If `f(x)=abs(1/(abs(absx-2))+1/(abs(absx-3)))` and `g(x)=sin^(-1)(2sqrtx)`, then f(x)=g(x) has

Options

A. no solution
B. less than 2 solutions
C. atleast 1 solution
D. none of these

Answer

To solve the equation \( f(x) = g(x) \) where \( f(x) = \left| \frac{1}{| |x| - 2 |} + \frac{1}{| |x| - 3 |} \right| \) and \( g(x) = \sin^{-1}(2\sqrt{x}) \), we will follow these steps:

### Step 1: Determine the domain of \( g(x) \)

The function \( g(x) = \sin^{-1}(2\sqrt{x}) \) is defined when the argument \( 2\sqrt{x} \) lies within the range \([-1, 1]\).

- This gives us the inequalities:
\[ -1 \leq 2\sqrt{x} \leq 1 \]
- From \( 2\sqrt{x} \geq 0 \), we have \( x \geq 0 \).
- From \( 2\sqrt{x} \leq 1 \), we have \( \sqrt{x} \leq \frac{1}{2} \) which implies \( x \leq \frac{1}{4} \).

Thus, the domain of \( g(x) \) is:
\[ x \in [0, \frac{1}{4}] \]

### Step 2: Analyze \( f(x) \)

Next, we simplify \( f(x) \):
\[ f(x) = \frac{1}{| |x| - 2 |} + \frac{1}{| |x| - 3 |} \]
Since \( |x| \) is always non-negative, we can analyze the function based on the critical points \( x = 2 \) and \( x = 3 \).

- For \( x < 2 \):
\[ f(x) = \frac{1}{2 - x} + \frac{1}{3 - x} \]
- For \( 2 \leq x < 3 \):
\[ f(x) = \frac{1}{x - 2} + \frac{1}{3 - x} \]
- For \( x \geq 3 \):
\[ f(x) = \frac{1}{x - 2} + \frac{1}{x - 3} \]

### Step 3: Evaluate \( f(x) \) at the endpoints of the domain

We will evaluate \( f(x) \) at the endpoints of the domain \( [0, \frac{1}{4}] \):

1. **At \( x = 0 \)**:
\[ f(0) = \frac{1}{2 - 0} + \frac{1}{3 - 0} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

2. **At \( x = \frac{1}{4} \)**:
\[ f\left(\frac{1}{4}\right) = \frac{1}{2 - \frac{1}{4}} + \frac{1}{3 - \frac{1}{4}} = \frac{1}{\frac{7}{4}} + \frac{1}{\frac{11}{4}} = \frac{4}{7} + \frac{4}{11} \]
To combine these fractions, find a common denominator:
\[ \frac{4}{7} + \frac{4}{11} = \frac{4 \cdot 11 + 4 \cdot 7}{77} = \frac{44 + 28}{77} = \frac{72}{77} \]

### Step 4: Compare \( f(x) \) and \( g(x) \)

Now, we need to find \( g(0) \) and \( g\left(\frac{1}{4}\right) \):
- **At \( x = 0 \)**:
\[ g(0) = \sin^{-1}(2 \cdot 0) = 0 \]
- **At \( x = \frac{1}{4} \)**:
\[ g\left(\frac{1}{4}\right) = \sin^{-1}(2 \cdot \sqrt{\frac{1}{4}}) = \sin^{-1}(1) = \frac{\pi}{2} \]

### Step 5: Analyze the intersection

- At \( x = 0 \), \( f(0) = \frac{5}{6} \) and \( g(0) = 0 \).
- At \( x = \frac{1}{4} \), \( f\left(\frac{1}{4}\right) = \frac{72}{77} \) and \( g\left(\frac{1}{4}\right) = \frac{\pi}{2} \).

Since \( f(x) \) is continuous and decreases from \( \frac{5}{6} \) to \( \frac{72}{77} \) while \( g(x) \) increases from \( 0 \) to \( \frac{\pi}{2} \), by the Intermediate Value Theorem, there is at least one intersection point in the interval \( [0, \frac{1}{4}] \).

### Conclusion

Thus, the equation \( f(x) = g(x) \) has at least one solution.