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 \( \sec^{-1}\left(\frac{7 - 5(x^2 + 3)}{2(x^2 + 2)}\right) \), we will follow these steps:

### Step 1: Simplify the expression inside the secant inverse function

We start with the expression:

\[ \sec^{-1}\left(\frac{7 - 5(x^2 + 3)}{2(x^2 + 2)}\right) \]

This can be rewritten as:

\[ \sec^{-1}\left(\frac{7 - 5x^2 - 15}{2x^2 + 4}\right) = \sec^{-1}\left(\frac{-5x^2 - 8}{2x^2 + 4}\right) \]

### Step 2: Analyze the expression

To find the maximum value of \( \sec^{-1}(y) \), we need to find the minimum value of \( y \) since the secant function is decreasing in the range where it is defined.

We need to analyze:

\[ y = \frac{-5x^2 - 8}{2x^2 + 4} \]

### Step 3: Find the critical points

To find the critical points, we will differentiate \( y \) with respect to \( x \) and set the derivative to zero.

Let \( y = \frac{-5x^2 - 8}{2x^2 + 4} \).

Using the quotient rule:

\[ y' = \frac{(2x^2 + 4)(-10x) - (-5x^2 - 8)(4x)}{(2x^2 + 4)^2} \]

Setting the numerator equal to zero:

\[ (2x^2 + 4)(-10x) + (5x^2 + 8)(4x) = 0 \]

### Step 4: Solve the equation

This simplifies to:

\[ -20x^3 - 40x + 20x^3 + 32x = 0 \]

Combining like terms:

\[ -8x + 32 = 0 \implies x = 4 \]

### Step 5: Evaluate \( y \) at critical points and endpoints

We will evaluate \( y \) at \( x = 4 \):

\[ y = \frac{-5(4^2) - 8}{2(4^2) + 4} = \frac{-80 - 8}{32 + 4} = \frac{-88}{36} = -\frac{22}{9} \]

### Step 6: Find the maximum value of \( \sec^{-1}(y) \)

The maximum value of \( \sec^{-1}(y) \) occurs when \( y \) is at its minimum. Since \( y \) can approach values close to -2, we find:

\[ \sec^{-1}(-2) = \frac{2\pi}{3} \]

### Conclusion

Thus, the maximum value of \( \sec^{-1}\left(\frac{7 - 5(x^2 + 3)}{2(x^2 + 2)}\right) \) is:

\[ \frac{2\pi}{3} \]