**Q12 ** Let A = {9,10,11,12,13} and let f : A N be defined by f (n) = the highest prime

factor of n. Find the range of f.

It is given that
A = {9,10,11,12,13}
And
f : A → N be defined by f(n) = the highest prime factor of n.
Now,
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
f(n) = the highest prime factor of n.
Hence,
f(9) = the highest prime factor of 9 = 3
f(10) = the highest prime factor of 10 = 5
f(11) = the highest prime factor...

**Q11** Let f be the subset of defined by . Is f a

function from Z to Z? Justify your answer.

It is given that
Now, we know that relation f from a set A to a set B is said to be a function only if every element of set A has a unique image in set B
Now, for value 2, 6, -2, -6
Now, we can observe that same first element i.e. 12 corresponds to two different images that are 8 and -8.
Thus, f is not a function

**Q10 (2) ** Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true? f is a function from A to B

It is given that
and
Now,
As we can observe that same first element i.e. 2 corresponds to two different images that is 9 and 11.
Hence f is not a function from A to B
Therefore, given statement is FALSE

**Q10 (1)** Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}

Are the following true? f is a relation from A to B

It is given that
and
Now,
Now, a relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B
And we can see that f is a subset of
Hence f is a relation from A to B
Therefore, given statement is TRUE

**Q9 (3)** Let R be a relation from N to N defined by . Are

the following true?

(a,b) R, (b,c) R implies (a,c) R.

It is given that
And
implies
Now, it can be seen that because and , But
Therefore,
Therefore, the given statement is FALSE

**Q9 (2)** Let R be a relation from N to N defined by . Are

the following true?

implies (b,a) R

It is given that
And
implies
Now , it can be seen that and , But
Therefore,
Therefore, given statement is FALSE

**Q9 (1)** Let R be a relation from N to N defined by . Are

the following true?

for all

It is given that
And
for all
Now, it can be seen that But,
Therefore, this statement is FALSE

**Q8** Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by

f(x) = ax + b, for some integers a, b. Determine a, b.

It is given that
And
Now,
At x = 1 ,
Similarly,
At ,
Now, put this value of b in equation (i)
we will get,
Therefore, values of a and b are 2 and -1 respectively

**Q7** Let f, g : R R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find

f + g, f – g and f/g

It is given that
Now,
Therefore,
Now,
Therefore,
Now,
Therefore, values of are respectively

**Q6** Let

R be a function from R into R. Determine the range

of f.

Given function is
Range of any function is the set of values obtained after the mapping is done in the domain of the function. So every value of the codomain that is being mapped is Range of the function.
Let's take
Now, 1 - y should be greater than zero and y should be greater than and equal to zero for x to exist because other than those values the x will be...

**Q5** Find the domain and the range of the real function f defined by

Given function is
As the given function is defined of all real number
The domain of the function is R
Now, as we know that the mod function always gives only positive values
Therefore,
Range of function is all non-negative real numbers i.e.

**Q4** Find the domain and the range of the real function f defined by

Given function is
We can clearly see that f(x) is only defined for the values of x ,
Therefore,
The domain of the function is
Now, as
take square root on both sides
Therefore,
Range of function is

3. Find the domain of the function

Given function is
Now, we will simplify it into
Now, we can clearly see that
Therefore, the Domain of f(x) is

**Q2** If find

**1**. The relation f is defined by

The relation g is defined by

Show that f is a function and g is not a function.

It is given that
Now,
And
At x = 3,
Also, at x = 3,
We can see that for , f(x) has unique images.
Therefore, By definition of a function, the given relation is function.
Now,
It is given that
Now,
And
At x = 2,
Also, at x = 2,
We can clearly see that element 2 of the domain of relation g(x) corresponds to two different images i.e. 4 and 6. Thus, f(x) does not have unique...

**Q5 (3)** Find the range of each of the following functions.

f (x) = x, x is a real number

**Q5 (2) ** Find the range of each of the following functions

, x is a real number.

Given function is
It is given that x is a real number
Now,
Add 2 on both the sides
Therefore,
Range of function is

**Q5 (1) ** Find the range of each of the following functions.

Given function is
It is given that
Now,
Add 2 on both the sides
Therefore,
Range of function is

**Q4(4) ** The function ‘t’ which maps temperature in degree Celsius into temperature in

degree Fahrenheit is defined by

The value of C, when t(C) = 212.

**Q4(3) ** The function ‘t’ which maps temperature in degree Celsius into temperature in

degree Fahrenheit is defined by

t (-10)

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