Let `f(x) = sin^(-1) x + cos^(-1) x ". Then " pi/2 ` is equal to

Options

A. `f(- 1/2)`
B. `f(k^(2) - 2k + 3), k in R`
C. `f(1/(1+ k^(2))), k in R`
D. `f(-2)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1}(x) + \cos^{-1}(x) \) and determine which of the given options is equal to \( \frac{\pi}{2} \).

### Step 1: Understand the function \( f(x) \)

The function \( f(x) \) is defined as:
\[ f(x) = \sin^{-1}(x) + \cos^{-1}(x) \]
We know a fundamental identity in inverse trigonometric functions:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \quad \text{for } x \in [-1, 1] \]

### Step 2: Evaluate the options

Now we will evaluate each option to see if they equal \( \frac{\pi}{2} \).

#### Option A: \( f\left(-\frac{1}{2}\right) \)

Since \( -\frac{1}{2} \) is in the interval \([-1, 1]\):
\[ f\left(-\frac{1}{2}\right) = \sin^{-1}\left(-\frac{1}{2}\right) + \cos^{-1}\left(-\frac{1}{2}\right) \]
Calculating the values:
\[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}, \quad \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \]
Thus,
\[ f\left(-\frac{1}{2}\right) = -\frac{\pi}{6} + \frac{2\pi}{3} = -\frac{\pi}{6} + \frac{4\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]
So, Option A is valid.

#### Option B: \( f(k^2 - 2k + 3) \)

Next, we analyze \( k^2 - 2k + 3 \):
\[ k^2 - 2k + 3 = (k-1)^2 + 2 \]
This expression is always greater than or equal to 2 for all real \( k \). Since \( 2 > 1 \), this value does not lie in the interval \([-1, 1]\). Therefore, Option B is invalid.

#### Option C: \( f\left(\frac{1}{1+k^2}\right) \)

We check if \( \frac{1}{1+k^2} \) is in \([-1, 1]\):
\[ \frac{1}{1+k^2} > 0 \quad \text{and} \quad \frac{1}{1+k^2} < 1 \quad \text{for all } k \in \mathbb{R} \]
Thus, it lies in the interval \([-1, 1]\). Therefore, Option C is valid.

#### Option D: \( f(-2) \)

Since \(-2\) is not in the interval \([-1, 1]\), Option D is invalid.

### Conclusion

The valid options that equal \( \frac{\pi}{2} \) are:
- Option A: \( f\left(-\frac{1}{2}\right) = \frac{\pi}{2} \)
- Option C: \( f\left(\frac{1}{1+k^2}\right) = \frac{\pi}{2} \)

Thus, the final answer is that \( \frac{\pi}{2} \) is equal to options A and C.

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