Solveeit Logo
Login

Question

Question: Let a function is given as \(f(x) = \dfrac{{x + 1}}{{x - 1}}\) for all \(x \ne 1.\) Let \({f^1}...

Let a function is given as f(x)=x+1x−1f(x) = \dfrac{{x + 1}}{{x - 1}} for all x≠1.x \ne 1.
Let
f1(x)=f(x),f2(x)=f(f(x)) {f^1}(x) = f(x),{f^2}(x) = f(f(x)){\text{ }} and generally
fn(x)=f(fn−1(x)) for n > 1 Let P=f1(2)f2(3)f3(4)f4(5)  {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?
A. 125 B. 375 C. 250 D. 147  {\text{A}}{\text{. 125}} \\\ {\text{B}}{\text{. 375}} \\\ {\text{C}}{\text{. 250}} \\\ {\text{D}}{\text{. 147}} \\\

Explanation

Solution

Hint: In this question first we examine how the function fn(x){f^n}(x) varies with even or odd values by solving the relations f1(x)=f(x),f2(x)=f(f(x)){f^1}(x) = f(x),{f^2}(x) = f(f(x)). After that we calculate values of f1(2),f2(3),f3(4),f4(5){f^1}(2),{f^2}(3),{f^3}(4),{f^4}(5) and put them in expression for PP.

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

f1(x)=f(x){f^1}(x) = f(x) …………….. eq.(2)
f2(x)=f(f(x)){f^2}(x) = f(f(x)) …………… eq.(3)

fn(x)=f(fn−1(x)) for n > 1{f^n}(x) = f({f^{n - 1}}(x)){\text{ for }}n{\text{ > 1}} …………………. eq.4
Considering eq.3
⇒f2(x)=f(f(x)) ⇒f2(x)=f(x)+1f(x)−1 ⇒f2(x)=(x+1x−1)+1(x+1x−1)−1 \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}} ……………..   from eq.1 {\text{ \\{ from eq}}{\text{.1\\} }} \\\

On solving above equation, we get
⇒f2(x)=x\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 f1(x){f^1}(x). Now, using eq.2 and eq.5 we can generalise the eq.4 as
if n=odd ⇒ fn(x)=f(x){\text{if }}n = {\text{odd }} \Rightarrow {\text{ }}{f^n}(x) = f(x) ………….. eq.(6)
if n=even ⇒ fn(x)=x{\text{if }}n = {\text{even }} \Rightarrow {\text{ }}{f^n}(x) = x …………...eq.(7)

Now considering P
⇒P=f1(2)f2(3)f3(4)f4(5)\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,
⇒f1(2)=2+12−1\Rightarrow {f^1}(2) = \dfrac{{2 + 1}}{{2 - 1}}

From  eq. 6\text{From}\ \text{ eq. 6}
⇒f1(2)=3\Rightarrow {f^1}(2) = 3.............eq.(9)

From  eq. 7\text{From}\ \text{ eq. 7}
⇒f2(3)=3\Rightarrow {f^2}(3) = 3............. eq.(10)

From  eq. 6\text{From}\ \text{ eq. 6}
⇒f3(4)=53\Rightarrow {f^3}(4) = \dfrac{5}{3}............ eq(11)

From  eq. 7\text{From}\ \text{ eq. 7}
⇒f4(5)=5\Rightarrow {f^4}(5) = 5............ eq.(12)

Now, on putting values of f1(2),f2(3),f3(4),f4(5){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
⇒P=3×3×53×5 ⇒P=75  \Rightarrow P = 3 \times 3 \times \dfrac{5}{3} \times 5 \\\ \Rightarrow P = 75 \\\
Therefore, PP is 75 and 375 is multiple of PP.
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 fn(x){f^n}(x) as in this question for n=even⇒fn(x)=x n = {\text{even}} \Rightarrow {f^n}(x) = x and for n = odd⇒fn(x)=f(x){\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.