Question-1 :- Show that the function f : R∗ → R∗ defined by f(x) = 1/x is one-one and onto, where R∗ is the set of all non-zero real numbers. Is the result true, if the domain R∗ is replaced by N with co-domain being same as R∗
Solution :- Given that f : R∗ → R∗ defined by f(x) = 1/x.
⇒ f(x) = f(y)
⇒ 1/x = 1/y
⇒ x = y
∴f is one-one.
Now, f(x) = 1/x
⇒ y = 1/x
⇒ x = 1/y
⇒ f(1/y) = x
∴f is on-to.
Again, g : R∗ → R∗ defined by g(x) = 1/x.
⇒ g(x₁) = g(x₂)
⇒ 1/x₁ = 1/x₂
⇒ x₁ = x₂ and (x₁, x₂) ∉ N
∴g is one-one.
Now, g(x) = 1/x
But every real number belonging to co-domain may not have a pre-image in N. e.g., 1/3 = 3/2 ∉ N.
∴g is not on-to.
Question-2 :- Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : R → R given by f(x) = x²
(iv) f : N → N given by f(x) = x³
(v) f : Z → Z given by f(x) = x³
(i) f : N → N given by f(x) = x²
⇒ f(x₁) = f(x₂)
⇒ x₁² = x₂²
⇒ x₁ = x₂
∴f is one-one.
Now, f(x) = x²
There are such numbers of co-domain which have no image in domain N.
e.g., 3 ∈ co-domain N, but there is no pre-image in domain of f.
∴f is not on-to.
Hence, f is injective but not surjective.
(ii) f : Z → Z given by f(x) = x²
since, Z = {0, ±1, ±2, ±3...}
Therefore, f(-1) = f(1) = 1
⇒ -1 and 1 have same image.
∴f is not one-one.
Now, f(x) = x²
There are such numbers of co-domain which have no image in domain Z.
e.g., 3 ∈ co-domain Z, but there is no √3 ∉ Z domain of f.
∴f is not on-to.
Hence, f is neither injective nor surjective.
(iii) f : R → R given by f(x) = x²
since, R = {0, ±1, ±2, ±3...}
Therefore, f(-1) = f(1) = 1
⇒ -1 and 1 have same image.
∴f is not one-one.
e.g., 3 ∈ co-domain R, but there is no √-3 ∉ R domain of f.
∴f is not on-to.
Hence, f is neither injective nor surjective.
(iv) f : N → N given by f(x) = x³
⇒ f(x₁) = f(x₂)
⇒ x₁³ = x₂³
⇒ x₁ = x₂
for every x ∈ N, has a unique image in its co-domain.
∴f is one-one.
There are many such members of co-domain of f which do not have pre-image in its domain. e.g., 2, 3 etc.
∴f is not on-to.
Hence, f is injective but not surjective.
(v) f : Z → Z given by f(x) = x³
⇒ f(x₁) = f(x₂)
⇒ x₁³ = x₂³
⇒ x₁ = x₂
for every x ∈ Z, has a unique image in its co-domain.
∴f is one-one.
There are many such members of co-domain of f which do not have pre-image in its domain.
∴f is not on-to.
Hence, f is injective but not surjective.
Question-3 :- Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution :- f: → R → R is given by,
f(x) = [x]
It is seen that f(1.4) = [1.4] = 1, f(1.8) = [1.8] = 1.
∴ f(1.4) = f(1.8), but 1.4 ≠ 1.8.
∴ f is not one-one.
Now, consider 0.8 ∈ R.
It is known that f(x) = [x] is always an integer.
Thus, there does not exist any element x ∈ R such that f(x) = 0.8.
∴ f is not on-to.
Hence, the greatest integer function is neither one-one nor on-to.
Question-4 :- Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where |x| is x, if x is positive or 0 and |x| is – x, if x is negative.
Solution :- Modulus Function f : R → R, given by f(x) = |x|
f contains (-1,1),(1,1),(-3,3),(3,3)
Thus negative integers are not images of any element.
∴f is not one-one.
Also second set R contains some negative numbers which are not images of any real number.
i.e., f(x) = |x| = -1
∴f is not on-to.
Hence, f is neither one-one nor on-to.
Question-5 :- Show that the Signum Function f : R → R, given by
is neither one-one nor onto.
Signum Function f : R → R, given by⇒ f(1) = f(2) = 1 ⇒ f(x₁) = f(x₂) = 1 for x>1 ⇒ x₁ ≠ x₂ ∴f is not one-one. Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain, there does not exist any x in domain such that f(x) = −2. ∴f is not onto. Hence, f is neither one-one nor on-to.
Question-6 :- Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
Solution :- Given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴f (1) = 4, f (2) = 5, f (3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.
Question-7 :- In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x²
(i) f : R → R defined by f(x) = 3 – 4x
⇒ f(x₁) = f(x₂)
⇒ 3 – 4x₁ = 3 – 4x₂
⇒ x₁ = x₂
∴f is one-one.
Now, f(x) = 3 – 4x, Let f(x) = y
⇒ y = 3 – 4x
⇒ x = (3 - y)/4 ∈ R
⇒ f((3 - y)/4 ) = 3 - 4[(3 - y)/4 ] = 3 - 3 + y = y
∴f is on-to.
Hence, f is injective and surjective or f is bijective function.
(ii) f : R → R defined by f(x) = 1 + x²
⇒ f(x₁) = f(x₂)
⇒ 1 + x₁² = 1 + x₂²
⇒ x₁² = x₂²
⇒ x₁ = ± x₂
∴f is not one-one.
Now, f(x) = 1 + x², Let f(x) = y
⇒ y = 1 + x²
⇒ x = ± √y - 1
⇒ f(√y - 1) = 1 + (1-y) = 2 - y ≠ y
∴f is not on-to.
Hence, f is neither injective nor surjective so, f is not bijective function.
Question-8 :- Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function
Solution :- f : A × B → B × A such that f(a, b) = (b, a)
⇒ Let (a₁, b₁) & (a₂, b₂) ∈ A x B such that f(a₁, b₁) = f(a₂, b₂)
⇒ (b₁, a₁) = (b₂, a₂)
⇒ a₁ = a₂, b₁ = b₂
⇒ (a₁, b₁) = (a₂, b₂)
⇒ a₁ = a₂, b₁ = b₂
So, for all (a₁, b₁),(a₂, b₂) ∈ A x B
∴f is one-one.
Now, let (b, a) ∈ B × A be any element.
Then, there exists (a, b) ∈ A × B such that f(a, b) = (b, a).
∴f is on-to.
Hence, f is injective and surjective so, f is bijective function.
Question-9 :- Let f : N → N be defined by
State whether the function f is bijective. Justify your answer.
Let f : N → N be defined by(a) f(1) = (n+1)/2 = (1+1)/2 = 2/2 = 1 and f(2) = n/2 = 2/2 = 1 The elements 1, 2, belonging to domain of f have the same image 1 in its co-domain. ∴f is one-one. (b) Every number of co-domain has pre-image in its domain e.g., 1 has two pre-images 1 and 2. ∴f is on-to. Hence, f is not injective but surjective. So, f is not bijective function.
Question-10 :- Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2)/(x-3). Is f one-one and onto? Justify your answer
Solution :- Let A = R – {3} and B = R – {1} and the function f : A → B defined by f(x) = (x-2)/(x-3).
⇒ f(x₁) = f(x₂)
⇒ (x₁-2)/(x₁-3) = (x₂-2)/(x₂-3)
⇒ (x₁-2)(x₂-3) = (x₁-3)(x₂-2)
⇒ x₁x₂ - 3x₁ - 2x₂ + 6 = x₁x₂ - 2x₁ - 3x₂ + 6
⇒ - 3x₁ - 2x₂ = - 2x₁ - 3x₂
⇒ - 3x₁ + 2x₁ = - 3x₂ + 2x₂
⇒ -x₁ = -x₂
⇒ x₁ = x₂
∴f is one-one.
Now, f(x) = (x-2)/(x-3)
Let f(x) = y
Then y = (x-2)/(x-3)
⇒ y(x-3) = x - 2
⇒ xy - 3y = x - 2
⇒ xy - x = 3y - 2
⇒ x(y-1) = 3y - 2
⇒ x = (3y-2)/(y-1)
⇒ f(x) = y
∴f is on-to.
Hence, f is injective and surjective so, f is bijective function.
Question-11 :- Let f : R → R be defined as f(x) = x⁴. Choose the correct answer.
(A) f is one-one onto,
(B) f is many-one onto,
(C) f is one-one but not onto,
(D) f is neither one-one nor onto.
Let f : R → R be defined as f(x) = x⁴
⇒ f(x₁) = f(x₂)
⇒ x₁⁴ = x₂⁴
⇒ ± x₁ = ± x₂
∴f is not one-one.
Now, f(x) = x⁴
Let f(x) = y
Then y = x⁴
⇒ x = ± (y)1/4
f(x) = (- y1/4)⁴
⇒ f(x) = y
∴f is not on-to.
Hence, f is neither one-one nor on-to.
Therefore, option (D) is correct.
Question-12 :- Let f : R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto.
Let f : R → R be defined as f(x) = 3x
⇒ f(x₁) = f(x₂)
⇒ 3x₁ = 3x₂
⇒ x₁ = x₂
∴f is one-one.
Now, f(x) = 3x
Let f(x) = y
Then y = 3x
x = y/3
f(x) = 3 x (y/3) = y
∴f is not on-to.
Hence, f is one-one and on-to.
Therefore, option (A) is correct.
