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Let $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}+1}{\mathrm{x}-1}$ for all $\mathrm{x} \neq 1$. Let
$\mathrm{f}^{1}(\mathrm{x})=\mathrm{f}(\mathrm{x}), \mathrm{f}^{2}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{x}))$ and generally
$f^{n}(x)=f\left(f^{n-1}(x)\right)$ for $n>1$
Let $\mathrm{P}=\mathrm{f}^{1}(2) \mathrm{f}^{2}(3) \mathrm{f}^{3}(4) \mathrm{f}^{4}(5)$
Which of the following is a multiple of P -
Options:
$\mathrm{f}^{1}(\mathrm{x})=\mathrm{f}(\mathrm{x}), \mathrm{f}^{2}(\mathrm{x})=\mathrm{f}(\mathrm{f}(\mathrm{x}))$ and generally
$f^{n}(x)=f\left(f^{n-1}(x)\right)$ for $n>1$
Let $\mathrm{P}=\mathrm{f}^{1}(2) \mathrm{f}^{2}(3) \mathrm{f}^{3}(4) \mathrm{f}^{4}(5)$
Which of the following is a multiple of P -
Solution:
2172 Upvotes
Verified Answer
The correct answer is:
375
$\begin{array}{l}
f(x)=\frac{x+1}{x-1} \\
f^{2}(x)=f(f(x))=f\left(\frac{x+1}{x-1}\right)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}=x \\
f^{3}(x)=f(x)=\frac{x+1}{x-1} \\
f^{4}(x)=x \\
P=f(2) \cdot f^{3}(3) f^{3}(4) f^{4}(5) \\
P=3 \times 3 \times \frac{5}{3} \times 5=75
\end{array}$
Multiple of $\mathrm{P}$ is 375
f(x)=\frac{x+1}{x-1} \\
f^{2}(x)=f(f(x))=f\left(\frac{x+1}{x-1}\right)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}=x \\
f^{3}(x)=f(x)=\frac{x+1}{x-1} \\
f^{4}(x)=x \\
P=f(2) \cdot f^{3}(3) f^{3}(4) f^{4}(5) \\
P=3 \times 3 \times \frac{5}{3} \times 5=75
\end{array}$
Multiple of $\mathrm{P}$ is 375
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