Let f(x) be a continuous function for all `x in R and f'(0) =1` then `g(x) = f(|x|)-sqrt((1-cos 2x)/(2)), at x=0,`

Options

A. is differentiable at x=0 and its value is 1
B. is disfferentiable at x=0 and its value is 0
C. is non-differentaible at x=0 as its graph has sharp turn at x=o
D. is non - differentiable at x=0 as its graph has veritical tanent at x=0

Answer

To solve the problem, we need to determine the differentiability of the function \( g(x) = f(|x|) - \sqrt{\frac{1 - \cos(2x)}{2}} \) at \( x = 0 \), given that \( f(x) \) is continuous for all \( x \in \mathbb{R} \) and \( f'(0) = 1 \).

### Step 1: Simplify the function \( g(x) \)

We start with the expression for \( g(x) \):
\[ g(x) = f(|x|) - \sqrt{\frac{1 - \cos(2x)}{2}} \]

Using the trigonometric identity \( 1 - \cos(2x) = 2\sin^2(x) \), we can rewrite the square root term:
\[ g(x) = f(|x|) - \sqrt{\frac{2\sin^2(x)}{2}} = f(|x|) - |\sin(x)| \]

### Step 2: Evaluate \( g(0) \)

Now, we calculate \( g(0) \):
\[ g(0) = f(0) - |\sin(0)| = f(0} - 0 = f(0) \]

### Step 3: Calculate the Left-Hand Derivative (LHD) at \( x = 0 \)

The left-hand derivative at \( x = 0 \) is given by:
\[ g'(0^-) = \lim_{h \to 0^-} \frac{g(0 + h) - g(0)}{h} \]
Substituting \( g(0) \):
\[ g'(0^-) = \lim_{h \to 0^-} \frac{g(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{f(-h) - |\sin(h)| - f(0)}{h} \]
Since \( |h| = -h \) when \( h \) is negative:
\[ g'(0^-) = \lim_{h \to 0^-} \frac{f(-h) - |\sin(h)| - f(0)}{h} \]

### Step 4: Separate the terms in the limit

We can separate the terms in the limit:
\[ g'(0^-) = \lim_{h \to 0^-} \left( \frac{f(-h) - f(0)}{h} - \frac{|\sin(h)|}{h} \right) \]

### Step 5: Evaluate each limit

1. The first term, using the definition of the derivative:
\[ \lim_{h \to 0^-} \frac{f(-h) - f(0)}{h} = -f'(0) = -1 \quad \text{(since \( f'(0) = 1 \))} \]

2. The second term, using the standard limit:
\[ \lim_{h \to 0^-} \frac{|\sin(h)|}{h} = 1 \]

Thus, we have:
\[ g'(0^-) = -1 - 1 = -2 \]

### Step 6: Calculate the Right-Hand Derivative (RHD) at \( x = 0 \)

Now we calculate the right-hand derivative:
\[ g'(0^+) = \lim_{h \to 0^+} \frac{g(0 + h) - g(0)}{h} \]
Substituting \( g(0) \):
\[ g'(0^+) = \lim_{h \to 0^+} \frac{g(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h) - |\sin(h)| - f(0)}{h} \]

### Step 7: Separate the terms in the limit for RHD

Similar to LHD:
\[ g'(0^+) = \lim_{h \to 0^+} \left( \frac{f(h) - f(0)}{h} - \frac{|\sin(h)|}{h} \right) \]

### Step 8: Evaluate each limit for RHD

1. The first term:
\[ \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = f'(0) = 1 \]

2. The second term:
\[ \lim_{h \to 0^+} \frac{|\sin(h)|}{h} = 1 \]

Thus, we have:
\[ g'(0^+) = 1 - 1 = 0 \]

### Step 9: Conclusion on differentiability

Since \( g'(0^-) = -2 \) and \( g'(0^+) = 0 \), the left-hand derivative and right-hand derivative at \( x = 0 \) are not equal. Therefore, \( g(x) \) is not differentiable at \( x = 0 \).

### Final Answer

The function \( g(x) \) is not differentiable at \( x = 0 \).