RD Chapter 2 Functions Ex 2.4 Solutions
Question - 1 : - State with reason whether the 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 - 1 : -
(i) Given f: {1, 2, 3,4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
We have:
f (1) = f (2) = f (3) = f (4) = 10
⇒ f is notone-one.
⇒ f is not abijection.
So, f does not have an inverse.
(ii) Given g: {5, 6,7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}
from the question it is clear that g (5) = g (7) = 4
⇒ f is notone-one.
⇒ f is not abijection.
So, f does not have an inverse.
(iii) Given h: {2, 3,4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}
Here, different elements of the domain have different images in the co-domain.
⇒ h is one-one.
Also, each element in the co-domain has a pre-image in the domain.
⇒ h is onto.
⇒ h is abijection.
Therefore h inverseexists.
⇒ h has an inverseand it is given by
h-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}
Question - 2 : - Find f −1 if it exists: f: A → B, where
(i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) =3 x.
(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81}and f(x) = x2
Answer - 2 : -
(i) Given A = {0,−1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}
Here, differentelements of the domain have different images in the co-domain.
Clearly, this is one-one.
Range of f = Range of f = B
so, f is a bijection and,
Thus, f -1 exists.
Hence, f -1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}
(ii) Given A ={1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2
So, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}
Here, differentelements of the domain have different images in the co-domain.
Clearly, f is one-one.
But this is not onto because the element 0 in the co-domain (B) has nopre-image in the domain (A)
⇒ f is not abijection.
So, f -1does not exist.
Question - 3 : - Consider f: {1, 2, 3} → {a, b, c} and g:{a, b, c} → {apple, ball, cat} defined as f (1)= a, f (2) = b, f (3) = c, g (a) =apple, g (b) = ball and g (c) = cat. Showthat f, g and gof are invertible. Find f−1, g−1 and gof−1andshow that (gof)−1 = f −1o g−1
Answer - 3 : -
Given f = {(1, a), (2,b), (c , 3)} and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and gare bijections.
So, f and g areinvertible.
Now,
f -1 ={(a ,1) , (b , 2) , (3,c)} and g-1 = {(apple, a), (ball , b),(cat , c)}
So, f-1 og-1= {apple, 1), (ball, 2), (cat, 3)}……… (1)
f: {1,2,3,} → {a,b, c} and g: {a, b, c} → {apple, ball, cat}
So, gof: {1, 2,3} → {apple, ball, cat}
⇒ (gof) (1) = g (f (1))= g (a) = apple
(gof) (2) = g (f (2))
= g (b)
= ball,
And (gof) (3) = g (f(3))
= g (c)
= cat
∴ gof = {(1, apple),(2, ball), (3, cat)}
Clearly, gof is abijection.
So, gof isinvertible.
(gof)-1 = {(apple,1), (ball, 2), (cat, 3)}……. (2)
Form (1) and (2), weget
(gof)-1 =f-1 o g -1
Question - 4 : - Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C ={7, 23, 47, 79}and f: A → B, g: B → C be definedas f(x) = 2x + 1 and g(x) = x2 − 2. Express(gof)−1 and f−1 og−1 asthe sets of ordered pairs and verify that (gof)−1 = f−1 og−1.
Answer - 4 : -
Given that f (x) = 2x+ 1
⇒ f= {(1, 2(1)+ 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) +1)}
={(1, 3), (2, 5), (3, 7), (4, 9)}
Also given that g(x) =x2−2
⇒ g= {(3, 32−2), (5, 52−2), (7, 72−2), (9, 92−2)}
={(3, 7), (5, 23), (7, 47), (9, 79)}
Clearly f and g are bijections and, hence, f−1: B→A and g−1: C→ B exist.
So, f−1= {(3, 1), (5, 2), (7, 3), (9, 4)}
And g−1= {(7, 3), (23, 5), (47, 7), (79, 9)}
Now, (f−1 o g−1): C→A
f−1 o g−1 ={(7, 1), (23, 2), (47, 3), (79, 4)}……….(1)
Also, f: A→B and g: B → C,
⇒ gof: A → C, (gof) −1 : C→A
So, f−1 o g−1and (gof)−1 have same domains.
(gof) (x) =g (f (x))
=g (2x + 1)
=(2x +1 )2−2
⇒ (gof) (x) = 4x2 + 4x +1 − 2
⇒ (gof) (x) = 4x2+ 4x −1
Then, (gof) (1) = g (f (1))
= 4 + 4 − 1
=7,
(gof) (2) =g (f (2))
= 4(2)2 +4(2) – 1 = 23,
(gof) (3) =g (f (3))
= 4(3)2 +4(3) – 1 = 47 and
(gof) (4) =g (f (4))
= 4(4)2 +4(4) − 1 = 79
So, gof ={(1, 7), (2, 23), (3, 47), (4, 79)}
⇒ (gof)– 1 ={(7, 1), (23, 2), (47, 3), (79, 4)}…… (2)
From (1) and (2), we get:
(gof)−1 = f−1 o g−1
Question - 5 : - Show that the function f: Q → Q, definedby f(x) = 3x + 5, is invertible. Also, find f−1
Answer - 5 : -
Givenfunction f: Q → Q, defined by f(x) = 3x + 5
Now we have to showthat the given function is invertible.
Injection of f:
Let x and y be two elements of the domain (Q),
Such that f(x) = f(y)
⇒ 3x + 5 =3y + 5
⇒ 3x = 3y
⇒ x = y
so, f is one-one.
Surjection of f:
Let y be in the co-domain (Q),
Such that f(x) =y
⇒ 3x +5 = y
⇒ 3x = y – 5
⇒ x = (y -5)/3 belongsto Q domain
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have tofind f-1:
Let f-1(x)= y…… (1)
⇒ x = f(y)
⇒ x = 3y + 5
⇒ x −5 = 3y
⇒ y = (x – 5)/3
Now substituting thisvalue in (1) we get
So, f-1(x)= (x – 5)/3
Question - 6 : - Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Answer - 6 : -
Givenf: R → R given by f(x) = 4x + 3
Now we have to showthat the given function is invertible.
Consider injectionof f:
Let x and y be two elements of domain (R),
Such that f(x) = f(y)
⇒ 4x + 3 =4y + 3
⇒ 4x = 4y
⇒ x = y
So, f is one-one.
Now surjectionof f:
Let y be in the co-domain (R),
Such that f(x) =y.
⇒ 4x + 3 = y
⇒ 4x = y -3
⇒ x = (y-3)/4 inR (domain)
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have tofind f -1
Let f-1(x)= y……. (1)
⇒ x = f (y)
⇒ x = 4y + 3
⇒ x − 3 = 4y
⇒ y = (x -3)/4
Now substituting thisvalue in (1) we get
So, f-1(x)= (x-3)/4
Question - 7 : - Consider f: R → R+ → [4, ∞) givenby f(x) = x2 + 4. Show that f is invertiblewith inverse f−1 of f given by f−1(x)= √ (x-4) where R+ is the set of all non-negative real numbers.
Answer - 7 : -
Givenf: R → R+ → [4, ∞) given by f(x) = x2 +4.
Now we have to showthat f is invertible,
Consider injectionof f:
Let x and y be two elements of the domain (Q),
Such that f(x) =f(y)
⇒ x2 +4 = y2 + 4
⇒ x2 =y2
⇒ x = y (as co-domain as R+)
So, f isone-one
Now surjectionof f:
Let y be in the co-domain (Q),
Such that f(x) =y
⇒ x2 +4 = y
⇒ x2 =y – 4
⇒ x = √ (y-4) in R
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Now we have tofind f-1:
Let f−1 (x) = y……(1)
⇒ x = f (y)
⇒ x = y2 + 4
⇒ x − 4 = y2
⇒ y = √ (x-4)
So, f-1(x)= √ (x-4)
Now substituting thisvalue in (1) we get,
So, f-1(x)= √ (x-4)
Question - 8 : - 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 - 8 : -
It is given that f(x)= (4x + 3)/ (6x – 4), x ≠ 2/3
Now we have to showfof(x) = x
(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
Therefore, fof(x) = xfor all x ≠ 2/3
=> fof = 1
Hence, the givenfunction f is invertible and the inverseof f is f itself.
Question - 9 : - Consider f: R+ → [−5, ∞) given by f(x) =9x2 + 6x − 5. Show that f is invertible with
f-1(x) = (√(x +6)-1)/3
Answer - 9 : -
Given f: R+ →[−5, ∞) given by f(x) = 9x2 + 6x – 5
We have to show that fis invertible.
Injectivity of f:
Let x and y be two elements of domain (R+),
Such that f(x) = f(y)
⇒ 9x2 +6x – 5 = 9y2 + 6y − 5
⇒ 9x2 +6x = 9y2 + 6y
⇒ x = y (As, x, y ∈ R+)
So, f isone-one.
Surjectivityof f:
Let y is in the co domain (Q)
Such that f(x) =y
⇒ 9x2 +6x – 5 = y
⇒ 9x2 +6x = y + 5
⇒ 9x2 +6x +1 = y + 6 (By adding 1 on both sides)
⇒ (3x + 1)2 =y + 6
⇒ 3x + 1 = √(y + 6)
⇒ 3x = √ (y + 6) – 1
⇒ x = (√ (y + 6)-1)/3in R+ (domain)
f is onto.
So, f is a bijection and hence, it is invertible.
Now we have to find f-1
Let f−1(x) = y…..(1)
⇒ x = f (y)
⇒ x = 9y2 + 6y − 5
⇒ x + 5 = 9y2 +6y
⇒ x + 6= 9y2+ 6y + 1 (adding 1 on both sides)
⇒ x + 6 = (3y + 1)2
⇒ 3y + 1 = √ (x + 6)
⇒ 3y =√(x +6) -1
⇒ y = (√ (x+6)-1)/3
Now substituting this valuein (1) we get,
So, f-1(x) =(√ (x+6)-1)/3
Question - 10 : - If f: R → R be defined by f(x) = x3 −3,then prove that f−1 exists and find a formula for f−1.Hence, find f−1 (24) and f−1 (5).
Answer - 10 : -
Givenf: R → R be defined by f(x) = x3 −3
Now we have to prove thatf−1 exists
Injectivity of f:
Let x and y be two elements in domain (R),
Such that, x3 − 3 = y3 − 3
⇒ x3 = y3
⇒ x = y
So, f isone-one.
Surjectivityof f:
Let y be in the co-domain (R)
Such that f(x) =y
⇒ x3 –3 = y
⇒ x3 =y + 3
⇒ x = ∛(y+3) in R
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Finding f -1:
Let f-1(x) = y……..(1)
⇒ x= f(y)
⇒ x = y3 −3
⇒ x + 3 = y3
⇒ y = ∛(x + 3) = f-1(x) [from (1)]
So, f-1(x)= ∛(x + 3)
Now, f-1(24)= ∛ (24 + 3)
= ∛27
= ∛33
= 3
And f-1(5)=∛ (5 + 3)
= ∛8
= ∛23
= 2