If the maximum possible value of `(sin^(- 1)x)^2+(tan^(- 1)y)^2+pisin^(- 1)x+pitan^(- 1)y+(pi^2)/2` is k, and [] represents the greatest integer function then the value of `[k/(2pi^2)]` is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the maximum possible value of the expression:

\[ f(x, y) = (\sin^{-1} x)^2 + (\tan^{-1} y)^2 + \pi \sin^{-1} x + \pi \tan^{-1} y + \frac{\pi^2}{2} \]

### Step 1: Understanding the Ranges
First, we need to determine the ranges of \(\sin^{-1} x\) and \(\tan^{-1} y\):
- The range of \(\sin^{-1} x\) is \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
- The range of \(\tan^{-1} y\) is also \([- \frac{\pi}{2}, \frac{\pi}{2}]\).

### Step 2: Rewrite the Function
We can rewrite the function \(f\) as follows:

\[ f(x, y) = \left(\sin^{-1} x\right)^2 + \pi \sin^{-1} x + \left(\tan^{-1} y\right)^2 + \pi \tan^{-1} y + \frac{\pi^2}{2} \]

### Step 3: Maximizing Each Component
To maximize \(f\), we need to maximize the two components separately:
1. Let \(u = \sin^{-1} x\) and \(v = \tan^{-1} y\). Then we need to maximize:

\[ g(u) = u^2 + \pi u \quad \text{for } u \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

2. Similarly, we need to maximize:

\[ h(v) = v^2 + \pi v \quad \text{for } v \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

### Step 4: Finding the Maximum of \(g(u)\)
To find the maximum of \(g(u)\), we can differentiate:

\[ g'(u) = 2u + \pi \]

Setting \(g'(u) = 0\):

\[ 2u + \pi = 0 \implies u = -\frac{\pi}{2} \]

Now we check the endpoints:
- At \(u = -\frac{\pi}{2}\):

\[ g\left(-\frac{\pi}{2}\right) = \left(-\frac{\pi}{2}\right)^2 + \pi \left(-\frac{\pi}{2}\right) = \frac{\pi^2}{4} - \frac{\pi^2}{2} = -\frac{\pi^2}{4} \]

- At \(u = \frac{\pi}{2}\):

\[ g\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 + \pi \left(\frac{\pi}{2}\right) = \frac{\pi^2}{4} + \frac{\pi^2}{2} = \frac{3\pi^2}{4} \]

Thus, the maximum value of \(g(u)\) is \(\frac{3\pi^2}{4}\).

### Step 5: Finding the Maximum of \(h(v)\)
Using the same method for \(h(v)\):

\[ h'(v) = 2v + \pi \]

Setting \(h'(v) = 0\):

\[ 2v + \pi = 0 \implies v = -\frac{\pi}{2} \]

Checking the endpoints:
- At \(v = -\frac{\pi}{2}\):

\[ h\left(-\frac{\pi}{2}\right) = -\frac{\pi^2}{4} \]

- At \(v = \frac{\pi}{2}\):

\[ h\left(\frac{\pi}{2}\right) = \frac{3\pi^2}{4} \]

Thus, the maximum value of \(h(v)\) is also \(\frac{3\pi^2}{4}\).

### Step 6: Combine the Maximum Values
Now, we combine the maximum values of \(g(u)\) and \(h(v)\):

\[ f_{\text{max}} = g(u) + h(v) + \frac{\pi^2}{2} = \frac{3\pi^2}{4} + \frac{3\pi^2}{4} + \frac{\pi^2}{2} \]

Calculating this:

\[ f_{\text{max}} = \frac{3\pi^2}{4} + \frac{3\pi^2}{4} + \frac{2\pi^2}{4} = \frac{8\pi^2}{4} = 2\pi^2 \]

### Step 7: Determine \(k\) and the Greatest Integer Function
Thus, the maximum value \(k\) is \(2\pi^2\). We need to find:

\[ \left\lfloor \frac{k}{2\pi^2} \right\rfloor = \left\lfloor \frac{2\pi^2}{2\pi^2} \right\rfloor = \left\lfloor 1 \right\rfloor = 1 \]

### Final Answer
The value of \(\left\lfloor \frac{k}{2\pi^2} \right\rfloor\) is:

\[ \boxed{1} \]