Question 1

# 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}

(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}

Answer

**(i)** *A* = {1, 2, 3 … 13, 14}

R = {(*x*, *y*): 3*x* − *y* = 0}

**∴ **R = {(1, 3), (2, 6), (3, 9), (4, 12)}

R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.

Also, R is not symmetric as (1, 3) ∈R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]

Also, R is not transitive as (1, 3), (3, 9) ∈R, but (1, 9) ∉ R.

[3(1) − 9 ≠ 0]

Hence, R is neither reflexive, nor symmetric, nor transitive.

**(ii)** R = {(*x*, *y*): *y* = *x* + 5 and *x* < 4} = {(1, 6), (2, 7), (3, 8)}

It is seen that (1, 1) ∉ R.

**∴ **R is not reflexive.

(1, 6) ∈R

But,

(1, 6) ∉ R.

**∴** R is not symmetric.

Now, since there is no pair in R such that (*x*, *y*) and (*y*, *z*) ∈R, then (*x*, *z*) cannot belong to R.

**∴** R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

**(iii)** *A* = {1, 2, 3, 4, 5, 6}

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

We know that any number (*x)* is divisible by itself.

(*x*, *x*) ∈R

**∴** R is reflexive.

Now,

(2, 4) ∈R [as 4 is divisible by 2]

But,

(4, 2) ∉ R. [as 2 is not divisible by 4]

**∴ **R is not symmetric.

Let (*x*, *y*), (*y*, *z*) ∈ R. Then, *y* is divisible by *x* and *z* is divisible by *y*.

**∴ ***z* is divisible by *x*.

⇒ (*x*, *z*) ∈R

**∴ **R is transitive.

Hence, R is reflexive and transitive but not symmetric.

**(iv)** R = {(*x*, *y*): *x* − *y* is an integer}

Now, for every *x* ∈ **Z**, (*x*, *x*) ∈R as *x* − *x* = 0 is an integer.

**∴ **R is reflexive.

Now, for every *x*, *y* ∈ **Z** if (*x*, *y*) ∈ R, then *x* − *y* is an integer.

⇒ −(*x* − *y*) is also an integer.

⇒ (*y* − *x*) is an integer.

**∴** (*y*, *x*) ∈ R

**∴ **R is symmetric.

Now,

Let (*x*, *y*) and (*y*, *z*) ∈R, where *x*, *y*, *z* ∈ **Z**.

⇒ (*x* − *y*) and (*y* − *z*) are integers.

⇒ *x *− *z* = (*x* − *y*) + (*y* − *z*) is an integer.

**∴** (*x*, *z*) ∈R

**∴** R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

(*x*, *x*) ∈ R

**∴** R is reflexive.

If (*x*, *y*) ∈ R, then *x* and *y* work at the same place.

⇒ *y* and *x* work at the same place.

⇒ (*y*, *x*) ∈ R.

**∴ **R is symmetric.

Now, let (*x*, *y*), (*y*, *z*) ∈ R

⇒ *x* and *y* work at the same place and *y* and *z* work at the same place.

⇒ *x* and *z* work at the same place.

⇒ (*x*, *z*) ∈R

**∴** R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

Clearly (*x*, *x*) ∈ R as *x* and *x* is the same human being.

**∴** R is reflexive.

If (*x*, *y*) ∈R, then *x* and *y* live in the same locality.

⇒ *y* and *x* live in the same locality.

⇒ (*y*, *x*) ∈ R

**∴ **R is symmetric.

Now, let (*x*, *y*) ∈ R and (*y*, *z*) ∈ R.

⇒ *x* and *y* live in the same locality and *y* and *z* live in the same locality.

⇒ *x* and *z* live in the same locality.

⇒ (*x,* *z*) ∈ R

**∴ **R is transitive.

Hence, R is reflexive, symmetric, and transitive.

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

Now, (*x*, *x*) ∉ R

Since human being *x *cannot be taller than himself.

**∴ **R is not reflexive.

Now, let (*x*, *y*) ∈R.

⇒ *x* is exactly 7 cm taller than *y*.

Then, *y* is not taller than *x*.

**∴** (*y*, *x*) ∉R

Indeed if *x* is exactly 7 cm taller than *y*, then *y* is exactly 7 cm shorter than *x*.

∴R is not symmetric.

Now,

Let (*x*, *y*), (*y*, *z*) ∈ R.

⇒ *x* is exactly 7 cm taller than* y *and *y* is exactly 7 cm taller than z.

⇒ *x* is exactly 14 cm taller than *z *.

**∴ **(*x*, *z*) ∉R

**∴ **R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

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

Now, (*x*, *x*) ∉ R

Since *x* cannot be the wife of herself.

∴R is not reflexive.

Now, let (*x*, *y*) ∈ R

⇒ *x* is the wife of *y.*

Clearly *y* is not the wife of *x*.

**∴** (*y*, *x*) ∉ R

Indeed if *x* is the wife of *y*, then *y* is the husband of *x*.

**∴ **R is not transitive.

Let (*x*, *y*), (*y*, *z*) ∈ R

⇒ *x* is the wife of *y* and *y* is the wife of *z*.

This case is not possible. Also, this does not imply that *x* is the wife of *z*.

**∴** (*x*, *z*) ∉ R

**∴ **R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

**(e)** R = {(*x*, *y*): *x* is the father of *y*}

Now (*x*, *x*) ∉ R

As *x* cannot be the father of himself.

**∴**R is not reflexive.

Now, let (*x*, *y*) ∈R.

⇒ *x* is the father of *y.*

⇒ *y* cannot be the father of *y.*

Indeed, *y* is the son or the daughter of *y.*

**∴ **(*y*, *x*) ∉ R

**∴** R is not symmetric.

Now, let (*x*, *y*) ∈ R and (*y*, *z*) ∈ R.

⇒ *x* is the father of *y* and *y* is the father of *z*.

⇒ *x* is not the father of *z*.

Indeed *x* is the grandfather of *z*.

∴ (*x*, *z*) ∉ R

**∴**R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

- 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:- 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:- 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:-
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:- 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:-
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:- Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
- Q:-
Let

*f*: R → R be defined as f(x) = 3x. Choose the correct answer.(A)

*f*is one-one onto(B)

*f*is many-one onto(C)

*f*is one-one but not onto(D)

*f*is neither one-one nor onto.

- 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:- If A = \(\begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}\), Find |A|
- Q:-
Consider

*f*: R_{+}→ [4, ∞) given by f(x) = x^{2}+ 4. Show that*f*is invertible with the inverse*f*^{–1}of f given by_{}, where R_{+}is the set of all non-negative real numbers. - 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:- Integrals cos3x
- Q:-
Determine order and degree(if defined) of differential equation \begin{align}\left(\frac{ds}{dt}\right)^4\;+\;3s\frac{d^2s}{dt^2}\;=\;0\end{align}

- 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:- The anti derivative of \begin{align} \left(\sqrt x + \frac {1}{\sqrt x}\right)\end{align} equals to \begin{align} (A) \frac{1}{3}.x^\frac{1}{3} + 2.x^\frac{1}{2} +C \;\;\;\; (B) \frac{2}{3}.x^\frac{2}{3} + \frac{1}{2}.x^2 +C \end{align}

\begin{align} (C) \frac{2}{3}.x^\frac{3}{2} +2 x^\frac{1}{2} +C \;\;\;\;(D) \frac{3}{2}.x^\frac{3}{2} +\frac{1}{2}. x^\frac{1}{2} +C \end{align} - 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:- 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.

Priyanka
2020-08-02 10:31:32

Very helpful

Priyanka
2020-08-02 10:29:50

Thanks for answer for Question no.1

Priyanka
2020-08-02 10:28:37

Thanks

Priyanka
2020-08-02 10:27:16

Thanks

Sant Kumar Hooda
2019-05-10 18:42:42

Thanks , Sant Kumar Hooda

2018-09-12 16:36:21

1 - ii is not correct

Manish Kumar
2018-05-01 09:10:13

2 is not correct

pankti naik
2017-09-01 15:03:11

solution of A={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15} S={(x,y)/y=5x ; xâA, yâA} find S is equivalence relation or not?

Hitesh
2017-03-26 20:54:15

Very helpful

- NCERT Chapter