Question

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

• A. \(g(y)=\frac {3y} {3-4y}\)
• B. \(g(y) = \frac {4y} {4-3y}\)
• C. \(g(y)= \frac {4y} {3-4y}\)
• D. \(g(y)= \frac {3y} {4-3y}\)

Answer 1

The Correct Option is B

It is given that \(f : R - \{ - \frac {-4} {3} \}\) → \(R\) be a function defined as \(f (x) = \frac {4x} {3x + 4}\)
Let \(y\) be an arbitrary element of Range \(f\).
Then, there exists \(x \in R - \{ \frac {-4} {3} \}\) such that \(y = f (x).\)
=> \(y = \frac {4x} {3x+4}\)
=> \(3xy + 4y = 4x\)
=> \(x = \frac {4y} {4-3y}\).
Let us define \(g:\) Range \(f\)---> \(R - \{ \frac {-4} {3} \}  \,as\  g (y) = \frac {4y} {4-3y} .\)
Now, \((gof) (x) = g (f (x)) = g (\frac {4x} {3x+4})\)
= \(\frac{4(\frac {4x}{3x+ 4})}{4-3\frac{4x}{3x+4}}\)=\(\frac {16x} {12x + 16 - 12x} = \frac {16x} {16} = x\)
And \(fog (y) = f \bigg(\frac {4y} {4-3y}\bigg) = \frac {4 \bigg( \frac {4y} {4-3y }\bigg )} { 3 \bigg (\frac {4y} {4-3y} \bigg ) +4 } = \frac {16y} {12y =16 -12y} = \frac {16y} {16} = y \)
∴ gof=I_R-\(gof = I_{R-\bigg (\frac {4} {3} \bigg )}\) and \(fog = I_{Range f}\)
Thus, g is the inverse of f i.e.. f ^-1 = g

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