Question

If f: R—>R, f(x)=2x—3, then......,Select Proper option from the given options.
(a) f⁻¹(x)=1/2x-3
(b) f⁻¹(x)= x+3/2
(c) f⁻¹ does not exist
(d) f⁻¹(x)=3x-2

Answer 1

Answer:

option (b) is correct

Step-by-step explanation:

Given:

f:R-->R, f(x)=2x-3

clearly f is a bijective function.

So, we can find inverse of f

$2x-3=y(say)$

$\implies\:2x=y+3$

$\implies\:x=\frac{y+3}{2}$

Hence, inverse of f(x) is

$\bf\:f^{-1}(x)=\frac{x+3}{2}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

DEFINITION

INJECTIVE FUNCTION :

$\sf{A \: function \: f : A \to B \: is \: said \: to \: be \: injective \: if}$

$\sf{For \: \: x_1 \ne x_2 \: \: we \: have \: \: f(x_1) \ne f(x_2)}$

SURJECTIVE FUNCTION

$\sf{A \: function \: f : A \to B \: is \: said \: to \: be \: surjective}$

if for every element y in the co-domain B there exists a pre-image x in domain set A such that y = f(x)

BIJECTIVE FUNCTION

$\sf{A \: function \: f : A \to B \: is \: said \: to \: be \: bijective}$

if f is both injective and surjective

CALCULATION

$\sf{ f : \mathbb{R} \to \mathbb{R} \: } \: defined \: by \: f(x) = 2x - 3$

CHECKING FOR INJECTIVE

$\sf{Let \: \: x_1, x_2 \in \mathbb{ R} \: \: with \: \: x_1 \ne \: x_2}$

Now

$\sf{ f(x_1) \ne f(x_2)} \: \: \: \: gives$

$\sf {2x_1 - 3\ne 2x_2 - 3}$

$\implies \: \sf {2x_1 \ne 2x_2 }$

$\implies \: \sf {x_1 \: \ne \: x_2 }$

Hence f is injective

CHECKING FOR SURJECTIVE

$\sf{y \in \: \mathbb{R} \: \: ( Co - domain \: \: set) }$

If possible let there exists an element x in domain set such that

$\sf{ \: y = f(x)\: }$

$\implies \sf{ \: y = 2x - 3\: }$

$\displaystyle \: \implies \sf{ \:x = \frac{y + 3}{2} \: }$

$\displaystyle \: \sf{As \: y \in \mathbb{R} \: \: we \: have \: \: \: \frac{y + 3}{2} \: \in \mathbb{R} \: }$

Hence f is surjective

Therefore f is bijective

$\sf{Hence \: \: {f}^{ - 1} \: \: exists}$

$\sf{let \: \: {f}^{ - 1} (x) = y \: \: }$

$\implies \: \sf{x = f(y)\: }$

$\implies \: \sf{x = 2y - 3\: }$

$\displaystyle \: \implies \sf{ \:y = \frac{x + 3}{2} \: }$

$\displaystyle \: \implies \sf{ \: {f}^{ - 1} (x) = \frac{x + 3}{2} \: }$

RESULT

$\sf{Hence \: \: {f}^{ - 1} \: \: exists} \: \: and \: \: \displaystyle \: \sf{ \: {f}^{ - 1} (x) = \frac{x + 3}{2} \: }$

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