Question

Let $f:[a,\infty )\to [a,\infty )$ be defined by $f\left(x\right)=x^2-2ax+a(a+1)$. If one of the solutions of the equation $f\left(x\right)=f^{-1}\left(x\right)$ is $5049$, then the other solution can be

• A. $5051$
• B. $5048$
• C. $5052$
• D. $5050$

Answer 1

Answer: (B), (D)

$5050$

$f\left(x\right)=x^2-2ax+a(a+1)$
$\Rightarrow (x-a{)}^2+a,x\in [a,\infty )$
Let $y=(x-a{)}^2+a$. Clearly, $y\ge a$.
Thus, $(x-a{)}^2=y-a$
or, $x=a+\sqrt{y-a}$
$\therefore f^{-1}\left(x\right)=a+\sqrt{x-a}$
Now, $f\left(x\right)=f^{-1}\left(x\right)$
$\Rightarrow (x-a{)}^2+a=a+\sqrt{x-a}$
$\Rightarrow (x-a{)}^2=\sqrt{x-a}$
$\Rightarrow (x-a{)}^4=x-a$
$\Rightarrow x=a$ or $(x-a{)}^3=1$
$\therefore x=a$ or $a+1$
If $a=5049$, then $a+1=5050$.
If $a+1=5049$, then $a=5048$.