If f(x) = ((tan 1° )x + log ₑ(123))/(xlog ₑ(1234) - (tan 1° )),x > 0, then the least value of f(f(x)) + f( f( 4/x ) ) is :

Solution

Maths sir · 4 steps

Do not touch those ugly constants: this f undoes itself, so f(f(x)) is just x.

1. Name the constants. Write a = tan 1°, b = logₑ 123, c = logₑ 1234. Then f(x) = (ax + b)/(cx - a). The x-coefficient on top is a and the constant at the bottom is -a, and they add to zero. That pattern makes f its own inverse.
2. Check f(f(x)). Substitute f(x) into f and clear the inner denominator cx - a. The top becomes a(ax+b) + b(cx-a) = (a² + bc)x and the bottom becomes c(ax+b) - a(cx-a) = a² + bc.

   f(f(x)) = ((a² + bc) x)/(a² + bc) = x

3. Apply AM-GM. So f(f(x)) + f(f(4/x)) = x + 4/x. For x > 0, AM-GM gives x + 4/x ≥ 2√(x × 4/x) = 4, with equality when x = 4/x, that is x = 2.
4. So. The least value is 4, reached at x = 2. That is option (b).

The idea to keep. For f(x) = (ax+b)/(cx-d), if the x-coefficient on top equals the constant subtracted at the bottom (a = d), then f(f(x)) = x. Spot this before expanding anything.

Answer from the NTA final answer key. Solution by PrepWiser.