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Question: If \[f(x) = \dfrac{{3x + 2}}{{5x - 3}}\], then which of the following option is true? A) \({f^{ - ...

If f(x)=3x+25x−3f(x) = \dfrac{{3x + 2}}{{5x - 3}}, then which of the following option is true?
A) f−1(x)=f(x){f^{ - 1}}(x) = f(x)
B) f−1(x)=−f(x){f^{ - 1}}(x) = - f(x)
C) f−1(x)=119f(x){f^{ - 1}}(x) = \dfrac{1}{{19}}f(x)
D) (f∘f)(x)=−x(f \circ f)(x) = - x

Explanation

Solution

From the options we can understand that we are asked to find the relation between the function f(x)f(x) and its inverse f−1(x){f^{ - 1}}(x). To extract xx from the given expression, introduce another variable say yy dependent of xx, such that, f(x)=yf(x) = y. This gives x=f=1(y)x = {f^{ = 1}}(y) and so we can calculate f−1(x){f^{ - 1}}(x).

Formula Used:
For two variables x,yx,y and a function ff,
y=f(x)⇒f−1(y)=f−1f(x)=xy = f(x) \Rightarrow {f^{ - 1}}(y) = {f^{ - 1}}f(x) = x
Also, for any two constants a,ba,b,
f(ax+b)=af(x)+bf(ax + b) = af(x) + b

Complete step by step answer:
Given, f(x)=3x+25x−3f(x) = \dfrac{{3x + 2}}{{5x - 3}}
We need to find the relation between f(x)f(x) and its inverse f−1(x){f^{ - 1}}(x).
Let f(x)=yf(x) = y
⇒3x+25x−3=y\Rightarrow \dfrac{{3x + 2}}{{5x - 3}} = y
Using this we can solve for xx.
Cross multiplying the above equation we get,
⇒3x+2=y(5x−3)\Rightarrow 3x + 2 = y(5x - 3)
Simplifying we get,
⇒3x+2=5xy−3y\Rightarrow 3x + 2 = 5xy - 3y
To find xx, collect the terms with xx to one side of the equation.
⇒3x−5xy=−3y−2\Rightarrow 3x - 5xy = - 3y - 2
Taking xx common on the left side we get,
⇒x(3−5y)=−3y−2\Rightarrow x(3 - 5y) = - 3y - 2
Simplifying we get,
⇒x=−3y−23−5y\Rightarrow x = \dfrac{{ - 3y - 2}}{{3 - 5y}}
Multiplying the numerator and denominator of the RHS by −1 - 1 we have,
⇒x=3y+25y−3\Rightarrow x = \dfrac{{3y + 2}}{{5y - 3}}
To find f−1(x){f^{ - 1}}(x), using above equation we can take f−1{f^{ - 1}} on both sides,
⇒f−1(x)=f−1(3y+25y−3)\Rightarrow {f^{ - 1}}(x) = {f^{ - 1}}(\dfrac{{3y + 2}}{{5y - 3}})
Also for any two constants a,ba,b, we have the result,
f(ax+b)=af(x)+bf(ax + b) = af(x) + b
Using this we get,
⇒f−1(x)=3f−1(y)+25f−1(y)−3\Rightarrow {f^{ - 1}}(x) = \dfrac{{3{f^{ - 1}}(y) + 2}}{{5{f^{ - 1}}(y) - 3}}
We can observe that in the RHS, y=f(x)⇒f−1(y)=f−1f(x)=xy = f(x) \Rightarrow {f^{ - 1}}(y) = {f^{ - 1}}f(x) = x
⇒f−1(x)=3x+25x−3\Rightarrow {f^{ - 1}}(x) = \dfrac{{3x + 2}}{{5x - 3}}
But we have f(x)=3x+25x−3{f^{}}(x) = \dfrac{{3x + 2}}{{5x - 3}}
So f−1(x)=f(x){f^{ - 1}}(x) = f(x)

∴\therefore Option A is correct.

Note:
In this question, the inverse of the given function is the function itself. But in general it is not true. A function need not have inverse as well. For the existence of inverse, the function must be bijective that is, injective (one to one) as well as surjective (onto). Also the inverse may be negative of the function or a constant multiple like the other options.