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# 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

Ris replaced by N with co-domain being same as R?

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

(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.

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.

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(ðÂ‘¥) = is neither one-one nor onto.

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.

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.

(i) f: R → R defined by f(x) = 3 − 4x

(ii) f: R → R defined by f(x) = 1 + x2

(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.

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.

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.

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

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