Let `f(x) = (1 - x)^2 sin^2 x + x^2 ` for all x ∈ R, and let `g(x) = ∫((2(t - 1))/(t + 1) - ln t)f(t)dt` for t ∈ [1, x] for all x ∈ (1, ∞).Which of the following is true ?

Options

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

Answer

To solve the problem step by step, we need to analyze the functions given and derive the necessary conclusions about \( g(x) \).

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

### Step 2: Analyze \( f(x) \)
We need to check if \( f(x) \) is always positive:
- The term \( (1 - x)^2 \sin^2 x \) is always non-negative since it is a square term multiplied by \( \sin^2 x \).
- The term \( x^2 \) is also always non-negative.
- Therefore, \( f(x) \geq 0 \) for all \( x \in \mathbb{R} \).

### Step 3: Define \( h(x) \)
Let’s define:
\[ h(x) = \frac{2(x - 1)}{x + 1} - \ln x \]
We will analyze the behavior of \( h(x) \) to understand the nature of \( g(x) \).

### Step 4: Find the derivative \( h'(x) \)
To find if \( h(x) \) is increasing or decreasing, we compute its derivative:
\[ h'(x) = \frac{d}{dx} \left( \frac{2(x - 1)}{x + 1} \right) - \frac{d}{dx} (\ln x) \]

Using the quotient rule for the first term:
\[ h'(x) = \frac{(2)(x + 1) - 2(x - 1)(1)}{(x + 1)^2} - \frac{1}{x} \]
\[ = \frac{2x + 2 - 2x + 2}{(x + 1)^2} - \frac{1}{x} \]
\[ = \frac{4}{(x + 1)^2} - \frac{1}{x} \]

### Step 5: Simplify \( h'(x) \)
To combine the terms, we find a common denominator:
\[ h'(x) = \frac{4x - (x + 1)^2}{x(x + 1)^2} \]
\[ = \frac{4x - (x^2 + 2x + 1)}{x(x + 1)^2} \]
\[ = \frac{4x - x^2 - 2x - 1}{x(x + 1)^2} \]
\[ = \frac{-x^2 + 2x - 1}{x(x + 1)^2} \]

### Step 6: Analyze the sign of \( h'(x) \)
The quadratic \( -x^2 + 2x - 1 \) can be factored or analyzed using the quadratic formula:
\[ = -(x^2 - 2x + 1) = -(x - 1)^2 \]
This is always less than or equal to zero for all \( x \).

### Step 7: Conclusion about \( h(x) \)
Since \( h'(x) \leq 0 \) for \( x > 1 \), it follows that \( h(x) \) is a decreasing function for \( x > 1 \).

### Step 8: Conclusion about \( g(x) \)
Since \( g(x) = \int_{1}^{x} h(t) f(t) \, dt \) and \( f(t) \geq 0 \), if \( h(x) < 0 \) for \( x > 1 \), then \( g(x) \) is decreasing.

### Final Answer
Thus, the correct statement is that \( g(x) \) is decreasing for \( x \in (1, \infty) \).

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