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let $f( x + \frac{ 1}{ x } ) = x ^ 2 + \frac{ 1}{ x ^ 2} $ then $f(x)$ equals

options are

$( A ) x ^ 2 - 2$

$( B ) x ^ 2 - 1$

$( C ) x ^ 2$

I don't know how to solve this type of problems

Mohit
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  • What have you tried? A very simple way to attempt this is just plug in all the answer choices for $f(x)$ in order to find the right function. If you do not know how to solve the problem, you have to check again your textbook (if available) or ask your teacher. A guide to ask a good question can be found here and here. – user061703 May 25 '18 at 11:28
  • Taking $x=1$ we see that $f(2)=2$ so... – lulu May 25 '18 at 11:32
  • @TrầnThúcMinhTrí can these type of questions ask in subjective papers I mean without options? – Mohit May 25 '18 at 11:49
  • I actually don't know much about functions, but by looking at your multiple choice questions I can still get some ideas. – user061703 May 25 '18 at 11:51

4 Answers4

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Hint. Note that $(a+b)^2 =a^2 +2ab +b^2$. Now let $a=x$ and $b=1/x$. Then relate $f(x+\frac1x)$ with $(x+\frac1x)^2$ somehow.

Rócherz
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The option is A. Just substitute $f(x+\frac 1x)$ in option A and verify the outcome

$$f(x+\frac 1x)=(x+\frac 1x)^2-2=x^2+\frac {1}{x^2}$$

So this outcome is only possible if $f(x)=x^2-2$

tien lee
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$$f(x+\frac1x)=x^2+\frac{1}{x^2}$$

Let's run through the options.

First, let's try $A$. $f(x)=x^2-2$

Thus $$f(x+\frac 1x)=(x+\frac 1x)^2-2$$ Evaluating the $RHS$ we get: $$(x+\frac1x)^2-2=x^2+\frac{1}{x^2}+2-2=x^2+\frac{1}{x^2}$$ as required. No need, therefore, to check $B$ and $C$.

Rhys Hughes
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Lets to this the mathematical way. Let $a=x+\frac 1x$. Simply $f(a) = a^2-2$. Now by obvious bijection and no known constraints, one to one mapping will obviously show $f(x) = x^2-2$.

So choose options A.

This seems a more complete answer, as OP requires if no options were provided.

Tony Hellmuth
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