FunctionshardFree

Functions — JEE Maths practice question

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

Question

f(x) = \dfrac{2010x+163}{165x-2010}

x > 0

x \neq \dfrac{2010}{165}

, then the least value of

f(f(x)) + f\!\left(f\!\left(\dfrac{4}{x}\right)\right)

is.

Solution

Answer: 4

Step 1: Show

f

is an involution For

f(x) = \dfrac{ax+b}{cx+d}

f(f(x)) = x

iff

a+d = 0

. Here

2010+(-2010) = 0

, so

f(f(x)) = x

. Step 2: Simplify and minimize

f(f(x))+f\!\left(f\!\left(\frac{4}{x}\right)\right) = x + \frac{4}{x}

By AM-GM for

x > 0

x + \frac{4}{x} \geq 2\sqrt{x \cdot \frac{4}{x}} = 2\sqrt{4} = 4

Equality holds at

x = 2

. Answer: 4

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