Sol. It is given that f : R* → R* is defined by f(x) = 1/x
For one-one, f(x) = f(y) → 1/x = 1/y → x = y
Therefore, f is one-one
For onto,
It is clear that for y ϵ R* , there exists x = ϵ R* . R*(Exists as y ≠ 0) such that
f(x) = 1/1/y = y
Therefore, 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
Therefore, g is one-one.
Further, it is clear that g is not onto as for 1.2 ϵ R*, there does not exist any x in N such that g(x) =1/1.2
Hence, function g is one-one but not onto
Q 2. Check the injectivity and surjectivity of the following functions:
- f : N → N given by f(x) = x2
- f : Z → Z given by f(x) = x2
- f : R → R given by f(x) = x3
- f : N → N given by f(x) = x3
- f : Z → Z given by f(x) = x3
Sol. (i) f : N→N is given by f(x) = x2
It is seen that for x, y ϵ N, f(x) = f(y)
→ x2 = y2 → x = y (x and y are positive numbers)
Therefore, f is injective
Now, 2 ϵ N but there does not exist any x in N such that f(x) = x2 = 2
It means there is some element in co-domain in which do not have any images. Therefore, 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
Therefore, f is not injective
Now, -2 ϵ Z. But there does not exist any element x ϵ Z such that f(x) = x2 = -2
Therefore, 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
Therefore, f is not injective
Now, -2 ϵ R. But there does not exist any element x ϵ R such that f(x) = x2 = -2. Therefore, 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 ϵ N, f(x) = f(y) → x3 = y3 → x = y
Therefore, f is injective. Now 2 ϵ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2
Therefore, f is not surjective
Hence, function f is injective but not surjective
(v) f : Z→Z given by f(x) = x3
It is seen that for x, y ϵ Z, f(x) = f(y) → x3 = y3 → x = y
Therefore, f is injective.
Now, 2 ϵ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2. Therefore, f is not surjective
Hence, function f is injective but not surjective
Sol. Here, 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
Therefore, f(1.2) = f(1.9), but 1.2 ≠ 1.9
Therefore, f is not one-one
Now, consider 0.7 ϵ R. It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ϵ R such that f(x) = 0.7
Therefore, f is not onto
Hence, the greatest integer function is neither one-one nor onto
Sol. Here, f : R →R is given by
It is seen that f(-1) = |-1| = 1, f(1) = |1| = 1
Therefore, f(-1) = f(1) but -1 ≠ 1
Therefore, f is not one-one
Now, Consider -1 ϵ 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
Therefore, f is not onto
Hence, the modulus functions is neither one-one nor onto
Q 5. Show that the Signum function f : R → R, given by , Is neither one-one nor onto

Now, as f(x) takes only three 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. Therefore, f is not onto.
Hence, the Signum function is neither one-one nor onto.
Sol 6.
Given that A = {1, 2, 3} and b = {4, 5, 6, 7}
Now, f : A→B is defined as f = {(1, 4), (2, 5), (3, 6)}
Therefore, 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
Q 7. In each of the following cases, state whether the functions is one-one, onto or bijective. Justify answer.
i. f : R →R defined by f(x) = 3 – 4x
ii. f : R →R defined by f(x) = 1 + x2
Sol. (i) Here, f : R → R is defined by f(x) = 3-4x
Let x1, x2 ϵ R such that f(x1) = f(x2)
→ 3 – 4x1 = 3 – 4x2 → – 4x1 = – 4x2 → x1= x2
Therefore, f is one-one.
For any real number y in R, there exists 3-y/4 in R such that
Therefore, f is onto. Hence, f is bijective.
(ii) Here, f : R →R is defined as
f(x) = 1 + x2
Let x1, x2 ϵ R such that f(x1) = f(x2)
→
For instance, f(1) = f(-1)= 2
Therefore, f(x1) = f(x2) does not imply that x1= x2
Therefore, 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 ϵ R
Thus, there does not exist any x in domain R such that f(x) = -2
Therefore, f is not onto. Hence, f is neither one-one nor onto.
Sol. Here, f : A X B → B X A is defined as f(a, b) = (b, a)
Let (a1, b1), (a2, b2) ϵ A X B
Such that f(a1, b1) = f(a2, b2) → (b1, a1) = (b2, a2)
→ b1 = b2 and a1 = a2 → (a1, b1) = (a2, b2)
Therefore, f is one-one. Now, let (b, a) ϵ B X A of any element
Then, there exist (a, b) ϵ A X B such that f(a, b) = (b, a) [definition of f]
Therefore, f is onto. Hence, f is bijective.
Q 9. Let f : N →N be defined by f(n) =
For all n ϵ N state whether the function f is onto, one-one or bijective. Justify your answer.
Sol. Here, f :N→N is defined as, f(n),
for all n ϵ N. It can be observed that
f(1)= 1+1/2 = 1 and f(2) = 2/2 = 1 [By definition of f]
f(1) = f(2), where 1 ≠ 2
Therefore, f is not one-one. Consider a natural number n in co-domain N.
Case 1 When n is odd.
Therefore, n = 2r +1 for some r ϵ N
Then, there exists 4r + 1 ϵ N such that f(4r +1) = 4r+1+1/2 = 2r + 1
Therefore, f is onto.
Case 2 When n is even
Therefore, n = 2r for some r ϵ N
Then, there exists 4r ϵ N such that f(4r) = 4r/2 = 2r
Therefore, f is onto. Hence, f is not a bijective function.
Note If f is one-one and onto, then we say that f is bijective functive.
Sol. Here, A = R – {3}, B = R – {1}
And f : A → B is defined as f(x) = (x-2/x-3)
Let x, y ϵ 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 – 3y – 2x + 6
→ -3x – 2y = -3y -2x
→ 3x – 2x = 3y – 2y → x = y
Therefore, f is one-one. Let y ϵ B = R – {1}. Then y ≠ 1
The function f is onto if there exists x ϵ 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 ϵ A [y ≠ 1]
Thus, for any y ϵ B, there exists (2-3y/1-y) ϵ A such that
Therefore, f is onto. Hence, function f is one-one and onto.
Q 11. Let f : R →R be defined as f(x) = x4. Choose the correct answer.
i. f is one-one ontp
ii. f is many-one onto
iii. f is one-one but not onto
iv. f is neither one-one nor onto
Sol. Function f : R →R is defined as f(x) = x4
Let x, y ϵ R such that f(x) =f(y)
→ x4 = y4
→ x = ±y
Therefore, f(x1) = f(x2) does not imply that x1 = x2
For instance, f(1) = f(-1) = 1
Therefore, 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.
Therefore, f is not onto. Hence, function f is neither one-one nor onto.
The correct answer is (d).
Q 12. Let f : R→R be defined as f(x) = 3x. Choose the correct answer.
- f is one-one onto
- f is many-one onto
- f is one-one but not onto
- f is neither one-one nor onto
Sol. f : R →R is defined as f(x) = 3x
Let x, y ϵ R such that f(x) =f(y)
→ 3x = 3y → x = y
Therefore, f is one-one
Also, for any real number (y) in co-domain R, there exists y/3 in R such that
f (y/3) = 3 (y/3) = y
Therefore, f is onto. Hence, function f is one-one and onto.
The correct answer is (a).