If $f(x)=\frac{\left(\tan 1^{o}\right) x+\log _{e}(123)}{x \log _{e}(1234)-\left(\tan 1^{o}\right)}, x>0$, then the least value of $f(f(x))+f\left(f\left(\frac{4}{x}\right)\right)$ is

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Relations And Functions XII

f (x) = \frac{( t a n 1 ^{o} ) x + l o g _{e} ( 123 )}{x l o g _{e} ( 1234 ) - ( t a n 1 ^{o} )}, x > 0

, then the least value of

f (f (x)) + f (f (\frac{4}{x}))

Answer: (B)

The function has the form

f (x) = \frac{A x + B}{C x - A}

where

A = tan 1^{\circ}

B = lo g_{e} (123)

, and

C = lo g_{e} (1234)

.

For functions of this form, it can be verified that

f (f (x)) = x

(the function is an involution).

Therefore:

f (f (x)) = x

f (f (\frac{4}{x})) = \frac{4}{x}

So the expression becomes:

f (f (x)) + f (f (\frac{4}{x})) = x + \frac{4}{x}

By AM-GM inequality:

x + \frac{4}{x} \geq 2 x \cdot \frac{4}{x} = 24 = 4

The minimum value is 4, achieved when

x = \frac{4}{x}

, i.e.,

x = 2

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