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Question: Which of the following is inverse of itself? A.\[f(x) = \dfrac{{1 - x}}{{1 + x}}\] B.\[g(x) = 5\...

Which of the following is inverse of itself?
A.f(x)=1−x1+xf(x) = \dfrac{{1 - x}}{{1 + x}}
B.g(x)=5log⁡xg(x) = 5\log x
C.h(x)=2x(x−1)h(x) = 2x(x - 1)
D.ϕ(x)=sin⁡x\phi (x) = \sin x

Explanation

Solution

First we have to define what the terms we need to solve the problem are.We want to find whether the given options are inverse by themselves or not. By applying the condition we can check. The option which satisfies the condition is inverse by itself. The inverse of the function is defined as the function that reverses another function. Not all the functions have an inverse function.
Formula to be used:
The function ff is the inverse by itself when f(x)=f−1(x)f(x) = {f^{ - 1}}(x)
A function gg is inverse of the function ff if whenever y=f(x)y = f(x)andx=g(y)x = g(y)
Complete step by step answer:
To find inverse by itself function from the given options
The given options A) f(x)=1−x1+xf(x) = \dfrac{{1 - x}}{{1 + x}}
We take f(x)=y=1−x1+xf(x) = y = \dfrac{{1 - x}}{{1 + x}}
Then it changes to
1y=1−x1+x\dfrac{1}{y} = \dfrac{{1 - x}}{{1 + x}}
Substituting y=1−x1+xy = \dfrac{{1 - x}}{{1 + x}} we get,
1−y1+y=1−1−x1+x1+1−x1+x\dfrac{{1 - y}}{{1 + y}} = \dfrac{{1 - \dfrac{{1 - x}}{{1 + x}}}}{{1 + \dfrac{{1 - x}}{{1 + x}}}}
Adding the terms in the numerator and adding the terms in denominator, we get
1−y1+y=1+x−(1−x)1+x1+x+1−x1+x\dfrac{{1 - y}}{{1 + y}} = \dfrac{{\dfrac{{1 + x - (1 - x)}}{{1 + x}}}}{{\dfrac{{1 + x + 1 - x}}{{1 + x}}}}
By simplifying we get,
1−y1+y=2x1+x21+x\dfrac{{1 - y}}{{1 + y}} = \dfrac{{\dfrac{{2x}}{{1 + x}}}}{{\dfrac{2}{{1 + x}}}}
1−y1+y=2x1+x×1+x2\dfrac{{1 - y}}{{1 + y}} = \dfrac{{2x}}{{1 + x}} \times \dfrac{{1 + x}}{2}
By canceling 1+x1 + xwe get,
1−y1+y=2x2\dfrac{{1 - y}}{{1 + y}} = \dfrac{{2x}}{2}
Then,
1−y1+y=x\dfrac{{1 - y}}{{1 + y}} = x
f(y)=x=1−y1+yf(y) = x = \dfrac{{1 - y}}{{1 + y}}
Finally,
f−1(x)=1−x1+x{f^{ - 1}}(x) = \dfrac{{1 - x}}{{1 + x}}Whereas f−1(y)=1−y1+y{f^{ - 1}}(y) = \dfrac{{1 - y}}{{1 + y}} which is inverse by itself.
So option A is correct
Then,
Given option B) g(x)=5log⁡xg(x) = 5\log x
y=5log⁡xy = 5\log x
Bringing 55 from R.H.S to L.H.S
y5=log⁡x\dfrac{y}{5} = \log x
Applying exponents on both sides,
ey5=x{e^{\dfrac{y}{5}}} = x
Finally,
g−1(y)=ey5{g^{ - 1}}(y) = {e^{\dfrac{y}{5}}} Whereas g−1(x)=ex5{g^{ - 1}}(x) = {e^{\dfrac{x}{5}}},which is not inverse by itself.
Then,
The given option C) h(x)=2x(x−1)h(x) = 2x(x - 1)
y=2x(x−1)y = 2x(x - 1)
Multiplying the terms, we get
2x2−2x−y=02{x^2} - 2x - y = 0
Finding roots−b±b2−4ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
a=2,b=−2,c=−1a = 2,b = - 2,c = - 1
Substituting a,b,c in the formula
x=2±4+8y4x = \dfrac{{2 \pm \sqrt {4 + 8y} }}{4}
By substituting, we get
x=1±1+2y2x = \dfrac{{1 \pm \sqrt {1 + 2y} }}{2}
Finally,
h−1(y)=1±1+2y2{h^{ - 1}}(y) = \dfrac{{1 \pm \sqrt {1 + 2y} }}{2} whereas h−1(x)=1±1+2x2{h^{ - 1}}(x) = \dfrac{{1 \pm \sqrt {1 + 2x} }}{2} ,which is not inverse by itself
Then,
Given option D) ϕ(x)=sin⁡x\phi (x) = \sin x
y=sin⁡xy = \sin x
We can also write it as,
x=sin⁡−1yx = {\sin ^{ - 1}}y
Finally,
ϕ−1(y)=sin⁡−1y{\phi ^{ - 1}}(y) = {\sin ^{ - 1}}y Whereas
ϕ−1(x)=sin⁡−1x{\phi ^{ - 1}}(x) = {\sin ^{ - 1}}x ,which is not inverse by itself.

Hence, the correct option is A)f(x)=1−x1+xf(x) = \dfrac{{1 - x}}{{1 + x}}.

Note:
The functionff is inverse by itself when f(x)=f−1(x)f(x) = {f^{ - 1}}(x)
We should check the function by substituting in the formula, if it satisfies the condition it is inverse by itself ,if not it is not inverse by itself .
Inverse by itself is also called as self inverse.
The word inverse refers to the opposite of another operation.