**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:**

**(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,

**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) = y ^{2}, \(0\leq y\leq 5\)**

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

**The relation ‘n’ is defined by**

**n (y) = y ^{2}, \(0\leq y\leq 4\)**

** = 5y, \(4\leq y\leq 30\)**

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

** **

**Sol:**

Here,

m (y) = y^{2}, \(0\leq y\leq 5\)

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

Now, \(0\leq y\leq 5\), m(y) = y^{2}

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

Now, at **y = 5**, m(y) = 5^{2} = 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) = y^{2}, \(0\leq y\leq 4\)

= 5y, \(4\leq y\leq 30\)

Now, at y = 4, m (y) = 4^{2} = 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) = y ^{2} then, Find \(\frac{g(1.2) – g(1)}{(1.2 – 1)}\)**

** **

**Sol:**

Here, g(y) = y^{2}

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:**

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

Thus, the **range of ‘g’ is the set of all 2 ^{nd} 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**

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**

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 = y ^{2}}. 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 = y ^{2}}**

**(1).** Now, take \((4, 2)\epsilon f, (25, 5)\epsilon f\;because\; 4,2,25,5\epsilon N\) and 4 = 2^{2} and 16 = 4^{2}.

**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 = 4^{2}.

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 1^{st} 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 1^{st} 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}**

## Leave a Reply