The area of region bounded by the lines y=x,y=0 and `x=sin^-1(a^4+1)+cos^-1(a^4+1)-tan^-1(a^4+1)` is

Options

A. `(pi)/(8) - (a^(2))/(4)`
B. `(pi^(2))/(8) - (a^(2))/(2)`
C. `(pi^(2))/(16)`
D. `(pi^(2))/(32)`

Explanation

To find the area of the region bounded by the lines \(y = x\), \(y = 0\), and \(x = \sin^{-1}(a^4 + 1) + \cos^{-1}(a^4 + 1) - \tan^{-1}(a^4 + 1)\), we will follow these steps:

### Step 1: Determine the domain of the expression
The expression \(a^4 + 1\) must satisfy the condition for the inverse sine function, which means:
\[ a^4 + 1 \leq 1 \]
This implies:
\[ a^4 \leq 0 \]
Since \(a^4\) is a non-negative quantity (as it is raised to an even power), the only solution is:
\[ a^4 = 0 \implies a = 0 \]

### Step 2: Substitute \(a = 0\) into the expression
Now, substituting \(a = 0\) into the expression gives:
\[ x = \sin^{-1}(0 + 1) + \cos^{-1}(0 + 1) - \tan^{-1}(0 + 1) \]
Calculating each term:
- \(\sin^{-1}(1) = \frac{\pi}{2}\)
- \(\cos^{-1}(1) = 0\)
- \(\tan^{-1}(1) = \frac{\pi}{4}\)

Thus, we have:
\[ x = \frac{\pi}{2} + 0 - \frac{\pi}{4} = \frac{\pi}{4} \]

### Step 3: Identify the bounded region
The lines that bound the region are:
1. \(y = x\)
2. \(y = 0\) (the x-axis)
3. \(x = \frac{\pi}{4}\)

### Step 4: Sketch the region
The region bounded by these lines forms a right triangle with vertices at:
- Origin \(O(0, 0)\)
- Point \(A\left(\frac{\pi}{4}, 0\right)\)
- Point \(B\left(\frac{\pi}{4}, \frac{\pi}{4}\right)\)

### Step 5: Calculate the area of the triangle
The area \(A\) of triangle \(OAB\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, both the base and height are equal to \(\frac{\pi}{4}\):
\[ \text{Area} = \frac{1}{2} \times \frac{\pi}{4} \times \frac{\pi}{4} = \frac{1}{2} \times \frac{\pi^2}{16} = \frac{\pi^2}{32} \]

### Conclusion
The area of the region bounded by the lines is:
\[ \frac{\pi^2}{32} \]

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