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Question
Mark the correct alternative in the following question:
Let f : R \[-\] \[\left\{ \frac{3}{5} \right\}\] \[\to\] R be defined by f(x) = \[\frac{3x + 2}{5x - 3}\] Then,
Options
f-1 (x) = f (x)
`f^-1 (x) = - f(x)`
fo f(x) = - x
`f^-1(x) = 1/19f(x)`
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Solution
We have,
f : R \[-\] \[\left\{ \frac{3}{5} \right\}\] \[\to\] R be defined by f(x) = \[\frac{3x + 2}{5x - 3}\]
\[fof\left( x \right) = f\left( f\left( x \right) \right)\]
\[ = f\left( \frac{3x + 2}{5x - 3} \right)\]
\[ = \frac{3\left( \frac{3x + 2}{5x - 3} \right) + 2}{5\left( \frac{3x + 2}{5x - 3} \right) - 3}\]
\[ = \frac{\left( \frac{9x + 6}{5x - 3} \right) + 2}{\left( \frac{15x + 10}{5x - 3} \right) - 3}\]
\[ = \frac{\left( \frac{9x + 6 + 10x - 6}{5x - 3} \right)}{\left( \frac{15x + 10 - 15x + 9}{5x - 3} \right)}\]
\[ = \frac{19x}{19}\]
\[ = x\]
\[\text{Let } y = \frac{3x + 2}{5x - 3}\]
\[ \Rightarrow 5xy - 3y = 3x + 2\]
\[ \Rightarrow 5xy - 3x = 3y + 2\]
\[ \Rightarrow x\left( 5y - 3 \right) = 3y + 2\]
\[ \Rightarrow x = \frac{3y + 2}{5y - 3}\]
\[ \Rightarrow f^{- 1} \left( y \right) = \frac{3y + 2}{5y - 3}\]
\[So, f^{- 1} \left( x \right) = \frac{3x + 2}{5x - 3} = f\left( x \right)\]
Hence, the correct alternative is option (a).
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