Let `f(x) = (1-x)^(2) sin^(2)x+ x^(2)` for all `x in IR` and let `g(x) = int_(1)^(x)((2(t-1))/(t+1)-lnt) f(t)` dt for all `x in (1,oo)`.
Which of the following is true ?

Options

A. g is increasing on `(1,oo)`
B. g is decreasing on `(1,oo)`
C. g is increasing on `(1,2)` and decreasing on `(2,oo)`
D. g is increasing on `(1,2)` and decreasing on `(2,oo)`

Answer

To solve the problem, we need to analyze the functions given and determine the behavior of \( g(x) \). Here is the step-by-step solution:

### Step 1: Define the functions
We have:
\[ f(x) = (1-x)^2 \sin^2 x + x^2 \]
for all \( x \in \mathbb{R} \), and
\[ g(x) = \int_{1}^{x} \left( \frac{2(t-1)}{t+1} - \ln t \right) f(t) \, dt \]
for all \( x \in (1, \infty) \).

### Step 2: Differentiate \( g(x) \)
Using the Fundamental Theorem of Calculus, we differentiate \( g(x) \):
\[ g'(x) = \left( \frac{2(x-1)}{x+1} - \ln x \right) f(x) \]

### Step 3: Analyze \( f(x) \)
Next, we need to analyze \( f(x) \):
\[ f(x) = (1-x)^2 \sin^2 x + x^2 \]
Since \( (1-x)^2 \sin^2 x \geq 0 \) and \( x^2 \geq 0 \), we conclude that \( f(x) \geq 0 \) for all \( x \).

### Step 4: Analyze \( g'(x) \)
Now, we need to analyze the expression for \( g'(x) \):
\[ g'(x) = \left( \frac{2(x-1)}{x+1} - \ln x \right) f(x) \]
Since \( f(x) \geq 0 \), the sign of \( g'(x) \) depends on the term \( \frac{2(x-1)}{x+1} - \ln x \).

### Step 5: Define \( h(x) \)
Let:
\[ h(x) = \frac{2(x-1)}{x+1} - \ln x \]

### Step 6: Differentiate \( h(x) \)
Now we differentiate \( h(x) \):
\[ h'(x) = \frac{4}{(x+1)^2} - \frac{1}{x} \]
To analyze \( h'(x) \), we can find a common denominator:
\[ h'(x) = \frac{4 - (x+1)}{x(x+1)^2} = \frac{3 - x}{x(x+1)^2} \]

### Step 7: Determine the sign of \( h'(x) \)
- For \( x > 3 \), \( h'(x) < 0 \) (function is decreasing).
- For \( 1 < x < 3 \), \( h'(x) > 0 \) (function is increasing).

### Step 8: Evaluate \( h(1) \)
Now, we evaluate \( h(1) \):
\[ h(1) = \frac{2(1-1)}{1+1} - \ln 1 = 0 - 0 = 0 \]

### Step 9: Conclusion about \( h(x) \)
Since \( h(1) = 0 \) and \( h(x) \) is increasing in \( (1, 3) \) and decreasing in \( (3, \infty) \), we conclude that \( h(x) < 0 \) for \( x > 3 \).

### Step 10: Conclusion about \( g(x) \)
Since \( g'(x) = h(x) f(x) \) and \( f(x) > 0 \), we conclude that \( g'(x) < 0 \) for \( x > 3 \). Thus, \( g(x) \) is decreasing for \( x > 1 \).

### Final Answer
The correct option is that \( g(x) \) is decreasing for \( x > 1 \).

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