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Q.
If an invertible function $y=f(x)$ defined implicity by the equation $x^3-3 x^2+3 x-y^2+2 y-2$ $=0$ where $x, y \geq 1$ then
Relations and Functions - Part 2
Solution:
$ x^3-3 x^2+3 x-y^2+2 y-2=0$
$( x -1)^3-( y -1)^2=0 $
$( x -1)^3=( y -1)^2 $
$( y -1)=( x -1)^{3 / 2} $
$y =1+( x -1)^{3 / 2}= f ( x ), x \geq 1 $
$\therefore f ^{-1}( x )=1+( x -1)^{2 / 3}, x \geq 1$
Now, number of solution of $f(x)=f^{-1}(x)$
$\Rightarrow( x -1)^{3 / 2}=( x -1)^{2 / 3} $
$\therefore x =1 \text { or } 2 \Rightarrow 2 \text { solution } \Rightarrow \text { (A) }$
Domain of $f(x)=$ Domain of $f^{-1}(x)$ is $[1, \infty) \Rightarrow(C)$
$f ^{-1}( x )> f ( x )$
$(x-1)^{2 / 3}>(x-1)^{3 / 2} $
$(x-1)^{3 / 2}-(x-1)^{2 / 3}<0$
$(x-1)^{2 / 3}\left((x-1)^{5 / 6}-1\right)<0 \Rightarrow x<2$