If the function `f(x)=((1-x))/(2)tan.(pix)/(2)` is continuous at x = 1, then `f(1)` is equal to

Options

A. `(1)/(pi)`
B. `(pi)/(2)`
C. 0
D. `pi`

Answer

To find the value of \( f(1) \) for the function \( f(x) = \frac{1-x}{2 \tan\left(\frac{\pi x}{2}\right)} \) such that it is continuous at \( x = 1 \), we follow these steps:

### Step 1: Check the continuity condition
A function is continuous at a point \( x = a \) if:
\[ \lim_{x \to a} f(x) = f(a) \]
In this case, we need to check:
\[ \lim_{x \to 1} f(x) = f(1) \]

### Step 2: Substitute \( x = 1 \) into \( f(x) \)
First, we find \( f(1) \):
\[ f(1) = \frac{1-1}{2 \tan\left(\frac{\pi \cdot 1}{2}\right)} = \frac{0}{2 \tan\left(\frac{\pi}{2}\right)} \]
Since \( \tan\left(\frac{\pi}{2}\right) \) is undefined, \( f(1) \) is not directly computable.

### Step 3: Find the limit as \( x \) approaches 1
Next, we compute the limit:
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{1-x}{2 \tan\left(\frac{\pi x}{2}\right)} \]
Substituting \( x = 1 \) directly gives us the indeterminate form \( \frac{0}{0} \).

### Step 4: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:
- The derivative of the numerator \( 1 - x \) is \( -1 \).
- The derivative of the denominator \( 2 \tan\left(\frac{\pi x}{2}\right) \) is \( 2 \cdot \frac{\pi}{2} \sec^2\left(\frac{\pi x}{2}\right) = \pi \sec^2\left(\frac{\pi x}{2}\right) \).

Now we can rewrite the limit:
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{-1}{\pi \sec^2\left(\frac{\pi x}{2}\right)} \]

### Step 5: Evaluate the limit
As \( x \) approaches 1:
\[ \sec^2\left(\frac{\pi x}{2}\right) \to \sec^2\left(\frac{\pi}{2}\right) \to \infty \]
Thus, the limit becomes:
\[ \lim_{x \to 1} f(x) = \frac{-1}{\pi \cdot \infty} = 0 \]

### Step 6: Set the limit equal to \( f(1) \)
For the function to be continuous at \( x = 1 \):
\[ f(1) = \lim_{x \to 1} f(x) = 0 \]

### Conclusion
Thus, \( f(1) \) is equal to \( 0 \).