Step-by-step Solution:

Correct Answer: D (at $x = \cos x$)

To find the condition under which the curve attains a maximum, we need to find the first derivative of the function, set it to zero to find the critical points, and then test these points.

Step 1: Find the First Derivative of the Function

The given function is $y = \frac{x}{1+x\tan x}$. We use the quotient rule for differentiation, which states that $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$.

• Let $u = x$, so $u' = 1$.
• Let $v = 1+x\tan x$. Using the product rule, $v' = (1)(\tan x) + (x)(\sec^2 x) = \tan x + x\sec^2 x$.

Applying the quotient rule:

$$ \frac{dy}{dx} = \frac{(1)(1+x\tan x) - (x)(\tan x + x\sec^2 x)}{(1+x\tan x)^2} $$

$$ \frac{dy}{dx} = \frac{1 + x\tan x - x\tan x - x^2\sec^2 x}{(1+x\tan x)^2} $$

$$ \frac{dy}{dx} = \frac{1 - x^2\sec^2 x}{(1+x\tan x)^2} $$

Step 2: Find the Critical Points

For a maximum or minimum, we set the first derivative to zero, $\frac{dy}{dx} = 0$. This occurs when the numerator is zero.

$1 - x^2\sec^2 x = 0$

$1 = x^2\sec^2 x$

Since $\sec^2 x = \frac{1}{\cos^2 x}$, we have:

$1 = \frac{x^2}{\cos^2 x} \implies x^2 = \cos^2 x$

Taking the square root of both sides gives the conditions for the critical points:

$|x| = |\cos x| \implies x = \cos x \quad \text{or} \quad x = -\cos x$

Step 3: Determine the Condition for a Maximum

To determine which condition corresponds to a maximum, we can use the First Derivative Test. The sign of $\frac{dy}{dx}$ is determined by the sign of its numerator, $1 - x^2\sec^2 x$, which can be rewritten as $\frac{\cos^2 x - x^2}{\cos^2 x}$. Since $\cos^2 x$ is non-negative, the sign depends on the expression $g(x) = \cos^2 x - x^2$.

• Case 1: $x = \cos x$ This equality holds for a unique positive value of $x$ (approximately 0.739).
  • For a value slightly less than this root, $|x| < |\cos x|$, so $x^2 < \cos^2 x$. This makes $g(x)$ positive ($\frac{dy}{dx} > 0$).
  • For a value slightly greater than this root, $|x| > |\cos x|$, so $x^2 > \cos^2 x$. This makes $g(x)$ negative ($\frac{dy}{dx} < 0$).
  Since the derivative changes from positive to negative, the function has a local maximum at $x = \cos x$.
• Case 2: $x = -\cos x$ This equality holds for a unique negative value of $x$ (approximately -0.739).
  • For a value slightly less than this root, $|x| > |\cos x|$, so $x^2 > \cos^2 x$. This makes $g(x)$ negative ($\frac{dy}{dx} < 0$).
  • For a value slightly greater than this root, $|x| < |\cos x|$, so $x^2 < \cos^2 x$. This makes $g(x)$ positive ($\frac{dy}{dx} > 0$).
  Since the derivative changes from negative to positive, the function has a local minimum at $x = -\cos x$.

Therefore, the curve attains a maximum at $x = \cos x$.