If `f(x)=sin^(2)x, g(x)=sqrtx and h(x)=cos^(-1)x, 0 le xle 1,` then

Options

A. hogof=fogoh
B. gofoh=fohog
C. fohog=hogof
D. none of these

Answer

To solve the problem, we need to evaluate the functions \( f(x) = \sin^2 x \), \( g(x) = \sqrt{x} \), and \( h(x) = \cos^{-1} x \) for \( 0 \leq x \leq 1 \) and check the validity of the given options step by step.

### Step 1: Evaluate \( h(g(f(x))) \)

1. **Calculate \( f(x) \)**:
\[ f(x) = \sin^2 x \]

2. **Calculate \( g(f(x)) \)**:
\[ g(f(x)) = g(\sin^2 x) = \sqrt{\sin^2 x} = |\sin x| \]
Since \( \sin x \) is non-negative in the interval \( [0, 1] \), we have:
\[ g(f(x)) = \sin x \]

3. **Calculate \( h(g(f(x))) \)**:
\[ h(g(f(x))) = h(\sin x) = \cos^{-1}(\sin x) \]
Using the identity \( \cos^{-1}(\sin x) = \frac{\pi}{2} - x \), we get:
\[ h(g(f(x))) = \frac{\pi}{2} - x \]

### Step 2: Evaluate \( f(g(h(x))) \)

1. **Calculate \( h(x) \)**:
\[ h(x) = \cos^{-1} x \]

2. **Calculate \( g(h(x)) \)**:
\[ g(h(x)) = g(\cos^{-1} x) = \sqrt{\cos^{-1} x} \]

3. **Calculate \( f(g(h(x))) \)**:
\[ f(g(h(x))) = f(\sqrt{\cos^{-1} x}) = \sin^2(\sqrt{\cos^{-1} x}) \]

### Step 3: Compare the results

Now we have:
- \( h(g(f(x))) = \frac{\pi}{2} - x \)
- \( f(g(h(x))) = \sin^2(\sqrt{\cos^{-1} x}) \)

Since \( \frac{\pi}{2} - x \) is a linear function and \( \sin^2(\sqrt{\cos^{-1} x}) \) is a non-linear function, they cannot be equal.

### Conclusion

None of the options provided in the question are correct based on our calculations.

### Final Answer
The correct answer is **Option D: None of these**.

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