FunctionshardFree

Functions: 2010x 165x Least Value

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

Question
If f(x)=2010x+163165x−2010f(x) = \dfrac{2010x+163}{165x-2010}, x>0x > 0, x≠2010165x \neq \dfrac{2010}{165}, then the least value of f(f(x))+f ⁣(f ⁣(4x))f(f(x)) + f\!\left(f\!\left(\dfrac{4}{x}\right)\right) is.
Solution
Answer: 4
Step 1: Show ff is an involution For f(x)=ax+bcx+df(x) = \dfrac{ax+b}{cx+d}, f(f(x))=xf(f(x)) = x iff a+d=0a+d = 0. Here 2010+(−2010)=02010+(-2010) = 0, so f(f(x))=xf(f(x)) = x. Step 2: Simplify and minimize
f(f(x))+f ⁣(f ⁣(4x))=x+4xf(f(x))+f\!\left(f\!\left(\frac{4}{x}\right)\right) = x + \frac{4}{x}
 By AM-GM for x>0x > 0:
x+4x≥2x⋅4x=24=4x + \frac{4}{x} \geq 2\sqrt{x \cdot \frac{4}{x}} = 2\sqrt{4} = 4
 Equality holds at x=2x = 2. Answer: 4
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