1. Give an example of a function
(i) Which is one-one but not onto.
(ii) Which is not one-one but onto.
(iii) Which is neither one-one nor onto.
Solution:
(i) Let f: Z → Z given by f(x) = 3x + 2
Let us check one-one condition on f(x) = 3x + 2
Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f (x) = f(y)
⇒ 3x + 2 =3y + 2
⇒ 3x = 3y
⇒ x = y
⇒ f(x) = f(y)
⇒ x = y
So, f is one-one.
Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z(domain).
Let f(x) = y
⇒ 3x + 2 = y
⇒ 3x = y – 2
⇒ x = (y – 2)/3. It may not be in the domain (Z)
Because if we take y = 3,
x = (y – 2)/3 = (3-2)/3 = 1/3 ∉ domain Z.
So, for every element in the co domain there need not be any element in the domain such that f(x) = y.
Thus, f is not onto.
(ii) Example for the function which is not one-one but onto
Let f: Z → N ∪ {0} given by f(x) = |x|
Injectivity:
Let x and y be any two elements in the domain (Z),
Such that f(x) = f(y).
⇒ |x| = |y|
⇒ x = ± y
So, different elements of domain f may give the same image.
So, f is not one-one.
Surjectivity:
Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
⇒ |x| = y
⇒ x = ± y
Which is an element in Z (domain).
So, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f is onto.
(iii) Example for the function which is neither one-one nor onto.
Let f: Z → Z given by f(x) = 2x2 + 1
Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)
⇒ 2x2+1 = 2y2+1
⇒ 2x2 = 2y2
⇒ x2 = y2
⇒ x = ± y
So, different elements of domain f may give the same image.
Thus, f is not one-one.
Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f (x) = y
⇒ 2x2+1=y
⇒ 2x2= y − 1
⇒ x2 = (y-1)/2
⇒ x = √ ((y-1)/2) ∉ Z always.
For example, if we take, y = 4,
x = ± √ ((y-1)/2)
= ± √ ((4-1)/2)
= ± √ (3/2) ∉ Z
So, x may not be in Z (domain).
Thus, f is not onto.
2. Which of the following functions from A to B are one-one and onto?
(i) f1 = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}
(ii) f2 = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}
(iii) f3 = {(a, x), (b, x), (c, z), (d, z)}; A = {a, b, c, d,}, B = {x, y, z}.
Solution:
(i) Consider f1 = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}
Injectivity:
f1 (1) = 3
f1 (2) = 5
f1 (3) = 7
⇒ Every element of A has different images in B.
So, f1 is one-one.
Surjectivity:
Co-domain of f1 = {3, 5, 7}
Range of f1 =set of images = {3, 5, 7}
⇒ Co-domain = range
So, f1 is onto.
(ii) Consider f2 = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}
Injectivity:
f2 (2) = a
f2 (3) = b
f2 (4) = c
⇒ Every element of A has different images in B.
So, f2 is one-one.
Surjectivity:
Co-domain of f2 = {a, b, c}
Range of f2 = set of images = {a, b, c}
⇒ Co-domain = range
So, f2 is onto.
(iii) Consider f3 = {(a, x), (b, x), (c, z), (d, z)} ; A = {a, b, c, d,}, B = {x, y, z}
Injectivity:
f3 (a) = x
f3 (b) = x
f3 (c) = z
f3 (d) = z
⇒ a and b have the same image x.
Also c and d have the same image z
So, f3 is not one-one.
Surjectivity:
Co-domain of f3 ={x, y, z}
Range of f3 =set of images = {x, z}
So, the co-domain is not same as the range.
So, f3 is not onto.
3. Prove that the function f: N → N, defined by f(x) = x2 + x + 1, is one-one but not onto
Solution:
Given f: N → N, defined by f(x) = x2 + x + 1
Now we have to prove that given function is one-one
Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
⇒ x2 + x + 1 = y2 + y + 1
⇒ (x2 – y2) + (x – y) = 0 `
⇒ (x + y) (x- y ) + (x – y ) = 0
⇒ (x – y) (x + y + 1) = 0
⇒ x – y = 0 [x + y + 1 cannot be zero because x and y are natural numbers
⇒ x = y
So, f is one-one.
Surjectivity:
When x = 1
x2 + x + 1 = 1 + 1 + 1 = 3
⇒ x2 + x +1 ≥ 3, for every x in N.
⇒ f(x) will not assume the values 1 and 2.
So, f is not onto.
4. Let A = {−1, 0, 1} and f = {(x, x2) : x ∈ A}. Show that f : A → A is neither one-one nor onto.
Solution:
Given A = {−1, 0, 1} and f = {(x, x2): x ∈ A}
Also given that, f(x) = x2
Now we have to prove that given function neither one-one or nor onto.
Injectivity:
Let x = 1
Therefore f(1) = 12=1 and
f(-1)=(-1)2=1
⇒ 1 and -1 have the same images.
So, f is not one-one.
Surjectivity:
Co-domain of f = {-1, 0, 1}
f(1) = 12 = 1,
f(-1) = (-1)2 = 1 and
f(0) = 0
⇒ Range of f = {0, 1}
So, both are not same.
Hence, f is not onto
5. Classify the following function as injection, surjection or bijection:
(i) f: N → N given by f(x) = x2
(ii) f: Z → Z given by f(x) = x2
(iii) f: N → N given by f(x) = x3
(iv) f: Z → Z given by f(x) = x3
(v) f: R → R, defined by f(x) = |x|
(vi) f: Z → Z, defined by f(x) = x2 + x
(vii) f: Z → Z, defined by f(x) = x − 5
(viii) f: R → R, defined by f(x) = sin x
(ix) f: R → R, defined by f(x) = x3 + 1
(x) f: R → R, defined by f(x) = x3 − x
(xi) f: R → R, defined by f(x) = sin2x + cos2x
(xii) f: Q − {3} → Q, defined by f (x) = (2x +3)/(x-3)
(xiii) f: Q → Q, defined by f(x) = x3 + 1
(xiv) f: R → R, defined by f(x) = 5x3 + 4
(xv) f: R → R, defined by f(x) = 5x3 + 4
(xvi) f: R → R, defined by f(x) = 1 + x2
(xvii) f: R → R, defined by f(x) = x/(x2 + 1)
Solution:
(i) Given f: N → N, given by f(x) = x2
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
f(x) = f(y)
x2 = y2
x = y (We do not get ± because x and y are in N that is natural numbers)
So, f is an injection.
Surjection condition:
Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).
f(x) = y
x2= y
x = √y, which may not be in N.
For example, if y = 3,
x = √3 is not in N.
So, f is not a surjection.
Also f is not a bijection.
(ii) Given f: Z → Z, given by f(x) = x2
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)
x2 = y2
x = ±y
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
x2 = y
x = ± √y which may not be in Z.
For example, if y = 3,
x = ± √ 3 is not in Z.
So, f is not a surjection.
Also f is not bijection.
(iii) Given f: N → N given by f(x) = x3
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
f(x) = f(y)
x3 = y3
x = y
So, f is an injection
Surjection condition:
Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).
f(x) = y
x3= y
x = ∛y which may not be in N.
For example, if y = 3,
X = ∛3 is not in N.
So, f is not a surjection and f is not a bijection.
(iv) Given f: Z → Z given by f(x) = x3
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y)
f(x) = f(y)
x3 = y3
x = y
So, f is an injection.
Surjection condition:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
x3 = y
x = ∛y which may not be in Z.
For example, if y = 3,
x = ∛3 is not in Z.
So, f is not a surjection and f is not a bijection.
(v) Given f: R → R, defined by f(x) = |x|
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
f(x) = f(y)
|x|=|y|
x = ±y
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
|x|=y
x = ± y ∈ Z
So, f is a surjection and f is not a bijection.
(vi) Given f: Z → Z, defined by f(x) = x2 + x
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)
x2+ x = y2 + y
Here, we cannot say that x = y.
For example, x = 2 and y = – 3
Then,
x2 + x = 22 + 2 = 6
y2 + y = (−3)2 – 3 = 6
So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (Z),
such that f(x) = y for some element x in Z (domain).
f(x) = y
x2 + x = y
Here, we cannot say x ∈ Z.
For example, y = – 4.
x2 + x = − 4
x2 + x + 4 = 0
x = (-1 ± √-5)/2 = (-1 ± i √5)/2 which is not in Z.
So, f is not a surjection and f is not a bijection.
(vii) Given f: Z → Z, defined by f(x) = x – 5
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)
x – 5 = y – 5
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
x – 5 = y
x = y + 5, which is in Z.
So, f is a surjection and f is a bijection
(viii) Given f: R → R, defined by f(x) = sin x
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
Sin x = sin y
Here, x may not be equal to y because sin 0 = sin π.
So, 0 and π have the same image 0.
So, f is not an injection.
Surjection test:
Range of f = [-1, 1]
Co-domain of f = R
Both are not same.
So, f is not a surjection and f is not a bijection.
(ix) Given f: R → R, defined by f(x) = x3 + 1
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
x3+1 = y3+ 1
x3= y3
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
x3+1=y
x = ∛ (y – 1) ∈ R
So, f is a surjection.
So, f is a bijection.
(x) Given f: R → R, defined by f(x) = x3 − x
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
x3 – x = y3 − y
Here, we cannot say x = y.
For example, x = 1 and y = -1
x3 − x = 1 − 1 = 0
y3 – y = (−1)3− (−1) – 1 + 1 = 0
So, 1 and -1 have the same image 0.
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
x3 − x = y
By observation we can say that there exist some x in R, such that x3 – x = y.
So, f is a surjection and f is not a bijection.
(xi) Given f: R → R, defined by f(x) = sin2x + cos2x
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
f(x) = sin2x + cos2x
We know that sin2x + cos2x = 1
So, f(x) = 1 for every x in R.
So, for all elements in the domain, the image is 1.
So, f is not an injection.
Surjection condition:
Range of f = {1}
Co-domain of f = R
Both are not same.
So, f is not a surjection and f is not a bijection.
(xii) Given f: Q − {3} → Q, defined by f (x) = (2x +3)/(x-3)
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y).
f(x) = f(y)
(2x + 3)/(x – 3) = (2y + 3)/(y – 3)
(2x + 3) (y − 3) = (2y + 3) (x − 3)
2xy − 6x + 3y − 9 = 2xy − 6y + 3x − 9
9x = 9y
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain).
f(x) = y
(2x + 3)/(x – 3) = y
2x + 3 = x y − 3y
2x – x y = −3y − 3
x (2−y) = −3 (y + 1)
x = -3(y + 1)/(2 – y) which is not defined at y = 2.
So, f is not a surjection and f is not a bijection.
(xiii) Given f: Q → Q, defined by f(x) = x3 + 1
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (Q), such that f(x) = f(y).
f(x) = f(y)
x3 + 1 = y3 + 1
x3 = y3
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q (domain).
f(x) = y
x3+ 1 = y
x = ∛(y-1), which may not be in Q.
For example, if y= 8,
x3+ 1 = 8
x3= 7
x = ∛7, which is not in Q.
So, f is not a surjection and f is not a bijection.
(xiv) Given f: R → R, defined by f(x) = 5x3 + 4
Now we have to check for the given function is injection, surjection and bijection condition.
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
5x3 + 4 = 5y3 + 4
5x3= 5y3
x3 = y3
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
5x3+ 4 = y
x3 = (y – 4)/5 ∈ R
So, f is a surjection and f is a bijection.
(xv) Given f: R → R, defined by f(x) = 5x3 + 4
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
5x3 + 4 = 5y3 + 4
5x3 = 5y3
x3 = y3
x = y
So, f is an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
5x3 + 4 = y
x3 = (y – 4)/5 ∈ R
So, f is a surjection and f is a bijection.
(xvi) Given f: R → R, defined by f(x) = 1 + x2
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
1 + x2 = 1 + y2
x2 = y2
x = ± y
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
1 + x2 = y
x2 = y − 1
x = ± √-1 = ± i` is not in R.
So, f is not a surjection and f is not a bijection.
(xvii) Given f: R → R, defined by f(x) = x/(x2 + 1)
Now we have to check for the given function is injection, surjection and bijection condition.
Injection condition:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
x /(x2 + 1) = y /(y2 + 1)
x y2+ x = x2y + y
xy2 − x2y + x − y = 0
−x y (−y + x) + 1 (x − y) = 0
(x − y) (1 – x y) = 0
x = y or x = 1/y
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
x /(x2 + 1) = y
y x2 – x + y = 0
x = (-(-1) ± √ (1-4y2))/(2y) if y ≠ 0
= (1 ± √ (1-4y2))/ (2y), which may not be in R
For example, if y=1, then
(1 ± √ (1-4)) / (2y) = (1 ± i √3)/2, which is not in R
So, f is not surjection and f is not bijection.
6. If f: A → B is an injection, such that range of f = {a}, determine the number of elements in A.
Solution:
Given f: A → B is an injection
And also given that range of f = {a}
So, the number of images of f = 1
Since, f is an injection, there will be exactly one image for each element of f .
So, number of elements in A = 1.
7. Show that the function f: R − {3} → R − {2} given by f(x) = (x-2)/(x-3) is a bijection.
Solution:
Given that f: R − {3} → R − {2} given by f (x) = (x-2)/(x-3)
Now we have to show that the given function is one-one and on-to
Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).
f(x) = f(y)
⇒ (x – 2) /(x – 3) = (y – 2) /(y – 3)
⇒ (x – 2) (y – 3) = (y – 2) (x – 3)
⇒ x y – 3 x – 2 y + 6 = x y – 3y – 2x + 6
⇒ x = y
So, f is one-one.
Surjectivity:
Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element x in R − {3} (domain).
f(x) = y
⇒ (x – 2) /(x – 3) = y
⇒ x – 2 = x y – 3y
⇒ x y – x = 3y – 2
⇒ x ( y – 1 ) = 3y – 2
⇒ x = (3y – 2)/ (y – 1), which is in R – {3}
So, for every element in the co-domain, there exists some pre-image in the domain.
⇒ f is onto.
Since, f is both one-one and onto, it is a bijection.
8. Let A = [-1, 1]. Then, discuss whether the following function from A to itself is one-one, onto or bijective:
(i) f (x) = x/2
(ii) g (x) = |x|
(iii) h (x) = x2
Solution:
(i) Given f: A → A, given by f (x) = x/2
Now we have to show that the given function is one-one and on-to
Injection test:
Let x and y be any two elements in the domain (A), such that f(x) = f(y).
f(x) = f(y)
x/2 = y/2
x = y
So, f is one-one.
Surjection test:
Let y be any element in the co-domain (A), such that f(x) = y for some element x in A (domain)
f(x) = y
x/2 = y
x = 2y, which may not be in A.
For example, if y = 1, then
x = 2, which is not in A.
So, f is not onto.
So, f is not bijective.
(ii) Given g: A → A, given by g (x) = |x|
Now we have to show that the given function is one-one and on-to
Injection test:
Let x and y be any two elements in the domain (A), such that f(x) = f(y).
g(x) = g(y)
|x| = |y|
x = ± y
So, f is not one-one.
Surjection test:
For y = -1, there is no value of x in A.
So, g is not onto.
So, g is not bijective.
(iii) Given h: A → A, given by h (x) = x2
Now we have to show that the given function is one-one and on-to
Injection test:
Let x and y be any two elements in the domain (A), such that h(x) = h(y).
h(x) = h(y)
x2 = y2
x = ±y
So, f is not one-one.
Surjection test:
For y = – 1, there is no value of x in A.
So, h is not onto.
So, h is not bijective.
9. Are the following set of ordered pair of a function? If so, examine whether the mapping is injective or surjective:
(i) {(x, y): x is a person, y is the mother of x}
(ii) {(a, b): a is a person, b is an ancestor of a}
Solution:
Let f = {(x, y): x is a person, y is the mother of x}
As, for each element x in domain set, there is a unique related element y in co-domain set.
So, f is the function.
Injection test:
As, y can be mother of two or more persons
So, f is not injective.
Surjection test:
For every mother y defined by (x, y), there exists a person x for whom y is mother.
So, f is surjective.
Therefore, f is surjective function.
(ii) Let g = {(a, b): a is a person, b is an ancestor of a}
Since, the ordered map (a, b) does not map ‘a’ – a person to a living person.
So, g is not a function.
10. Let A = {1, 2, 3}. Write all one-one from A to itself.
Solution:
Given A = {1, 2, 3}
Number of elements in A = 3
Number of one-one functions = number of ways of arranging 3 elements = 3! = 6
(i) {(1, 1), (2, 2), (3, 3)}
(ii) {(1, 1), (2, 3), (3, 2)}
(iii) {(1, 2 ), (2, 2), (3, 3 )}
(iv) {(1, 2), (2, 1), (3, 3)}
(v) {(1, 3), (2, 2), (3, 1)}
(vi) {(1, 3), (2, 1), (3,2 )}
11. If f: R → R be the function defined by f(x) = 4x3 + 7, show that f is a bijection.
Solution:
Given f: R → R is a function defined by f(x) = 4x3 + 7
Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
⇒ 4x3 + 7 = 4y3 + 7
⇒ 4x3 = 4y3
⇒ x3 = y3
⇒ x = y
So, f is one-one.
Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain)
f(x) = y
⇒ 4x3 + 7 = y
⇒ 4x3 = y − 7
⇒ x3 = (y – 7)/4
⇒ x = ∛(y-7)/4 in R
So, for every element in the co-domain, there exists some pre-image in the domain. f is onto.
Since, f is both one-to-one and onto, it is a bijection.
1. Find gof and fog when f: R → R and g : R → R is defined by
(i) f(x) = 2x + 3 and g(x) = x2 + 5.
(ii) f(x) = 2x + x2 and g(x) = x3
(iii) f (x) = x2 + 8 and g(x) = 3x3 + 1
(iv) f (x) = x and g(x) = |x|
(v) f(x) = x2 + 2x − 3 and g(x) = 3x − 4
(vi) f(x) = 8x3 and g(x) = x1/3
Solution:
(i) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
Also given that f(x) = 2x + 3 and g(x) = x2 + 5
Now, (gof) (x) = g (f (x))
= g (2x +3)
= (2x + 3)2 + 5
= 4x2+ 9 + 12x +5
=4x2+ 12x + 14
Now, (fog) (x) = f (g (x))
= f (x2 + 5)
= 2 (x2 + 5) +3
= 2 x2+ 10 + 3
= 2x2 + 13
(ii) Given, f: R → R and g: R → R
so, gof: R → R and fog: R → R
f(x) = 2x + x2 and g(x) = x3
(gof) (x)= g (f (x))
= g (2x+x2)
= (2x+x2)3
Now, (fog) (x) = f (g (x))
= f (x3)
= 2 (x3) + (x3)2
= 2x3 + x6
(iii) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = x2 + 8 and g(x) = 3x3 + 1
(gof) (x) = g (f (x))
= g (x2 + 8)
= 3 (x2+8)3 + 1
Now, (fog) (x) = f (g (x))
= f (3x3 + 1)
= (3x3+1)2 + 8
= 9x6 + 6x3 + 1 + 8
= 9x6 + 6x3 + 9
(iv) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = x and g(x) = |x|
(gof) (x) = g (f (x))
= g (x)
= |x|
Now (fog) (x) = f (g (x))
= f (|x|)
= |x|
(v) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = x2 + 2x − 3 and g(x) = 3x − 4
(gof) (x) = g (f(x))
= g (x2 + 2x − 3)
= 3 (x2 + 2x − 3) − 4
= 3x2 + 6x − 9 − 4
= 3x2 + 6x − 13
Now, (fog) (x) = f (g (x))
= f (3x − 4)
= (3x − 4)2 + 2 (3x − 4) −3
= 9x2 + 16 − 24x + 6x – 8 − 3
= 9x2 − 18x + 5
(vi) Given, f: R → R and g: R → R
So, gof: R → R and fog: R → R
f(x) = 8x3 and g(x) = x1/3
(gof) (x) = g (f (x))
= g (8x3)
= (8x3)1/3
= [(2x)3]1/3
= 2x
Now, (fog) (x) = f (g (x))
= f (x1/3)
= 8 (x1/3)3
= 8x
2. Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.
Solution:
Given f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}
f : {3, 9, 12} → {1, 3, 4} and g : {1, 3, 4, 5} → {3, 9}
Co-domain of f is a subset of the domain of g.
So, gof exists and gof: {3, 9, 12} → {3, 9}
(gof) (3) = g (f (3)) = g (1) = 3
(gof) (9) = g (f (9)) = g (3) = 3
(gof) (12) = g (f (12)) = g (4) = 9
⇒ gof = {(3, 3), (9, 3), (12, 9)}
Co-domain of g is a subset of the domain of f.
So, fog exists and fog: {1, 3, 4, 5} → {3, 9, 12}
(fog) (1) = f (g (1)) = f (3) = 1
(fog) (3) = f (g (3)) = f (3) = 1
(fog) (4) = f (g (4)) = f (9) = 3
(fog) (5) = f (g (5)) = f (9) = 3
⇒ fog = {(1, 1), (3, 1), (4, 3), (5, 3)}
3. Let f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.
Solution:
Given f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}
f: {1, 4, 9, 16} → {-1, -2, -3, 4} and g: {-1, -2, -3, 4} → {-2, -4, -6, 8}
Co-domain of f = domain of g
So, gof exists and gof: {1, 4, 9, 16} → {-2, -4, -6, 8}
(gof) (1) = g (f (1)) = g (−1) = −2
(gof) (4) = g (f (4)) = g (−2) = −4
(gof) (9) = g (f (9)) = g (−3) = −6
(gof) (16) = g (f (16)) = g (4) = 8
So, gof = {(1, −2), (4, −4), (9, −6), (16, 8)}
But the co-domain of g is not same as the domain of f.
So, fog does not exist.
4. Let A = {a, b, c}, B = {u, v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as: f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.
Show that f and g both are bijections and find fog and gof.
Solution:
Given f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.
Also given that A = {a, b, c}, B = {u, v, w}
Now we have to show f and g both are bijective.
Consider f = {(a, v), (b, u), (c, w)} and f: A → B
Injectivity of f: No two elements of A have the same image in B.
So, f is one-one.
Surjectivity of f: Co-domain of f = {u, v, w}
Range of f = {u, v, w}
Both are same.
So, f is onto.
Hence, f is a bijection.
Now consider g = {(u, b), (v, a), (w, c)} and g: B → A
Injectivity of g: No two elements of B have the same image in A.
So, g is one-one.
Surjectivity of g: Co-domain of g = {a, b, c}
Range of g = {a, b, c}
Both are the same.
So, g is onto.
Hence, g is a bijection.
Now we have to find fog,
we know that Co-domain of g is same as the domain of f.
So, fog exists and fog: {u v, w} → {u, v, w}
(fog) (u) = f (g (u)) = f (b) = u
(fog) (v) = f (g (v)) = f (a) = v
(fog) (w) = f (g (w)) = f (c) = w
So, fog = {(u, u), (v, v), (w, w)}
Now we have to find gof,
Co-domain of f is same as the domain of g.
So, fog exists and gof: {a, b, c} → {a, b, c}
(gof) (a) = g (f (a)) = g (v) = a
(gof) (b) = g (f (b)) = g (u) = b
(gof) (c) = g (f (c)) = g (w) = c
So, gof = {(a, a), (b, b), (c, c)}
5. Find fog (2) and gof (1) when f: R → R; f(x) = x2 + 8 and g: R → R; g(x) = 3x3 + 1.
Solution:
Given f: R → R; f(x) = x2 + 8 and g: R → R; g(x) = 3x3 + 1.
Consider (fog) (2) = f (g (2))
= f (3 × 23 + 1)
= f(3 × 8 + 1)
= f (25)
= 252 + 8
= 633
(gof) (1) = g (f (1))
= g (12 + 8)
= g (9)
= 3 × 93 + 1
= 2188
6. Let R+ be the set of all non-negative real numbers. If f: R+ → R+ and g : R+ → R+ are defined as f(x)=x2 and g(x)=+ √x, find fog and gof. Are they equal functions.
Solution:
Given f: R+ → R+ and g: R+ → R+
So, fog: R+ → R+ and gof: R+ → R+
Domains of fog and gof are the same.
Now we have to find fog and gof also we have to check whether they are equal or not,
Consider (fog) (x) = f (g (x))
= f (√x)
= √x2
= x
Now consider (gof) (x) = g (f (x))
= g (x2)
= √x2
= x
So, (fog) (x) = (gof) (x), ∀x ∈ R+
Hence, fog = gof
7. Let f: R → R and g: R → R be defined by f(x) = x2 and g(x) = x + 1. Show that fog ≠ gof.
Solution:
Given f: R → R and g: R → R.
So, the domains of f and g are the same.
Consider (fog) (x) = f (g (x))
= f (x + 1) = (x + 1)2
= x2 + 1 + 2x
Again consider (gof) (x) = g (f (x))
= g (x2) = x2 + 1
So, fog ≠ gof
1. Find fog and gof, if
(i) f (x) = ex, g (x) = loge x
(ii) f (x) = x2, g (x) = cos x
(iii) f (x) = |x|, g (x) = sin x
(iv) f (x) = x+1, g(x) = ex
(v) f (x) = sin−1 x, g(x) = x2
(vi) f (x) = x+1, g (x) = sin x
(vii) f(x)= x + 1, g (x) = 2x + 3
(viii) f(x) = c, c ∈ R, g(x) = sin x2
(ix) f(x) = x2 + 2 , g (x) = 1 − 1/ (1-x)
Solution:
(i) Given f (x) = ex, g(x) = loge x
Let f: R → (0, ∞); and g: (0, ∞) → R
Now we have to calculate fog,
Clearly, the range of g is a subset of the domain of f.
fog: ( 0, ∞) → R
(fog) (x) = f (g (x))
= f (loge x)
= loge ex
= x
Now we have to calculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→ R
(gof) (x) = g (f (x))
= g (ex)
= loge ex
= x
(ii) f (x) = x2, g(x) = cos x
f: R→ [0, ∞) ; g: R→[−1, 1]
Now we have to calculate fog,
Clearly, the range of g is not a subset of the domain of f.
⇒ Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}
⇒ Domain (fog) = x: x ∈ R and cos x ∈ R}
⇒ Domain of (fog) = R
(fog): R→ R
(fog) (x) = f (g (x))
= f (cos x)
= cos2 x
Now we have to calculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→R
(gof) (x) = g (f (x))
= g (x2)
= cos x2
(iii) Given f (x) = |x|, g(x) = sin x
f: R → (0, ∞) ; g : R→[−1, 1]
Now we have to calculate fog,
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R→R
(fog) (x) = f (g (x))
= f (sin x)
= |sin x|
Now we have to calculate gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog : R→ R
(gof) (x) = g (f (x))
= g (|x|)
= sin |x|
(iv) Given f (x) = x + 1, g(x) = ex
f: R→R ; g: R → [ 1, ∞)
Now we have calculate fog:
Clearly, range of g is a subset of domain of f.
⇒ fog: R→R
(fog) (x) = f (g (x))
= f (ex)
= ex + 1
Now we have to compute gof,
Clearly, range of f is a subset of domain of g.
⇒ fog: R→R
(gof) (x) = g (f (x))
= g (x+1)
= ex+1
(v) Given f (x) = sin −1 x, g(x) = x2
f: [−1,1]→ [(-π)/2 ,π/2]; g : R → [0, ∞)
Now we have to compute fog:
Clearly, the range of g is not a subset of the domain of f.
Domain (fog) = {x: x ∈ domain of g and g (x) ∈ domain of f}
Domain (fog) = {x: x ∈ R and x2 ∈ [−1, 1]}
Domain (fog) = {x: x ∈ R and x ∈ [−1, 1]}
Domain of (fog) = [−1, 1]
fog: [−1,1] → R
(fog) (x) = f (g (x))
= f (x2)
= sin−1 (x2)
Now we have to compute gof:
Clearly, the range of f is a subset of the domain of g.
fog: [−1, 1] → R
(gof) (x) = g (f (x))
= g (sin−1 x)
= (sin−1 x)2
(vi) Given f(x) = x+1, g(x) = sin x
f: R→R ; g: R→[−1, 1]
Now we have to compute fog
Clearly, the range of g is a subset of the domain of f.
Set of the domain of f.
⇒ fog: R→ R
(fog) (x) = f (g (x))
= f (sin x)
= sin x + 1
Now we have to compute gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R → R
(gof) (x) = g (f (x))
= g (x+1)
= sin (x+1)
(vii) Given f (x) = x+1, g (x) = 2x + 3
f: R→R ; g: R → R
Now we have to compute fog
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R→ R
(fog) (x) = f (g (x))
= f (2x+3)
= 2x + 3 + 1
= 2x + 4
Now we have to compute gof
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R → R
(gof) (x) = g (f (x))
= g (x+1)
= 2 (x + 1) + 3
= 2x + 5
(viii) Given f (x) = c, g (x) = sin x2
f: R → {c} ; g: R→ [ 0, 1 ]
Now we have to compute fog
Clearly, the range of g is a subset of the domain of f.
fog: R→R
(fog) (x) = f (g (x))
= f (sin x2)
= c
Now we have to compute gof,
Clearly, the range of f is a subset of the domain of g.
⇒ fog: R→ R
(gof) (x) = g (f (x))
= g (c)
= sin c2
(ix) Given f (x) = x2+ 2 and g (x) = 1 – 1 / (1 – x)
f: R → [ 2, ∞ )
For domain of g: 1− x ≠ 0
⇒ x ≠ 1
⇒ Domain of g = R − {1}
g (x )= 1 – [1/(1 – x)] = (1 – x – 1)/ (1 – x) = -x/(1 – x)
For range of g
y = (- x)/ (1 – x)
⇒ y – x y = − x
⇒ y = x y − x
⇒ y = x (y−1)
⇒ x = y/(y – 1)
Range of g = R − {1}
So, g: R − {1} → R − {1}
Now we have to compute fog
Clearly, the range of g is a subset of the domain of f.
⇒ fog: R − {1} → R
(fog) (x) = f (g (x))
= f (-x/ (1 – x))
= ((-x)/ (1 – x))2 + 2
= (x2 + 2x2 + 2 – 4x) / (1 – x)2
= (3x2 – 4x + 2)/ (1 – x)2
Now we have to compute gof
Clearly, the range of f is a subset of the domain of g.
⇒ gof: R→R
(gof) (x) = g (f (x))
= g (x2 + 2)
= 1 – 1 / (1 – (x2 + 2))
= – 1/ (1 – (x2 + 2))
= (x2 + 2)/ (x2 + 1)
2. Let f(x) = x2 + x + 1 and g(x) = sin x. Show that fog ≠ gof.
Solution:
Given f(x) = x2 + x + 1 and g(x) = sin x
Now we have to prove fog ≠ gof
(fog) (x) = f (g (x))
= f (sin x)
= sin2 x + sin x + 1
And (gof) (x) = g (f (x))
= g (x2+ x + 1)
= sin (x2+ x + 1)
So, fog ≠ gof.
3. If f(x) = |x|, prove that fof = f.
Solution:
Given f(x) = |x|,
Now we have to prove that fof = f.
Consider (fof) (x) = f (f (x))
= f (|x|)
= ||x||
= |x|
= f (x)
So,
(fof) (x) = f (x), ∀x ∈ R
Hence, fof = f
4. If f(x) = 2x + 5 and g(x) = x2 + 1 be two real functions, then describe each of the following functions:
(i) fog
(ii) gof
(iii) fof
(iv) f2
Also, show that fof ≠ f2
Solution:
f(x) and g(x) are polynomials.
⇒ f: R → R and g: R → R.
So, fog: R → R and gof: R → R.
(i) (fog) (x) = f (g (x))
= f (x2 + 1)
= 2 (x2 + 1) + 5
=2x2 + 2 + 5
= 2x2 +7
(ii) (gof) (x) = g (f (x))
= g (2x +5)
= (2x + 5)2 + 1
= 4x2 + 20x + 26
(iii) (fof) (x) = f (f (x))
= f (2x +5)
= 2 (2x + 5) + 5
= 4x + 10 + 5
= 4x + 15
(iv) f2 (x) = f (x) x f (x)
= (2x + 5) (2x + 5)
= (2x + 5)2
= 4x2 + 20x +25
Hence, from (iii) and (iv) clearly fof ≠ f2
5. If f(x) = sin x and g(x) = 2x be two real functions, then describe gof and fog. Are these equal functions?
Solution:
Given f(x) = sin x and g(x) = 2x
We know that
f: R→ [−1, 1] and g: R→ R
Clearly, the range of f is a subset of the domain of g.
gof: R→ R
(gof) (x) = g (f (x))
= g (sin x)
= 2 sin x
Clearly, the range of g is a subset of the domain of f.
fog: R → R
So, (fog) (x) = f (g (x))
= f (2x)
= sin (2x)
Clearly, fog ≠ gof
Hence they are not equal functions.
6. Let f, g, h be real functions given by f(x) = sin x, g (x) = 2x and h (x) = cos x. Prove that fog = go (f h).
Solution:
Given that f(x) = sin x, g (x) = 2x and h (x) = cos x
We know that f: R→ [−1, 1] and g: R→ R
Clearly, the range of g is a subset of the domain of f.
fog: R → R
Now, (f h) (x) = f (x) h (x) = (sin x) (cos x) = ½ sin (2x)
Domain of f h is R.
Since range of sin x is [-1, 1], −1 ≤ sin 2x ≤ 1
⇒ -1/2 ≤ sin x/2 ≤ 1/2
Range of f h = [-1/2, 1/2]
So, (f h): R → [(-1)/2, 1/2]
Clearly, range of f h is a subset of g.
⇒ go (f h): R → R
⇒ Domains of fog and go (f h) are the same.
So, (fog) (x) = f (g (x))
= f (2x)
= sin (2x)
And (go (f h)) (x) = g ((f(x). h(x))
= g (sin x cos x)
= 2sin x cos x
= sin (2x)
⇒ (fog) (x) = (go (f h)) (x), ∀x ∈ R
Hence, fog = go (f h)
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)}
Solution:
(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 not one-one.
⇒ f is not a bijection.
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 not one-one.
⇒ f is not a bijection.
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 a bijection.
Therefore h inverse exists.
⇒ h has an inverse and it is given by
h-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}
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
Solution:
(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, different elements 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, different elements 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 no pre-image in the domain (A)
⇒ f is not a bijection.
So, f -1does not exist.
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. Show that f, g and gof are invertible. Find f−1, g−1 and gof−1and show that (gof)−1 = f −1o g−1
Solution:
Given f = {(1, a), (2, b), (c , 3)} and g = {(a , apple) , (b , ball) , (c , cat)} Clearly , f and g are bijections.
So, f and g are invertible.
Now,
f -1 = {(a ,1) , (b , 2) , (3,c)} and g-1 = {(apple, a), (ball , b), (cat , c)}
So, f-1 o g-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 a bijection.
So, gof is invertible.
(gof)-1 = {(apple, 1), (ball, 2), (cat, 3)}……. (2)
Form (1) and (2), we get
(gof)-1 = f-1 o g -1
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 defined as f(x) = 2x + 1 and g(x) = x2 − 2. Express (gof)−1 and f−1 og−1 as the sets of ordered pairs and verify that (gof)−1 = f−1 og−1.
Solution:
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
5. Show that the function f: Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f−1
Solution:
Given function f: Q → Q, defined by f(x) = 3x + 5
Now we have to show that 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 belongs to Q 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 = 3y + 5
⇒ x −5 = 3y
⇒ y = (x – 5)/3
Now substituting this value in (1) we get
So, f-1(x) = (x – 5)/3
6. Consider f: R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Solution:
Given f: R → R given by f(x) = 4x + 3
Now we have to show that the given function is invertible.
Consider injection of 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 surjection of f:
Let y be in the co-domain (R),
Such that f(x) = y.
⇒ 4x + 3 = y
⇒ 4x = y -3
⇒ x = (y-3)/4 in 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 = 4y + 3
⇒ x − 3 = 4y
⇒ y = (x -3)/4
Now substituting this value in (1) we get
So, f-1(x) = (x-3)/4
7. Consider f: R → R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with inverse f−1 of f given by f−1(x) = √ (x-4) where R+ is the set of all non-negative real numbers.
Solution:
Given f: R → R+ → [4, ∞) given by f(x) = x2 + 4.
Now we have to show that f is invertible,
Consider injection of 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 is one-one
Now surjection of 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 to find 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 this value in (1) we get,
So, f-1(x) = √ (x-4)
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?
Solution:
It is given that f(x) = (4x + 3)/ (6x – 4), x ≠ 2/3
Now we have to show fof(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) = x for all x ≠ 2/3
=> fof = 1
Hence, the given function f is invertible and the inverse of f is f itself.
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
Solution:
Given f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x – 5
We have to show that f is 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 is one-one.
Surjectivity of 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)/3 in 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 value in (1) we get,
So, f-1(x) = (√ (x+6)-1)/3
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).
Solution:
Given f: R → R be defined by f(x) = x3 −3
Now we have to prove that f−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 is one-one.
Surjectivity of 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
11. A function f: R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f−1 (3).
Solution:
Given that f: R → R is defined as f(x) = x3 + 4
Injectivity of f:
Let x and y be two elements of domain (R),
Such that f (x) = f (y)
⇒ x3 + 4 = y3 + 4
⇒ x3 = y3
⇒ x = y
So, f is one-one.
Surjectivity of f:
Let y be in the co-domain (R),
Such that f(x) = y.
⇒ x3 + 4 = y
⇒ x3 = y – 4
⇒ x = ∛ (y – 4) in R (domain)
⇒ 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 + 4
⇒ x − 4 = y3
⇒ y =∛ (x-4)
So, f-1(x) =∛ (x-4) [from (1)]
f-1 (3) = ∛(3 – 4)
= ∛-1
= -1