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

Answer

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

### Step 1: Determine the limit as \( x \) approaches 1
Since we want \( f(1) \) to be continuous, we need to find:
\[ f(1) = \lim_{x \to 1} f(x) \]
This means we need to evaluate:
\[ \lim_{x \to 1} \frac{(1-x)}{2} \tan\left(\frac{\pi x}{2}\right) \]

### Step 2: Substitute \( x = 1 \) into the function
Substituting \( x = 1 \) directly into \( f(x) \):
\[ f(1) = \frac{(1-1)}{2} \tan\left(\frac{\pi \cdot 1}{2}\right) = \frac{0}{2} \tan\left(\frac{\pi}{2}\right) \]
Since \( \tan\left(\frac{\pi}{2}\right) \) is undefined, we get an indeterminate form \( 0 \cdot \text{undefined} \).

### Step 3: Rewrite the limit using sine and cosine
To resolve this, we can rewrite the tangent function:
\[ \tan\left(\frac{\pi x}{2}\right) = \frac{\sin\left(\frac{\pi x}{2}\right)}{\cos\left(\frac{\pi x}{2}\right)} \]
Thus, we have:
\[ \lim_{x \to 1} \frac{(1-x)}{2} \cdot \frac{\sin\left(\frac{\pi x}{2}\right)}{\cos\left(\frac{\pi x}{2}\right)} \]

### Step 4: Apply L'Hôpital's Rule
As \( x \) approaches 1, both the numerator and denominator approach 0, resulting in a \( \frac{0}{0} \) form. We can apply L'Hôpital's Rule:
\[ \lim_{x \to 1} \frac{(1-x) \sin\left(\frac{\pi x}{2}\right)}{2 \cos\left(\frac{\pi x}{2}\right)} \]
Differentiating the numerator and denominator separately:
- The derivative of the numerator \( (1-x) \sin\left(\frac{\pi x}{2}\right) \) using the product rule:
\[ \frac{d}{dx}[(1-x) \sin\left(\frac{\pi x}{2}\right)] = -\sin\left(\frac{\pi x}{2}\right) + (1-x) \cdot \frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right) \]
- The derivative of the denominator \( 2 \cos\left(\frac{\pi x}{2}\right) \) is:
\[ -\pi \sin\left(\frac{\pi x}{2}\right) \]

### Step 5: Evaluate the limit again
Now we evaluate:
\[ \lim_{x \to 1} \frac{-\sin\left(\frac{\pi x}{2}\right) + (1-x) \cdot \frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right)}{-\pi \sin\left(\frac{\pi x}{2}\right)} \]
Substituting \( x = 1 \):
\[ = \frac{-1 + 0 \cdot \frac{\pi}{2} \cdot 0}{-\pi \cdot 1} = \frac{-1}{-\pi} = \frac{1}{\pi} \]

### Conclusion
Thus, the value of \( f(1) \) is:
\[ f(1) = \frac{1}{\pi} \]