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)`[(3pi)/(4),(1)/(2)]`
B. (b)`[(5pi)/(6),3]`
C. (c)`[(3pi)/(4),3]`
D. (d)`(3,oo)`

Answer

To solve the problem step by step, we will first tackle the inequation and then the equation, followed by defining the function \( f: A \to B \).

### Step 1: Solve the Inequation

We need to solve the inequation:

\[ n(\sin x - \frac{1}{2})(\sin x - \frac{1}{\sqrt{2}}) < 0 \]

for \( x \) in the interval \( [\frac{\pi}{2}, \pi] \).

#### Step 1.1: Identify the critical points

The critical points occur when either factor is zero:

1. \( \sin x - \frac{1}{2} = 0 \)
- This gives \( \sin x = \frac{1}{2} \) which corresponds to \( x = \frac{\pi}{6} \) (not in the interval) and \( x = \frac{5\pi}{6} \) (in the interval).

2. \( \sin x - \frac{1}{\sqrt{2}} = 0 \)
- This gives \( \sin x = \frac{1}{\sqrt{2}} \) which corresponds to \( x = \frac{\pi}{4} \) (not in the interval) and \( x = \frac{3\pi}{4} \) (in the interval).

Thus, the critical points in the interval \( [\frac{\pi}{2}, \pi] \) are \( x = \frac{3\pi}{4} \) and \( x = \frac{5\pi}{6} \).

#### Step 1.2: Test the intervals

We will test the intervals determined by the critical points:

1. Interval \( [\frac{\pi}{2}, \frac{3\pi}{4}) \)
2. Interval \( [\frac{3\pi}{4}, \frac{5\pi}{6}) \)
3. Interval \( [\frac{5\pi}{6}, \pi] \)

- For \( x = \frac{\pi}{2} \):
\[ n(\sin(\frac{\pi}{2}) - \frac{1}{2})(\sin(\frac{\pi}{2}) - \frac{1}{\sqrt{2}}) = n(1 - \frac{1}{2})(1 - \frac{1}{\sqrt{2}}) > 0 \]

- For \( x = \frac{3\pi}{4} \):
\[ n(\sin(\frac{3\pi}{4}) - \frac{1}{2})(\sin(\frac{3\pi}{4}) - \frac{1}{\sqrt{2}}) = n(\frac{\sqrt{2}}{2} - \frac{1}{2})(\frac{\sqrt{2}}{2} - \frac{1}{\sqrt{2}}) < 0 \]

- For \( x = \frac{5\pi}{6} \):
\[ n(\sin(\frac{5\pi}{6}) - \frac{1}{2})(\sin(\frac{5\pi}{6}) - \frac{1}{\sqrt{2}}) = n(\frac{1}{2} - \frac{1}{2})(\frac{1}{2} - \frac{1}{\sqrt{2}}) = 0 \]

- For \( x = \pi \):
\[ n(\sin(\pi) - \frac{1}{2})(\sin(\pi) - \frac{1}{\sqrt{2}}) = n(0 - \frac{1}{2})(0 - \frac{1}{\sqrt{2}}) < 0 \]

### Conclusion for Step 1

The solution set \( A \) is:

\[ A = \left( \frac{3\pi}{4}, \frac{5\pi}{6} \right) \cup \{ \pi \} \]

### Step 2: Solve the Equation

Next, we solve the equation:

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

#### Step 2.1: Use the identity

Using the identity \( \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) \):

Let \( y = \sin^{-1}(x) \). Then \( x = \sin(y) \) and we can rewrite the equation as:

\[ 3\sin(y) - 4\sin^3(y) = \sin(3y) \]

This is satisfied when:

\[ 3\sin(y) - 4\sin^3(y) = 3\sin(y) \]

This simplifies to:

\[ -4\sin^3(y) = 0 \implies \sin^3(y) = 0 \implies \sin(y) = 0 \]

Thus, \( y = 0 \) which gives \( x = 0 \).

#### Step 2.2: Check the range

We need to check the range of \( x \):

The equation \( \sin^{-1}(3x - 4x^3) \) is defined for \( x \in [-1, 1] \).

### Conclusion for Step 2

The solution set \( B \) is:

\[ B = \{ 0 \} \]

### Step 3: Define the Function \( f: A \to B \)

Now we define the function \( f: A \to B \).

Since \( A \) contains values in the interval \( \left( \frac{3\pi}{4}, \frac{5\pi}{6} \right) \cup \{ \pi \} \) and \( B = \{ 0 \} \), we can define:

\[ f(x) = 0 \quad \text{for all } x \in A \]

### Final Conclusion

The function \( f: A \to B \) is defined as \( f(x) = 0 \) for all \( x \in A \).