} - 1}}$ = { 1, if  x – 1 ≥ 0 and –1, if x –1 < 0

Thus, the range R(f) = {–1, 1}

 

9.

Solution:

Test of one – one.

Let, x, yϵ R (domain)

And f(x) = f(y)

Then, cx + d = cy + d.

Then, x= y

So, f is one–one.

Test of onto.

Let yϵ R(range) and let y = cx + d.

i.e. x = $\frac{{{\rm{y}} - {\rm{d}}}}{{\rm{c}}}$ϵ R (domain)

So, for everyvalue yϵ R (range) there exist x ϵ R (domain), therefore given function is onto.

Hence, the given function is bijective, so there exits f^–1.

For this, let y = cx + d then x = $\frac{{\left( {{\rm{y}} - {\rm{d}}} \right)}}{{\rm{c}}}$.

I.e. f^–1(y)= $\frac{{\left( {{\rm{y}} - {\rm{d}}} \right)}}{{\rm{c}}}$

Since, y is a dummy variable, so interchanging y to x, we get,

f^–1(x)= $\frac{{\left( {{\rm{x}} - {\rm{d}}} \right)}}{{\rm{c}}}$

Now, f(f^–1(x)) = f$\left( {\frac{{{\rm{x}} - {\rm{d}}}}{{\rm{c}}}} \right)$ = c . $\frac{{\left( {{\rm{x}} - {\rm{d}}} \right)}}{{\rm{c}}} + {\rm{\: d}}$ = x

Again, f^–1(f(x)) = f^–1(cx+d) = $\frac{{\left( {{\rm{cx}} + {\rm{d}}} \right) - {\rm{d}}}}{{\rm{c}}}$ = x.

Hence, f(f^–1(x)) =,f^–1(f(x)).

 

10. (i)

Solution:

Test of one–one

Let x, y ϵ R – {2} and let f(x) = f(y).

Or, $\frac{{3{\rm{x}}}}{{{\rm{x}} - 2}}$ = $\frac{{3{\rm{y}}}}{{{\rm{y}} - 2}}$

Or, xy – 2x = xy – 2y

Or, x = y

So, the given function is one–one.

Test of onto.

Let y ϵ R – {3} and let y = $\frac{{3{\rm{x}}}}{{{\rm{x}} - 2}}$

Ie. xy – 2y = 3x

Or, x – 3x = 2y

Or, x (y – 3) = 2y                                                                   

Or, x = $\frac{{2{\rm{y}}}}{{{\rm{y}} - 3}}$ϵ R – {2}

So, the given function is onto.

Hence, the given function is bijective function. So, there exists inverse function:

For this, let y = $\frac{{3{\rm{x}}}}{{{\rm{x}} - 2}}$ then x = $\frac{{2{\rm{y}}}}{{{\rm{y}} - 3}}$.

I.e. f^–1(y) = $\frac{{2{\rm{y}}}}{{{\rm{y}} - 3}}$

Since, y is a dummy variable, so interchanging y to x,

f^–1(x) = $\frac{{2{\rm{x}}}}{{{\rm{x}} - 3}}$