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Mathmatics - Chapter No. : 1

Relation Functions

NCERTworld ncert class 12 textbook maths

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) = 8x3 and g(x) = x1/3

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) = 8x3 and g(x) = x1/3
So, gof(x) = g(f(x)) = g(8x3) = (8x3)1/3 = 2x
fog(x) = f(g(x)) = f(x1/3) = 8(x1/3)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 = Ix
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 = IRange 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 = IR
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) = x2 + 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) = x2 + 4
For one – one:
Let f(x) = f(y)
⇒ x2 + 4 = y2 + 4
⇒ x2 = y2
⇒ x = y                [as x = y ∈ R+]
So, f is a one – one function.
For onto:
For y ∈ [4, ∞), let y = x2 + 4
⇒ x2 = 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(x2 + 4) = √{x2 + 4 – 4} = √x2 = x
and
(fog)(y) = f(g(y)) = f(√(y - 4)) = (√(y – 4))2 + 4 = y - 4 + 4 = y
So, gof = fog = IR
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) = 9x2 + 6x − 5. Show that f is invertible with  f−1(y) = [{√(y + 6) − 1}/3]

Answer :
f: R+ → [−5, ∞) is given as f(x) = 9x2 + 6x − 5.
Let y be an arbitrary element of [−5, ∞).
Let y = 9x2 + 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(9x2 + 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 = IR and fog = y = IRange 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 g1 and g2 are two inverses of f.
Then for all y ∈ Y, fog1(y) = IY(y) = fog2(y). Use one – one ness of f).

Answer :
Let f : X → Y be an invertible function.
Also, suppose f has two inverses (say g1 and g2)
Then, for all y ∈Y, we have
     fog1(y) = IY(y) = fog2(y)
⇒ f(g1(y)) = f(g2(y))
⇒ g1(y) = g2(y)                  [as f is invertible ⇒ f is one – one]
⇒ g1 = g2                          [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 = IX and fog = IY, 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 = IX and hog = IY, 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 = IX and fog = IY.
Here, f−1 = g.
Now, gof = IX and fog = IY
⇒ f−1of = IX and fof−1 = IY
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 − x3)1/3, then fof(x) is
(A) 1/x3                      (B) x3                        (C) x                         (D) (3 − x3)

Answer :
Given, f : R → R be given as f(x) = (3 - x3)1/3
So, fof(x) = f(f(x)) = f ((3 - x3)1/3) = [3 -  ((3 - x3)1/3)3]1/3
                               = [3 - (3 - x3)]1/3
                               = (x3)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 = IR-{-4/3} and fog = IRange 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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