The maximum value of `sec^(-1)((7-5(x^(2)+3))/(2(x^(2)+2)))` is:

Options

A. `(5pi)/(6)`
B. `(5pi)/(12)`
C. `(7pi)/(12)`
D. `(2pi)/(3)`

Answer

To find the maximum value of the function \( y = \sec^{-1}\left(\frac{7 - 5(x^2) + 3}{2(x^2) + 2}\right) \), we will follow these steps:

### Step 1: Simplify the expression inside the secant inverse
We start by simplifying the expression:
\[ y = \sec^{-1}\left(\frac{(7 + 3) - 5x^2}{2x^2 + 2}\right) = \sec^{-1}\left(\frac{10 - 5x^2}{2x^2 + 2}\right) \]

### Step 2: Rewrite the expression
Now, we can factor out constants from the numerator and denominator:
\[ y = \sec^{-1}\left(\frac{5(2 - x^2)}{2(x^2 + 1)}\right) \]

### Step 3: Set up the inequality for maximum value
To find the maximum value of \( y \), we need to ensure that the argument of the secant inverse function is valid. The secant function is defined for values greater than or equal to 1 or less than or equal to -1. Therefore, we need:
\[ \left|\frac{5(2 - x^2)}{2(x^2 + 1)}\right| \geq 1 \]

### Step 4: Solve the inequality
This leads us to two cases:
1. \( \frac{5(2 - x^2)}{2(x^2 + 1)} \geq 1 \)
2. \( \frac{5(2 - x^2)}{2(x^2 + 1)} \leq -1 \)

#### Case 1:
\[ 5(2 - x^2) \geq 2(x^2 + 1) \]
Expanding and rearranging gives:
\[ 10 - 5x^2 \geq 2x^2 + 2 \implies 10 - 2 \geq 7x^2 \implies 8 \geq 7x^2 \implies x^2 \leq \frac{8}{7} \]

#### Case 2:
\[ 5(2 - x^2) \leq -2(x^2 + 1) \]
Expanding and rearranging gives:
\[ 10 - 5x^2 \leq -2x^2 - 2 \implies 10 + 2 \leq 3x^2 \implies 12 \leq 3x^2 \implies x^2 \geq 4 \]

### Step 5: Analyze the results
From the two cases, we have:
1. \( x^2 \leq \frac{8}{7} \)
2. \( x^2 \geq 4 \)

Since \( x^2 \) cannot satisfy both conditions simultaneously, we focus on the first case for maximum values.

### Step 6: Find the maximum value of \( y \)
To find the maximum value of \( y \), we substitute \( x^2 = \frac{8}{7} \) into the expression:
\[ y = \sec^{-1}\left(\frac{5(2 - \frac{8}{7})}{2(\frac{8}{7} + 1)}\right) \]
Calculating this gives:
\[ y = \sec^{-1}\left(\frac{5\left(\frac{14 - 8}{7}\right)}{2\left(\frac{15}{7}\right)}\right) = \sec^{-1}\left(\frac{5\left(\frac{6}{7}\right)}{\frac{30}{7}}\right) = \sec^{-1}\left(\frac{30}{30}\right) = \sec^{-1}(1) \]
Thus, the maximum value of \( y \) is:
\[ y = 0 \]

### Conclusion
The maximum value of \( \sec^{-1}\left(\frac{7 - 5(x^2) + 3}{2(x^2) + 2}\right) \) is \( 0 \).