1. **Statement of Problem**: Given two propositions $P$ and $Q$: - $(P): \exists x \in \mathbb{R} : x^2 + x - 2 = 0$ - $(Q): \forall x \in \mathbb{R} : x^2 + x - 2 = 0 \Rightarrow x = 1$ Verify $P$, analyze $Q$, then solve given equations and prove inequalities. 2. **Check proposition P**: Solve the quadratic equation: $$x^2 + x - 2 = 0$$ Using the discriminant: $$\Delta = 1^2 - 4 \times 1 \times (-2) = 1 + 8 = 9$$ Roots: $$x = \frac{-1 \pm \sqrt{9}}{2} = \frac{-1 \pm 3}{2}$$ So, $$x_1 = 1, \quad x_2 = -2$$ Since solutions exist, $P$ is **true**. 3. **Analyze Q and deduce it is false**: $Q$ claims **all** real $x$ satisfying $x^2+x-2=0$ equal 1. But from step 2, $x=-2$ is also a root. Therefore, $Q$ is **false**. 4. **Solve equation using cases**: $$x^2 + |x-1| - 1 = 0$$ **Case 1: $x \geq 1$** Then $|x-1|=x-1$. Equation becomes: $$x^2 + x - 1 -1 = 0 \Rightarrow x^2 + x - 2 = 0$$ Roots from step 2: $x=1$ or $x=-2$. Only $x=1 \geq 1$ valid. **Case 2: $x < 1$** Then $|x-1|=1-x$. Equation becomes: $$x^2 + 1 - x - 1 = 0 \Rightarrow x^2 - x = 0$$ Factor: $$x(x-1)=0$$ Roots: $$x=0,\, x=1$$ Only $x=0 <1$ valid. **Solution set:** $\boxed{\{0,1\}}$. 5. **Show $\forall x \in \mathbb{R}: \frac{x^2 +3}{x^2 +4} \neq 1$ by contradiction**: Assume: $$\frac{x^2 + 3}{x^2 + 4} = 1$$ Multiply both sides: $$x^2 + 3 = x^2 + 4 \Rightarrow 3 = 4$$ Contradiction! Hence $$\boxed{\forall x \in \mathbb{R}, \frac{x^2 + 3}{x^2 + 4} \neq 1}$$ 6. **Prove the implication:** $$\frac{3x - 7}{x - 1} \neq 1 \Rightarrow x \neq 3$$ Rewrite: $$\frac{3x - 7}{x - 1} = 1 \Rightarrow 3x - 7 = x - 1 \Rightarrow 2x = 6 \Rightarrow x = 3$$ So if $\frac{3x - 7}{x - 1} \neq 1$, then $x \neq 3$. 7. **By induction, prove for $n \in \mathbb{N}$:** $$1 + 4 + 4^2 + 4^3 + \dots + 4^n = \frac{4^{n+1} - 1}{3}$$ **Base case $(n=0)$:** $$1 = \frac{4^{1} -1}{3} = \frac{4-1}{3} = 1$$ True. **Inductive step:** assume true for $n$, prove for $n+1$: $$1 + 4 + ... + 4^n + 4^{n+1} = \frac{4^{n+1} - 1}{3} + 4^{n+1} = \frac{4^{n+1} - 1}{3} + \frac{3 \times 4^{n+1}}{3} = \frac{4^{n+1} -1 + 3 \times 4^{n+1}}{3} = \frac{4^{n+1} (1 + 3) -1}{3} = \frac{4^{n+2} - 1}{3}$$ Hence the formula holds for $n+1$. 8. **Define functions:** $$f(x) = -x^2 + 2x + 3, \quad g(x) = \sqrt{x + 7}$$ 9. **Domains $D_f$ and $D_g$:** $f$ is polynomial: domain all real numbers $$D_f = \mathbb{R}$$ For $g$, radicand $x+7 \geq 0 \Rightarrow x \geq -7$ $$D_g = [-7, +\infty)$$ 10. **Show $f$ is neither even nor odd:** Check $f(-x)$: $$f(-x) = -(-x)^2 + 2(-x) + 3 = -x^2 - 2x + 3$$ Not equal to $f(x)$ or $-f(x)$, so $f$ is neither even nor odd. 11. **Evaluate $f(2)$ and $g(2)$:** $$f(2) = -(2)^2 + 2\times 2 + 3 = -4 + 4 + 3 = 3$$ $$g(2) = \sqrt{2 + 7} = \sqrt{9} = 3$$ Both equal 3, a coincidence at $x=2$. 12. **Solve quadratic $f(x)=0$:** $$-x^2 + 2x + 3 = 0 \Rightarrow x^2 - 2x - 3 = 0$$ Discriminant: $$\Delta = (-2)^2 - 4(1)(-3) = 4 + 12 = 16$$ Roots: $$x = \frac{2 \pm 4}{2}$$ So, $$x_1 = 3, \quad x_2 = -1$$ **Sign table:** Since $a=-1 < 0$, $f$ is positive between roots: $$\begin{cases} f(x)>0 & \text{if } -1 < x < 3 \\ f(x)<0 & \text{if } x < -1 \text{ or } x > 3 \end{cases}$$ 13. **Find intercepts of $C_f$:** - $x$-intercepts are roots found: $x=-1, 3$ - $y$-intercept: $f(0) = 3$ 14. **Variation tables for $f$ and $g$:** - $f$ is quadratic opening downward with vertex: $$x_v = -\frac{b}{2a} = -\frac{2}{2(-1)}=1$$ $$f(1) = -1 + 2 + 3 = 4$$ Function $f$ increases on $(-\infty, 1]$ and decreases on $[1, +\infty)$. - $g$ is increasing square root from $[-7, +\infty)$. 15. **Sketch curves $C_f$ and $C_g$ together**. 16. **Determine maximum of $f$: $4$ at $x=1$**. Values on intervals: - $f([0;1])$ between $f(0) =3$ and $f(1)=4$. - $f([1;3])$ between $f(1)=4$ and $f(3)=0$. 17. **Solve inequalities graphically:** - $g(x) < 3 \Rightarrow \sqrt{x+7} < 3 \Rightarrow x+7 < 9 \Rightarrow x < 2$ - $f(x) \leq 3$ from sign and vertex info: $f(x)$ maximum is 4, equals 3 at $x=0$ and at two symmetric points. (Exact intervals found from $f(x) = 3$ equation). 18. **Expression and domain of composition $f \circ g$:** $$f(g(x)) = f(\sqrt{x+7}) = - (\sqrt{x+7})^2 + 2 \sqrt{x+7} + 3 = -(x+7) + 2 \sqrt{x+7} + 3 = -x -4 + 2\sqrt{x+7}$$ Domain intersection: $$D_g = [ -7 , +\infty ), \quad f \text{ defined on } \mathbb{R}$$ So, $$D_{f\circ g} = D_g = [ -7 , +\infty )$$ 19. **Determine monotonicity of $f\circ g$ on $[0,1]$ and $[1,3]$**. Check values or derivative behavior (sketch or compute). **Final answers summary:** - $P$ is true. - $Q$ is false. - Solutions of $x^2 + |x-1| -1=0$ are $\{0,1\}$. - For all real $x, \frac{x^2 +3}{x^2 +4} \neq 1$. - $\frac{3x -7}{x-1} \neq 1 \Rightarrow x \neq 3$. - Induction formula for geometric sum with ratio 4 holds. - $D_f = \mathbb{R}$ and $D_g = [-7, +\infty)$. - $f$ neither even nor odd. - $f(2) = g(2) = 3$. - Roots and sign table of $f$ found. - Monotonicity and max values of $f$, $g$ explored. - Composition $f\circ g$ expression and domain verified.