If the function f(x) = (1-x) ` tan ""(pix)/2` is continuous at x =1 then f (1) =

Options

A. `2/pi`
B. ` pi/2`
C. 0
D. `2pi`

Answer

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

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

### Step 2: Rewrite the function
To resolve the indeterminate form, we can rewrite the function:
\[ f(x) = (1-x) \tan\left(\frac{\pi x}{2}\right) = (1-x) \frac{\sin\left(\frac{\pi x}{2}\right)}{\cos\left(\frac{\pi x}{2}\right)} \]

### Step 3: Find the limit as \( x \) approaches 1
We need to find:
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} (1-x) \tan\left(\frac{\pi x}{2}\right) \]
This can be rewritten as:
\[ \lim_{x \to 1} \frac{(1-x) \sin\left(\frac{\pi x}{2}\right)}{\cos\left(\frac{\pi x}{2}\right)} \]
As \( x \to 1 \), both the numerator and denominator approach 0, leading to a \( \frac{0}{0} \) form.

### Step 4: Apply L'Hôpital's Rule
We can apply L'Hôpital's Rule, which states that if we have a \( \frac{0}{0} \) form, we can differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}[(1-x) \sin\left(\frac{\pi x}{2}\right)] = -\sin\left(\frac{\pi x}{2}\right) + (1-x) \cdot \frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right) \]
\[ \text{Denominator: } \frac{d}{dx}[\cos\left(\frac{\pi x}{2}\right)] = -\frac{\pi}{2} \sin\left(\frac{\pi x}{2}\right) \]

### Step 5: Evaluate the limit again
Substituting \( x = 1 \) into the derivatives:
\[ \text{Numerator at } x = 1: -\sin\left(\frac{\pi}{2}\right) + (1-1) \cdot \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) = -1 + 0 = -1 \]
\[ \text{Denominator at } x = 1: -\frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) = -\frac{\pi}{2} \cdot 1 = -\frac{\pi}{2} \]
Thus, the limit becomes:
\[ \lim_{x \to 1} f(x) = \frac{-1}{-\frac{\pi}{2}} = \frac{2}{\pi} \]

### Conclusion
Since \( f(x) \) is continuous at \( x = 1 \), we have:
\[ f(1) = \frac{2}{\pi} \]

### Final Answer
\[ f(1) = \frac{2}{\pi} \]