Question

please answer this question ​

Answer 1

Given:

A function y= f(x) such that:

$f\bigg(x+\dfrac{1}{x}\bigg)=x^{2}+\dfrac{1}{x^{2}}$

To Find:

Definition of given function

Solution:

Here, definition of given function has to be determined by hit and trial method

Case-1 Let 2-x² be the definition of given function

Then,

f(x)= 2-x² ----------------(1)

And it should satisfy the condition

$f\bigg(x+\dfrac{1}{x}\bigg)=x^{2}+\dfrac{1}{x^{2}}$

Now,

On substituting x with $x+\dfrac{1}{x}$ in (1), we get

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

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

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

But it is not satisfying the condition

So, 2-x² is not the correct defintion of given function

Case-2 Let x²-2 be the definition of given function

Then,

f(x)= x² -2 ----------------(2)

And it should satisfy the condition

$f\bigg(x+\dfrac{1}{x}\bigg)=x^{2}+\dfrac{1}{x^{2}}$

Now,

On substituting x with $x+\dfrac{1}{x}$ in (2), we get

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

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

$f\bigg(x+\dfrac{1}{x}\bigg)=x^{2}+\dfrac{1}{x^{2}}$

Hence, it is satisfying the condition.

So, the given function y=f(x) is defined by x² -2.