**Q-1: The relation ‘m’ is defined by:**

**m (y) = y ^{2}, 0≤y≤5**

** = 5y, 5≤y≤30**

**The relation ‘n’ is defined by**

**n (y) = y ^{2}, 0≤y≤4**

** = 5y, 4≤y≤30**

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

** **

**Sol:**

Here,

m (y) = y^{2},

= 5y,

Now, ^{2}

And,

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 **Thus, the given relation is a function.**

Now, n(y) = y^{2},

= 5y,

Now, at y = 4, m (y) = 4^{2} = 16 or m(y) = 5 × 4 = 20

Thus, element 4 of the domain ** 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 g(1.2)–g(1)(1.2–1)**

** **

**Sol:**

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

Therefore,

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

**Sol:**

Here,

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:**

**Sol:**

Here, ** function.**

So, it is clear that the **function** is defined for

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**

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

**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,∞).**

**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:

**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’.**

** **

**Sol:**

=

Thus, the **range of ‘g’ is the set of all 2 ^{nd} elements**. It can be seen that all these elements are:

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

**Q-7: Assume that function ‘m’ and ‘n’ is defined from: R →R.**

**m (y) = y + 2, n(y) = 3y – 2**

**Find m + n, m – n and mn**

** **

**Sol:**

Here, **m(y) = y + 2** and **n(y) = 3y – 2** are defined from **R →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,

**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,yϵN and x = y ^{2}}. Find out which of the following is true and which one is false.**

**1. (x,y)ϵf,(y,z)ϵf⇒(x,z)ϵf.**

**2. (x,x)ϵf,forallxϵN**

**3. (x,y)ϵf⇒(y,x)ϵf**

**Also justify your answer.**

** **

**Sol:**

**f = {(x,y): x,yϵN and x = y ^{2}}**

**(1).** Now, take ^{2} and 16 = 4^{2}.

**Therefore, the given statement is true**.

**(2).** Now, let

**Therefore, the statement is false.**

**(3).** ^{2}.

Now,

**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,yϵZ}. Is ‘g’ a function from ‘Z to Z’, also justify your answer.**

** **

**Sol:**

Here,

**f = {(xy, x + y): x,yϵZ}**

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,

i.e. [ (48, 16), (48, -16) ]

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 →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

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

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