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

Relation and Function Exercise 1.2 Solutions

17. Show that the function f: R* → R* defined by f(x) = 1/x is one-one and onto, where: R* is the set of all non-zero real numbers. Is the result true, if the domain: R* is replaced by N with the co-domain being same as: R*?

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

The $f x$ n is $f (x) = \frac{1}{x}$ , which is a $f : R_{*}$ $\to$ $R_{*}$ and $R_{*}$ is set of all non-zero real numbers

For, $x_{1}, x_{2} \in R_{*}, f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}}$

$\Rightarrow x_{1} = x_{2}$ So, $f$ is one-one

For, $y \in R_{*},$ $x = \frac{1}{f (x)} = \frac{1}{y}$ such that

So, $f (x) = y$

So, every element in the co-domain has a pre-image in $f$

So, $f$ is onto

If $f : N \to R_{*}$ such that $f (x) = \frac{1}{x}$

For, $x_{1}, x_{2} \in N,$ $f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{1}{x_{1}} = \frac{1}{x_{2}}$

$\Rightarrow x_{1} = x_{2}$ So, $f$ is one-one

For, $y \in R_{*}$ and $f (x) = y$ we have $x = \frac{1}{y} \notin N$

Eg., $3 \in R_{*}$ so $x = \frac{1}{3} \notin N$

So, $f$ is not onto

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

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

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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

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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

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

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

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

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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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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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Relation and Function Exercise 1.2 Solutions

17. Show that the function f: R* → R* defined by f(x) = 1/x is one-one and onto, where: R* is the set of all non-zero real numbers. Is the result true, if the domain: R* is replaced by N with the co-domain being same as: R*?