Let the solution set of the inequation `n(sinx-1/2)(sinx-1/(sqrt(2)))lt=0 in [pi/2,pi]` be `A` and let solution set of equation `sin^-1(3x-4x^3)=3sin^-1x` be `B.` Now define a function `f:A->B.`

Options

A. (a)`(12x)/(pi)-(19)/(2)`
B. (b)`-(12x)/(pi)-(19)/(2)`
C. (c)`(12x)/(pi)+(19)/(2)`
D. (d)`12pi+(19)/(2)x `

Answer

To solve the given problem, we will break it down into two parts: finding the solution set \( A \) for the inequality and finding the solution set \( B \) for the equation. Then, we will define the function \( f: A \to B \).

### Step 1: Solve the Inequation
We need to solve the inequation:

\[ n(\sin x - \frac{1}{2})(\sin x - \frac{1}{\sqrt{2}}) < 0 \quad \text{for } x \in \left[\frac{\pi}{2}, \pi\right] \]

#### Step 1.1: Identify the critical points
The critical points occur when each factor is equal to zero:

1. \(\sin x - \frac{1}{2} = 0 \Rightarrow \sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}\) (not in the interval)
2. \(\sin x - \frac{1}{\sqrt{2}} = 0 \Rightarrow \sin x = \frac{1}{\sqrt{2}} \Rightarrow x = \frac{\pi}{4}\) (not in the interval)

Since both critical points are outside the interval \(\left[\frac{\pi}{2}, \pi\right]\), we will evaluate the sign of the expression at the endpoints of the interval.

#### Step 1.2: Evaluate the expression at the endpoints
- At \( x = \frac{\pi}{2} \):
\[ \sin\left(\frac{\pi}{2}\right) = 1 \Rightarrow n(1 - \frac{1}{2})(1 - \frac{1}{\sqrt{2}}) = n(\frac{1}{2})(1 - \frac{1}{\sqrt{2}}) \]
This is positive since \( n > 0 \).

- At \( x = \pi \):
\[ \sin(\pi) = 0 \Rightarrow n(0 - \frac{1}{2})(0 - \frac{1}{\sqrt{2}}) = n(-\frac{1}{2})(-\frac{1}{\sqrt{2}}) = n\frac{1}{2\sqrt{2}} \]
This is also positive.

#### Step 1.3: Determine the sign in the interval
Since both endpoints yield positive values and there are no critical points in the interval, the expression does not change sign. Therefore, the solution set \( A \) is empty:

\[ A = \emptyset \]

### Step 2: Solve the Equation
Now we solve the equation:

\[ \sin^{-1}(3x - 4x^3) = 3\sin^{-1}(x) \]

#### Step 2.1: Find the domain
The function \( \sin^{-1}(y) \) is defined for \( y \in [-1, 1] \). Therefore, we need:

\[ -1 \leq 3x - 4x^3 \leq 1 \]

#### Step 2.2: Solve \( 3x - 4x^3 = 1 \)
\[ 4x^3 - 3x + 1 = 0 \]

Using numerical methods or graphing, we can find the roots of this polynomial. Let's denote the roots as \( r_1, r_2, r_3 \).

#### Step 2.3: Solve \( 3x - 4x^3 = -1 \)
\[ 4x^3 - 3x - 1 = 0 \]

Again, using numerical methods or graphing, we can find the roots of this polynomial. Let's denote these roots as \( s_1, s_2, s_3 \).

#### Step 2.4: Combine the results
The solution set \( B \) will be the union of the intervals defined by the roots found in the previous steps.

### Step 3: Define the Function \( f: A \to B \)
Since \( A = \emptyset \), the function \( f \) is not defined for any element. Therefore, we can say:

\[ f: A \to B \quad \text{is not defined.} \]

### Summary
- The solution set \( A \) for the inequation is empty: \( A = \emptyset \).
- The solution set \( B \) is determined by solving the cubic equations.
- The function \( f: A \to B \) is not defined.