If `f:R rarr (0,oo)`be a differentiable function `f(x)` satisfying
`f(x+y)-f(x-y)=f(x)*{f(y)-f(y)-y}, AA x, y in R, (f(y)!= f(-y) " for all " y in R)` and `f'(0)=2010`.
Now, answer the following questions.
let `g(x)=log_(e)(sin x)`, and `int f(g(x))cos x dx=h(x)+c`, (where c is constant of integration), then `h(pi/2)` is equal to

Options

A. 0
B. `(1)/(2010)`
C. 1
D. `(1)/(2011)`

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the solution systematically.

### Step 1: Define the function \( g(x) \)
We are given that \( g(x) = \log_e(\sin x) \).

### Step 2: Set up the integral
We need to evaluate the integral:
\[ \int f(g(x)) \cos x \, dx = h(x) + c \]
Substituting \( g(x) \) into the integral, we have:
\[ \int f(\log_e(\sin x)) \cos x \, dx \]

### Step 3: Use the information about \( f \)
From the problem, we know that \( f'(0) = 2010 \). This implies that \( f(x) \) can be approximated around \( x = 0 \) as:
\[ f(x) \approx f(0) + f'(0)x = f(0) + 2010x \]
However, we will use the given functional equation to express \( f(g(x)) \).

### Step 4: Substitute \( f(g(x)) \)
Using the information provided and the fact that \( f(g(x)) \) can be expressed in terms of \( e^{2010 \log(\sin x)} \):
\[ f(g(x)) = e^{2010 \log(\sin x)} = (\sin x)^{2010} \]

### Step 5: Rewrite the integral
Now, we can rewrite the integral:
\[ \int f(g(x)) \cos x \, dx = \int (\sin x)^{2010} \cos x \, dx \]

### Step 6: Integrate
To integrate \( \int (\sin x)^{2010} \cos x \, dx \), we can use the substitution \( u = \sin x \), which gives \( du = \cos x \, dx \):
\[ \int (\sin x)^{2010} \cos x \, dx = \int u^{2010} \, du = \frac{u^{2011}}{2011} + C = \frac{(\sin x)^{2011}}{2011} + C \]

### Step 7: Define \( h(x) \)
From the integration, we can identify:
\[ h(x) = \frac{(\sin x)^{2011}}{2011} \]

### Step 8: Evaluate \( h\left(\frac{\pi}{2}\right) \)
Now, we need to find \( h\left(\frac{\pi}{2}\right) \):
\[ h\left(\frac{\pi}{2}\right) = \frac{(\sin(\frac{\pi}{2}))^{2011}}{2011} = \frac{1^{2011}}{2011} = \frac{1}{2011} \]

### Final Answer
Thus, the value of \( h\left(\frac{\pi}{2}\right) \) is:
\[ \boxed{\frac{1}{2011}} \]