Let `f(x) = {(x "sin" (pi)/(x)",",0 lt x le 1),(0,x =0):}` . Then f'(x) = 0 for

Options

A. exactly two value of x
B. no value of x
C. infinitely many values of x
D. exactly one value of x

Answer

To find the values of \( x \) for which \( f'(x) = 0 \) given the function

\[ f(x) = \begin{cases} x \sin\left(\frac{\pi}{x}\right) & \text{for } 0 < x \leq 1 \\ 0 & \text{for } x = 0 \end{cases} \]

we will differentiate \( f(x) \) for \( 0 < x \leq 1 \).

### Step 1: Differentiate \( f(x) \)

We need to differentiate \( f(x) = x \sin\left(\frac{\pi}{x}\right) \). We will use the product rule for differentiation, which states that if \( u(x) = x \) and \( v(x) = \sin\left(\frac{\pi}{x}\right) \), then:

\[ f'(x) = u'v + uv' \]

where \( u' = 1 \) and \( v' \) can be found using the chain rule.

### Step 2: Find \( v' \)

To differentiate \( v(x) = \sin\left(\frac{\pi}{x}\right) \):

1. Let \( w = \frac{\pi}{x} \), then \( v = \sin(w) \).
2. The derivative \( v' = \cos(w) \cdot w' \).
3. Now, \( w' = -\frac{\pi}{x^2} \) (using the power rule).

Thus,

\[ v' = \cos\left(\frac{\pi}{x}\right) \cdot \left(-\frac{\pi}{x^2}\right) = -\frac{\pi \cos\left(\frac{\pi}{x}\right)}{x^2} \]

### Step 3: Substitute \( u, u', v, v' \) into the product rule

Now substituting back into the product rule:

\[ f'(x) = 1 \cdot \sin\left(\frac{\pi}{x}\right) + x \cdot \left(-\frac{\pi \cos\left(\frac{\pi}{x}\right)}{x^2}\right) \]

This simplifies to:

\[ f'(x) = \sin\left(\frac{\pi}{x}\right) - \frac{\pi \cos\left(\frac{\pi}{x}\right)}{x} \]

### Step 4: Set \( f'(x) = 0 \)

Now we set the derivative equal to zero:

\[ \sin\left(\frac{\pi}{x}\right) - \frac{\pi \cos\left(\frac{\pi}{x}\right)}{x} = 0 \]

Rearranging gives:

\[ \sin\left(\frac{\pi}{x}\right) = \frac{\pi \cos\left(\frac{\pi}{x}\right)}{x} \]

### Step 5: Divide both sides by \( \cos\left(\frac{\pi}{x}\right) \)

Assuming \( \cos\left(\frac{\pi}{x}\right) \neq 0 \):

\[ \tan\left(\frac{\pi}{x}\right) = \frac{\pi}{x} \]

### Step 6: Analyze the equation

The equation \( \tan\left(\frac{\pi}{x}\right) = \frac{\pi}{x} \) will have infinitely many solutions as \( x \) varies from 0 to 1. The function \( \tan\left(\frac{\pi}{x}\right) \) oscillates and intersects the line \( y = \frac{\pi}{x} \) multiple times in this interval.

### Step 7: Consider the case when \( x = 0 \)

At \( x = 0 \), \( f'(0) = 0 \) since the function is defined to be 0 at that point.

### Conclusion

Thus, \( f'(x) = 0 \) for infinitely many values of \( x \) in the interval \( (0, 1] \) and also at \( x = 0 \).

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