Let g(x) be a polynomial of degree one and f(x) be defined by `f(x)=-g(x), x<=0 and |x|^sinx, x>0` If f(x) is continuous satisfying `f'(1)=f(-1)`, then g(x) is

Options

A. (1+ sin1 )x+1
B. (1-sin 1)x + 1
C. (sin 1 - 1)x-1
D. none of these

Answer

To solve the problem, we need to find the polynomial \( g(x) \) of degree one, given the function \( f(x) \) defined piecewise and the condition that \( f'(1) = f(-1) \).

### Step-by-step Solution:

1. **Define \( g(x) \)**: Since \( g(x) \) is a polynomial of degree one, we can express it as:
\[ g(x) = ax + b \]

2. **Define \( f(x) \)**: The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} -g(x) & \text{if } x \leq 0 \\ |x|^{\sin x} & \text{if } x > 0 \end{cases} \]
For \( x > 0 \), since \( |x| = x \), we have:
\[ f(x) = x^{\sin x} \]

3. **Check continuity at \( x = 0 \)**: For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ f(0) = \lim_{x \to 0^+} f(x) \]
We calculate \( f(0) \):
\[ f(0) = -g(0) = -b \]
Now, we find \( \lim_{x \to 0^+} f(x) \):
\[ \lim_{x \to 0^+} x^{\sin x} = \lim_{x \to 0^+} e^{\sin x \ln x} \]
We need to evaluate \( \lim_{x \to 0^+} \sin x \ln x \). Using L'Hôpital's Rule:
\[ \lim_{x \to 0^+} \sin x \ln x = \lim_{x \to 0^+} \frac{\sin x}{1/\ln x} \]
Both the numerator and denominator approach 0, so we differentiate:
\[ \lim_{x \to 0^+} \frac{\cos x}{-\frac{1}{x \ln^2 x}} = \lim_{x \to 0^+} -x \cos x \ln^2 x \]
As \( x \to 0^+ \), this limit approaches 0. Therefore:
\[ \lim_{x \to 0^+} x^{\sin x} = e^0 = 1 \]
Setting the limits equal for continuity:
\[ -b = 1 \implies b = -1 \]

4. **Differentiate \( f(x) \)**: We need to find \( f'(1) \):
\[ f'(x) = \frac{d}{dx}(x^{\sin x}) \text{ for } x > 0 \]
Using the product rule:
\[ f'(x) = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \ln x \right) \]
Evaluating at \( x = 1 \):
\[ f'(1) = 1^{\sin 1} \left( \frac{\sin 1}{1} + \cos 1 \cdot 0 \right) = \sin 1 \]

5. **Calculate \( f(-1) \)**: We find \( f(-1) \):
\[ f(-1) = -g(-1) = -(-a + b) = a - b \]
Substituting \( b = -1 \):
\[ f(-1) = a - (-1) = a + 1 \]

6. **Set the condition \( f'(1) = f(-1) \)**:
\[ \sin 1 = a + 1 \implies a = \sin 1 - 1 \]

7. **Final expression for \( g(x) \)**:
\[ g(x) = ax + b = (\sin 1 - 1)x - 1 \]

### Conclusion:
Thus, the polynomial \( g(x) \) is:
\[ g(x) = (\sin 1 - 1)x - 1 \]