Inverse Functions - Solved Exercises | Class 10 Readmore Optional Mathematics

4 - Inverse Functions

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Before starting, you must check:

1. Notes of Inverse Functions - Class 10

Remember the following points:

1. The function obtained by inversing a function may not necessarily be a function.
2. For the inverse of a function to be a function, the initial function should be a one to one onto function.

Here are more solutions from Algebra:

1. Composite Function
2. Inverse Function
3. Remainder Theorem
4. Factor Theorem

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1. Find the inverse function of the following function:

a) f={(x,y): y=x and x = {1,2,3}}

Solution:

Given,

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

So,

Inverse of function f = {(1,1),(2,2),(3,3)}

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2a) If f={(1,2),(2,3),(4,5)}, find $f^{-1}$ and the domain and range of  $f^{-1}$.

Solution:

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

Domain of $f^{-1}$ = {2,3,5}

Range of $f^{-1}$ = {1,2,4}

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3a) If f={(1,5),(2,4),(3,6)} and fog={(5,6),(4,5),(6,4)} then find $g^{-1}$.

Solution:

fog={(5,6),(4,5),(6,4)}

f = {(1,5),(2,4),(3,6)}

We know, in fog, f is the range and g is the domain.

So, g = {(5,3),(4,1),(6,2)}

And, $g^{-1}$ = {(1,4),(2,6),(3,5)}

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4. Find the inverse functions of the following functions:

a) f(x) = 2x + 3

Solution:

Let, f(x) = y = 2x + 3

Interchanging the roles of x and y, we get,

x = 2y + 3

or, x - 3 = 2y

or, y = (x - 3)/2

So, $f^{-1} = \dfrac{x - 3}{2}$

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c) f(x) = (x + 5)/3

Solution:

Let f(x) = y = (x + 5)/3

Interchanging the roles of x and y, we get,

x = (y + 5)/3

or, 3x = y + 5

or, 3x - 5 = y

or, y = 3x - 5

So, $f^{-1} = 3x - 5$

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e) f:x - > (2x + 5)/6

Solution:

Let f(x) = y = (2x+5)/6

Interchanging the roles of x and y, we get,

x = (2y +5)/6

or, 6x = 2y + 5

or, 6x - 5 = 2y

or, y = (6x -5)/2

So, $f^{-1} = \dfrac{6x - 5}{2}$

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g) f:x -> (2x +3)/(x -3); x ≠ 3

Solution:

Let f(x) = y = (2x + 3)/(x - 3)

Interchanging the roles of x and y, we get,

x = (2y + 3)/(y - 3)

or, x(y - 3) = 2y + 3

or, xy - 3x = 2y + 3

or, xy - 2y - 3x = 3

or, xy - 2y = 3 + 3x

or, y(x - 2) = 3x + 3

or, y = (3x + 3)/(x - 2)

So, $f^{-1} = \dfrac{3x + 3}{x - 2}$

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i) f:x -> (4x + 7)/(5x + 3)

Solution:

Let f(x) = y = (4x + 7)/(5x +3)

Interchanging the roles of x and y, we get,

x = (4y +7)/(5y +3)

or, x(5y + 3) = 4y +7

or, 5xy + 3x = 4y + 7

or, 3x - 7 = 4y - 5xy

or, 3x - 7 = y(4 - 5x)

or, y(4 -5x) = 3x - 7

or, y = (3x - 7)/(4 - 5x)

So, $f^{-1} = \dfrac{3x - 7}{4 - 5x}$

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5a) If f(x) = (x -5)/4 and f-¹(x) = 25, find the value of x.

Solution:

Let f(x) = y = (x -5)/4

Interchanging roles of x and y, we get,

x = (y -5)/4

or, 4x = y - 5

or, y = 4x + 5

So, f-¹(x) = 4x + 5

Given,

f-¹(x) = 25

or, 4x + 5 = 25

or, 4x = 25 - 5

or, 4x = 20

So, value of x = 5

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6a) If f(x) = 8 - 3x, evaluate f-¹(-4) and ff(2).

Solution:

Let f(x) = y = 8 - 3x

Interchanging the roles of x and y, we get,

x = 8 - 3y

or, 3y + x = 8

or, 3y = 8 - x

or, y = (8-x)/3

So, f-¹(x) = (8 - x)/3

So, f-¹(-4) = {8 - (-4)}/3 = (8+4)/3 = 12/3 = 4

And,

ff(x) = fof(x) = f[f(x)]

= f(8 - 3x)

= 8 - 3(8 - 3x)

= 8 - 24 + 9x

= 9x - 16

So, ff(2) = 9×2 - 16 = 18 - 16 = 2

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7a) If f(4x+5) = 12x + 18 then find f-¹(x).

Solution:

Here,

f(4x + 5) = 12x + 18

or, f(4x + 5) = 12x + 15 + 3

or, f(4x + 5) = 3(4x + 5) + 3

or, f(x) = 3x + 3

Let f(x) = y = 3x + 3

Interchanging the roles of x and y, we get,

x = 3y + 3

or, x - 3 = 3y

or, (x - 3)/3 = y

So, f-¹(x) = (x -3)/3

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8a) If f(x) = 2x +7, find f(x+2) and $f^{-}$(x+2).

Solution:

Given,

f(x) = 2x +7

Now,

f(x+2) = 2(x +2) + 7

= 2x + 4 + 7

= 2x + 11

And,

Let f(x) = y = 2x +11

Interchanging the roles of x and y, we get,

or, x = 2y + 11

or, x -11 = 2y

or, 2y = x -11

or, y = $\frac{x -11}{2}$

So, $f^{-1}$(x+2) = $\frac{x -11}{2}$

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9a) If f(x) = 2x +1, g(x) = $\frac{x-5}{2}$ then find the value of $f^{-1}og^{-1}$(3).

Solution:

Given,

f(x) = 2x +1

Let, f(x) = y = 2x +1

or, y = 2x +1

Interchanging the roles of x and y, we get,

or, x = 2y +1

or, x -1 = 2y

or, 2y = x -1

or, y = $\frac{x -1}{2}$

So, $f^{-1}(x) = \frac{x-1}{2}$

And,

g(x) = $\frac{x-5}{2}$

Let, g(x) = y = $\frac{x-5}{2}$

or, y = $\frac{x-5}{2}$

Interchanging the roles of x and y, we get,

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

or, 2x = y -5

or, 2x +5 = y

So, $g^{-1}(x) = 2x+5$

Now,

$f^{-1}g{-1}(x) = f^{-1}[g^{-1}(x)]$

= $f^{-1}[2x +5]$

= $\frac{2x + 5-1}{2}$

= $\frac{2x+4}{2}$

= $\frac{2(x+2)}{2}$

= $x+2$

So,

$f^{-1}g{-1}(3) = (3)+2 = 5$

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10a) If f(x) = 1 +2x and g(x) = $\frac{1}{1-x}$, where x ≠ 1, find the value of $g^{-1}\frac{1}{2}$ and fog(-1).

Solution:

Given,

f(x) = 1 + 2x

g(x) = $\frac{1}{1-x}$, x≠1

Let g(x) = y = $\frac{1}{1-x}$

or, y = $\frac{1}{1-x}$

Interchanging the roles of x and y, we get,

or, x = $\frac{1}{1 - y}$

or, x(1 - y) = 1

or, x - xy = 1

or, x -1 = xy

or, $\frac{x-1}{x}$ = y

So, $g^{-1}(x) = \frac{x-1}{x}$

And,

$g^{-1}(\frac{1}{2}) = \dfrac{ \frac{1}{2} - 1}{\frac{1}{2}}$

= $\dfrac{1 - 2}{2} × \dfrac{2}{1}$

= $-1$

Now,

fog(x) = f[g(x)]

= f[$\frac{1}{1-x}$]

= 1 + $2 × \frac{1}{1 - x}$

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

Again,

fog(-1) = 1 + $\frac{2}{1-(-1)}$

= 1 + $\frac{2}{2}$

= 1 + 1

= 2

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11a) If f(x) = 3x + 4; g(x) = 2(x +1), prove that fog = gof and find the value of $f^{-1}$(2).

Solution:

Given,

f(x) = 3x + 4

g(x) = 2(x + 1)

To prove: fog = gof

Taking LHS

fog(x) = f[g(x)]

= f[2(x +1)]

= f[2x + 2]

= 3(2x + 2) + 4

= 6x + 6 + 4

= 6x + 10

Taking RHS

gof(x) = g[f(x)]

= g[3x + 4]

= 2(3x + 4 + 1)

= 2(3x + 5)

= 6x + 10

Since, LHS = 6x + 10 = RHS

fog(x) = gof(x) #proved.

Again,

Let f(x) = y = 3x + 4

or, y = 3x + 4

Interchanging the roles of x and y

or, x = 3y + 4

or, x - 4 = 3y

or, 3y = x - 4

or, y = $\frac{x - 4}{3}$

$\therefore f^{-1}$(x) = $\frac{x -4}{3}$

Put x = 2 in $f^{-1}(x)$, we get,

$f^{-1}(2) = \dfrac{2-4}{3}$

= \dfrac{-2}{3}$

= - \dfrac{2}{3}$

Hence, the required value of $f^{-1}$(2) is -2/3.

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12a) Given that function f(x) = 3x + a. If ff (6) = 10, find the value of a and $f^{-1}$(4).

Solution:

Given,

f(x) = 3x + a

We know,

fof (6) = f[f(x)]

or, 10 = f[3x + a]

or, 10 = 3(3x + a) + a

or, 10 = 9(6) + 3a + a

or, 10 = 9(6) + 4a

or, 10 = 54 + 4a

or, 4a = 10 - 54

or, 4a = - 44

or, 4 × a = 4 × (-11)

So, a = -11

Now,

f(x) = 3x + (-11) = 3x -11

Let y = f(x) = 3x - 11

or, y = 3x - 11

Interchanging the roles of x and y

or, x = 3y - 11

or, x + 11 = 3y

or, y = $\frac{x + 11}{3}$

$\therefore f^{-1}(x) = \frac{x +11}{3}$

And,

$f^{-1}$(4) = $\frac{4 +11}{3}$

$= \frac{15}{3}$

$= 5$

So, $f^{-1}$(-4) = 5.

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13a) Given that f(x)=4x +7 and g(x) = 3x -5. If $fg^{-1}$ = 15, find the value of x.

Solution:

Given,

f(x) = 4x +7

g(x) = 3x -4

Let g(x) = y = 3x - 4

or, y = 3x - 4

Interchanging the roles of x and y, we get,

or, x = 3y - 4

or, x - 4 = 3y

or, y = $\frac{x - 4}{3}$

$\therefore g^{-1}(x) = \frac{x - 4}{3}$

Now,

$fg^{-1}$(x) = f[$g^{-1}$(x)]

or, 15 = f[$\frac{x -4}{3}$]

or, 15 = 4 × $\frac{x - 4}{3}$ + 7

or, 15 - 7 = $\frac{4x - 16}{3}$

or, 8×3 = 4x - 16

or, 24 + 16 = 4x

or, 40 = 4x

$\therefore$ x = 10

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14a) If f(x) = 5 - $\frac{6}{x}$ then what is the value of x at $f(x) = - f^{-1}(-1)$?

Solution:

Here,

f(x) = 5 - $\frac{6}{x}$

Let y = f(x) = 5 - $\frac{6}{x}$

$or, y = 5 -$\frac{6}{x}$

Interchanging the roles of x and y, we get,

or, x = 5 - $\frac{6}{y}$

or, $\frac{6}{y}$ = (5 - x)

or, y = $\frac{6}{5-x}$

So, $f^{-1}$(x) = $\frac{6}{5-x}$

Also, $f^{-1}$(-1) = $\frac{6}{5-(-1)}$

= $\frac{6}{6}$

= 1

Given,

$f(x) = - f^{-1}(-1)$

$or, 5 - \frac{6}{x} = - (1)$

$or, \frac{5x -6}{x} = -1$

$or, 5x - 6 = -x$

$or, 5x + x = 6$

$or, 6x = 6$

$\therefore x = 1$

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15a) If f(x) = 4x + 5 and g(x) = 8x + 9, prove that $f^{-1}$og(x) is a linear function.

Solution:

A function in the form of f(x) = mx + c, is said to be a linear function.

Given,

f(x) = 4x + 5

g(x) = 8x + 9

Let y = f(x) = 4x + 5

or, y = 4x + 5

Interchanging the roles of x and y, we get,

or, x = 4y + 5

or, x - 5 = 4y

or, y = $\frac{x - 5}{4}$

$\therefore f^{-1}(x) = \frac{x - 5}{4}$

Now,

$f^{-1}$og(x) = $f^{-1}$[g(x)]

$= f^{-1}[ 8x + 9]$

$= \frac{8x + 9 - 5}{4}$

$= \frac{8x + 4}{4}$

$= \frac{4(2x + 1)}{4}$

$= 2x +1$

Hence, $f^{-1}og(x)$ is a linear function. #proved.

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16a) Let f(x) = $\frac{3x +5}{x -2}$ and g(x) = 7 be two functions. Prove that $f^{-1}$og(x) is a constant function.

Solution:

A function in the form of f(x) = c where c is a constant number is said to be a constant function.

Here,

f(x) = $\frac{3x +5}{x -2}$

g(x) = 7

Let y = f(x) = $\frac{3x + 5}{x -2}$

or, y = $\frac{3x +5}{x -2}$

Interchanging the roles of x and y, we get,

$or, x = \frac{3y + 5}{y -2}$

$or, x(y - 2) = 3y  + 5$

$or, xy - 2x = 3y +5$

$or, xy - 3y - 2x = 5$

$or, y(x - 3) = 2x +5$

$or, y = \frac{2x +5}{x -3}$

$\therefore f^{-1}(x) = \frac{2x +5}{x -3}$

Now,

$f^{-1}og(x) = f^{-1}[g(x)]$

$= f^{-1}[9]$

$= \frac{2×9+5}{9-3}$

$= \frac{18+5}{6}$

$= \frac{23}{6}$

Hence, $f^{-1}og(x)$ is a constant function. #proved.

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17a) Let, f(x) = 2x +3 and g(x) =x²+2x +1 be two functions. Prove that $f^{-1}$og(x) is a quadratic function.

Solution:

Given,

f(x) = 2x + 3

g(x) = x² +2x +1 = (x +1)²

Let y = f(x) = 2x + 3

or, y = 2x + 3

Interchanging the roles of x and y, we get,

or, x = 2y + 3

or, x - 3 = 2y

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

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

Now,

$f^{-1}og(x) = f^{-1}[g(x)]$

$= f^{-1}[(x +1)²]$

$= 2(x +1)² + 3$

It is the equation of a parabola which is definitely a quadratic function.

$= 2(x²+2x+1)+3$

$= 2x² + 4x + 2 + 3$

$= 2x² + 4x + 5$ #proved

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18 a) Given that f(x) = x³ + x² + 1 and g(x) = x -1. Prove that $fog^{-1}$ is a cubic function.

Solution:

Let y = g(x) = x - 1

or, y = x -1

Interchanging the roles of x and y, we get,

or, x = y - 1

or, y = x + 1

$\therefore g^{-1}(x) = x + 1$

Now,

$fog^{-1}(x)$= $f[g^{-1}(x)]$

$= f[x + 1]$

$= (x +1)^3 + (x +1)^2 + 1$

$= x^3 + 3x^2 + 3x + 1 + x^2 + 2x + 1 + 1$

$= x^3 + 4x^2 + 5x + 3$

Since, the above function has its highest degree equal to 3, it is proved that $fog^{-1}(x)$ is a cubic function.

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19 a) Given that f(x) = 3x + 5 and g(x) = $\dfrac{3x +2}{4}. What value of x makes$fog^{-1}(x)$ an identity function?

Solution:

Let y = g(x) = $\dfrac{3x +2}{4}$

or, y = $\dfrac{3x +2}{4}$

Interchanging the roles of x and y, we get,

$or, x = \dfrac{3y +2}{4}$

$or, 4x = 3y +2$

$or, 4x - 2 = 3y$

$\therefore y = \dfrac{4x -2}{3}$

Given,

$fog^{-1}(x) = x$

$or, f[g^{-1}(x)] = x$

$or, f[ \frac{4x -2}{3} ] = x$

$or, 3 × \dfrac{4x -2}{3} + 5 = x$

$or, 4x - 2 + 5 = x$

$or, 4x - x + 3 = 0$

$or, 3x = - 3$

$\therefore x = -1$

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About Readmore Optional Mathematics Book 10

Author: D. R. Simkhada

Editor: I. R. Simkhada

Readmore Publishers and Distributors

T.U. Road, Kuleshwor - 14, Kathmandu, Nepal

Phone: 4672071, 5187211, 5187226

readmorenepal6@gmail.com

About this page:

Inverse Functions - Solved Exercises | Class 10 Readmore Optional Mathematics is a collection of the solutions related to exercises of Inverse Functions from the Algebra chapter for Nepal's Secondary Education Examination (SEE) appearing students. 

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