If f(x) = ((tan 1°)x + loge(123)) / (x loge(1234) - (tan 1°)), x 0, then the least value of f(f(x)

Q: Why can't we just plug in a value for $x$?

While plugging in values might give you a hint, the JEE Advanced examiner tests your ability to prove the property for all $x$. The algebraic proof is the only way to be certain that the function behaves this way universally.

Q: How did you know $f(f(x)) = x$ so quickly?

It is a property of functions where the coefficient of $x$ in the numerator is the negative of the constant term in the denominator. This makes the function an involution, meaning it is its own inverse. Whenever you see this structure, you can anticipate this beautiful cancellation.

Q: Why is the minimum value 4?

We used the AM-GM inequality on the expression $x + \frac{4}{x}$. Since $x > 0$, the sum of a number and its reciprocal is minimized when the two terms are equal, leading to $2 + 2 = 4$.

Question

If $f(x) = \frac{(	an 1^\circ)^x + \log_e(123)}{x \log_e(1234) - (	an 1^\circ)}$, $x > 0$, then the least value of $f(f(x)) + f(f(\frac{4}{x}))$ is

Options:

(A) 0

(B) 8

(C) 2

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you first look at this problem, your heart might skip a beat. You see $	an 1^\circ$, $\log_e(123)$, and $\log_e(1234)$ all jumbled together in a rational function. It looks like a nightmare of calculation, but complexity is often a mask.

The examiner is testing your ability to see through the noise. Let us strip away the mask by defining the following constants:
$a = 	an 1^\circ$
$b = \log_e(123)$
$c = \log_e(1234)$

Suddenly, the function transforms into something elegant:
$$f(x) = \frac{ax+b}{cx-a}$$

### The Dance of Composition

Now, we must find $f(f(x))$. We are not just evaluating a function; we are nesting it by substituting the simplified form back into itself.

We write the composition as:
$$f(f(x)) = \frac{a\left(\frac{ax+b}{cx-a}ight) + b}{c\left(\frac{ax+b}{cx-a}ight) - a}$$

To simplify this, we clear the denominator by multiplying the entire numerator and denominator by $(cx-a)$. Watch the magic happen as we expand the terms:

Numerator: $a(ax+b) + b(cx-a) = a^2x + ab + bcx - ab$

Denominator: $c(ax+b) - a(cx-a) = acx + bc - acx + a^2$

### The Revelation

Look closely at the expansion. In the numerator, the $+ab$ and $-ab$ terms vanish into thin air. In the denominator, the $+acx$ and $-acx$ terms cancel out perfectly.

We are left with:
$$f(f(x)) = \frac{(a^2+bc)x}{a^2+bc}$$

The entire constant term $(a^2+bc)$ cancels out, leaving us with the most beautiful result:
$$f(f(x)) = x$$

This function is an **involution**; it is its own inverse. This realization is the turning point of the problem, transforming a terrifying algebraic expression into a simple variable.

### The Final Ascent

The problem asks for the minimum value of $f(f(x)) + f(f(\frac{4}{x}))$. Since we have proven that $f(f(x)) = x$, this expression simplifies instantly to:
$$x + \frac{4}{x}$$

We are now in the realm of the **AM-GM inequality**, a favorite tool of the JEE. For any positive number $x$, the Arithmetic Mean is greater than or equal to the Geometric Mean:
$$\frac{x + \frac{4}{x}}{2} \ge \sqrt{x \cdot \frac{4}{x}}$$

The $x$ terms cancel, leaving $\sqrt{4} = 2$. Multiplying by $2$, we find:
$$x + \frac{4}{x} \ge 4$$

The **minimum value is 4**. You have conquered the monster by seeing the pattern beneath the skin.

Animated Solution for Mathematics - Functions: If $f (x) = \frac{( t a n 1 ^{\circ} ) ^{x} + l o g _{e} ( 123 )}{x l o g _{e} ( 1234 ) - ( t a n 1 ^{\circ} )}$ , $x > 0$ , then the least value of $f (f (x)) + f (f (\frac{4}{x}))$ is

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Visualized Solution

$A na l y z in g t h e F u n c t i o n f (x)$

$S im pl i f i e d F or m o f f (x)$

$S e tt in g u p f (f (x))$

$A l g e b r ai c E x p an s i o n$

$S im pl i f y in g f (f (x))$

$E v a l u a t in g t h e T a r g e tE x p r ess i o n$

$T h e F ina l E x p r ess i o n$

$A ppl y in g A M - GM I n e q u a l i t y$

$F in d in g t h e M inim u mV a l u e$

The Sigma Insight: Composite Functions

Analyzing the Setup

The Dance of Composition

The Revelation

The Final Ascent

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=xloge​(1234)−(tan1∘)(tan1∘)x+loge​(123)​, x>0, then the least value of f(f(x))+f(f(x4​)) is

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Visualized Solution

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