Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If $f\left( x \right) = {\left( {p - {x^n}} \right)^{\dfrac{1}{n}}},p > 0$ and n is a positive integer, then $f\left( {f\left( x \right)} \right)$=?A. xB. ${x^n}$C. ${p^{\dfrac{1}{n}}}$D. $p - {x^n}$

Last updated date: 17th Jun 2024
Total views: 403.2k
Views today: 5.03k
Verified
403.2k+ views
Hint: Start by writing the given function f(x) and find out $f\left( {f\left( x \right)} \right)$by substituting the value of f(x) in place of x , Simplify the new expression formed by solving the exponents and get the most simplified form , the value obtained is the desired answer.

Given,
$f\left( x \right) = {\left( {p - {x^n}} \right)^{\dfrac{1}{n}}},p > 0$
Let us find out the value of $f\left( {f\left( x \right)} \right)$
Substituting the value of f(x), by replacing the equation f(x) in place of x variable , we get
$f\left( {f\left( x \right)} \right) = {\left( {p - {{\left( {{{\left( {p - {x^n}} \right)}^{\dfrac{1}{n}}}} \right)}^n}} \right)^{\dfrac{1}{n}}}$
Here , The inside powers $\dfrac{1}{n}$ and n gets cancelled and on simplification, we have
$f\left( {f\left( x \right)} \right) = {\left( {p - \left( {p - {x^n}} \right)} \right)^{\dfrac{1}{n}}} \\ \Rightarrow f\left( {f\left( x \right)} \right) = {\left( {p - p + {x^n}} \right)^{\dfrac{1}{n}}} \\$
So now p will be cancelled out with -p and we are left with
$f\left( {f\left( x \right)} \right) = {\left( {{x^n}} \right)^{\dfrac{1}{n}}}$
Now, Again the powers of n and $\dfrac{1}{n}$ will be cancelled out and hence we have
$f\left( {f\left( x \right)} \right) = x$

So, the correct answer is “Option A”.

Note: Similar questions can be asked with multiple iteration of f(x) ,for .e.g. $f\left[ {f\left( {f\left( x \right)} \right)} \right]$, follow the same procedure as above. Attention must be given while substituting and simplifying as any missed sign or wrong interpretation may lead to wrong answers.