# NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2

Here you will get NCERT Solutions for Class 12 Maths Chapter 1 – Exercise 1.2 in the text as well as pdf format. As we all know that NCERT books are very helpful in Class 12 CBSE Board. Here we have given question from Chapter 1 of Class 12th Maths subject along with their respective answers.

NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 Overview
Book National Council of Educational Research and Training (NCERT)
Class 12th Class
Subject NCERT Solution Class 12th Maths
Chapter Chapter 1 – Relations and Functions

## Chapter 1: Relations and Functions – Exercise 1.2

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Q-1: Prove that the function f: RR which is defined by f(a) = 1a which is one- one and onto, where R is the set of all of the non- zero real numbers. Check whether the result is true or not, if the domain, say, R is replaced by M having co- domain as same as R?

Solutions

Solution

As per the data given in the question,

RR which is defined by f(a) = 1/a

For one- one condition:

Let, a and b ϵR such that, f(a) = f(b)

1/a=1/b

a = b

Hence, the function f is one- one.

For onto condition:

Clearly, for bϵR, there must exists a = 1/ϵR [since, b ≠ 0], so that

f(x) = 1/1./b = y

Hence, the function f is onto.

Therefore, the given function f(a) is one- one as well as onto.

Let us consider a another function g: M → R which is defined by g(a) = 1/a.

g(x1) = g(x2)

1/a1=1/a2

x1 = x2

Hence, the another function g is also one- one.

It is obvious that g is not onto as for 1.2 ϵR, there doesn’t exist any a in M so that:

g(a) = 1/1.2

Therefore, the function ‘g’ is one- one but, it is not onto.

Q-2: Check the following functions for their injectivity and surjectivity:

(i) f: N →N which is given by f(a) = a2

(ii) f: Z →Z which is given by f(a) = a2

(iii) f: R → R which is given by f(a) = a2

(iv) f: N →N which is given by f(a) = a3

(v) f: Z → Z which is given by f(a) = a3

Solutions

Solution

(i) f: N → N which is given by f(a) = a2

We can see that for a, b ɛ N

f(a) = f(y)

a= b2

a = b

Hence, the function f is injective.

2 ɛ N

But, there doesn’t exist any of the a in N, so that f(a) = b= 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but, it is injective.

(ii) f: Z → Z which is given by f(a) = a2

We can see that for a, b ɛ Z

f(-1) = f(1) = 1

But, -1≠ 1.

Hence, the function f is not injective.

-2 ɛ Z

But, there doesn’t exist any of the a ɛ Z, so that f(a) = – 2 or a= -2

Hence, the function f is not surjective.

Therefore, the function f is neither surjective nor it is injective.

(iii) f: R → R which is given by f(a) = a2

We can see that for a, b ɛ R

f(-1) = f(1) = 1

But, -1≠ 1.

Hence, the function f is not injective.

-2 ɛ R

But, there doesn’t exist any of the a ɛ R, so that f(a) = – 2 or a= -2

Hence, the function f is not surjective.

Therefore, the function f is neither surjective nor it is injective.

(iv) f: N → N which is given by f(a) = a3

We can see that for a, b ɛ N

f(a) = f(y)

a= b3

a = b

Hence, the function f is injective.

2 ɛ N

But, there doesn’t exist any of the element of a ɛ Z, so that f(a) = 2 or a= 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but it is injective.

(v) f: Z → Z which is given by f(a) = a3

We can see that for a, b ɛ Z

f(a) = f(y)

a= b3

a = b

Hence, the function f is injective.

2 ɛ Z

But, there doesn’t exist any of the element of a ɛ Z, so that f(a) = 2 or a= 2

Hence, the function f is not surjective.

Therefore, the function f is not surjective but it is injective.

Q-3: Show that the GIF (Greatest Integer Function) f: R →R which is given by f(a) = [a], which is neither one- one nor onto, where [a] notifies the greatest integer which must be less than or equal to a.

Solutions

Solution

: R → R which is given by f(a) = [a]

f(1.4) = [1.4] = 1, f(1.7) = [1.7] = 1

f(1.4) = f(1.7) = 1, but 1.4 ≠ 1.7

Hence, the function f is not one- one.

Let us consider that, 0.9 ɛ R.

We know that, f(a) = [a] will always be any integer.

So, there doesn’t exist any of the element a ɛ R, such that f(a) = 0.9

Hence, the function f is not onto.

Therefore, the GIF (Greatest Integer Function) is neither onto nor one- one.

Q-4: Prove that the Modulus Function f: R → R which is given by f(a) = [a], which is neither one- one nor it is onto, where |a| is a, if a will be positive or 0 and |a| is –a, if a will be negative.

Solutions

Solution

f : R → R which is given by

f(a) = |a| = a, if a ≥ 0 and (-a), if a ≤ 0

It is obvious that f(-2) = |-2| = 2 and f(2) = |2| = 2

So, f(-2) = f(2), but -2 ≠ 2

Hence, f is not one- one

Let us consider -2 ɛ R.

We know that, f (a) = |a| will always be non- negative.

Therefore, neither of any element a exist in the domain R, so that

f(a) = |a| = -2.

Hence, f is not onto.

Hence, the modulus function is neither onto nor one- one.

Q-5: Prove that the Signum function f: R → R which is given by f(a) =

1, if a > 0

0, if a = 0

-1, if x < 0 is neither onto nor one- one.

Solutions

Solution

f: R → R which is given by f(a) =

1, if a > 0

0, if a = 0

-1, if x < 0

Since, f(a) can only take 3 values (i.e., 1, 0, -1) for the elements -3 in the co- domain R, there does not exist any of the a in domain R, so that f(a) = -3.

Hence, f is not onto.

Here, we can see that:

f(2) = f(3) = 1, but 2 ≠ 3

Hence, f is not one- one.

Therefore, the Signum function in this case is neither onto nor one- one.

Q-6: Let us take P as {2, 3, 4}, Q as {5, 6, 7, 8} and let f = {(2, 5), (3, 6), (4, 7)} which are the function from P and Q. Prove that the function f is one- one.

Solution

Solution

As per the data given in the question, we have

P = {2, 3, 4} and Q = {5, 6, 7, 8}

f : P → Q is defines as f = {(2, 5), (3, 6), (4, 7)}

Thus, f (2) = 5, f(3) = 6 and f(4) = 7

Now,

Here we can see that the images of the distinct elements of P under f are also distinct.

Therefore, the function f is not one- one.

Q-7. Check whether the function in each of the following conditions is one- one, onto or bijective. Justify your answer in each cases.

(i) f : R → R which is defines by f(a) = 4 – 5a

(ii) f : R → R which is defined by f(a) = 2 + a2

Solution

Solution

(i) f : R → R which is defined as f(a) = 4 – 5a

Let, a1, a2 ɛ so that, f(a1) = f(a2)

4 – 5a1 = 4 – 5a2

– 5a1 = – 5a2

a1 = a2

Hence, the function f is one- one.

Now,

For any of the real number (b) in R, there exists 4b5 in R, so that

f (4b5) = 45(4b5)

f (4b5) = 4 – (4 – b)

f (4b5) = b

Hence, the function f is onto.

Therefore, the function f is bijective.

(ii) f : R → R which is defined as f(a) = 2 + a2

Let, a1, a2 ɛ so that, f(a1) = f(a2)

2 + a12 = 2 + a22

a12 = a22

a1 = ± a2

Thus, f(a1) = f(a2) which doesn’t imply that a1 = a2

Let us illustrate an example, f(2) = f(-2) = 3

Hence, the function f is not one- one.

Now,

Let us take an element -3 in the co- domain R.

We can see that, f(a) = 2 + a2 which is positive for all of the x ɛ R.

So, there doesn’t exist any a in the domain R so that f(a) = -3.

Hence, the function f is not onto.

Therefore, the function f is neither one- one nor onto so, it is not bijective.

Q-8: Let us consider P and Q be the two sets. Prove that f : P × Q → Q × P so that (x, y) = (y, x) is bijective function.

Solution

Solution

f : P × Q → Q × P which is defined as f(x, y) = y, x.

Let, (x1, y1), (x2, y2) ɛ P × Q so that, f(x1, y1)= f(x2, y2)

(y1, x1) = (y2, x2)

y= yand x= x2

(x1, y1) = (x2, y2)

Hence, the function f is one- one.

Let us consider that (y, x) ɛ Q × P be any element.

Thus,

There will exist (x, y) ɛ P × Q so that,

f(x, y) = (y, x).

Hence, f is onto.

Therefore, the function f is bijective.

Q-9: Let f : N → N which will be defined by f(p) =

p+1/2, if p is odd

p2, if p is even, for all p ɛ N.

Check and justify your answer that whether the function f is bijective or not?

Solution

Solution

f : N → N which is defined as f(p) =

p+1/2, if p is odd

p/2, if p is even

for all n ɛ N

We can see that:

f(3) = 3+1/2 = 2 and f(4) = 4/2 = 2

f(3) = f(4), but 3 ≠ 4

Hence, f is not one- one.

Let us consider any natural number (p) in the co- domain N.

Case-1: [ p is odd ]

p = 2r + 1 for few r ɛ N.

Thus, there exists 4r + 1 ɛ N so that,

f(4r + 1) = 4r+1+1/2 = 4r+2/2 = 2r + 1

Case-2: [p is even]

p = 2r for few r ɛ N.

Thus, there exists 4r ɛ N so that,

f(4r) = 4r/2 = 2r

Hence, the function f is onto.

Therefore, the function f is not a bijective function.

Q-10: Let P = A – {3} and Q = A – {1}. Let us consider the function f : P → Q which defined by f(a) =(a3/a4). Check and Justify your answer, whether the function f is onto and one- one?

Solution

Solution

P = A – {3}, Q = A – {1} and f : P → Q which is defined by

f(a) = (a3 / a4)

Let, a, b ɛ P so that f(a) = f(b)

a3 / a4=b3 / b4

(a – 3)(b – 4) = (b – 3)(a – 4)

ab – 4a – 3b + 12 = ab – 4b – 3a + 12

a = b

Hence, the function f is one- one.

Let, b ɛ Q = A – {1}.

Thus, b ≠ 1

Then, the function f will be onto if a ɛ P so that, f(a) = b.

f(a) = b

a3/a4 = b

(a – 3) = b(a – 4)

a – 3 = ab – 4b

a(1 – b) = 3 – 4b

a = 34b  /  1b   ɛ [b ≠ 1]

Then,

For any value of b ɛ Q, there will exist 34b1b ɛ P, so that

f (34b / 1b) = (34b/1b)3   /   (34b/1b)4

(34b)3(1b) / (34b)4(1b)

34b3+3b / 34b4+4b

b/1 = b

Hence, the function f is onto.

Therefore, the function f is onto as well as it is one- one.

Q-11: Let us assume, f : R → R will be defined as g(a) = a5. Select the correct answer from the following:

(a) f is many- one onto (b) f is one- one onto

(c) f is neither one- one nor onto (d) f is one- one but, not onto

Solution

Solution

f : R → R which is defined as g(a) = a4.

Let, a, b ɛ R so that f(a) = f(b).

a4 = b4

a = ± b

Thus, f(a) = f(b) doesn’t imply that a = b.

Example:

f(3) = f(-3) = 3

Hence, the function f is not one- one.

Let us consider an element 3 in the co- domain R. It’s obvious that there doesn’t exist any a in the domain R so that, f(a) = 3.

Hence, the function f is not onto.

Therefore, the function f is neither one- one nor it is onto.

Thus, the correct answer is option (c).

Q-12: Let us assume, f : R → R will be defined as g(a) = 4a . Select the correct answer from the following:

(a) f is many- one onto (b) f is one- one onto

(c) f is neither one- one nor onto (d) f is one- one but, not onto

Solution

Solution

f : R → R which is defined as g(a) = 4a.

Let, a, b ɛ R so that f(a) = f(b).

4a = 4b

a = b

Hence, the function f is one- one.

For any of the real number (b) in the co- domain R, there must exist b3 in R so that, f(b3=3(b3)=y)

Hence, the function f is onto.

Therefore, the function f is one- one and also, it is onto.

Thus, the correct answer is option (b).