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

11. Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

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Vishal Baghel | Contributor-Level 10

5 months ago

The given relation in set A of points in a plane is

R= ${(P, Q) :$ distance of point P from origin=distance of point Q from $o r i g i n}$

If O is the point of origin

R= ${(P, Q) : P O = Q O}$

Then, for $P \in A$ we have PO=PO

So, $(P, P) \in R$

i.e., P is reflexive

for, $P, Q \in A$ and $(P, Q) \in R$ we have

PO=QO

QO=PO i.e., $(Q, P) \in R$

i.e., R is symmetric

for $P, Q, S \in A$ and $(P, Q) & (Q, S) \in R$

PO=QO and QO=SO

PO=SO

i.e., $(P, S) \in R$

so, R is transitive

Hence, R is an equivalence relation

For a point $P \neq (o, o)$ the set of all points related to P i.e., distance from origin to the points are equal is a circle with center at origin (o, o) by the definition of circle

The given relation in set A of points in a plane isR= <math><mrow><mrow><mo> {</mo><mrow><mrow><mo> (</mo><mrow><mi>P</mi><mo>, </mo><mi>Q</mi></mrow><mo>)</mo></mrow><mo>:</mo></mrow></mrow></mrow></math> distance of point P from origin=distance of point Q from <math><mrow><mrow><mrow><mi>o</mi><mi>r</mi><mi>i</mi><mi>g</mi><mi>i</mi><mi>n</mi></mrow><mo>}</mo></mrow></mrow></math>If O is the point of originR= <math><mrow><mrow><mo> {</mo><mrow><mrow><mo> (</mo><mrow><mi>P</mi><mo>, </mo><mi>Q</mi></mrow><mo>)</mo></mrow><mo>:</mo><mi>P</mi><mi>O</mi><mo>=</mo><mi>Q</mi><mi>O</mi></mrow><mo>}</mo></mrow></mrow></math>Then, for <math><mrow><mi>P</mi><mo>∈</mo><mi>A</mi></mrow></math> we have PO=POSo,  <math><mrow><mrow><mo> (</mo><mrow><mi>P</mi><mo>, </mo><mi>P</mi></mrow><mo>)</mo></mrow><mo>∈</mo><mi>R</mi></mrow></math>i.e., P is reflexivefor,  <math><mrow><mi>P</mi><mo>, </mo><mi>Q</mi><mo>∈</mo><mi>A</mi></mrow></math> and <math><mrow><mrow><mo> (</mo><mrow><mi>P</mi><mo>, </mo><mi>Q</mi></mrow><mo>)</mo></mrow><mo>∈</mo><mi>R</mi></mrow></math> we havePO=QOQO=PO i.e.,  <math><mrow><mrow><mo> (</mo><mrow><mi>Q</mi><mo>, </mo><mi>P</mi></mrow><mo>)</mo></mrow><mo>∈</mo><mi>R</mi></mrow></math>i.e., R is symmetricfor <math><mrow><mi>P</mi><mo>, </mo><mi>Q</mi><mo>, </mo><mi>S</mi><mo>∈</mo><mi>A</mi></mrow></math> and <math><mrow><mrow><mo> (</mo><mrow><mi>P</mi><mo>, </mo><mi>Q</mi></mrow><mo>)</mo></mrow><mo>&</mo><mrow><mo> (</mo><mrow><mi>Q</mi><mo>, </mo><mi>S</mi></mrow><mo>)</mo></mrow><mo>∈</mo><mi>R</mi></mrow></math>PO=QO and QO=SOPO=SOi.e.,  <math><mrow><mrow><mo> (</mo><mrow><mi>P</mi><mo>, </mo><mi>S</mi></mrow><mo>)</mo></mrow><mo>∈</mo><mi>R</mi></mrow></math>so, R is transitiveHence, R is an equivalence relationFor a point <math><mrow><mi>P</mi><mo>≠</mo><mrow><mo> (</mo><mrow><mi>o</mi><mo>, </mo><mi>o</mi></mrow><mo>)</mo></mrow></mrow></math> the set of all points related to P i.e., distance from origin to the points are equal is a circle with center at origin (o, o) by the definition of circle

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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

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$?$ 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

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

Then the number of possible functions g : A ® A such that gof = f is:

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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)

Case I : If x is even then g (x) = x . (i)

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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