Question

Let a function is given as $f(x) = \dfrac{{x + 1}}{{x - 1}}$ for all $x \ne 1.$
Let
${f^1}(x) = f(x),{f^2}(x) = f(f(x)){\text{ }}$ and generally
$
  {f^n}(x) = f({f^{n - 1}}(x)){\text{ for }}n{\text{ > 1}} \\
  {\text{Let }}P = {f^1}(2){f^2}(3){f^3}(4){f^4}(5) \\
$
Which of the following is a multiple of P?
$
  {\text{A}}{\text{. 125}} \\
  {\text{B}}{\text{. 375}} \\
  {\text{C}}{\text{. 250}} \\
  {\text{D}}{\text{. 147}} \\
 $

Answer 1

Hint: In this question first we examine how the function ${f^n}(x)$ varies with even or odd values by solving the relations ${f^1}(x) = f(x),{f^2}(x) = f(f(x))$. After that we calculate values of ${f^1}(2),{f^2}(3),{f^3}(4),{f^4}(5)$ and put them in expression for $P$.

Complete step-by-step solution -
Given:
$f(x) = \dfrac{{x + 1}}{{x - 1}}$ for all $x \ne 1.$ …………….. eq.1

  $ {f^1}(x) = f(x) $ …………….. eq.(2)
  $ {f^2}(x) = f(f(x)) $ …………… eq.(3)

${f^n}(x) = f({f^{n - 1}}(x)){\text{ for }}n{\text{ > 1}}$ …………………. eq.4
Considering eq.3
$
   \Rightarrow {f^2}(x) = f(f(x)) \\
   \Rightarrow {f^2}(x) = \dfrac{{f(x) + 1}}{{f(x) - 1}} \\
   \Rightarrow {f^2}(x) = \dfrac{{(\dfrac{{x + 1}}{{x - 1}}) + 1}}{{(\dfrac{{x + 1}}{{x - 1}}) - 1}} $ …………….. $ {\text{ \{ from eq}}{\text{.1\} }} \\
$

On solving above equation, we get
\[ \Rightarrow {f^2}(x) = x\] ………………………... eq.5
Since, 2 is an even number. So, eq.4 varies the same for all even values and for all odd numbers it varies the same as \[{f^1}(x)\]. Now, using eq.2 and eq.5 we can generalise the eq.4 as
$ {\text{if }}n = {\text{odd }} \Rightarrow {\text{ }}{f^n}(x) = f(x) $ ………….. eq.(6)
$ {\text{if }}n = {\text{even }} \Rightarrow {\text{ }}{f^n}(x) = x $ …………...eq.(7)

Now considering P
$ \Rightarrow P = {f^1}(2){f^2}(3){f^3}(4){f^4}(5)$ …………………. eq.8
For finding value of P we need to find
So,
 $ \Rightarrow {f^1}(2) = \dfrac{{2 + 1}}{{2 - 1}} $

  $ \text{From}\ \text{ eq. 6} $
  $ \Rightarrow {f^1}(2) = 3 $.............eq.(9)

  $ \text{From}\ \text{ eq. 7} $
  $ \Rightarrow {f^2}(3) = 3 $............. eq.(10)

  $ \text{From}\ \text{ eq. 6} $
  $ \Rightarrow {f^3}(4) = \dfrac{5}{3} $............ eq(11)

$ \text{From}\ \text{ eq. 7} $
$ \Rightarrow {f^4}(5) = 5 $............ eq.(12)

Now, on putting values of ${f^1}(2),{f^2}(3),{f^3}(4),{f^4}(5)$ from eq.9, eq.10, eq.11, eq. 12 respectively into eq.8 we get
$
   \Rightarrow P = 3 \times 3 \times \dfrac{5}{3} \times 5 \\
   \Rightarrow P = 75 \\
$
Therefore, $P$ is 75 and 375 is multiple of $P$.
Hence option B. is correct.

Note:- Whenever you get this type of question the key concept to solve this is to find pattern in values of function ${f^n}(x)$ as in this question for $ n = {\text{even}} \Rightarrow {f^n}(x) = x$ and for ${\text{n = odd}} \Rightarrow {{\text{f}}^n}{\text{(}}x) = f(x)$.And one more thing to remember is that Multiple of P means any number that P divides evenly or any number for which P is a factor. So anything in the P times table is going to be a multiple of P.