1. Problem 24: Find the spanning set for the null space of matrix $$A=\begin{pmatrix}1 & 3 & -2 \\ 2 & 1 & -1 \\ -4 & -2 & 2\end{pmatrix}$$ Step 1: Write the homogeneous system $A\mathbf{x} = \mathbf{0}$ where $\mathbf{x} = \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}$. Step 2: Solve the system: \[ \begin{cases} 1x_1 + 3x_2 - 2x_3 = 0 \\ 2x_1 + 1x_2 - 1x_3 = 0 \\ -4x_1 - 2x_2 + 2x_3 = 0 \end{cases} \] Step 3: Row reduce the augmented matrix or substitute: From first, $x_1 = -3x_2 + 2x_3$. Substitute in second: $2(-3x_2 + 2x_3) + x_2 - x_3 = 0 \Rightarrow -6x_2 + 4x_3 + x_2 - x_3 =0 \Rightarrow -5x_2 + 3x_3=0 \Rightarrow 5x_2=3x_3 \Rightarrow x_2=\frac{3}{5}x_3$. Then $x_1 = -3\cdot\frac{3}{5}x_3 + 2x_3 = -\frac{9}{5}x_3 + 2x_3 = \frac{1}{5}x_3$. Step 4: The null space vector is $$\mathbf{x} = \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix} = x_3 \begin{pmatrix} \frac{1}{5} \\ \frac{3}{5} \\ 1 \end{pmatrix} = t \begin{pmatrix} 1 \\ 3 \\ 5 \end{pmatrix}$$ where $t = \frac{x_3}{5}$. Step 5: Hence the null space is spanned by $\begin{pmatrix}1 \\ 3 \\ 5\end{pmatrix}$ i.e., option (B). --- 2. Problem 25: Given $f(x) = A \cos(\omega x - \phi)$ with a cosine wave plotted from $0$ to $2\pi$ and amplitude between $-3$ and $3$. Step 1: The phase shift $\phi$ determines the horizontal shift. Step 2: Cosine normally starts at maximum ($x=0$, $f(0)=A$ if no phase shift). Step 3: The graph shows behavior shifted right by $\frac{\pi}{4}$, so possible $\phi = \frac{\pi}{4}$. Step 4: Thus, the phase shift $\phi = \frac{\pi}{4}$ matches option (B). --- 3. Problem 26: Find $\int_{-2}^2 f(x) \, dx$ where $$f(x) = \max \{ \sqrt{4 - x^2}, x+2 \}, \quad -2 \leq x \leq 2.$$ Step 1: Find where $\sqrt{4 - x^2} = x+2$. Square both sides: $$4 - x^2 = (x+2)^2 = x^2 + 4x + 4,$$ which simplifies to $$4 - x^2 = x^2 + 4x + 4 \implies 0 = 2x^2 + 4x,$$ or $$x^2 + 2x = 0 \implies x(x+2)=0.$$ Step 2: The roots of equality are $x=0$ and $x=-2$. Step 3: For $x$ in [-2,0], compare values: At $x=-2$, $\sqrt{4-4} = 0$ and $x+2=0$ equal. At $x=-1$, $\sqrt{4-1}=\sqrt{3}\approx1.732$, $-1+2=1$ so max is $\sqrt{4-x^2}$. At $x=0$, $\sqrt{4-0}=2$, $0+2=2$ equal. Step 4: For $x$ in [0,2], check: At $x=1$, $\sqrt{4-1}=\sqrt{3}\approx1.732$, $1+2=3$ max is $x+2$. Step 5: Integral splits as $$\int_{-2}^2 f(x) dx = \int_{-2}^0 \sqrt{4 - x^2} \, dx + \int_0^2 (x+2) \, dx.$$ Step 6: Calculate integrals: First integral (semicircle area with radius 2): $$\int_{-2}^0 \sqrt{4 - x^2} \, dx = \frac{1}{4} \pi \cdot 2^2 = \pi$$ But this is the area from -2 to 2, so from -2 to 0 is half semicircle area, i.e. $\frac{\pi \cdot 2^2}{2 \cdot 2} = \frac{\pi \cdot 4}{4} = \pi$ actually half semicircle area is $ \frac{\pi 2^2}{2} = 2\pi$. Therefore, calculation: Use formula for circle sector area with integral: $$\int \sqrt{r^2 - x^2} dx = \frac{x}{2} \sqrt{r^2 - x^2} + \frac{r^2}{2} \arcsin \frac{x}{r} + C,$$ so $$\int_{-2}^0 \sqrt{4 - x^2} \, dx = \Big[ \frac{x}{2} \sqrt{4-x^2} + 2 \arcsin(\frac{x}{2}) \Big]_{-2}^0.$$ Evaluate at 0: $0 + 2 \arcsin(0) = 0$. At -2: $$\frac{-2}{2} \cdot 0 + 2 \arcsin(-1) = 0 + 2(-\frac{\pi}{2}) = -\pi.$$ So result is $0 - (-\pi) = \pi$. Step 7: For second integral: $$\int_0^2 (x+2) dx = \Big[ \frac{x^2}{2} + 2x \Big]_0^2 = \Big( \frac{4}{2} + 4 \Big) - 0 = 2 + 4 = 6.$$ Step 8: Sum results: $$\int_{-2}^2 f(x) dx = \pi + 6.$$ Step 9: Corresponds to option (C) $6 + \pi$. --- 4. Problem 27: Negate "If $P$ is true, then $Q$ is true and $R$ is true." Step 1: Original statement is $P \to (Q \wedge R)$. Step 2: Negation is $P \wedge \neg (Q \wedge R)$. Step 3: Using De Morgan, $\neg(Q \wedge R) = \neg Q \vee \neg R$. Step 4: Negation is $P \wedge (\neg Q \vee \neg R)$ meaning: "$P$ is true and either $Q$ is false or $R$ is false." Step 5: Matches option (A). --- 5. Problem 28: Evaluate $$\int_0^\infty x^2 e^{-x^2} dx.$$ Step 1: Use substitution $u = x^2 \implies du = 2x dx$ so $x dx = \frac{du}{2}$. However substitution is tricky here; better use gamma function: Step 2: Recall Gamma function: $$\Gamma(s) = \int_0^\infty t^{s-1} e^{-t} dt.$$ Step 3: Express integral in Gamma form by letting $t = x^2$, so $x = \sqrt{t}$, $dx = \frac{1}{2 \sqrt{t}} dt$. Step 4: Change variable: $$\int_0^\infty x^2 e^{-x^2} dx = \int_0^\infty t e^{-t} \frac{1}{2 \sqrt{t}} dt = \frac{1}{2} \int_0^\infty t^{1 - \frac{1}{2}} e^{-t} dt = \frac{1}{2} \int_0^\infty t^{\frac{1}{2}} e^{-t} dt.$$ Step 5: This is $$\frac{1}{2} \Gamma(\frac{3}{2}) = \frac{1}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{\sqrt{\pi}}{4}.$$ (Since $\Gamma(3/2) = \frac{1}{2} \sqrt{\pi}$). Step 6: Answer is option (A) $\frac{\sqrt{\pi}}{4}$. --- Final answers: 24: (B), 25: (B), 26: (C), 27: (A), 28: (A)