1. Problem P1 states: For all positive real numbers $x$, prove that $\sqrt{x} \leq x$. Step 1: Note that $x > 0$. Consider two cases: - If $0 < x \leq 1$, then $\sqrt{x} \leq 1$ and $x \leq 1$, and since $\sqrt{x} = x^{1/2}$, and $x < 1$, we have $x^{1/2} > x$, so $\sqrt{x} > x$ in this case, so P1 is false in general. - If $x \geq 1$, then $\sqrt{x} = x^{1/2} \leq x$ since $x^{1/2 - 1} = x^{-1/2} \leq 1$. Thus, P1 is false for all $x \in \mathbb{R}^+$. 2. Problem P2: There exists $x \in \mathbb{R}$ such that $\cos x = 0$ and $\sin x = 0$. Step 1: Recall that for all real numbers $x$, $\sin^2 x + \cos^2 x = 1$. Step 2: If $\cos x = 0$ and $\sin x = 0$, then $0^2 + 0^2 = 0 \neq 1$. Conclusion: No such real $x$ exists. P2 is false. 3. Problem P3: There exists $x \in \mathbb{R}$ such that $\cos x = 0$, and there exists $x \in \mathbb{R}$ such that $\sin x = 0$. Step 1: Recall $\cos x = 0$ when $x = \frac{\pi}{2} + k\pi$, $k \in \mathbb{Z}$. Step 2: Recall $\sin x = 0$ when $x = k\pi$, $k \in \mathbb{Z}$. Step 3: Therefore, such $x$ exist separately, so P3 is true. 4. Problem P4: For all $x \in \mathbb{R}$, prove $\sqrt{x^2 + 2x + 2} > x + 1$. Step 1: Simplify the radicand: $x^2 + 2x + 2 = (x+1)^2 + 1$. Step 2: Then $\sqrt{(x+1)^2 + 1} > x + 1$. Step 3: Since $\sqrt{(x+1)^2 + 1} > |x+1|$, and $|x+1| \geq x + 1$ only if $x + 1 \geq 0$, but we must compare carefully. New approach: Square both sides (both sides $\geq 0$ or ensure inequality direction): $$ (x+1)^2 + 1 > (x+1)^2 \implies 1 > 0$$ True for all $x$. Also, note $\sqrt{(x+1)^2 + 1} \geq |x+1|$, and since $x+1 \leq |x+1|$, we have: $$ \sqrt{x^2 + 2x + 2} = \sqrt{(x+1)^2 + 1} > |x+1| \geq x + 1$$ Thus, P4 is true. 5. Problem P5: There exists positive integer $n$ such that for all real $x$, $\frac{x^{2n}}{1 + x^2} > 4$. Step 1: For $x=0$, the expression is 0, which is not greater than 4. Therefore, no such $n$ exists for the given statement. P5 is false. 6. Problem P6: For all $x \in \mathbb{R}$, there exists $y \in \mathbb{R}$ such that $x^2 + y^2 + xy + 3 = 0$. Step 1: For fixed $x$, consider the quadratic equation in $y$: $$ y^2 + x y + (x^2 + 3) = 0 $$ Step 2: The discriminant $\Delta$ must be nonnegative for real $y$: $$ \Delta = x^2 - 4(x^2 + 3) = x^2 - 4x^2 - 12 = -3x^2 - 12 $$ Step 3: Since $\Delta = -3x^2 - 12 < 0$ for all real $x$, no real $y$ satisfies the equation. Therefore, P6 is false. Final answers: - P1 is false. - P2 is false. - P3 is true. - P4 is true. - P5 is false. - P6 is false.