NCERT Solutions For Class 11 Maths Chapter 2

NCERT Solutions Class 11 Maths Relations and Functions

Ncert Solutions For Class 11 Maths Chapter 2 PDF Download

NCERT Solutions For Class 11 Maths Chapter 2 relations and functions are provided here in a detailed and easy to understand way. Here, the complete exercise-wise solutions for NCERT class 11 maths solution for chapter 2 (relations and functions) are given which the students can check and clear their doubts instantly. These solutions are available for free and any student can refer to these for clearing doubts and for knowing the best answers to the respective questions in chapter 2 of NCERT class 11 book.

In the class 11 maths book of NCERT, several practice questions are included in which the students often face doubts and eventually they end up piling those up. At this point, the NCERT Solutions For Class 11 Maths comes in handy which can help the students to clear their doubts instantly. The solutions are also available in PDF format which students can download and check offline. Similarly, for the relations and functions chapter, the detailed solutions are given here in a very detailed but easily understandable way. Check out the complete NCERT Solutions Class 11 Maths Relations and Functions below.

NCERT Solutions Class 11 Maths Chapter 2 Exercises

Exercise 2.1

Q.1: If \((\frac{ a }{ 3 } \; + \; 1, \; b \; – \; \frac{ 2 }{ 3 })\) = \((\frac{ 5 }{ 3 }, \; \frac{ 1 }{ 3 })\), what is the value of a and b?

 

Sol:

\((\frac{ a }{ 3 } \; + \; 1, \; b \; – \; \frac{ 2 }{ 3 })\) = \((\frac{ 5 }{ 3 }, \; \frac{ 1 }{ 3 })\)

As the ordered pairs are equal, the corresponding elements will also be equal.

Therefore, \(\frac{ a }{ 3 } \; + \; 1 \; = \; \frac{ 5 }{ 3 }\\\)

\(\frac{ a }{ 3 } \; = \; \frac{ 5 }{ 3 } \; – \; 1\)

\(\frac{ a }{ 3 } \; = \; \frac{ 2 }{ 3 }\)

Therefore,   a = 2

Now, \(b \; – \; \frac{ 2 }{ 3 } \; = \; \frac{ 1 }{ 3 }\)

\(b \; = \; \frac{ 1 }{ 3 } \; + \; \frac{ 2 }{ 3 }\)

Therefore,   b = 1

Hence, a = 2 and b = 1

 

 

Q:2. If the set X has 4 elements and the set Y = {2, 3, 4, 5}, then find the number of elements in X × Y

 

Sol:

There are 4 elements in set X and the elements of set X are 2, 3, 4, and 5.

No. of elements in X × Y = (No. of elements in X) × (No. of elements in Y)

= 4 × 4

= 16

Therefore, the no. of elements in \((X \; \times \; Y)\) is 16.

 

 

Q.3: If A = {8, 9} and B = {4, 5, 2}, what is the value of A × B and B × A?

 

Sol:

A = {8, 9}

B = {4, 5, 2}

As we know that Cartesian product ‘P × Q’ of two non-empty sets ‘P’ and ‘Q’ is defined as P × Q = {(p, q): p \(\in\) P, q \(\in\) Q}

Therefore,

A × B = {(8, 4), (8, 5), (8, 2), (9, 4), (9, 5), (9, 2)}

B × A = {(4, 8), (4, 9), (5, 8), (5, 9), (2, 8), (2, 9)}

 

 

Q.4: State whether the given statements are True or False. If the statement is false, write that statement correctly.

(i). If X = {a, b} and Y = {b, a}, then X × Y = {(a, b), (b, a)}

(ii). If P and Q are non – empty sets, then P × Q is a non – empty set of ordered pairs (a, b) such that x \(\in\) P and b \(\in\) Q.

(iii). If M = {2, 3}, N = {4, 5}, then M × (N \(\cap\)Ø ) = Ø.

 

Sol:

(i). If X = {a, b} and Y = {b, a}, then X × Y = {(a, b), (b, a)}

The given statement is False.

X = {a, b}

Y = {b, a}

Therefore, X × Y = {(a, b), (a, a), (b, b), (b, a)}

 

(ii). If P and Q are non – empty sets, then P × Q is a non – empty set of ordered pairs (a, b) such that x \(\in\) P and b \(\in\) Q.

The given statement is True.

 

(iii). If M = {2, 3}, N = {4, 5}, then M × (N \(\cap\)Ø ) = Ø.

The given statement is True.

 

 

Q.5: If M = {-2, 2}, then find M × M × M.

 

Sol:

For any non – empty set ‘M’, M × M × M is defined as:

M × M × M = {(x, y, z): x, y, z \(\in\) M}

Since, M = {-2, 2}             [ Given]

Therefore, M × M × M = {(–2, –2, –2), (–2, –2, 2), (–2, 2, –2), (–2, 2, 2), (2, –2, –2), (2, –2, 2), (2, 2, –2), (2, 2, 2)}

 

Q.6: If X × Y = {(a, m), (a, n), (b, m), (b, n)}. Find X and Y.

 

Sol:

X × Y = {(a, m), (a, n), (b, m), (b, n)}

As we know, that Cartesian product P × Q of two non-empty sets P and Q is defined as P × Q = {(p, q): p \(\in\) P, q \(\in\) Q}

Therefore, ‘X’ is the set of all the first elements and ‘Y’ is the set of all the second elements.

Therefore, X = {a, b} and Y = {m, n}

 

 

Q.7: Let P = {2, 3}, Q = {2, 3, 4, 5}, R = {6, 7} and S = {6, 7, 8, 9}. Verify the following:

(i). \(P \; \times \; (Q \; \cap \; R)\) = \((P \; \times \; Q) \; \cap \; (P \; \times \; R)\)

(ii). P × R is a subset of Q × S

 

Answer:

(i). \(P \; \times \; (Q \; \cap \; R)\) = \((P \; \times \; Q) \; \cap \; (P \; \times \; R)\)

Taking LHS:

\((Q \; \cap \; R)\) = {2, 3, 4, 5} \(\cap\) {6, 7}

= Ø

\(P \; \times \; (Q \; \cap \; R)\) = P × Ø = Ø

Now Taking RHS:

\(P \; \times \; Q\) = {(2, 2), (2, 3), (2, 4), (2, 5), (3, 2), (3, 3), (3, 4), (3, 5)}

\(P \; \times \; R\) = {(2, 6), (2, 7), (3, 6), (3, 7)}

\((P \; \times \; Q) \; \cap \; (P \; \times \; R)\) = Ø

Therefore, LHS = RHS

\(P \; \times \; (Q \; \cap \; R)\) = \((P \; \times \; Q) \; \cap \; (P \; \times \; R)\)

(ii).   P × R is a subset of Q × S

P × R = {(2, 6), (2, 7), (3, 6), (3, 7)}

Q × S = {(2, 6), (2, 7), (2, 8), (2, 9), (3, 6), (3, 7), (3, 8), (3, 9), (4, 6), (4, 7), (4, 8), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9)}

We can see that all the elements of set P × R are the elements of the set Q × S.

Therefore, P × R is a subset of Q × S.

 

 

Q.8: Let P = {2, 3} and Q = {4, 5}. Find P × Q and then find how many subsets will P × Q have? List them.

 

Sol:

P = {2, 3}

Q = {4, 5}

P × Q = {(2, 4), (2, 5), (3, 4), (3, 5)}

n (P × Q) = 4

As we know, that If ‘A’ is a set with n(A) = m,

Then, n[P(A)] = \(2^{ m }\)

Therefore,

For the set P × Q = \(2^{ 4 }\)

= 16 subsets

The subsets are as following:

Ø, {(2, 4)}, {(2, 5)}, {(3, 4)}, {(3, 5)}, {(2, 4), (2, 5)}, {(2, 4), (3, 4)}, {(2, 4), (3, 5)}, {(2, 5), (3, 4)}, {(2, 5), (3, 5)}, {(3, 4), (3, 5)}, {(2, 4), (2, 5), (3, 4)}, {(2, 4), (2, 5), (3, 5)}, {(2, 4), (3, 4), (3, 5)}, {(2, 5), (3, 4), (3, 5)}, {(2, 4), (2, 5), (3, 4), (3, 5)}

 

 

Q.9: Let M and N be two sets where n (M) = 3 and n (N) = 2. If (a, 1), (b, 2), (c, 1) are in M × N, find M and N, where a, b and c are different elements.

 

Sol:

n(M) = 3

n(N) = 2

Since, (a, 1), (b, 2), (c, 1) are in M × N

[M = Set of first elements of the ordered pair elements of M × N]

[N = Set of second elements of the ordered pair elements of M × N]

Therefore, a, b and c are the elements of M

And, 1 and 2 are the elements of N

As n(M) = 3 and n(N) = 2, hence M = {a, b, c} and N = {1, 2}

 

 

Q.10: The Cartesian product Z × Z has 9 elements among which are found (-2, 0) and (0, 2). Find the set Z and also the remaining elements of Z × Z.

 

Sol:

As we know that, If n(M) = p and n(N) = q, then n(M × N) = pq.

Now,

n (Z × Z) = n(Z) × n(Z)

But, it is given that, n(Z × Z) = 9

Therefore, n(Z) × n(Z) = 9

n(Z) = 3

The pairs (-2, 0) and (0, 2) are two of the nine elements of Z × Z

As we know, Z × Z = {(x, x): x \(\in\) Z}

Therefore, –2, 0, and 2 are elements of Z

Since, n(Z) = 3, we can see that Z = {–2, 0, 2}

Therefore, the remaining elements of the set Z × Z are (–2, –2), (–2, 2), (0, –2), (0, 0), (2, –2), (2, 0), and (2, 2).

 

 

Exercise 2.2 

Q.1: Let X = {1, 2, 3, 4, . . . . . 14}. Define a relation Z from X to X by Z= {(a, b): 3a – b = 0, where a, b \(\in\) X}. Find its co – domain, domain and range.

 

Sol:

The relation ‘Z’ from ‘X to X’ is:

Z = {(a, b): 3a – b = 0, where a, b \(\in\) X}

Z = {(a, b): 3a = b, where a, b \(\in\) X}

Z = {(1, 3), (2, 6), (3, 9), (4, 12)}

The domain of Z is the set of all the first elements of the ordered pairs in the relation.

Domain of Z = {1, 2, 3, 4}

The set X is the co – domain of the relation Z.

Therefore, co – domain of Z = X = {1, 2, 3, 4, . . . . . . 14}

The range of Z is the set of the second elements of the ordered pairs in the relation.

Therefore, Range of Z = {3, 6, 9, 12}

 

 

Q.2: Define a relation Z on the set N of natural no. by Z = {(a, b): b = a + 5, a is a natural no less than 4; a, b \(\in\) N}. Give this relationship in the roaster form. Find the domain and the range.

 

Sol:

Z = {(a, b): b = a + 5, a is a natural number less than 4; a, b \(\in\) N}.

Natural numbers less than 4 are 1, 2 and 3.

Z = {(1, 6), (2, 7), (3, 8)}

The domain of Z is the set of all the first elements of the ordered pairs in the relation.

Domain of Z = {1, 2, 3}

The range of Z is the set of the second elements of the ordered pairs in the relation.

Therefore, Range of Z = {6, 7, 8}

 

 

Q.3: M = {1, 2, 3, 5} and N = {4, 6, 9}. Define a relation Z from M to N by Z = {(a, b): the difference between a and b is odd; a \(\in\) M, b \(\in\) N}. Find Z in roster form.

 

Sol:

M = {1, 2, 3, 5}

N = {4, 6, 9}

Z = {(a, b): the difference between a and b is odd; a \(\in\) M, b \(\in\) N}

Therefore, Z = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

 

 

Q.4: The figure given below shows a relationship between the sets A and B. Find the following relation:

sets A and B

(i) In set-builder form

(ii) In roster form.

What is its range and domain?

 

Sol:

According to the information given in the figure:

A = {5, 6, 7}

B = {3, 4, 5}

(i).   Z = {(a, b): b = a – 2; a \(\in\) A}

(or),  Z = {(a, b): b = a – 2 for a = 5, 6, 7}

(ii).   Z = {(5, 3), (6, 4), (7, 5)}

Domain of Z = {5, 6, 7}

Range of Z = {3, 4, 5}

 

 

Q.5: Let X = {1, 2, 3, 4, 6}. Let Z be the relation on X defined by {(p, q): p, q \(\in\) X, q is divisible by p}.

(i) Write Z in the roster form

(ii) Find domain of Z

(iii) Find range of Z

 

Sol:

X = {1, 2, 3, 4, 6}

Z = {(p, q): p, q \(\in\) X, q is divisible by p}

(i) Z = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

(ii) Domain of Z = {1, 2, 3, 4, 6}

(iii) Range of Z = {1, 2, 3, 4, 6}

 

 

Q.6: Find the range and domain of the relation Z defined by Z = {(a, a + 5): a \(\in\) {0, 1, 2, 3, 4, 5}}.

 

Sol:

Z = {(a, a + 5): a \(\in\) {0, 1, 2, 3, 4, 5}}

Therefore, Z = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

Domain = {0, 1, 2, 3, 4, 5}

Range = {5, 6, 7, 8, 9, 10}

 

 

Q.7: Find the relation Z = {(a, \(a^{3}\)): a is a prime number less than 10} in the roster form.

 

Sol:

Z = {(a, \(a^{3}\)): a is a prime no. less than 10}

The prime number less than 10 are 2, 3, 5, and 7.

Therefore, Z = {(2, 8), (3, 27), (5, 125), (7, 343)}

 

 

Q.8: Let X = {a, b, c} and Y = {11, 12}. Find the no. of relations from X to Y.

 

Sol:

It is given that X = {a, b, c} and Y = {11, 12}.

X × Y = {(a, 11), (a, 12), (b, 11), (b, 12), (c, 11), (c, 12)}

As n(X × Y) = 6, the no of subsets of X × Y = \(2^{6}\).

Therefore, the number of relations from X to Y is \(2^{6}\).

 

 

Q.9: Let Z be the relation on P defined by Z = {(x, y): x, y \(\in\) P, x – y is an integer}. Find the range and domain of Z.

 

Sol:

Z = {(x, y): x, y \(\in\) P, x – y is an integer}

As we know that the difference between any two integers is always an integer.

Domain of Z = P

Range of Z = P

 

 

Exercise 2.3 

Q.1: Which of the given relations are functions? Answer with reason. If it is a function, find its range and domain.

(i) {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}

(ii) {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}

(iii) {(11, 13), (11, 15), (12, 15)}

 

Sol:

(i).   {(12, 11), (15, 11), (18, 11), (1, 11), (4, 11), (7, 11)}

Since, 12, 15, 18, 1, 4 and 7 are the elements of the domain of the given relation which have unique images.

Therefore, the given relation is a function.

Domain = {12, 15, 18, 1, 4, 7}

Range = {11}

 

(ii).   {(12, 11), (14, 12), (16, 13), (18, 14), (0, 15), (2, 16), (4, 17)}

Since, 12, 14, 16, 18, 0, 2 and 4 are the elements of the domain of the given relation which have unique images.

Therefore, the given relation is a function.

Domain = {12, 14, 16, 18, 0, 2, 4}

Range = {11, 12, 13, 14, 15, 16, 17}

(iii).  {(11, 13), (11, 15), (12, 15)}

Since, the same first element that is 11 corresponds to two images that is 13 and 15.

Therefore, the given relation is not a function.

 

 

Q.2: Find the range and domain of the given real function:

(i) f(y) = -|y|

(ii) f(y) = \(\sqrt{9 – y^{2}}\)

 

Sol:

(i) f(y) = -|y|, y \(\in\) R

As we know:

|y| = \(\left\{\begin{matrix} y,\; i\! f \; \; y\geq 0\\ -y,\; i\! f \; \; y< 0 \end{matrix}\right.\\\)

Therefore,

f(y) = -|y| = \(\left\{\begin{matrix} -y,\; i\! f \; \; y\geq 0\\ y,\; i\! f \; \; y< 0 \end{matrix}\right.\\\)

Since, f(y) is defined for y \(\in\) R, the domain of ‘f’ is R.

The range of f(y) = -|y|is all the real number except positive real number.

Therefore, range of ‘f’ is \((-\infty , 0]\).

 

(ii) f(y) = \(\sqrt{9 – y^{2}}\\\)

\(\sqrt{9 – y^{2}}\) is defined for the real number which are greater than or equal to -3 and less than or equal to 3.

Therefore, Domain of f(y) = {y: -3 \(\leq\) y \(\leq\) 3} or [-3, 3].

For any value of ‘y’ such that \(-3 \; \leq \; y \; \leq 3\), the value of f(y) will lie between 0 and 3.

Therefore, the Range of f(y) = {y: 0 \(\leq\) y \(\leq\) 3} or [0, 3]

 

 

Q.3: A function f is f(y) = 3y – 6. Find the values of the following:

(i) f(1)

(ii) f(8)

(iii) f(-2)

Sol:

The function of ‘f’ is:

f(y) = 3y – 6

(i) f(1) = (3 × 1) – 6

= 3 – 6 = -3

 

(ii) f(8) = (3 × 8) – 6

= 24 – 6= 18

 

(iii) f(-2)  = (3 × -2) – 6

= -6 – 6= -12

 

 

Q.4: The function ‘f’ which shows temperature in degree Celsius into temperature in degree Fahrenheit is expressed as: \(\frac{9C}{5}\; + \; 32\).

Find for the following values:

(i) f(0)

(ii) f(28)

(iii) f(-10)

(iv) The value of C, when f(C) = 212

 

Sol:

F = \(\frac{9C}{5}\; + \; 32\)

(i) f(0):

= \(\frac{9 \times 0}{5}\; + \; 32\)

= 0 + 32= 32

 

(ii) f(28):

= \(\frac{9 \times 28}{5}\; + \; 32\)

= \(\frac{252 + 160}{5}\)

= \(\frac{412}{5}\)

 

(iii) f(-10):

= \(\frac{9 \times -10}{5}\; + \; 32\)

= \(9 \times (-2) \;+ \; 32\)

= -18 + 32= 14

 

(iv) The value of C, when f(C) = 212:

\(\Rightarrow\) \(212=\frac{9 \times C}{5}+32\\\)

\(\Rightarrow\) \(\\\frac{9 \times C}{5}=212-32\\\)

\(\Rightarrow\) \(\\\frac{9 \times C}{5} = 180\\\)

\(\Rightarrow\) \(\\9 \times C = 180 \times 5\\\)

\(\Rightarrow\) \(\\C=\frac{180\times5}{9}\\\)

i.e.  C=100

Therefore, The value of f, when f(C) = 212 is 100.

 

 

Q.5: Calculate range of the given functions:

(i) f(y) = 2 – 3y, y \(\in\) R, y > 0.

(ii) f(y) = \(y^{2}\; + \;2\), is a real no.

(iii) f(y)  = y, y is a real no.

Sol:

(i)  \(f\left ( y \right ) = 2 – 3y, \; y \in R, \; y > 0\)

We can write the value of f(y) for different real numbers x > 0 in tabular form as:

y 0.01 0.1 0.9 1 2 2.5 4 5
f(y) 1.97 1.7

0.7

1

4

5.5

10

13

Thus, we can clearly observe that the range of ‘f’ forms the set for all real numbers which are less than 2. 

i.e. Range of f = \(\left ( -\infty , 2 \right )\)

Alternative:

Let, y > 0

3y > 0

2 – 3y < 2

i.e f(y) < 2

Therefore, Range of f = \(\left ( -\infty , 2 \right )\)

 

(ii) \(f(y) = y^{2} + 2\), y, is a real number.

We can write the value of f(y) for different real numbers x, in tabular form as:

y 0 \(\pm\)0.3 \(\pm\)0.8 \(\pm\)1 \(\pm\)2 \(\pm\)3 . . .
f(y) 2 2.09 2.64 3 6 11 . . .

Thus, we can clearly observe that the range of f forms the set for all real numbers which are less than 2. 

i.e. Range of f = \(\left ( -\infty , 2 \right )\)

Alternative:

Let ‘y’ be any real number. Then,

\(\\ y^{2} \geq 0 \\ => y^{2} + 2 \geq 0 + 2 \\ => y^{2} + 2 \geq 2 \\ f(y) \geq 2\)

Therefore, Range of f = \(\left [ 2, \; \infty \right )\)

 

(iii) f(y) = y, where y is a real number

Here, we can see that the range of f is the set of all the real numbers.

Therefore, Range of f = R

 

 

Miscellaneous Exercise

Q-1:  The relation ‘m’ is defined by:

m (y) = y2,  \(0\leq y\leq 5\)

          = 5y,  \(5\leq y\leq 30\)

The relation ‘n’ is defined by

n (y) = y2,  \(0\leq y\leq 4\)

         = 5y,  \(4\leq y\leq 30\)

Now, prove that ‘m’ is a function and ‘n’ is not a function.

 

Sol:

Here,

m (y) = y2,  \(0\leq y\leq 5\)

= 5y, \(5\leq y\leq 30\)

Now, \(0\leq y\leq 5\),   m(y) = y2

And, \(5\leq y\leq 30\),   m(y) = 5y

Now, at y = 5, m(y) = 52 = 25 or m(y) = 5 × 5 = 25

i.e., at y = 5, m(y) = 25

Therefore, for \(0\leq y\leq 30\) , the images of m(y) are unique. Thus, the given relation is a function.

Now, n(y) = y2,   \(0\leq y\leq 4\)

= 5y,   \(4\leq y\leq 30\)

Now, at y = 4, m (y) = 42 = 16 or m(y) = 5 × 4 = 20

Thus, element 4 of the domain \(0\leq y\leq 30\) of relation ‘n’ has 2 different images i.e., 16 and 20

Therefore, this relation is not a function.

 

 

Q-2: If g(y) = y2 then, Find \(\frac{g(1.2) – g(1)}{(1.2 – 1)}\)

 

Sol:

Here, g(y) = y2

Therefore, \(\frac{g(1.2) – g(1)}{(1.2 – 1)}\\\)

\(= \frac{(1.2)^{2} – (1)^{2}}{(1.2 – 1)}= \frac{1.44 – 1}{0.2}\)

\(= \frac{0.44}{0.2} = 2.2\)

 

 

Q-3: Find the domain for the function given below:

\(g(y) = \frac{y^{2} – 2y + 1}{y^{2} – 9y + 20}\)

Sol:

Here, \(g(y) = \frac{y^{2} – 2y + 1}{y^{2} – 9y + 20}\\\)

\(= \frac{y^{2} – 2y + 1}{(y – 5)(y – 4)}\\\)

Now, it clear from above equation that the function ‘g’ is defined for all real numbers except ‘y = 4’ and ‘y = 5’.

Therefore, the required domain is R: {4, 5}

 

 

Q-4: Find the range and domain of the function given below:

\(g(y) = \sqrt{(y – 5)}\)

Sol:

Here, \(g(y) = \sqrt{(y – 5)}\) is the given function.

So, it is clear that the function is defined for \(y\geq 5\).

So, the domain will be the set of all real numbers greater than or equal to 5. i.e., The domain for g(y) is \([5,\infty )\)

Now, for range of the given function, we have:

\(y\geq 5\)

\(\Rightarrow (y – 5)\geq 0\)

\(\Rightarrow \sqrt{(y – 5)} \geq 0\\\)

Therefore, the range of g(y) is the set of all real numbers greater than or equal to 0.

i.e, the range of g(y) is \([0,\infty )\).

 

 

Q-5: Find the range and domain of the function:   g(y) = |y – 4|

 

Sol:

Here |y – 4| is the given function.

So, it is clear that the function is defined for all the real numbers.

The domain for g(y) is R

Now, for range of the given function, we have:

\(y\epsilon R ,\), |y – 4| assumes for all real numbers.

Therefore, the range of g(y) is the set of all non- negative real numbers.

 

 

Q-6: A function from ‘R into R’ is given below. Find the range of ‘g’.

\(g = \left [ \left ( y, \frac{y^{2}}{1 + y^{2}} \right ); y\epsilon R \right ]\)

 

Sol:

\(g = \left [ \left ( y, \frac{y^{2}}{1 + y^{2}} \right ); y\epsilon R \right ]\\\)

=\([(0,0), (\pm 0.5, \frac{1}{5}), (\pm 1, \frac{1}{2}),(\pm 1.5, \frac{9}{13}),(\pm 2, \frac{4}{5}), (3, \frac{9}{10}),…….]\\\)

Thus, the range of ‘g’ is the set of all 2nd elements. It can be seen that all these elements are:    \(\geq 0\;but\;< 1\).

Therefore, the range of ‘g’ is = [0,1)

 

 

Q-7: Assume that function ‘m’ and ‘n’ is defined from: R\(\rightarrow\)R.

m (y) = y + 2, n(y) = 3y – 2

Find m + n, m – n and \(\frac{m}{n}\)

 

Sol:

Here, m(y) = y + 2 and n(y) = 3y – 2 are defined from R\(\rightarrow\)R.

Now, (m + n) (y) = m(y) + n(y) = (y + 2) + (3y – 2) = 4y

Therefore, (m + n) (y) = 4y

(m – n) (y) = m(y) – n(y) = (y + 2) – (3y – 2) = -2y + 4

Therefore, (m – n) (y) = -2y + 4

\((\frac{m}{n})(y) = \frac{m(y)}{n(y)}, n(y)\neq 0, y\epsilon R\\\)

Therefore, \((\frac{m}{n})(y) = \frac{y + 2}{3y – 2}, 3y – 2 \neq 0, 3y\neq 2\\\)

\((\frac{m}{n})(y) = \frac{y + 2}{3y – 2}, y\neq \frac{2}{3}\)

 

 

Q-8: Let g = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a function from ‘Z to Z’ defined by g(y) = uy + v, for some integers u,v. Find u,v.

Sol:

g = {(1, 1), (2, 3), (0, -1), (-1, -3)} and g(y) = uy + v

\((1, 1) \;\epsilon g \Rightarrow g(1) = 1 \Rightarrow u\times 1 + v = 1\)

\(\Rightarrow u + v = 1\)

\((0,-1)\;\epsilon g \Rightarrow g(0) = -1 \Rightarrow u\times 0 + v = -1\)

\(\Rightarrow v = -1\)

By putting v = -1 in u + v = 1, we get

u = 2.

Therefore, u = 2 and v = -1.

 

 

Q-9: Let ‘f’ be a relation from ‘N to N’ defined by f = {(x,y): \(x,y \epsilon N\) and x = y2}. Find out which of the following is true and which one is false.

1.\((x,y) \epsilon f, (y,z)\epsilon f \Rightarrow (x,z)\epsilon f\).

2.\((x,x) \epsilon f, \;for\;all\;x\epsilon N\)

3.\((x,y) \epsilon f \Rightarrow (y,x)\epsilon f\)

Also justify your answer.

 

Sol:

f = {(x,y): \(x,y \epsilon N\) and x = y2}

(1). Now, take \((4, 2)\epsilon f, (25, 5)\epsilon f\;because\; 4,2,25,5\epsilon N\) and 4 = 22 and 16 = 42.

Therefore, the given statement is true.

(2). Now, let \(3\epsilon N\;but\;3\neq 3^{2} = 9\)

Therefore, the statement is false.

(3). \((16, 4)\epsilon N\;because\;16,4\epsilon N\) and 16 = 42.

Now, \(4\neq 16^{2} = 256;\) therefore (4, 16) does not belongs to N.

Therefore, the statement is false.

 

 

Q-10: Assume U = {1, 2, 3, 4}, V = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2,9), (3,1), (4, 5), (2, 11)}. Find out which of the following is true and which one is false.

(1). ‘f’ is a function from U to V.

(2). ‘f’ is a relation from U to V.

Justify your answer.

 

Sol:

Here, U = {1, 2, 3, 4} and V = {1, 5, 9, 11, 15, 16}

Therefore, U x V = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(1). As, the 1st member i.e. ‘2’ corresponds to 2 different images i.e. ‘9’ and ‘11’, Therefore, ‘f’ is not a function.

(2). A relation of non-empty set ‘U’ to a non-empty set ‘V’ is a subset of the Cartesian product U x V.

Also it can be seen that ‘f’ is a subset of U x V.

Therefore, ‘f’ is a relation from U to V.

 

 

Q-11: Assume ‘g’ be the subset of ‘Z to Z’ defined by f = {(xy, x + y): \(x, y \epsilon Z\)}. Is ‘g’ a function from ‘Z to Z’, also justify your answer.

 

Sol:

Here,

f = {(xy, x + y): \(x, y \epsilon Z\)}

As we know that a relation g from set X to a set Y is said to be a function only if every element of set X is having a unique images in set Y.

Since, \(4, 12, -4, -12 \;\epsilon \;Z\)

\([4\times 12, 4 + 12], [(-4)\times (-12),-4 + (-12)]\;\epsilon \;g\)

i.e.  [ (48, 16), (48, -16) ] \(\epsilon\; g\)

Here, the same 1st member ‘48’ is having 2 images ‘16’ and ‘-16’.

Therefore, relation ‘g’ is not a function from ‘Z to Z’.

 

 

Q-12: Assume ‘X’ = {5,7,9, 10, 11, 12, 13} and let ‘g’: X\(\rightarrow\)N be defined by g(n) = The highest prime factor of ‘n’. Find the range of ‘g’.

 

Sol:

X = {5, 7, 9, 10, 11, 12, 13, 14, 15};

g: X\(\rightarrow\)N be defined by g(n) = The highest prime factor of ‘n’

Prime factor of 5 = 5

Prime factor of 7 = 7

Prime factor of 9 = 3

Prime factor of 10 = 2,5

Prime factor of 11 = 11

Prime factor of 12 = 2,3

Prime factor of 13 = 13

Prime factor of 14 = 2, 7

Prime factor of 15 = 3, 5

Therefore,

g(5) = The highest prime factor of 5 = 5

g(7) = The highest prime factor of 7 = 7

g(9) = The highest prime factor of 9 = 3

g(10) = The highest prime factor of 10 = 5

g(11) = The highest prime factor of 11 = 11

g(12) = The highest prime factor of 12 = 3

g(13) = The highest prime factor of 13 = 13

g(14) = The highest prime factor of 14 = 7

g(15) = The highest prime factor of 15 = 5

Thus, the range of ‘g’ is the set of all ‘g(n)’, where \(n\epsilon X\).

Therefore, Range of g = {3, 5, 7, 11, 13}

The relations and functions chapter in class 11 is an introduction to some of the most important concepts in maths. These concepts are continued in class 12 and so, students are required to be completely thorough with all the important concepts to be able to comprehend the advanced concepts in CBSE class 12.

The chapter starts with the basic concepts about relations and functions and then introduces the important topics of Cartesian Products of Sets, concepts of relations and then about functions along with the graphs of some common functions and algebra of real functions.

In chapter 2 of CBSE class 11 book, several example questions based on each of the topic are given to help the students understand the concepts in a better way. Several exercise questions are also included in the book to help the students strengthen their skills from these concepts. Students are suggested to solve all the questions as it will help them develop their problem-solving abilities and help them to have a deeper understanding of the concepts.

Keep visiting BYJU’S to get science and maths NCERT Solutions for all the classes and for all the chapters. These solutions are extremely helpful for the students as they can clear all their doubts instantly and help them have a better understanding of the respective concepts.