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QUESTION: 1

If y is expressed in terms of a variable x as Y = ƒ(x), then y is called

Solution:

If y is expressed in terms of a variable x as y=f(x), then y is called explicit function.

QUESTION: 2

Let f : R → R defined as f(x) = x be an identity function. Then

Solution:

Here, we are given a function f ,which is defined from R to R. Also, f is said to be the Identity Function.

The Identity Function returns the same output as given in the input.

That is, if you input some x into f(x), you get the result as x itself. So, the Identity Function is:

As we can see, the Domain of the function is R , and the co-domain as well as range is also R.

QUESTION: 3

Let A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, 7), (3, 6)} then f is

Solution:

QUESTION: 4

In the mapping above, the function f_{4} is an:

Solution:

Clearly, mappings given in options (a),(b) and (c ) satisfy the given conditions and are one-one onto.

QUESTION: 5

Let *f : N →N* be defined by for all n ∈ N.

Then, the function f is

Solution:

F(n) = n+2 if n is odd

so if n=1, then f(1)= 3

if n=3 , f(3)=5

On the other hand

F(n)= n+3 if n is even

if n= 2,f(2)=5

if n=4 ,f(4)= 7

Clearly f(3) = f(2) = 5

but 3 does not equals to 3, so it is not one one i.e. injective also 1€N

i.e. codomain

but here 1€ range does not equals to codomain so it is not surjective.

QUESTION: 6

A function f: R → R defined as f(x) = x^{4} is

Solution:

f: R → R is defined as f(x) = x^{4}

Let x, y ∈ R such that f(x) = f(y).

=> x^{4} = y^{4}

=> x =+-y

Therefore, f(x^{1}) = f(x^{2}) does not imply that x^{1}=x^{2}.

For instance,f(1) = f(-1) = 1

∴ f is not one-one.

Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.

∴ f is not onto.

Hence, function f is neither one-one nor onto.

QUESTION: 7

A function f: X → Y is bijective if and only if

Solution:

QUESTION: 8

A function f: X → Y is injective if and only if

Solution:

Let X, Y be sets, and let f : X → Y be a function. We say that f is injective (sometimes called one-to-one) if ∀x_{1}, x_{2} ∈ X, f(x_{1}) = f(x_{2})

⇒ x_{1} = x_{2}.

QUESTION: 9

A function f: R → R defined as f(x) = 5x is

Solution:

Suppose x1 and x2 are real numbers such that f(x1) = f(x2). (We need to show x1 = x2 .)

5x1 = 5x2

Dividing by 5 on both sides gives

x1 = x2 (function is one - one)

Let y ∈R. (We need to show that x in R such that f(x) = y.)

If such a real number x exists, then 5x -2 = y and x = (y + 2)/5. x is a real number since sums and quotients (except for division by 0) of real numbers are real numbers. It follows that

f(x) = 5((y + 2)/5)

by the substitution and the definition of f

= y + 2

= y

by basic algebra

Hence, f is onto.

QUESTION: 10

A function f: X → Y is surjective if and only if

Solution:

A function f (from set X to Y) is surjective if and only if for every y in Y, there is at least one x in X such that f(x) = y, in other words f is surjective if and only if f(X) = Y.

QUESTION: 11

The range of function f : R → R defined by f(x) = x^{2} is

Solution:

As x is square of every real number, so the f(x) = x^{2} cannot be the negative.

QUESTION: 12

Let f : R → R defined as f(x) = 5 be a constant function. Then

Solution:

QUESTION: 13

The function, f(x) = 2x + 1 is

Solution:

Let the domain of x be R

For surjectivity-

Let f(x) = Y

=> 2x +1 = Y

=> x = (Y - 1)/2

It is clear that Y takes all the values from (-∞, +∞)

So the range of f(x) is the same as the domain of x.

So it is surjective.

QUESTION: 14

Let f: {1, 2, 3} → {1, 2, 3} be an onto function. Then, f is

Solution:

(i) f(1)=1,f(2)=2,f(3)=3

(ii) f(1)=1,f(2)=3,f(3)=2

(iii) f(1)=2,f(2)=3,f(3)=1

(iv) f(1)=2,f(2)=1,f(3)=3

(v) f(1)=3,f(2)=1,f(3)=2

(vi) f(1)=3,f(2)=2,f(3)=1

Since, f is onto, all elements of {1,2,3} have unique pre-image.

The above cases are possible.

Since, every element 1,2,3 has either of image 1,2,3 and that image is unique.

∴ f is one-one.

∴ Function f:A→A is one-one.

QUESTION: 15

A function f: Z → Z defined as f(x) = x^{3} is

Solution:

f(x) = x^{3}

Checking one-one f(x1) = (x1)^{3}

f(x2) = (x2)^{3}

Comparing f(x1) = f(x2)

(x1)^{3} = (x2)^{3}

x1 = x2 => x1 = -x2

Since, x1 & x2 are natural number, they are always positive.

Hence, x1 = x2 ( it is one - one)

Checking onto : f(x) = (x)^{3}

Let f(x) = y such that y implies N

x^{3} = y

x = (y)^⅓

x = 1.71, hence it is not a natural number

Thus function is not onto

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