Question

If an invertible function $y=f(x)$ defined implicity by the equation $x^3-3x^2+3x-y^2+2y-2$ $=0$ where $x,y\ge 1$ then

• A. number of solutions of the equation $f(x)=f^{-1}(x)$ is 2 .
• B. number of solutions of the equation $f(x)=f^{-1}(x)$ is 1 .
• C. domain of $f^{-1}(x)=$ domain of $f(x)$.
• D. solution set of the inequality $f^{-1}(x)>f(x)$ is $(2,\infty )$.

Answer 1

Answer: (A), (C)

$x^3-3x^2+3x-y^2+2y-2=0$
$(x-1{)}^3-(y-1{)}^2=0$
$(x-1{)}^3=(y-1{)}^2$
$(y-1)=(x-1{)}^{3/2}$
$y=1+(x-1{)}^{3/2}=f(x),x\ge 1$
$\therefore f^{-1}(x)=1+(x-1{)}^{2/3},x\ge 1$
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 $[1,\infty )\Rightarrow (C)$
$f^{-1}(x)>f(x)$
$(x-1{)}^{2/3}>(x-1{)}^{3/2}$
$(x-1{)}^{3/2}-(x-1{)}^{2/3}<0$
$(x-1{)}^{2/3}\left((x-1{)}^{5/6}-1\right)<0\Rightarrow x<2$