Let `f(x)` be a non-positive continuous function and `F(x)=int_(0)^(x)f(t)dt AA x ge0` and `f(x) ge cF(x)` where `c lt 0` and let `g:[0, infty) to R` be a function such that `(dg(x))/(dx) lt g(x) AA x gt 0` and `g(0)=0`
The solution set of inequation `g(x)(cos^(-1)x-sin^(-1)) le0`

Options

A. `[-1,1/sqrt(2)]`
B. `[1/sqrt(2),1]`
C. `[0,1/sqrt(2)]`
D. `[0,1/sqrt(2)]`

Explanation

To solve the inequality \( g(x)(\cos^{-1} x - \sin^{-1} x) \leq 0 \), we will follow these steps:

### Step 1: Analyze the inequality
We start with the inequality:
\[ g(x)(\cos^{-1} x - \sin^{-1} x) \leq 0 \]
Since \( g(x) \) is a non-negative function (as \( g(0) = 0 \) and \( g(x) \) is continuous), we can analyze the expression \( \cos^{-1} x - \sin^{-1} x \).

### Step 2: Simplify the expression
Using the identity:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
we can rewrite:
\[ \cos^{-1} x - \sin^{-1} x = \cos^{-1} x - \left(\frac{\pi}{2} - \cos^{-1} x\right) = 2\cos^{-1} x - \frac{\pi}{2} \]
Thus, our inequality becomes:
\[ g(x) \left(2\cos^{-1} x - \frac{\pi}{2}\right) \leq 0 \]

### Step 3: Determine the critical points
The expression \( 2\cos^{-1} x - \frac{\pi}{2} = 0 \) gives us:
\[ 2\cos^{-1} x = \frac{\pi}{2} \implies \cos^{-1} x = \frac{\pi}{4} \implies x = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 4: Analyze the sign of the expression
Now we need to analyze the sign of \( g(x) \) and \( 2\cos^{-1} x - \frac{\pi}{2} \):
1. For \( x < \frac{1}{\sqrt{2}} \), \( \cos^{-1} x > \frac{\pi}{4} \) implies \( 2\cos^{-1} x > \frac{\pi}{2} \) and thus \( 2\cos^{-1} x - \frac{\pi}{2} > 0 \).
2. For \( x = \frac{1}{\sqrt{2}} \), \( 2\cos^{-1} x - \frac{\pi}{2} = 0 \).
3. For \( x > \frac{1}{\sqrt{2}} \), \( \cos^{-1} x < \frac{\pi}{4} \) implies \( 2\cos^{-1} x < \frac{\pi}{2} \) and thus \( 2\cos^{-1} x - \frac{\pi}{2} < 0 \).

### Step 5: Combine results
Since \( g(x) \) is non-negative and \( g(0) = 0 \), the inequality \( g(x)(2\cos^{-1} x - \frac{\pi}{2}) \leq 0 \) holds when:
- \( x \in [0, \frac{1}{\sqrt{2}}] \) where \( g(x) \geq 0 \) and \( 2\cos^{-1} x - \frac{\pi}{2} \geq 0 \)
- \( x > \frac{1}{\sqrt{2}} \) where \( g(x) \geq 0 \) and \( 2\cos^{-1} x - \frac{\pi}{2} < 0 \)

### Final Solution
Thus, the solution set of the inequality \( g(x)(\cos^{-1} x - \sin^{-1} x) \leq 0 \) is:
\[ x \in [0, \frac{1}{\sqrt{2}}] \]