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

67. Consider the binary operations: R ×R → and o: R × R → R defined as a b = |a - b| and ao b = a, &mn For E; a, b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that &mn For E;a, b, c ∈ R, a(b o c) = (a b) o (a * c). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.

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Vishal Baghel | Contributor-Level 10

4 months ago

It is given that*: $R \times R \to$ and $o : R \times R \to R$ is defined as

$a * b = | a - b | a n d, a o b = a, & m n F o r E; a, b \in R .$

For $a, b \in R$ , we have:

$\begin{array}{l} a * b = | a - b | \\ b * a = | b - a | = | - (a - b) | = | a - b | \\ ∴ a * b = b * a \end{array}$

$∴$ The operation* is commutative.

It can be observed that,

$\begin{array}{l} (1.2) . 3 = (| 1 - 2 |) . 3 = 1.3 = | 1 - 3 | = 2 \\ 1 * (2 * 3) = 1 * (| 2 - 3 |) = 1 * 1 = | 1 - 1 | = 0 \\ ∴ (1 * 2) * 3 = 1 * (2 * 3) (w h e r e, 1, 2, 3 \in R) \end{array}$

$∴$ The operation* is not associative.

Now, consider the operation o:

It can be observed that $1 o 2 = 1, a n d, 2 o 1 = 2 .$

$∴ 1 o 2 \neq 2 o 1 (w h e r e, 1, 2 \in R)$

$∴$ The operation o is not commutative.

Let, $a, b, c \in R$ . Then we have:

$\begin{array}{l} (a ο b) ο c = a o c = a \\ a o (b ο c) = a o b = a \\ \Rightarrow (a ο b) ο c = a o (b ο c) \end{array}$

$∴$ The operation o is associative.

Now, $a, b, c \in R$ . Then we have:

$\begin{array}{l} a * (b ο c) = a * b = | a - b | \\ (a * b) o (a * c) = (| a - b |) o (| a - c |) = | a - b | \\ H e n c e, a * (b ο c) = (a * b) o (a * c) . \\ N o w, \\ 1 o (2 ο 3) = 1 o (| 2 - 3 |) = 1 o 1 = 1 \\ (1 o 2) * (1 o 3) = 1 * 1 = | 1 - 1 | = 0 \\ ∴ 1 o (2 ο 3) \neq (1 o 2) * (1 o 3) (w h e r e, 1, 2, 3 \in R) \end{array}$

$∴$ The operation o does not distribute over*.

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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) = ?

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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),
(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)

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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)

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

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

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

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