Let f : R - R be such that for all x ∈ R (2^(1+x)+2^(1-x)), f (x) and (3^x + 3^(-x)) are in A.P., t

Q: Wait, why can't we just differentiate $f(x)$ to find the minimum? Isn't that the standard way to find extrema?

You absolutely can! Differentiating $f(x)$ with respect to $x$ and setting $f'(x) = 0$ is a perfectly valid and rigorous method. However, in JEE Advanced, time is your most precious resource. Recognizing the structure of $a^x + a^{-x}$ allows you to use the AM-GM inequality, which provides the answer in seconds without the need for complex chain-rule differentiation. Both paths lead to the same truth, but AM-GM is the 'elegant' path.

Q: Why do we need to check if the minimums occur at the same $x$? Can't we just add the minimums of the two parts?

This is a brilliant question! You can only add the minimum values of two functions if they achieve those minimums at the exact same input value. If the first part reached its minimum at $x=1$ and the second at $x=5$, you could never reach both minimums simultaneously. In this problem, both $2^x + 2^{-x}$ and $3^x + 3^{-x}$ reach their minimum at $x=0$. Because they 'agree' on the location of the minimum, we can safely add them together.

Q: Is it possible for the minimum value to be negative? The terms look like they could be small.

Look closely at the base of the exponents. Since $2^x$ and $3^x$ are always positive for any real $x$, their sum must also be positive. The AM-GM inequality confirms this: the sum of a positive number and its reciprocal is always at least 2. Therefore, $f(x)$ is bounded below by a positive value, and it can never touch or cross into negative territory.

Question

Let $f : R \rightarrow R$ be such that for all $x \in R (2^{1+x}+2^{1-x}), f (x)$ and $(3^x + 3^{-x})$ are in A.P., then the minimum value of $f (x)$ is :

Options:

(A) 1

(B) 2

(C) 3

(D) 4

Sigma Percentile · Accepted Answer

### The Harmony of Arithmetic Progressions

Welcome, traveler of the mathematical landscape. Today, we are going to peel back the layers of a problem that, at first glance, might seem like a daunting collection of exponential terms.

But beneath the surface, it is a beautiful study of symmetry and balance. We are given three terms: $T_1 = (2^{1+x}+2^{1-x})$, $T_2 = f(x)$, and $T_3 = (3^x + 3^{-x})$, and we are told they exist in an Arithmetic Progression (A.P.).

### The Golden Key: The A.P. Property

What does it mean to be in an A.P.? It means the difference between consecutive terms is constant.

If $a, b, c$ are in A.P., then $b - a = c - b$, which simplifies to the elegant relation $2b = a + c$. This is our golden key. It tells us that the middle term, $f(x)$, is simply the average of its neighbors.

So, our equation becomes:

$$2f(x) = (2^{1+x}+2^{1-x}) + (3^x+3^{-x})$$

Imagine you are standing on a bridge, and $f(x)$ is the height of the center point, perfectly balanced between the two exponential 'pillars' on either side. Our goal is to find the lowest point this bridge ever reaches.

### Simplifying the Algebraic Dance

Before we rush into finding the minimum, let us tidy up our expression. The first term, $(2^{1+x}+2^{1-x})$, looks a bit cluttered.

Using the laws of exponents, specifically $a^{m+n} = a^m \cdot a^n$, we can rewrite this as:

$$2^{1+x} + 2^{1-x} = 2^1 \cdot 2^x + 2^1 \cdot 2^{-x} = 2(2^x + 2^{-x})$$

Now, substitute this back into our main equation and isolate $f(x)$ by dividing by 2:

$$f(x) = (2^x + 2^{-x}) + \frac{1}{2}(3^x + 3^{-x})$$

Look at that structure. It is a sum of two distinct parts, each involving a base raised to $x$ and its reciprocal. This is the signature of the AM-GM inequality.

### The AM-GM Revelation

Whenever you see a function of the form $g(x) = a^x + a^{-x}$, your intuition should immediately trigger the Arithmetic Mean-Geometric Mean (AM-GM) inequality. The inequality states that for any positive real numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$, or more simply, $a+b \ge 2\sqrt{ab}$.

Let us apply this to our first part, $(2^x + 2^{-x})$:

$$2^x + 2^{-x} \ge 2\sqrt{2^x \cdot 2^{-x}} = 2\sqrt{2^0} = 2$$

The minimum value is 2. Now, let us apply the same logic to the second part, $(3^x + 3^{-x})$:

$$3^x + 3^{-x} \ge 2\sqrt{3^x \cdot 3^{-x}} = 2\sqrt{3^0} = 2$$

### The Final Convergence

We have two minimums, but can we simply add them? Yes, because they occur at the same point!

For both $2^x + 2^{-x}$ and $3^x + 3^{-x}$, the minimum occurs when the two terms are equal, which is when $x=0$. Since the 'valley' of both functions aligns perfectly at $x=0$, we can combine them without hesitation:

$$f(x) \ge 2 + \frac{1}{2}(2) = 2 + 1 = 3$$

And there it is. The absolute minimum value of our function $f(x)$ is **3**.

It is a moment of pure mathematical harmony. We started with a complex-looking expression, applied the symmetry of an A.P., simplified the algebra, and used the power of AM-GM to find the lowest point of the curve. Remember, in JEE Advanced, the most complex problems often yield to the most fundamental principles if you approach them with patience and clarity.

Animated Solution for Mathematics - Sequence and Series: Let $f : R \to R$ be such that for all $x \in R (2^{1 + x} + 2^{1 - x}), f (x)$ and $(3^{x} + 3^{- x})$ are in A.P., then the minimum value of $f (x)$ is :

Select Answer:

Visualized Solution

The Given Sequence

Arithmetic Progression Property

Setting up the Equation

Simplifying the First Term

Isolating $f (x)$

The AM-GM Inequality

Minimum of $2^{x} + 2^{- x}$

Minimum of $3^{x} + 3^{- x}$

Combining the Minimums

Final Calculation

Conclusion

The Sigma Insight: Relation Between A.M., G.M., and H.M.

The Harmony of Arithmetic Progressions

The Golden Key

Simplifying the Algebraic Dance

The AM-GM Revelation

The Final Convergence

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