The function `f(x)=(x)/(1+x tanx )`

Options

A. one point of minimum in the interval `(0,pi//2)`
B. one point of maxmimum `(0,pi//2)`
C. no points of maximum , no point of minimum in the interval `(0,pi//2)`
D. two points of maxima in the interval `(0,pi//2)`

Explanation

To find the maxima and minima of the function \( f(x) = \frac{x}{1 + x \tan x} \), we will follow these steps:

### Step 1: Find the derivative \( f'(x) \)

To find the critical points, we first need to compute the derivative of \( f(x) \). We will use the quotient rule for differentiation, which states that if \( f(x) = \frac{g(x)}{h(x)} \), then:

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

Here, \( g(x) = x \) and \( h(x) = 1 + x \tan x \).

1. Compute \( g'(x) = 1 \).
2. Compute \( h'(x) \):
- \( h(x) = 1 + x \tan x \)
- Using the product rule, \( h'(x) = \tan x + x \sec^2 x \).

Now, applying the quotient rule:

\[ f'(x) = \frac{(1)(1 + x \tan x) - (x)(\tan x + x \sec^2 x)}{(1 + x \tan x)^2} \]

### Step 2: Simplify the derivative

Now, let's simplify the numerator:

\[ f'(x) = \frac{1 + x \tan x - x \tan x - x^2 \sec^2 x}{(1 + x \tan x)^2} \]

This simplifies to:

\[ f'(x) = \frac{1 - x^2 \sec^2 x}{(1 + x \tan x)^2} \]

### Step 3: Set the derivative to zero

To find critical points, set \( f'(x) = 0 \):

\[ 1 - x^2 \sec^2 x = 0 \]

This implies:

\[ x^2 \sec^2 x = 1 \quad \Rightarrow \quad \sec^2 x = \frac{1}{x^2} \]

### Step 4: Solve for \( x \)

Taking the reciprocal gives:

\[ \cos^2 x = x^2 \]

This equation can be solved for \( x \) to find the critical points.

### Step 5: Determine maxima or minima

To determine whether each critical point is a maximum or minimum, we can use the second derivative test or analyze the sign of \( f'(x) \) around the critical points.

### Step 6: Conclusion

After finding the critical points and determining their nature, we can conclude where the function has its maxima and minima.

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