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Mathematics Question on Relations and Functions

Let f:R -−−43\\{ - \frac {-4} {3} \\}→R be a function defined as f(x)=4x3x+4.f (x) = \frac {4x} {3x + 4}. The inverse of f is map g : Range f →R - −43\\{ \frac {- 4} {3}\\} given by

A

g(y)=3y3−4yg(y)=\frac {3y} {3-4y}

B

g(y)=4y4−3yg(y) = \frac {4y} {4-3y}

C

g(y)=4y3−4yg(y)= \frac {4y} {3-4y}

D

g(y)=3y4−3yg(y)= \frac {3y} {4-3y}

Answer

g(y)=4y4−3yg(y) = \frac {4y} {4-3y}

Explanation

Solution

It is given that f:R−−−43f : R - \\{ - \frac {-4} {3} \\} → RR be a function defined as f(x)=4x3x+4f (x) = \frac {4x} {3x + 4}
Let yy be an arbitrary element of Range ff.
Then, there exists x∈R−−43x \in R - \\{ \frac {-4} {3} \\} such that y=f(x).y = f (x).
=> y=4x3x+4y = \frac {4x} {3x+4}
=> 3xy+4y=4x3xy + 4y = 4x
=> x=4y4−3yx = \frac {4y} {4-3y}.
Let us define g:g: Range ff---> R−−43 as g(y)=4y4−3y.R - \\{ \frac {-4} {3} \\} \,as\ g (y) = \frac {4y} {4-3y} .
Now, (gof)(x)=g(f(x))=g(4x3x+4)(gof) (x) = g (f (x)) = g (\frac {4x} {3x+4})
= 4(4x3x+4)4−34x3x+4\frac{4(\frac {4x}{3x+ 4})}{4-3\frac{4x}{3x+4}}=16x12x+16−12x=16x16=x\frac {16x} {12x + 16 - 12x} = \frac {16x} {16} = x
And fog(y)=f(4y4−3y)=4(4y4−3y)3(4y4−3y)+4=16y12y=16−12y=16y16=yfog (y) = f \bigg(\frac {4y} {4-3y}\bigg) = \frac {4 \bigg( \frac {4y} {4-3y }\bigg )} { 3 \bigg (\frac {4y} {4-3y} \bigg ) +4 } = \frac {16y} {12y =16 -12y} = \frac {16y} {16} = y
∴ gof =IR-gof=IR−(43)gof = I_{R-\bigg (\frac {4} {3} \bigg )} and fog=IRangeffog = I_{Range f}
Thus, g is the inverse of f i.e..f -1 = g

Hence, the inverse of f is the map g : Range f --->R−−43R - \\{\frac {-4} {3}\\} which is given by g(y)=4y4−3yg (y)= \frac {4y} {4-3y}
The correct answer is B.