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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

Let

f(x) = \dfrac{x+1}{x-1}

for all

x \neq 1

. Let

f^{-1}(x) = f(x)

f^2(x) = f(f(x))

, and

f^n(x) = f(f^{n-1}(x))

for

n > 1

. Let

P = f^1(2)\cdot f^2(3)\cdot f^3(4)\cdot f^4(5)

. The number of divisors of

P

Solution

Answer: 6

Step 1: Show

f^2(x) = 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} = \frac{2x}{2} = x

f^{2k}(x) = x

and

f^{2k+1}(x) = f(x)

for all

k \geq 0

. Step 2: Evaluate each factor

f^1(2) = f(2) = \frac{3}{1} = 3

f^2(3) = 3 \quad (\text{since } f^2(x) = x)

f^3(4) = f(4) = \frac{5}{3} \quad (\text{since } f^3(4) = f(f^2(4)) = f(4))

f^4(5) = 5 \quad (\text{since } f^4(x) = x)

Step 3: Find divisors of

P

P = 3 \times 3 \times \frac{5}{3} \times 5 = 3^1 \times 5^2 = 75

Number of divisors

= (1+1)(2+1) = 6

. Answer: 6

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