`(cos^(- 1)x+sec^(- 1)x+tan^(- 1)x+cot^(- 1)x)_(max)+6((sin^(- 1)x)^2+pi sin^-1 x)_(min)` equals

Options

A. (a)`(5pi^2+3pi)/2`
B. (b)`(pi^2+3pi)/2`
C. (c)`(5pi-3pi^2)/2`
D. (d)`(pi+3pi^3)/2`

Answer

To solve the problem, we need to evaluate the expression:

\[ (\cos^{-1}x + \sec^{-1}x + \tan^{-1}x + \cot^{-1}x)_{\text{max}} + 6((\sin^{-1}x)^2 + \pi \sin^{-1}x)_{\text{min}} \]

### Step 1: Evaluate \((\cos^{-1}x + \sec^{-1}x + \tan^{-1}x + \cot^{-1}x)_{\text{max}}\)

Let \( t = \cos^{-1}x + \sec^{-1}x + \tan^{-1}x + \cot^{-1}x \).

Using the identity \(\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}\), we can rewrite \( t \):

\[ t = \cos^{-1}x + \sec^{-1}x + \frac{\pi}{2} \]

### Step 2: Determine the domain for \( t \)

The domain for \(\cos^{-1}x\) is \(x \in [-1, 1]\).

- At \(x = -1\):
\[ \cos^{-1}(-1) = \pi, \quad \sec^{-1}(-1) = \pi \]
Thus,
\[ t = \pi + \pi + \frac{\pi}{2} = \frac{5\pi}{2} \]

- At \(x = 1\):
\[ \cos^{-1}(1) = 0, \quad \sec^{-1}(1) = 0 \]
Thus,
\[ t = 0 + 0 + \frac{\pi}{2} = \frac{\pi}{2} \]

### Step 3: Find the maximum value of \( t \)

From the evaluations:
- The maximum value occurs at \(x = -1\):
\[ t_{\text{max}} = \frac{5\pi}{2} \]

### Step 4: Evaluate \(6((\sin^{-1}x)^2 + \pi \sin^{-1}x)_{\text{min}}\)

Let \( v = 6((\sin^{-1}x)^2 + \pi \sin^{-1}x) \).

### Step 5: Determine the domain for \( \sin^{-1}x \)

The domain for \(\sin^{-1}x\) is \(x \in [-1, 1]\).

Let \(y = \sin^{-1}x\), then \(y \in [-\frac{\pi}{2}, \frac{\pi}{2}]\).

### Step 6: Find the minimum of \( v = 6(y^2 + \pi y) \)

To find the minimum, we differentiate \(v\) with respect to \(y\):

\[ \frac{dv}{dy} = 6(2y + \pi) \]

Setting the derivative to zero for critical points:

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

### Step 7: Evaluate \(v\) at the critical point

Substituting \(y = -\frac{\pi}{2}\):

\[ v = 6\left(\left(-\frac{\pi}{2}\right)^2 + \pi\left(-\frac{\pi}{2}\right)\right) \]
\[ = 6\left(\frac{\pi^2}{4} - \frac{\pi^2}{2}\right) = 6\left(\frac{\pi^2}{4} - \frac{2\pi^2}{4}\right) = 6\left(-\frac{\pi^2}{4}\right) = -\frac{3\pi^2}{2} \]

### Step 8: Combine results

Now we combine the maximum and minimum values:

\[ \text{Final Result} = t_{\text{max}} + v_{\text{min}} = \frac{5\pi}{2} - \frac{3\pi^2}{2} \]

### Final Answer

Thus, the final answer is:

\[ \frac{5\pi - 3\pi^2}{2} \]