Statement I: The equation `("sin"^-1x)^3 +("cos"^-1x)^3 -api^(3)=0` has a solution for all a> `(1)/(32)`
Statement II: for any X `epsilon R` ,`"sin"^-1x+"cos"^(-1)x=(pi)/(2)` and `0<("sin" x-(pi)/(4))^2 < (9pi^2)/(16)`

Options

A. Both statements I and II are true
B. Both statements I and II are False
C. Statement I is true and statement II is false.
D. Statement I is true and statement II is true.

Answer

To solve the problem, we need to analyze both statements provided and verify their correctness step by step.

### Step 1: Analyze Statement I
The equation given is:
\[ (\sin^{-1} x)^3 + (\cos^{-1} x)^3 - a \pi^3 = 0 \]
We can rewrite this as:
\[ (\sin^{-1} x)^3 + (\cos^{-1} x)^3 = a \pi^3 \]

### Step 2: Use the identity for inverse trigonometric functions
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Let \( y = \sin^{-1} x \). Then, we have:
\[ \cos^{-1} x = \frac{\pi}{2} - y \]

### Step 3: Substitute into the equation
Substituting into our equation gives:
\[ y^3 + \left(\frac{\pi}{2} - y\right)^3 = a \pi^3 \]

### Step 4: Expand the cube
Expanding the second term:
\[ \left(\frac{\pi}{2} - y\right)^3 = \frac{\pi^3}{8} - \frac{3\pi^2}{4}y + \frac{3\pi}{2}y^2 - y^3 \]
Thus, we can rewrite the equation as:
\[ y^3 + \left(\frac{\pi^3}{8} - \frac{3\pi^2}{4}y + \frac{3\pi}{2}y^2 - y^3\right) = a \pi^3 \]
This simplifies to:
\[ \frac{\pi^3}{8} + \frac{3\pi}{2}y^2 - \frac{3\pi^2}{4}y = a \pi^3 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \frac{3\pi}{2}y^2 - \frac{3\pi^2}{4}y + \left(\frac{\pi^3}{8} - a \pi^3\right) = 0 \]

### Step 6: Analyze the quadratic equation
This is a quadratic equation in \( y \). For this equation to have real solutions, the discriminant must be non-negative:
\[ D = \left(-\frac{3\pi^2}{4}\right)^2 - 4 \cdot \frac{3\pi}{2} \cdot \left(\frac{\pi^3}{8} - a \pi^3\right) \geq 0 \]

### Step 7: Calculate the discriminant
Calculating the discriminant:
\[ D = \frac{9\pi^4}{16} - 6\pi\left(\frac{\pi^3}{8} - a \pi^3\right) \]
This simplifies to:
\[ D = \frac{9\pi^4}{16} - \frac{3\pi^4}{4} + 6a\pi^4 \]
Combining terms gives:
\[ D = \left(6a - \frac{3}{4} + \frac{9}{16}\right)\pi^4 \]

### Step 8: Set the discriminant to be non-negative
For \( D \geq 0 \):
\[ 6a - \frac{3}{4} + \frac{9}{16} \geq 0 \]
Solving this gives:
\[ 6a \geq \frac{3}{4} - \frac{9}{16} \]
Calculating the right-hand side:
\[ \frac{3}{4} = \frac{12}{16} \Rightarrow \frac{12}{16} - \frac{9}{16} = \frac{3}{16} \]
Thus:
\[ 6a \geq \frac{3}{16} \Rightarrow a \geq \frac{1}{32} \]

### Conclusion for Statement I
Thus, Statement I is correct for \( a > \frac{1}{32} \).

### Step 9: Analyze Statement II
The statement is:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
This is always true for \( x \in [-1, 1] \).

### Step 10: Analyze the second part of Statement II
We need to check:
\[ 0 < \left(\sin x - \frac{\pi}{4}\right)^2 < \frac{9\pi^2}{16} \]
This implies:
\[ \sin x \neq \frac{\pi}{4} \]
and the maximum value of \( \sin x \) is \( 1 \) when \( x = \frac{\pi}{2} \).

### Conclusion for Statement II
Thus, Statement II is also correct.

### Final Conclusion
Both Statement I and Statement II are correct.