Question

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

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

Answer 1

Answer: (B), (D)

$5050$

$f\left(x\right)=x^2-2ax+a(a+1)$
$y=f\left(x\right)=(x-a{)}^2+a$
As $x\in [a,\infty )$, then
$y\ge a(x-a{)}^2=y-a\Rightarrow x=a+\sqrt{y-a}\therefore f^{-1}(x)=a+\sqrt{x-a}$

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 $(x-a{)}^3=1$
$\Rightarrow x=a$ or $a+1$
If $a=5049,$ then $a+1=5050$
If $a+1=5049,$ then $a=5048$
$\therefore$ Other possible solutions are $5050$ or $5048.$