The local maximum value of the function f(x) = (2/x)^(x²), x 0, is (1) (2√e)^(1/e) (2) (4/√e)^(e/4

Q: Wait, why can't I just use the power rule? Isn't it just $x^2$ times something?

That is a very common trap! The power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ is only valid when the exponent $n$ is a constant. In our function $f(x) = (\frac{2}{x})^{x^2}$, the exponent is $x^2$, which changes as $x$ changes. Because both the base and the exponent are variables, you must use logarithmic differentiation to handle the relationship between them.

Q: When we set $f'(x) = 0$, why can we just ignore the $x$ and $f(x)$ terms?

Excellent question! We found that $f'(x) = x f(x) [2 \ln(\frac{2}{x}) - 1] = 0$. Since the problem defines the domain as $x > 0$, $x$ cannot be zero. Furthermore, $f(x) = (\frac{2}{x})^{x^2}$ is an exponential form, which is always positive for $x > 0$. Since neither $x$ nor $f(x)$ can be zero, the only way for the entire product to be zero is if the bracketed term $[2 \ln(\frac{2}{x}) - 1]$ equals zero.

Q: How do I know if the critical point I found is a maximum or a minimum?

In this specific problem, we are looking for the local maximum. Given the behavior of the function $f(x) = (\frac{2}{x})^{x^2}$ for $x > 0$, it rises from zero, hits a peak, and then decays. If you were unsure, you could use the second derivative test, but for JEE problems of this type, identifying the unique critical point within the domain is usually sufficient to confirm it as the local maximum.

Question

The local maximum value of the function $f(x) = (\frac{2}{x})^{x^2}, x > 0$, is

Options:

(A) $(2\sqrt{e})^{1/e}$

(B) $(\frac{4}{\sqrt{e}})^{e/4}$

(C) $(e)^{2/e}$

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to tackle a problem that often intimidates students at first glance: $f(x) = (\frac{2}{x})^{x^2}$.

At first, it looks like a standard power function, but look closer. The base is changing, and the exponent is changing. This is a classic 'variable-to-variable' trap.

If you try to apply the standard power rule, you will find yourself lost in a maze of incorrect derivatives. But fear not—we have the perfect key to unlock this mystery: **Logarithmic Differentiation**.

### Phase 1: Bringing the Exponent Down to Earth

When we face a function where both the base and the exponent are variables, our first instinct should be to simplify the structure. We want to turn that exponent into a coefficient using the natural logarithm.

Let us take the natural log of both sides:

$$\ln f(x) = \ln \left( \left(\frac{2}{x}\right)^{x^2} \right)$$

Now, we invoke the beautiful property of logarithms: $\ln(a^b) = b \ln a$. This property allows us to pull that $x^2$ down from its lofty position in the exponent.

Suddenly, our equation becomes much more manageable:

$$\ln f(x) = x^2 \ln\left(\frac{2}{x}\right)$$

See how the complexity has vanished? We have transformed a power function into a simple product of two functions. This is the elegance of calculus—finding the right tool to simplify the seemingly impossible.

### Phase 2: The Product Rule in Action

Now, we differentiate both sides with respect to $x$. On the left, we use the chain rule: the derivative of $\ln f(x)$ is $\frac{f'(x)}{f(x)}$.

On the right, we have a product of two functions: $u = x^2$ and $v = \ln(\frac{2}{x})$. Recall the product rule: $(uv)' = u'v + uv'$.

Let's execute this carefully:

$$\frac{f'(x)}{f(x)} = (2x)\ln\left(\frac{2}{x}\right) + x^2 \cdot \frac{d}{dx} \left[ \ln 2 - \ln x \right]$$

Since $\ln(\frac{2}{x}) = \ln 2 - \ln x$, the derivative of the second part is simply $-\frac{1}{x}$. Substituting this back, we get:

$$\frac{f'(x)}{f(x)} = 2x \ln\left(\frac{2}{x}\right) + x^2 \left(-\frac{1}{x}\right) = 2x \ln\left(\frac{2}{x}\right) - x$$

### Phase 3: The Critical Moment

To find the local maximum, we need to find where the slope of the tangent is zero. We set $f'(x) = 0$.

Isolating $f'(x)$, we have:

$$f'(x) = f(x) \left[ 2x \ln\left(\frac{2}{x}\right) - x \right] = x f(x) \left[ 2 \ln\left(\frac{2}{x}\right) - 1 \right]$$

Since $x > 0$ and $f(x) > 0$, the only way for $f'(x)$ to be zero is if the bracketed term is zero:

$$2 \ln\left(\frac{2}{x}\right) - 1 = 0 \implies \ln\left(\frac{2}{x}\right) = \frac{1}{2}$$

Converting this to exponential form, we get $\frac{2}{x} = e^{1/2} = \sqrt{e}$. Solving for $x$, we find our critical point: $x = \frac{2}{\sqrt{e}}$.

### Phase 4: The Final Reveal

We have found the location of the peak. Now, we must find the height of the peak by substituting $x = \frac{2}{\sqrt{e}}$ back into our original function $f(x)$:

$$f\left(\frac{2}{\sqrt{e}}\right) = \left( \frac{2}{\frac{2}{\sqrt{e}}} \right)^{\left(\frac{2}{\sqrt{e}}\right)^2}$$

Simplifying the base, the $2$s cancel, and $\sqrt{e}$ moves to the numerator. The exponent becomes $(\frac{2}{\sqrt{e}})^2 = \frac{4}{e}$.

Thus, we have:

$$f\left(\frac{2}{\sqrt{e}}\right) = (\sqrt{e})^{4/e} = (e^{1/2})^{4/e} = e^{(1/2) \cdot (4/e)} = e^{2/e}$$

And there it is! The local maximum value is **$e^{2/e}$**. It is a clean, beautiful result that emerges from the chaos of variables.

Animated Solution for Mathematics - Differentiation: The local maximum value of the function $f (x) = (\frac{2}{x})^{x^{2}}, x > 0$ , is

Select Answer:

Visualized Solution

Visualizing the Function $f (x)$

The Strategy: Logarithmic Differentiation

Simplifying the Logarithm

Differentiating Both Sides

Applying the Product Rule

Isolating the Derivative

Finding the Critical Point

Solving for $x$

Exponential Conversion

Calculating the Maximum Value

Final Simplification

The Final Answer

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The local maximum value of the function f(x)=(x2​)x2,x>0, is