Consider the statements : P : There exists some x IR such that f(x) + 2x = 2(1+x2) Q : There exists some x IR such that 2f(x) +1 = 2x(1+x) `f(x)=(1-x)^(2) sin^(2) x+x^(2)" "AA x in R` Then (A) both P and Q are true (B) P is true and Q is false (C) P is false and Q is true (D) both P and Q are false.

Options

A. both P and Q are true
B. P is true and Q is false
C. P is false and Q is true
D. both P and Q are false

Explanation

To solve the problem, we need to analyze the statements P and Q based on the function \( f(x) = (1 - x^2) \sin^2 x + x^2 \).

### Step 1: Analyze Statement P
We need to check if there exists some \( x \in \mathbb{R} \) such that:
\[ f(x) + 2x = 2(1 + x^2) \]
Substituting \( f(x) \):
\[ (1 - x^2) \sin^2 x + x^2 + 2x = 2(1 + x^2) \]
This simplifies to:
\[ (1 - x^2) \sin^2 x + x^2 + 2x = 2 + 2x^2 \]
Rearranging gives:
\[ (1 - x^2) \sin^2 x = 2 + 2x^2 - x^2 - 2x \]
\[ (1 - x^2) \sin^2 x = 2 - x^2 - 2x \]
Now, we can analyze the left side:
\[ 1 - x^2 \geq 0 \implies x^2 \leq 1 \implies -1 \leq x \leq 1 \]
The right side \( 2 - x^2 - 2x \) is a quadratic function. We can find its vertex to determine its maximum or minimum value.

### Step 2: Analyze the Quadratic Function
The quadratic \( 2 - x^2 - 2x \) can be rewritten as:
\[ -(x^2 + 2x - 2) \]
The vertex occurs at \( x = -b/(2a) = -2/(-2) = 1 \). Evaluating at \( x = 1 \):
\[ 2 - 1 - 2 = -1 \]
Evaluating at \( x = -1 \):
\[ 2 - 1 + 2 = 3 \]
Thus, the maximum value of \( 2 - x^2 - 2x \) in the interval \( [-1, 1] \) is 3.

### Step 3: Check for Real Solutions
Now we need to check if \( (1 - x^2) \sin^2 x = 2 - x^2 - 2x \) can have real solutions. Since \( (1 - x^2) \sin^2 x \) can take values from 0 to 1 (as \( \sin^2 x \) ranges from 0 to 1), it cannot equal 2 or any value greater than 1. Therefore, statement P is false.

### Step 4: Analyze Statement Q
Now we check if there exists some \( x \in \mathbb{R} \) such that:
\[ 2f(x) + 1 = 2x(1 + x) \]
Substituting \( f(x) \):
\[ 2((1 - x^2) \sin^2 x + x^2) + 1 = 2x(1 + x) \]
This simplifies to:
\[ 2(1 - x^2) \sin^2 x + 2x^2 + 1 = 2x + 2x^2 \]
Rearranging gives:
\[ 2(1 - x^2) \sin^2 x + 1 = 2x \]
This implies:
\[ 2(1 - x^2) \sin^2 x = 2x - 1 \]
Thus, we can define:
\[ h(x) = \frac{2x - 1}{1 - x^2} - 2 \sin^2 x \]
We need to check if this function has a root.

### Step 5: Evaluate \( h(x) \)
Evaluate \( h(0) \):
\[ h(0) = \frac{2(0) - 1}{1 - 0^2} - 2 \sin^2(0) = -1 \]
As \( x \) approaches 1 from the left, \( h(x) \) approaches infinity. By the Intermediate Value Theorem, since \( h(0) < 0 \) and \( h(x) \to \infty \) as \( x \to 1 \), there exists at least one solution in the interval \( (0, 1) \).

### Conclusion
- Statement P is false.
- Statement Q is true.

Thus, the correct answer is:
**(C) P is false and Q is true.**