An algebra problem by Aman Banka

Algebra Level 3

If $f \left(x+\dfrac 1x\right)=x^2+ \dfrac 1{x^2}$ for all $x \in \mathbb{C}$ , then find $f(x)$ in $\mathbb{C} \backslash \{ 0 \}$ .

x^2-1

x^2-2

x^2

None of the others

2 solutions

Chew-Seong Cheong
Mar 26, 2017

$\begin{aligned} f \left({\color{#3D99F6}x+\frac 1x}\right) & = x^2+\frac 1{x^2} \\ & = \left({\color{#3D99F6}x+\frac 1x}\right)^2 - 2 \\ \implies f \left({\color{#3D99F6}x}\right) & = \boxed{{\color{#3D99F6}x}^2 - 2} \end{aligned}$

Technically, we've only shown that $f(x) = x^2 - 2$ on $( - \infty, -2] \cup [2, \infty )$ .

Even if we allowed for $x$ to be a complex number, there is no way to determine $f(0)$ .

Calvin Lin Staff - 4 years, 2 months ago

Yes, $x$ cannot be 0.

Chew-Seong Cheong - 4 years, 2 months ago

Aman Banka
Mar 26, 2017

let x+1/x=z.Then, f(z)=f(x+1/x)=(x^2+1/x^2)=(x+1/x)^2 -2=(z^-2) So,f(x)=(x^2-2).

from where did you get this solution?

Ajinkya Rahane - 4 years, 2 months ago

An algebra problem by Aman Banka

2 solutions

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