13. Show that the relation R defined in the set A of all polygons as R = {(P1, P2): P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right-angle triangle T with sides 3, 4 and 5?

The given relation in set A of all polygons is defined asR= {(P1,P2):P1 and P2 have same number of sides }Let P1∈A ,As number of sides (P1) = number of sides (P1)(P1,P1)∈RSo, R is reflexive.Let P1,P2∈A and (P1,P2)∈RThen, number of sides of P1 = number of sides of P2Number of sides of P2 = number of sides of P1i.e., (P2,P1)∈Rso, R is symmetric.Let P1,P2,P3∈A and (P1,P2) and (P2,P3)∈RThen, number of sides (P1) = number of sides (P2)Number of sides (P2) = number of sides (P3)So, number of sides (P1) = number of sides (P3)I.e., (P1,P3)∈RSo, R is transitive.Hence, R is an equivalence relation.The set of all triangles will be the elements in A related to the right-angle triangle T with sides 3,4 and 5 as they all have three number of sides.

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Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th Relations and Functions

13. Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂): P_{1 and} P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right-angle triangle T with sides 3, 4 and 5?

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Vishal Baghel | Contributor-Level 10

4 months ago

The given relation in set A of all polygons is defined as

R= ${(P_{1}, P_{2}) : P_{1}$ and $P_{2}$ have same number of sides $}$

Let $P_{1} \in A$ ,

As number of sides $(P_{1})$ = number of sides $(P_{1})$

$(P_{1}, P_{1}) \in R$

So, R is reflexive.

Let $P_{1}, P_{2} \in A$ and $(P_{1}, P_{2}) \in R$

Then, number of sides of $P_{1}$ = number of sides of $P_{2}$

Number of sides of $P_{2}$ = number of sides of $P_{1}$

i.e., $(P_{2}, P_{1}) \in R$

so, R is symmetric.

Let $P_{1}, P_{2}, P_{3} \in A$ and $(P_{1}, P_{2})$ and $(P_{2}, P_{3}) \in R$

Then, number of sides $(P_{1})$ = number of sides $(P_{2})$

Number of sides $(P_{2})$ = number of sides $(P_{3})$

So, number of sides $(P_{1})$ = number of sides $(P_{3})$

I.e., $(P_{1}, P_{3}) \in R$

So, R is transitive.

Hence, R is an equivalence relation.

...more

The given relation in set A of all polygons is defined as

R= ${(P_{1}, P_{2}) : P_{1}$ and $P_{2}$ have same number of sides $}$

Let $P_{1} \in A$ ,

As number of sides $(P_{1})$ = number of sides $(P_{1})$

$(P_{1}, P_{1}) \in R$

So, R is reflexive.

Let $P_{1}, P_{2} \in A$ and $(P_{1}, P_{2}) \in R$

Then, number of sides of $P_{1}$ = number of sides of $P_{2}$

Number of sides of $P_{2}$ = number of sides of $P_{1}$

i.e., $(P_{2}, P_{1}) \in R$

so, R is symmetric.

Let $P_{1}, P_{2}, P_{3} \in A$ and $(P_{1}, P_{2})$ and $(P_{2}, P_{3}) \in R$

Then, number of sides $(P_{1})$ = number of sides $(P_{2})$

Number of sides $(P_{2})$ = number of sides $(P_{3})$

So, number of sides $(P_{1})$ = number of sides $(P_{3})$

I.e., $(P_{1}, P_{3}) \in R$

So, R is transitive.

Hence, R is an equivalence relation.

The set of all triangles will be the elements in A related to the right-angle triangle T with sides 3,4 and 5 as they all have three number of sides.

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Let S = {1, 2, 3 ,....,20}

R₁ = {(a, b) : a divide b}

R₂ = {(a, b) : a is integral multiple of b} a, b Î s

n(R₁ – R₂) = ?

alok kumar singh

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

n (R₁) = 66

R₂ = {a is integral multiple of b}

So n (R₁ – R₂) = 66 – 20 = 46

as R₁ Ç R₂ = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

$f (x) = {\begin{matrix} e^{- x}, & x < 0 \\ l n x, & x > 0 \end{matrix}$

$g (x) = {\begin{matrix} e^{x}, & x < 0 \\ x, & x > 0 \end{matrix}$

The gof : A->R is

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$g o f (x) = {\begin{matrix} f (x), & f (x) < 0 \\ f (x), & f (x) > 0 \end{matrix}$

$= {\begin{array}{l} e^{l n x} = x & (0, 1) & e^{- x} \\ (- \infty, 0) & l n x & (1, \infty) \end{array}$

If A = {1, 2, 3, … 100}, R = {(x, y) | 2x = 3y, x, y ∈ A} is symmetric relation on A and the number of elements in R is n, the smallest integer value of n is

alok kumar singh

$?$ R is symmetric relation

⇒ (y, x) ∈ R V (x, y) ∈ R

(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y

Which holds only for (0, 0)

Which does not belongs to R.

∴ Value of n = 0

Let : f(0, R)->∞ be increasing function such that $\underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)} = 1$ then $\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1)$ is equal to

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f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

$\frac{f (x)}{f (x)} < \frac{f (5 x)}{f (x)} < \frac{f (7 x)}{f (x)}$

$\underset{x \to \infty}{l i m} \frac{f (x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)}$

-> $1 < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < 1$ $\underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} = 1$

$\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1) = 0$

Let A = ${1, 2, 3, . . . . ., 10} a n d f : A \to A$ be defined as

$f (k) = {\begin{matrix} k + 1 & i f k i s o d d \\ k & i f k i s e v e n \end{matrix}$

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alok kumar singh

Given f (k) = ${\begin{cases} k + 1, k i s o d d \\ k, k i s e v e n \end{cases}$

$? g : A \to A$ such that g (f (x) = f (x)

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Case II : If x is odd then g (x + 1) = x + 1 . (ii)

From (i) & (ii), g (x) = x, when x is even

So total no. of functions = 10⁵ × 1 = 10⁵

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13. Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂): P_{1 and} P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right-angle triangle T with sides 3, 4 and 5?

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