Existence of Real Roots for Equations Involving Trigonometric and Polynomial Function

To determine the truth value of statements $\mathbf{P}$ and $\mathbf{Q}$, we analyze each statement individually.

The given function is: $f(x) = (1 - x)^2 \sin^2 x + x^2, \quad \forall x \in \mathbb{R}$

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Analysis of Statement $\mathbf{P}$:

The equation given in statement $\mathbf{P}$ is: $f(x) + 2x = 2(1 + x^2)$

Substitute the definition of $f(x)$ into the equation: $(1 - x)^2 \sin^2 x + x^2 + 2x = 2 + 2x^2$

Rearranging the terms: $(1 - x)^2 \sin^2 x = x^2 - 2x + 2$ $(x - 1)^2 \sin^2 x = (x - 1)^2 + 1$ $(x - 1)^2 (\sin^2 x - 1) = 1$ $-(x - 1)^2 \cos^2 x = 1$ $(x - 1)^2 \cos^2 x = -1$

For any real number $x$, $(x - 1)^2 \ge 0$ and $\cos^2 x \ge 0$, which implies that the left-hand side is always non-negative: $(x - 1)^2 \cos^2 x \ge 0$

Since the right-hand side is $-1 < 0$, there exists no real value of $x$ satisfying this equation.

Hence, Statement $\mathbf{P}$ is FALSE.

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Analysis of Statement $\mathbf{Q}$:

The equation given in statement $\mathbf{Q}$ is: $2f(x) + 1 = 2x(1 + x)$

Substitute the definition of $f(x)$ into the equation: $2\left((1 - x)^2 \sin^2 x + x^2\right) + 1 = 2x + 2x^2$ $2(1 - x)^2 \sin^2 x + 2x^2 + 1 = 2x + 2x^2$ $2(1 - x)^2 \sin^2 x + 1 - 2x = 0$

Define a continuous function $h(x)$ on $\mathbb{R}$ as: $h(x) = 2(1 - x)^2 \sin^2 x + 1 - 2x$

Let us evaluate $h(x)$ at $x = 0$ and $x = 1$:

• For $x = 0$: $h(0) = 2(1 - 0)^2 \sin^2(0) + 1 - 2(0) = 0 + 1 - 0 = 1 > 0$
• For $x = 1$: $h(1) = 2(1 - 1)^2 \sin^2(1) + 1 - 2(1) = 0 + 1 - 2 = -1 < 0$

Since $h(x)$ is continuous on the closed interval $[0, 1]$ and $h(0) \cdot h(1) < 0$, by the Intermediate Value Theorem, there exists at least one $c \in (0, 1)$ such that $h(c) = 0$.

Hence, there exists some $x \in \mathbb{R}$ such that $2f(x) + 1 = 2x(1 + x)$, which means Statement $\mathbf{Q}$ is TRUE.

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Conclusion:

$\mathbf{P}$ is false and $\mathbf{Q}$ is true, which corresponds to Option (C).