### Question 1. Let f : {1, 3, 4} -> {1, 2, 5} and g : {1, 2, 5} -> {1, 3} be given by f = {(1, 2), (3, 5), (4, 1) and g = {(1, 3), (2, 5), (5, 1)}. Write down gof.

**Solution: **

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

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

f(1)= 2, g(2) = 3 => gof(1) = 3

f(3) = 5, g(5) = 1 => gof(3) = 1

f(4) =1, g(1) = 3 => gof(4) = 3

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

### Question 2. Let f, g and h be functions from R to R. Show that (f+g) oh = foh + goh, (f * g) oh = (foh) * (goh).

**Solution:**

f: R-> R, g: R-> R, h: R-> R

(f+g) oh(x) = (f+g) oh(x)

= (f+g) [h(x)]

= f[h(x)] + g[h(x)]

= foh(x) + goh(x)

(f+g) oh = foh + goh

(f * g) oh(x) = (f * g) oh(x)

= (f * g) [h(x)]

= f[h(x)] * g[h(x)]

= foh(x) * goh (x)

(f * g) oh = (foh) * (goh)

### Question 3. Find gof and fog, if

**(i) f(x) = |x| and g(x) = |5x – 2|**

**(ii) f(x) = 8x**^{3 }and g(x) = x^{1/3}

^{3 }and g(x) = x

^{1/3}

**Solution: **

(i)We have,f(x) = |x| and g (x) = | 5x – 2 |

gof(x) = g(f(x)) = g(|x|)

=> gof(x) = | 5 |x|-2 |

fog(x) = f(g(x)) = f(|5x-2|)

=> fog(x) = || 5x-2|| = | 5x -2 |

(ii)We have,f(x) = 8x

^{3}and g(x) = x^{1/3}gof(x) = g(f(x)) = g(8x

^{3})=> gof(x) = (8x

^{3})^{1/3}= 2xfog(x) = f(g(x)) = f(x

^{1/3})=> fog(x) = 8(x

^{1/3})^{3}= 8x

### Question 4. If f(x) = , show that fof(x) = x for all . What is the inverse of f ?

**Solution: **

Given that,

Now,

fof(x) = f(f(x)) =

=

On simplifying by taking LCM = (6x-4)

fof(x) =

=> fof (x) = = x

=> fof(x) = I

_{A }(x) for all=> fof(x) = I

_{A}such that A = R – which is the domain of f=> f

^{-1}= fHence, proved.

### Question 5. State with reason whether the following functions have inverse. Find the inverse, if it exists.

**(i) f : {1, 2, 3, 4} -> {10}**

**with f = {(1, 10), (2, 10), (3,10), (4,10)}**

**(ii) g: {5, 6, 7, 8} -> {1, 2, 3, 4}**

**with g = {(5, 4), (6, 3), (7, 4), (8, 2)}**

**(iii) h : {2, 3, 4, 5} -> {7, 9, 11, 13}**

**with h : {(2, 7), (3, 9), (4, 11), (5, 13)}**

**Solution:**

(i)We have f(1) = f(2) = f(3) = f(4) = 10 which means that f is many-oneand not one-one, therefore inverse of f does not exist.

(ii)Here g(5) = g(7) =4 i.e. g is many-one, so inverse of g does not exist.

(iii)Since range of h = {7, 9, 11, 13} = co-domain, therefore h is onto,Also, each element of domain has a unique image in h, therefore h is one-one.

Now, since h is both one-one and onto,thus inverse of h exists.

h

^{-1 }= {(7, 2), (9, 3), (11, 4), (13, 5)}

### Question 6. Show that f : [-1, 1] -> R, given by f(x) = is one-one. Find the inverse of the function f : {-1,1} -> Range f.

**Solution: **

Let x, y

_{ }[-1, 1]f(x) =

f(y) =

Now,

Let f(x) = f(y)

=> x(x + 2) = y(x + 2)

=> x y + 2x = x y + 2y

=> 2x = 2y

=> x = y

=> f is one-one

Also,

X = [-1, 1] and,

Y = { } = range of f.

=> f is onto

Since f is one-one and onto, therefore inverse of f exists.

Let y = f(x) => x =f

^{-1}(y)=> y =

=> x y + 2y = x

=> 2y = x(1 – y)

=> x =

Therefore, f : Y-> X is defined by f(y) = .

### Question 7. Consider f : R -> R is given by f(x) = 4x + 3 . Show that f is invertible. Find the inverse of f.

**Solution:**

It is given that,

f(x) = 4x + 3 where f : R -> R

Let,

f(x) = f(y)

=> 4x + 3 = 4y + 3

=> 4x = 4y

=> x = y

=> f is one-one function

Also,

Let y = 4x + 3 where y R

=> x =

Since for any . there exists such that

f(x) = = 4 +3 = y

=> f is onto

Since f is both one-one and onto, therefore f

^{-1}exists=> f

^{-1}(y) =

### Question 8. Consider f : R_{+} -> [4, ) given by f(x) = x^{2} + 4. Show that f is invertible with the inverse f^{-1} of f given by f^{-1}(y) = , where R_{+} is the set of all non-negative real numbers.

**Solution: **

Let f(x) = f(y)

=> x + 4 = y + 4

=> x

^{2}= y^{2}=> x = y [ x,y R

_{+}]=> f is one-one

Let y = x

^{2}+ 4 where y=> x

^{2}= y – 4 4 [ y=> x =

Therefore, for any y , there exists x =

=> f is onto

Since, f is both one-one and onto, f

^{-1}exists for every ,=> f

^{-1}(y) =

### Question 9. Consider R_{+ }-> [ -5, ) given by f (x) = 9x^{2} + 6x -5. Show that f is invertible with f^{-1} (y) =

**Solution: **

Let f(x) = f(y)

=> 9x

^{2}+ 6x -5 = 9y^{2}+ 6y – 5=> 9x

^{2}+ 6x = 9y^{2}+ 6y=> 9(x

^{2}– y^{2}) + 6 (x – y) = 0=> (x – y) [9 (x + y) + 6] = 0

=> x – y =0

=> x = y

=> f is one-one

Now, let y = 9x

^{2}+ 6x – 5=> 9x

^{2}+ 6x – 5 (x + y) = 0=> x =

=> f(x) =

On simpliying, we have f (x) = y

=> f is onto

Since f is both one-one and onto. f

^{-1}existsf

^{-1}(y) =

### Question 10. Let f : X -> Y be an invertible function. Show that f has unique inverse.

**Solution:**

We have,

f : X -> Y is an invertible function

Let g and h be two distinct inverses of f.

Then, for all y Y,

fog (y) = I (y) = foh (y)

=> f g (y)) = f(h (y))

=> g(y) = h(y) [f is one-one]

=> g = h [g is one-one]

which contradicts our supposition.

Hence, f has a unique inverse.

### Question 11. Consider f : {1, 2, 3} -> {a, b, c} given by f (1) = a, f (2) = b, f (3) = c. Find f and show that (f^{-1})f^{-1} = f.

**Solution:**

Given that,

f(1) = a, f(2) = b, f(3) = c

We have,

f = {(1, a), (2, b), (c, 3)}

which shows that f is both one-one and onto and thus f is invertible.

Therefore,

f

^{-1}= {(a, 1), (b, 2), (c, 3)}Also,

(f

^{-1})^{-1}= {(1, a), (2,b), (3, c)}=> (f

^{-1})^{-1}= fHence proved.

### Question 12. Let f: X -> Y be an invertible function. Show that the inverse of f^{-1} is f, i.e., (f^{-1} )^{-1} = f.

**Solution:**

Since, f is an invertible function,

=> f is both one-one and onto

Also,

Let g : Y -> X , where g is a one-one and onto function such that

gof (x) = I

_{x}and fog (y) = I_{y}=> g = f^{-1}=> f

^{-1}o (f^{-1})^{-1}= I=> f o [f

^{-1}o (f^{-1})^{-1}] = f o I=> (f o f

^{-1}) o (f^{-1})^{-1}= f=> I o (f

^{-1})^{-1}= fHence, (f

^{-1})^{-1}= f

### Question 13. If f : R -> R given by f (x) = (3 – x^{3})^{1/3 },then fof (x) is :

### (A) x^{1/3 }(B) x^{3} (C) x. (D) (3 – x^{3})

**Solution:**

Answer: (C)We have,

f(x) = (3 – x

^{3})^{1/3}where f : R -> RNow,

fof(x) = f(f(x))

=> fof(x) = f((3 – x

^{3})^{1/3})=> fof(x) = [3 – ((3 – x

^{3})^{1/3})^{3}]^{1/3}=> fof(x) = [3 – (3 – x

^{3})]^{1/3}=> fof(x) = (x

^{3})^{1/3}=> fof(x) = x

Hence, option C is correct.

### Question 14. Let f : R -{ } -> R be a function defined as f(x) = . The inverse of f is the map g : Range f -> R – { } given by

**(A) g(y) = **** (B) g(y) = **

**(C) g(y) = **** (D) g(y) = **

**Solution:**

Answer: (B)Let y = f(x)

=> y =

=> 3xy + 4y = 4x

=> x( 4 – 3y) = 4y

=> x =

f

^{-1}(y) = g (y) =

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