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## Class - 12 : Mathmatics

### Mathmatics - Chapter No. : 1

**Relation Functions**

**Exercise 1.3**

**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, 3), (5, 1)}. Write down gof.**

**Answer :**

The functions f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} are defined as:

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

gof(1) = g[f(1)] = g(2) = 3 [Since f(1) = 2 and g(2) = 3]

gof(3) = g[f(3)] = g(5) = 1 [Since f(3) = 5 and g(5) = 1]

gof(4) = g[f(4)] = g(1) = 3 [Since f(4) = 1 and g(1) = 3]

So, 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)**

**Answer :**

To prove: (f + g)oh = foh + goh

LHS = [(f + g)oh](x)

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

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

= (foh)(x) + (goh)(x)

= {(foh)(x) + (goh)}(x)

= RHS

So, {(f + g)oh}(x) = {(foh)(x) + (goh)}(x) for all x ∈R

Hence, (f + g)oh = foh + goh

To Prove: (f.g)oh = (foh).(goh)

LHS = [(f.g)oh](x)

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

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

= (foh)(x) . (goh)(x)

= {(foh).(goh)}(x) = RHS

So, [(f.g)oh](x) = {(foh).(goh)}(x) for all x ∈R

Hence, (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}

**Answer :**

(i). f(x) = |x| and g(x) = |5x-2|

So, gof(x) = g(f(x)) = g(|x|) = |5|x|-2|

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

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

So, gof(x) = g(f(x)) = g(8x^{3}) = (8x^{3})^{1/3} = 2x

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

**Question : 4
If f(x) = (4x+3)/(6x−4), x ≠ 2/3, show that fof(x) = x, for all x ≠ 2/3. What is the inverse of f?**

**Answer :**

It is given that f(x) = (4x+3)/(6x-4), x ≠ 2/3

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

= f((4x + 3)/(6x – 4))

= {4(4x + 3)/(6x – 4) + 3}/{6(4x + 3)/(6x – 4) - 4}

= (16x + 12 + 18x – 12)/(24x + 18 - 24x + 16)

= 34x/34

= x

So, fof(x) = x, for all x ≠ 2/3

⇒fof = I_{x}

Hence, the given function f is invertible and the inverse of f is f itself.

**Question : 5
State with reason whether following functions have inverse
(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)}**

**Answer :**

(i) f: {1, 2, 3, 4} → {10} defined as f = {(1, 10), (2, 10), (3, 10), (4, 10)}

From the given definition of f, we can see that f is a many one function as

f(1) = f(2) = f(3) = f(4) = 10

So, f is not one – one.

Hence, function f does not have an inverse.

(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as

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

From the given definition of g, it is seen that g is a many one function as

g(5) = g(7) = 4

So, g is not one – one.

Hence, function g does not have an inverse.

(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as

h = {(2, 7), (3, 9), (4, 11), (5, 13)}

It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h.

So, the function h is one – one.

Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in

the set {2, 3, 4, 5}, such that h(x) = y

Thus, h is a one – one and onto function.

Hence, h has an inverse.

**Question : 6
Show that f : [−1, 1] → R, given byf(x) = x/(x+2) is one – one. Find the inverse of the function f : [−1, 1] → Range f
(Hint: For y ∈ Range f, y = f(x) = x/(x+2), for some x in [−1, 1], i.e., x = 2y/(1−y)**

**Answer :**

Given, f : [−1, 1] → R is given as f(x) = x/(x+2)

For one – one:

Let f(x) = f(y)

⇒ x/(x + 2) = y/(y + 2)

⇒xy +2x = xy +2y

⇒ 2x = 2y

⇒ x = y

So, f is a one – one function.

It is clear that f : [−1, 1] → Range f is onto.

So, f : [−1, 1] → Range f is one – one and onto and therefore, the inverse of the function

f : [−1, 1] → Range f exists.

Let g : Range f → [−1, 1] be the inverse of f.

Let y be an arbitrary element of range f.

Since f : [−1, 1] → Range f is onto, we have

y = f(x) for some x ∈ [−1, 1]

⇒ y = x/(x + 2)

⇒xy + 2y = x

⇒ x(1-y) = 2y

⇒ x = 2y/(1 – y), y ≠ 1

Now, let us define g : Range f → [−1, 1] as

g(y) = 2y/(1 – y), y ≠ 1

Now, (gof)(x) = g(f(x))

= g(x/(x + 2))

= 2x/(x + 2)/{1 - x/(x + 2)}

= 2x/(x + 2 - x)

= 2x/2

= x

and (fog)(x) = f(g(x))

= f(2y/(1- y))

= 2y/(1- y)/{ 2y/(1- y) + 2}

= 2y/(2y + 2 – 2y)

= 2y/2

= y

So, gof = x = I_{[-1, 1]} and fog = y = I_{Range f}

So, f^{-1} = g

=> f^{-1}(y) = 2y/(1 - y), y ≠ 1

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

**Answer :**

Given, f: R → R is given by, f(x) = 4x + 3

For one – one:

Let f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y

⇒ x = y

So, f is a one – one function.

For onto:

For y ∈ R, let y = 4x + 3.

⇒ x = (y – 3)/4 ∈ R

Therefore, for any y ∈ R, there exists x = (y – 3)/4 ∈ R, such that

f(x) = f((y – 3)/4) = 4((y – 3)/4)) + 3 = y

So, f is onto.

Thus, f is one – one and onto and therefore, f^{−1} exists.

Let us define g: R → R by g(x) = (y – 3)/4

Now,

(gof)(x) = g(f(x)) = g(4x + 3) = {(4x + 3) – 3}/4 = 4x/4 = x

and (fog)(y) = f(g(y)) = f((y - 3)/4) = 4(y - 3)/4 + 3 = y – 3 + 3 = y

So, gof = fog = I_{R}

Hence, f is invertible and the inverse of f is given by f^{-1} (y) = g(y) = (y – 3)/4

**Question : 8
Consider f: R _{+}→ [4, ∞) given by f(x) = x^{2} + 4. Show that f is invertible with the inverse f^{−1} of given f by f^{−1}(y) = √(y – 4),
where R+ is the set of all non-negative real numbers.**

**Answer :**

Given, f : R_{+}→ [4, ∞) is given as f(x) = x^{2} + 4

For one – one:

Let f(x) = f(y)

⇒ x^{2} + 4 = y^{2} + 4

⇒ x^{2} = y^{2}

⇒ x = y [as x = y ∈ R_{+}]

So, f is a one – one function.

For onto:

For y ∈ [4, ∞), let y = x^{2} + 4

⇒ x^{2} = y − 4 ≥ 0 [as y ≥ 4]

⇒ x = √(y – 4) ≥ 0

Therefore, for any y ∈ [4, ∞), there exists x = √(y – 4) ∈ R_{+}, such that

f(x) = f(√(y -4)) = (√(y -4))^{2} + 4 = y - 4 + 4 = y

So, f is onto.

Thus, f is one – one and onto and therefore, f^{−1} exists.

Let us define g: [4, ∞) → R+ by g(y) = √(y – 4)

Now,

(gof)(x) = g(f(x)) = g(x^{2} + 4) = √{x^{2} + 4 – 4} = √x^{2} = x

and

(fog)(y) = f(g(y)) = f(√(y - 4)) = (√(y – 4))^{2} + 4 = y - 4 + 4 = y

So, gof = fog = I_{R}

Hence, f is invertible and the inverse of f is given by

f^{-1}(y) = g(y) = √(y – 4)

**Question : 9
Consider f: R _{+}→ [−5, ∞) given by f(x) = 9x^{2} + 6x − 5. Show that f is invertible with f^{−1}(y) = [{√(y + 6) − 1}/3]**

**Answer :**

f: R+ → [−5, ∞) is given as f(x) = 9x^{2} + 6x − 5.

Let y be an arbitrary element of [−5, ∞).

Let y = 9x^{2} + 6x – 5

⇒ y = (3x + 1)^{2}– 1 - 5

⇒ y = (3x + 1)^{2} - 6

⇒ y + 6 = (3x + 1)^{2}

⇒ 3x + 1 = √(y + 6) [as y ≥ −5 ⟹ y + 6 > 0]

⇒ x =[{√(y + 6) − 1}/3]

So, f is onto, thereby range f = [−5, ∞).

Let us define g: [−5, ∞) → R+ as g(y) = [{√(y + 6) − 1}/3]

Now,

(gof)(x) = g(f(x)) = g(9x^{2} + 6x-5) = g((3x + 1)^{2} - 6)

= [√{(3x + 1)^{2} - 6 + 6} – 1]/3

= (3x + 1 – 1)/3

= 3x/3

= x

and (fog)(y) = f(g(y)) = f ({√(y + 6) − 1}/3)

= [3{√(y + 6) − 1}/3 + 1]^{2} - 6

= {√(y + 6)}^{2} - 1

= y + 6 - 6

= y

So, gof = x = I_{R} and fog = y = I_{Range f}

Hence, f is invertible and the inverse of f is given by

f^{−1}(y) = g(y) = {√(y + 6) – 1}/3

**Question : 10
Let f : X → Y be an invertible function. Show that f has unique inverse. (Hint: suppose g _{1} and g_{2} are two inverses of f.
Then for all y ∈ Y, fog_{1}(y) = I_{Y}(y) = fog_{2}(y). Use one – one ness of f).**

**Answer :**

Let f : X → Y be an invertible function.

Also, suppose f has two inverses (say g_{1} and g_{2})

Then, for all y ∈Y, we have

fog_{1}(y) = I_{Y}(y) = fog_{2}(y)

⇒ f(g_{1}(y)) = f(g_{2}(y))

⇒ g_{1}(y) = g_{2}(y) [as f is invertible ⇒ f is one – one]

⇒ g_{1} = g_{2} [as g is one – one]

Hence, f has a unique inverse.

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

**Answer :**

Function f: {1, 2, 3} → {a, b, c} is given by f(1) = a, f(2) = b, and f(3) = c

If we define g: {a, b, c} → {1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3.

We have

(fog)(a) = f(g(a)) = f(1) = a

(fog)(b) = f(g(b)) = f(2) = b

(fog)(c) = f(g(c)) = f(3) = c

and

(gof)(1) = g(f(1)) = f(a) = 1

(gof)(2) = g(f(2)) = f(b) = 2

(gof)(3) = g(f(3)) = f(c) = 3

So, gof = I_{X} and fog = I_{Y}, where X = {1, 2, 3} and Y= {a, b, c}.

Thus, the inverse of f exists and f^{−1} = g.

So, f^{−1} : {a, b, c} → {1, 2, 3} is given by f^{−1}(a) = 1, f^{−1}(b) = 2, f^{−1}(c) = 3

Let us now find the inverse of f^{−1 }i.e., find the inverse of g.

If we define h: {1, 2, 3} → {a, b, c} as h(1) = a, h(2) = b, h(3) = c

We have

(goh)(1) = g(h(1)) = g(a) = 1

(goh)(2) = g(h(2)) = g(b) = 2

(goh)(3) = g(h(3)) = g(c) = 3

and

(hog)(a) = h(g(a)) = h(1) = a

(hog)(b) = h(g(b)) = h(2) = b

(hog)(c) = h(g(c)) = h(3) = c

So, goh = I_{X} and hog = I_{Y}, where X = {1, 2, 3} and Y = {a, b, c}.

Thus, the inverse of g exists and g^{−1} = h

⇒ (f^{−1})^{−1} = h

It can be noted that h = f

Hence, (f^{−1})^{−1} = f

**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.**

**Answer :**

Let f: X → Y be an invertible function.

Then, there exists a function g: Y → X such that gof = I_{X} and fog = I_{Y}.

Here, f_{−1} = g.

Now, gof = I_{X} and fog = I_{Y}

⇒ f^{−1}of = I_{X} and fof^{−1} = I_{Y}

Hence, f^{−1} : Y → X is invertible and f is the inverse of f^{−1} i.e., (f^{−1})^{−1} = f.

**Question : 13
If f : R → R be given by f(x) = (3 − x ^{3})^{1/3}, then fof(x) is
(A) 1/x^{3} (B) x^{3} (C) x (D) (3 − x^{3})**

**Answer :**

Given, f : R → R be given as f(x) = (3 - x^{3})^{1/3}

So, fof(x) = f(f(x)) = f ((3 - x^{3})^{1/3}) = [3 - ((3 - x^{3})^{1/3})^{3}]^{1/3}

= [3 - (3 - x^{3})]^{1/3}

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

= x

So, fof(x) = x

Hence, the correct answer is option C.

**Question : 14
Let f: R- {-4/3} → R be a function as f(x) = 4x/(3x + 4). The inverse of f is map g : Range f → R- {-4/3} given by (A) g(y) = 3y/(3 - 4y) (B) g(y) = 4y/(4 - 3y) (C) g(y) = 4y/(3 -4y) (D) g(y) = 3y/(4 - 3y)**

**Answer :**

It is given that f : R − {−4/3} → R be a function as f(x) = 4x/(3x + 4)

Let y be an arbitrary element of Range f.

Then, there exists x ∈ R - {-4/3} such that y = f(x)

⇒ y = 4x/(3x + 4)

⇒ 3xy + 4y = 4x

⇒ x(4 - 3y) = 4y

⇒ x = 4y/(4 - 3y)

Let us define g: Range f → R- {-4/3} as g(y) = 4y/(4 - 3y)

Now,

gof(x) = g(f(x)) = g (4x/(3x + 4) = 4(4x/(3x + 4))/{4 − 3(4x/(3x + 4))}

= 16x/(12x + 16 − 12x) = 16x/16 = x

and fog(y) = f(g(y)) = f (4y/(4 − 3y)) = 4(4y/(4 − 3y))/{3(4y/(4 − 3y)) + 4}

= 16y/(12y + 16 − 12y)

= 16y/16

= y

So, gof = I_{R-{-4/3}} and fog = I_{Range f}

Thus, g is the inverse of f i.e., f^{−1} = g

Hence, the inverse of f is the map g: Range f → R - {-4/3}, which is given by g(y) = 4y/(4 - 3y)

Hence, the correct answer is option B.

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