Which of the following functions are not derivable at `x=0?`

Options

A. `f(x)=sin^(-1)2xsqrt(1-x^(2))`
B. `g(x)=sin^(-1)((2^(x)+1)/(1+4^(x)))`
C. `h(x)=sin^(-1)((1-x^(2))/(1+x^(2)))`
D. `k(x)=sin^(-1)(cosx)`

Answer

To determine which of the given functions are not derivable at \( x = 0 \), we will evaluate each function one by one.

### Step-by-Step Solution:

1. **Function 1: \( f(x) = \sin^{-1}(2x \sqrt{1 - x^2}) \)**

- Substitute \( x = 0 \):
\[ f(0) = \sin^{-1}(2 \cdot 0 \cdot \sqrt{1 - 0^2}) = \sin^{-1}(0) = 0 \]
- The function is continuous at \( x = 0 \). To check if it's derivable, we need to check the derivative:
\[ f'(x) = \frac{d}{dx} \sin^{-1}(2x \sqrt{1 - x^2}) \]
- Since \( \sin^{-1}(x) \) is derivable wherever its argument is continuous and differentiable, we can conclude that \( f(x) \) is derivable at \( x = 0 \).

2. **Function 2: \( g(x) = \sin^{-1}\left(\frac{2^x + 1}{1 + 4^x}\right) \)**

- Substitute \( x = 0 \):
\[ g(0) = \sin^{-1}\left(\frac{2^0 + 1}{1 + 4^0}\right) = \sin^{-1}\left(\frac{1 + 1}{1 + 1}\right) = \sin^{-1}(1) \]
- The value \( \sin^{-1}(1) \) is \( \frac{\pi}{2} \). The function \( g(x) \) is continuous at \( x = 0 \). However, we need to check the derivative:
\[ g'(x) = \frac{d}{dx} \sin^{-1}\left(\frac{2^x + 1}{1 + 4^x}\right) \]
- The derivative will involve evaluating the limit as \( x \) approaches 0. Since \( \sin^{-1}(1) \) is at the boundary of the function's domain, \( g(x) \) is not derivable at \( x = 0 \).

3. **Function 3: \( h(x) = \sin^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \)**

- Substitute \( x = 0 \):
\[ h(0) = \sin^{-1}\left(\frac{1 - 0^2}{1 + 0^2}\right) = \sin^{-1}(1) = \frac{\pi}{2} \]
- Similar to \( g(x) \), \( h(x) \) is continuous at \( x = 0 \) but we need to check the derivative:
\[ h'(x) = \frac{d}{dx} \sin^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]
- Since \( \sin^{-1}(1) \) is again at the boundary, \( h(x) \) is not derivable at \( x = 0 \).

4. **Function 4: \( k(x) = \sin^{-1}(\cos x) \)**

- Substitute \( x = 0 \):
\[ k(0) = \sin^{-1}(\cos(0)) = \sin^{-1}(1) = \frac{\pi}{2} \]
- The function \( k(x) \) is continuous at \( x = 0 \). Checking the derivative:
\[ k'(x) = \frac{d}{dx} \sin^{-1}(\cos x) \]
- Since \( \sin^{-1}(1) \) is also at the boundary, \( k(x) \) is not derivable at \( x = 0 \).

### Conclusion:
The functions that are not derivable at \( x = 0 \) are:
- \( g(x) = \sin^{-1}\left(\frac{2^x + 1}{1 + 4^x}\right) \)
- \( h(x) = \sin^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \)
- \( k(x) = \sin^{-1}(\cos x) \)

### Final Answer:
The functions that are not derivable at \( x = 0 \) are options 2, 3, and 4.