5. Check whether the relation R in R defined as R = {(a, b): a ≤ b3} is reflexive, symmetric or transitive.

We have,R= {(a,b):a≤b3} is a relation in R.For, (a,b)∈R and a=12 we can writea≤a3 => 12≤(12)3 => 12≤18 which is not true.So, R is not reflexive.For (a,b)=(1,2)∈R we have,a≤b3 => 1≤23 => 1≤8 is true.So, (1,2)∈RBut 2≤13 => 2≤1 is not trueSo, (2,1)∉R and (b,a)∉RHence, R is not symmetric.For, (a,b)=(10,4) and (b,c)=(4,2)∈R10≤43 => 10≤64 is true=> (10,4)∈R4≤23 => 4≤8 is true=> (4,2)∈RBut 10≤23 => 10≤8 is not true=> (10,2)∉RHence, for (a,b),(b,c)∈R,(a,c)∉RSo, R is not transitive.

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

5. Check whether the relation R in R defined as R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.

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Vishal Baghel | Contributor-Level 10

4 months ago

We have,

R= ${(a, b) : a \leq b^{3}}$ is a relation in R.

For, $(a, b) \in R$ and $a = \frac{1}{2}$ we can write

$a \leq a^{3}$ => $\frac{1}{2} \leq {(\frac{1}{2})}^{3}$ => $\frac{1}{2} \leq \frac{1}{8}$ which is not true.

So, R is not reflexive.

For $(a, b) = (1, 2) \in R$ we have,

$a \leq b^{3}$ => $1 \leq 2^{3}$ => $1 \leq 8$ is true.

So, $(1, 2) \in R$

But $2 \leq 1^{3}$ => $2 \leq 1$ is not true

So, $(2, 1) \notin R$ and $(b, a) \notin R$

Hence, R is not symmetric.

For, $(a, b) = (10, 4)$ and $(b, c) = (4, 2) \in R$

$10 \leq 4^{3}$ => $10 \leq 64$ is true=> $(10, 4) \in R$

$4 \leq 2^{3}$ => $4 \leq 8$ is true=> $(4, 2) \in R$

But $10 \leq 2^{3}$ => $10 \leq 8$ is not true=> $(10, 2) \notin R$

Hence, for $(a, b), (b, c) \in R, (a, c) \notin R$

So, R is not transitive.

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

R₁ = {(a, b) : a divide b}

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R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
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$f (x) = {\begin{matrix} e^{- x}, & x < 0 \\ l n x, & x > 0 \end{matrix}$

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

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

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f is increasing function

x < 5x < 7x

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

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

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

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

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

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