Question 2

Check the injectivity and surjectivity of the following functions:

(i)* f* : **N → N** given by* f(x*) = x^{2}

(ii)* f* : **Z → Z** given by *f(x)* = x^{2}

(iii)* f* : **R → R** given by* f(x)* = x^{2}

(iv)* f *: **N → N** given by *f(x)* = x^{3}

(v)* f* : **Z → Z** given by *f(x)* = x^{3 }

Answer

(i) *f*: **N** → **N** is given by,

*f*(*x*) = *x*^{2}

It is seen that for *x*, *y* ∈**N**, *f*(*x*) = *f*(*y*) ⇒ *x*2 = *y*2 ⇒ *x* = *y*.

∴*f* is injective.

Now, 2 ∈ **N**. But, there does not exist any *x* in **N** such that *f*(*x*) = *x*^{2} = 2.

∴ *f* is not surjective.

Hence, function *f* is injective but not surjective.

(ii) *f*: **Z** → **Z** is given by,

*f*(*x*) = *x*^{2}

It is seen that *f*(-1) = *f*(1) = 1, but -1 ≠ 1.

∴ *f* is not injective.

Now,-2 ∈ **Z**. But, there does not exist any element *x* ∈**Z** such that *f*(*x*) = *x*^{2} = -2.

∴ *f* is not surjective.

Hence, function *f* is neither injective nor surjective.

(iii) *f*: **R** → **R** is given by,

*f*(*x*) = *x*^{2}

It is seen that *f*(-1) = *f*(1) = 1, but -1 ≠ 1.

∴ *f* is not injective.

Now,-2 ∈ **R**. But, there does not exist any element *x* ∈ **R** such that *f*(*x*) = *x*^{2} = -2.

∴ *f* is not surjective.

Hence, function *f* is neither injective nor surjective.

(iv) *f*: **N** → **N** given by,

*f*(*x*) = *x*^{3}

It is seen that for *x*, *y* ∈**N**, *f*(*x*) = *f*(*y*) ⇒ *x*^{3} = *y*^{3} ⇒ *x* = *y*.

∴*f* is injective.

Now, 2 ∈ **N**. But, there does not exist any element *x* in domain **N** such that *f*(*x*) = *x*^{3 }= 2.

∴ *f* is not surjective

Hence, function *f* is injective but not surjective.

(v) *f*: **Z** → **Z** is given by,

*f*(*x*) = *x*^{3}

It is seen that for *x*, *y* ∈ **Z**, *f*(*x*) = *f*(*y*) ⇒ *x*^{3} = *y*^{3} ⇒ *x* = *y*.

∴ *f* is injective.

Now, 2 ∈ **Z**. But, there does not exist any element *x* in domain **Z** such that *f*(*x*) = *x*^{3} = 2.

∴ *f* is not surjective.

Hence, function *f* is injective but not surjective.

- Q:- Given an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive. - Q:- Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation R in the set A = {1, 2, 3,13, 14} defined as

R = {(x, y): 3x − y = 0}

(ii) Relation R in the set N of natural numbers defined as

R = {(x, y): y = x + 5 and x < 4}

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as

R = {(x, y): y is divisible by x}

(iv) Relation R in the set Z of all integers defined as

R = {(x, y): x − y is as integer}

(v) Relation R in the set A of human beings in a town at a particular time given by

(a) R = {(x, y): x and y work at the same place}

(b) R = {(x, y): x and y live in the same locality}

(c) R = {(x, y): x is exactly 7 cm taller than y}

(d) R = {(x, y): x is wife of y}

(e) R = {(x, y): x is father of y} - Q:- Show that each of the relation R in the set A = { x ∈Z: 0≤x≤12}, A={x} given by

(i) R = { (a,b) : |a - b| is a multiple of 4}

(ii) R = {(a,b):a = b} is an equivalence relation.

Find the set of all elements related to 1 in each case. - Q:- Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b
^{2}} is neither reflexive nor symmetric nor transitive. - Q:-
Prove that the Greatest Integer Function

*f*: R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x. - Q:- Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
- Q:-
Show that the Modulus Function

*f*: R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative. - Q:- Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
- Q:- If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elements?
- Q:- Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as

R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

- Q:-
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i)

*f*:**R → R**defined by*f(x)*= 3 – 4x(ii)

*f*:**R → R**defined by*f(x)*= 1 + x^{2 } - Q:-
Show that

*f*: [–1, 1] → R, given by is one-one. Find the inverse of the function*f*: [–1, 1] → Range*f*.**(Hint: For***y*∈ Range*f*,*y*=, for some*x*in [ - 1, 1], i.e.,) - Q:- Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b
^{2}} is neither reflexive nor symmetric nor transitive. - Q:- Find the rate of change of the area of a circle with respect to its radius r when

(a) r = 3 cm

(b) r = 4 cm - Q:- Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = { (a,b) ; |a - b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
- Q:- Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
- Q:- Given an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive. - Q:- Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
- Q:-
Consider

*f*: {1, 2, 3} → {a, b, c} given by*f(1)*= a,*f(2)*= b and*f(3)*= c. Find*f*and show that^{ –1}*(f*=^{ –1})^{–1}*f*. - Q:-
Show that the Modulus Function

*f*: R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.

- NCERT Chapter

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