NCRT Solution Exercise 2
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∗?
Answer 1:
It is given that f: R* → R* is defined by f(x) = (1/x)
For one – one:
Let x, y ∈ R* such that f(x) = f(y)
⇒1/x = 1/y
⇒ x = y
∴ f is one – one.
For onto:
It is clear that for y ∈ R*, there exists x =(1/y) ∈ R* [as y ≠ 0] such that
f(x) = (1 / (1/y)) = y
∴ f is onto.
Thus, the given function f is one – one and onto.
Now, consider function g: N → R* defined by g(x) = 1/x
We have,
g(x1) = g(x2) ⇒ (1/x1) = (1/x2) ⇒ x1 = x2
∴ g is one – one.
Further, it is clear that g is not onto as for 1.2 ∈ R* there does not exit any x in N
such that g(x) =(1/1.2).
Hence, function g is one-one but not onto.
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
Answer 2:
(i) f: N → N is given by f(x) = x2
It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x2 = y2 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
(ii) f: Z → Z is given by f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now, −2 ∈ Z. But, there does not exist any element x ∈ Z such that
f(x) = −2 or x2= −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iii) f: R → R is given by f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now, −2 ∈ R. But, there does not exist any element x ∈ R such that
f(x) = −2 or x2= −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iv) f: N → N given by f(x) = x3
It is seen that for x, y ∈ N, f(x) = f(y) ⇒ x3= y3 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ N. But, there does not exist any element x ∈ N such that
f(x) = 2 or x3 = 2.
∴ f is not surjective
Hence, function f is injective but not surjective.
(v) f: Z → Z is given by f(x) = x3
It is seen that for x, y ∈ Z, f(x) = f(y) ⇒ x3= y3 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x ∈ Z such that
f(x) = 2 or x3= 2.
∴ f is not surjective.
Hence, function 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.
Answer 3:
f: R → R is given by, f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one – one.
Now, consider 0.7 ∈ 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) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one – one nor onto.
Question 4: Show that the Modulus Function f: R → R given by f(x) = |x|, is neither one –
one nor onto, where |x| is x, if x is positive or 0 and |x| is − x, if x is negative.
Answer 4:
f: R → R is given by f(x) = |x| =
It is clear that f(-1) = |-1| = 1 and f(1) = |1| = 1
∴ f(−1) = f(1), but −1 ≠ 1.
∴ f is not one – one.
Now, consider −1 ∈ R.
It is known that f(x) = |x| is always non-negative. Thus, there does not exist any
element x in domain R such that f(x) = |x| = −1.
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.
Question 5: Show that the Signum Function f: R → R, given by f(ð‘¥) =
is neither one-one nor onto.
Answer 5:
f: R → R is given by f(x) =
It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴ f is not one – one.
Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain
R, there does not exist any x in domain R such that f(x) = −2.
∴ f is not onto.
Hence, the Signum function is neither one – one nor onto.
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.
Answer 6:
It is 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
Answer 7:
(i) f: R → R is defined as f(x) = 3 − 4x.
Let x1, x2 ∈ R such that f(x1) = f(x2)
⇒ 3-4x1 = 3-4x2
⇒ -4x1 = -4x2
⇒ x1 = x2
∴ f is one – one.
For any real number (y) in R, there exists (3-y/4) in R such that
f ((3-y)/4)) = 3-4 ((3-y)/4)) = y
∴ f is onto.
Hence, f is bijective.
(ii) f: R → R is defined as f(x) = 1 + x2
Let x1, x2 ∈ R such that f(x1) = f(x2)
⇒ 1+ x12 = 1 + x22
⇒ x12 = x22
⇒ x1 = ± x2
∴ f(x1) = f(x2) does not imply that x1 = x2
For example f(1) = f(-1) = 2
∴ f is not one – one.
Consider an element −2 in co-domain R.
It is seen that f(x) = 1 + x2 is positive for all x ∈ R.
Thus, there does not exist any x in domain R such that f(x) = −2.
∴ f is not onto.
Hence, f is neither one – one nor onto.
Question 8: Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is
bijective function.
Answer 8:
f: A × B → B × A is defined as f(a, b) = (b, a).
Let (a1, b1), (a2, b2) ∈ A × B such that f (a1, b1) = f(a2, b2)
⇒ (b1, a1) = (b2, a2)
⇒ b1 = b2 and a1 = a2
⇒ (a1, b1) = (a2, b2)
∴ 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). [By definition of f ]
∴ f is onto.
Hence, f is bijective.
Question 9:
Let f: N → N be defined by f(n) =
State whether the function f is bijective. Justify your answer.
Answer 9:
f: N → N is defined as f(n) =
It can be observed that:
f(1) =((1+1) / 2)= 1 and f(2) =(2/2)= 1 [By definition of f(n)]
f(1) = f(2), where 1 ≠ 2
∴ f is not one-one.
Consider a natural number (n) in co-domain N.
Case I: n is odd
∴ n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1∈ N such that
f(4r + 1) = ((4r + 1 + 1) /2) = 2r + 1
Case II: n is even
∴ n = 2r for some r ∈ N. Then, there exists 4r ∈ N such that
f(4r) =(4r/2) = 2r.
∴ f is onto.
Hence, f is not a bijective function.
Question 10:
Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by
(x) = ((x−2) / (x−3)) . Is f one-one and onto? Justify your answer.
Answer 10:
A = R − {3}, B = R − {1} and f: A → B defined by f(x) = ((x−2) / (x−3))
Let x, y ∈ A such that f(x) = f(y)
⇒ (x-2/x-3) = (y-2/y-3)
⇒ (x – 2)(y – 3) = (y – 2)(x – 3)
⇒ xy – 3x – 2y + 6 = xy – 2x – 3y + 6
⇒ – 3x – 2y = – 2x – 3y ⇒ x = y
∴ f is one-one.
Let y ∈ B = R − {1}. Then, y ≠ 1.
The function f is onto if there exists x ∈ A such that f(x) = y.
Now, f(x) = y
⇒ (x-2 /x-3) = y
⇒ x – 2 = xy – 3y ⇒ x(1 – y) = – 3y + 2
⇒ x = (2-3y/1-y) ∈ A [y ≠ 1]
Thus, for any y ∈ B, there exists (2-3y/1-y) ∈ A such that
∴ f is onto.
Hence, function f is one – one and onto.
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
Answer 11:
f: R → R is defined as f(x) = x4.
Let x, y ∈ R such that f(x) = f(y).
⇒ x4 = y4
⇒ x = ± y
∴ f(x) = f(y) does not imply that x = y.
For example f(1) = f(–1) = 1
∴ f is not one-one.
Consider an element 2 in co-domain R. It is clear that there does not exist any x
in domain R such that f(x) = 2.
∴ f is not onto.
Hence, function f is neither one – one nor onto.
The correct answer is D.
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
Answer 12:
f: R → R is defined as f(x) = 3x.
Let x, y ∈ R such that f(x) = f(y).