EE1M1_202526_Exam1_all

📝 EE1M1 Midterm Exam 1 - 2025/2026

🚀 Overview

This document contains the collected questions and solutions from the EE1M1 Midterm Exam 1.

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🔹 Exercise #78626: Vector Angles 📐

Question: Given are the vectors:

$$\mathbf{u}=\left(\begin{array}{c}-2\\ -1\\ -2\end{array}\right)\text{and}\mathbf{v}=\left(\begin{array}{c}-2\\ 1\\ 3\end{array}\right)$$

Let $\theta$ be the angle between $\mathbf{u}$ and $\mathbf{v}$. Find $\cos (\theta )$.

✅ Correct Answer:

$$\cos (\theta )=-\frac{\sqrt{14}}{14}$$

💡 Stepwise Solution:

1. Recall the formula: The dot product relates to the angle via $\mathbf{u}\cdot \mathbf{v}=\Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert \cos (\theta )$.
2. Calculate dot product: $\mathbf{u}\cdot \mathbf{v}=(-2)(-2)+(-1)(1)+(-2)(3)=4-1-6=-3$
3. Calculate magnitudes:
   • $\Vert \mathbf{u}\Vert =\sqrt{(-2{)}^2+(-1{)}^2+(-2{)}^2}=\sqrt{4+1+4}=\sqrt{9}=3$
   • $\Vert \mathbf{v}\Vert =\sqrt{(-2{)}^2+1^2+3^2}=\sqrt{4+1+9}=\sqrt{14}$
4. Isolate $\cos (\theta )$: $-3=3\cdot \sqrt{14}\cdot \cos (\theta )⟹\cos (\theta )=\frac{-3}{3\sqrt{14}}=-\frac{1}{\sqrt{14}}=-\frac{\sqrt{14}}{14}$

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🔹 Exercise #76107: Trigonometric Integral 🌀

Question: Evaluate the following integral. Use a capital $C$ as a constant of integration.

$$\int (\sin (x){)}^{3}dx=\dots$$

✅ Correct Answer:

$$-\cos (x)+\frac{1}{3}(\cos (x){)}^3+C$$

💡 Stepwise Solution:

1. Identify Strategy: Use the identity ${\sin }^2(x)=1-{\cos }^2(x)$ to set up a $u$-substitution.
2. Rewrite integral: $\int {\sin }^2(x)\sin (x)dx=\int (1-{\cos }^2(x))\sin (x)dx$
3. Substitute: Let $u=\cos (x)$, then $du=-\sin (x)dx$ (or $\sin (x)dx=-du$).
4. Integrate: $-\int (1-u^2)du=-(u-\frac{1}{3}{u}^3)+C=-u+\frac{1}{3}{u}^3+C$
5. Back-substitute: $-\cos (x)+\frac{1}{3}(\cos (x){)}^3+C$

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🔹 Exercise #95723: Implicit Differentiation 📉

Question: The curve $C$ is implicitly described by:

$$y^3-5e^y=-2x^{2}y$$

Determine $\frac{dy}{dx}$.

✅ Correct Answer:

$$\frac{dy}{dx}=\frac{-4xy}{2x^2+3y^2-5e^y}$$

💡 Stepwise Solution:

1. Differentiate both sides with respect to $x$: $\frac{d}{dx}(y^3-5e^y)=\frac{d}{dx}(-2x^{2}y)$
2. Apply Chain Rule and Product Rule:
   • Left side: $3y^2\frac{dy}{dx}-5e^y\frac{dy}{dx}=(3y^2-5e^y)\frac{dy}{dx}$
   • Right side: $-4xy-2x^2\frac{dy}{dx}$
3. Group $\frac{dy}{dx}$ terms: $(3y^2-5e^y)\frac{dy}{dx}+2x^2\frac{dy}{dx}=-4xy$ $(2x^2+3y^2-5e^y)\frac{dy}{dx}=-4xy$
4. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{-4xy}{2x^2+3y^2-5e^y}$

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🔹 Exercise #105507 & #105508: Inverse Functions 🔄

Question: Consider $f(x)=3x^2-9x+6$ with domain $(-\infty ,-2]$.

1. Find the inverse function $f^{-1}(x)$.
2. Give the domain of $f^{-1}$.

✅ Correct Answers:

1. $f^{-1}(x)=\frac{3}{2}-\frac{\sqrt{12x+9}}{6}$
2. Domain of $f^{-1}$: $[36,\infty )$

💡 Stepwise Solution:

1. Find Range of $f$:
   • $f(x)$ is a parabola opening upwards. Vertex is at $x=-b/2a=9/6=1.5$.
   • Since the domain is $(-\infty ,-2]$, the function is decreasing.
   • Lower bound of range: $f(-2)=3(-2{)}^2-9(-2)+6=12+18+6=36$.
   • As $x\to -\infty ,f(x)\to \infty$.
   • Range of $f$ = Domain of $f^{-1}=[36,\infty )$.
2. Solve for $x$: Set $y=3x^2-9x+6⟹3x^2-9x+(6-y)=0$. Using quadratic formula: $x=\frac{9\pm \sqrt{(-9{)}^2-4(3)(6-y)}}{2(3)}=\frac{9\pm \sqrt{81-72+12y}}{6}=\frac{9\pm \sqrt{9+12y}}{6}=\frac{3}{2}\pm \frac{\sqrt{12y+9}}{6}$
3. Choose sign: Since the domain is $x\le -2$, we must pick the negative root: $x=\frac{3}{2}-\frac{\sqrt{12y+9}}{6}$
4. Swap variables: $f^{-1}(x)=\frac{3}{2}-\frac{\sqrt{12x+9}}{6}$.

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🔹 Exercise #123403: Differential Equation 🧪

Question: Solve: $(x^2+2x+2)y^{\prime }=\frac{1}{y^2}$ a. Find the explicit general solution. b. Find the solution satisfying $y(0)=2$.

✅ Correct Solution:

1. Separate variables: $y^{2}dy=\frac{1}{x^2+2x+2}dx$
2. Integrate both sides:
   • Left: $\int y^{2}dy=\frac{1}{3}{y}^3$
   • Right: $\int \frac{1}{(x+1{)}^2+1}dx=\arctan (x+1)+C$ (completing the square)
3. General solution: $\frac{1}{3}{y}^3=\arctan (x+1)+C⟹y(x)=\sqrt[3]{3\arctan (x+1)+C}$
4. Initial condition $y(0)=2$: $2=\sqrt[3]{3\arctan (1)+C}⟹8=3(\frac{\pi }{4})+C⟹C=8-\frac{3\pi }{4}$
5. Specific solution: $y(x)=\sqrt[3]{3\arctan (x+1)+8-\frac{3\pi }{4}}$

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🔹 Exercise #123404: Taylor Approximation 📏

Question: Consider $f(x)=e^{2-2x}$. a. Use a 3rd order Taylor approximation around $x=1$ to approximate $e^{-1}$. b. Is the error less than $1/20$?

✅ Correct Solution:

1. Derivatives at $x=1$:
   • $f(x)=e^{2-2x}⟹f(1)=1$
   • $f^{\prime }(x)=-2e^{2-2x}⟹f^{\prime }(1)=-2$
   • $f^{\prime \prime }(x)=4e^{2-2x}⟹f^{\prime \prime }(1)=4$
   • $f^{\prime \prime \prime }(x)=-8e^{2-2x}⟹f^{\prime \prime \prime }(1)=-8$
2. Taylor Polynomial $T_3(x)$: $T_3(x)=1-2(x-1)+\frac{4}{2}(x-1{)}^2-\frac{8}{6}(x-1{)}^3=1-2(x-1)+2(x-1{)}^2-\frac{4}{3}(x-1{)}^3$
3. Approximate $e^{-1}$: Note $e^{-1}=f(3/2)$. Plug in $x-1=1/2$: $T_3(3/2)=1-2(1/2)+2(1/4)-\frac{4}{3}(1/8)=1-1+1/2-1/6=1/3$
4. Error check (Lagrange Remainder): $|R_3|=\frac{|f^{(4)}(s)|}{4!}|x-a{|}^4$. $f^{(4)}(x)=16e^{2-2x}$. On $[1,3/2]$, $f^{(4)}(s)$ is max at $s=1$: $16e^0=16$. $|R_3|\le \frac{16}{24}(\frac{1}{2}{)}^4=\frac{2}{3}\cdot \frac{1}{16}=\frac{1}{24}$ Since $1/24<1/20$, the error is indeed less than $1/20$.

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🔹 Exercise #123405: Limits & Squeeze Theorem 🔍

Question: a. Evaluate ${\lim }_{x\to -\infty }x+\sqrt{x^2+3x+1}$. b. Use Squeeze Theorem for $f(x)$ where $\frac{e^x-e}{x-1}\le f(x)\le x^{\frac{1}{x-1}}$ for $x\to 1^{-}$.

✅ Correct Solution:

1. Limit (Part a): Use conjugate multiplication. ${\lim }_{x\to -\infty }\frac{(x+\sqrt{\dots })(x-\sqrt{\dots })}{x-\sqrt{\dots }}=\lim \frac{x^2-(x^2+3x+1)}{x-\sqrt{x^2+3x+1}}=\lim \frac{-3x-1}{x-|x|\sqrt{1+3/x+1/x^2}}$ Since $x<0,|x|=-x$: $\lim \frac{-3x-1}{x+x\sqrt{1+3/x+1/x^2}}=\lim \frac{-3-1/x}{1+\sqrt{1+3/x+1/x^2}}=\frac{-3}{1+1}=-\frac{3}{2}$
2. Squeeze Theorem (Part b):
   • Left limit: ${\lim }_{x\to 1^{-}}\frac{e^x-e}{x-1}\stackrel{L^{\prime }H}{\to }\lim \frac{e^x}{1}=e$.
   • Right limit: ${\lim }_{x\to 1^{-}}{x}^{\frac{1}{x-1}}=\lim e^{\frac{\ln (x)}{x-1}}\stackrel{L^{\prime }H\text{ on exponent}}{\to }{e}^{\lim \frac{1/x}{1}}=e^1=e$.
   • Since both sides go to $e$, ${\lim }_{x\to 1^{-}}f(x)=e$.

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🔹 Exercise #123406: Partial Fractions 🧩

Question: Evaluate: $\int \frac{5x^2-11x+8}{(x-2{)}^2(x-1)}dx$

✅ Correct Solution:

1. Setup decomposition: $\frac{5x^2-11x+8}{(x-2{)}^2(x-1)}=\frac{A}{x-2}+\frac{B}{(x-2{)}^2}+\frac{C}{x-1}$
2. Solve for constants:
   • $5x^2-11x+8=A(x-2)(x-1)+B(x-1)+C(x-2{)}^2$
   • Let $x=1⟹5-11+8=C(-1{)}^2⟹2=C$.
   • Let $x=2⟹20-22+8=B(1)⟹6=B$.
   • Coefficient of $x^2$: $5=A+C⟹5=A+2⟹3=A$.
3. Integrate: $\int (\frac{3}{x-2}+\frac{6}{(x-2{)}^2}+\frac{2}{x-1})dx=3\ln |x-2|-\frac{6}{x-2}+2\ln |x-1|+C$

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🔹 Exercise #123407: Planes & Lines ✈️

Question: Planes $V:3x-2y+z=5$ and $W:-x+2y+z=9$ intersect in line $L_1$. a. Is $L_2:\mathbf{r}(t)=⟨4,5,-3⟩+t⟨1,1,-1⟩$ parallel to $L_1$? b. Find plane through $P(2,3,0)$ orthogonal to $L_1$.

✅ Correct Solution:

1. Direction of $L_1$: Perpendicular to both normals ${\mathbf{n}}_{\mathbf{V}}=⟨3,-2,1⟩$ and ${\mathbf{n}}_{\mathbf{W}}=⟨-1,2,1⟩$. ${\mathbf{d}}_1={\mathbf{n}}_{\mathbf{V}}\times {\mathbf{n}}_{\mathbf{W}}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 3& -2& 1\\ -1& 2& 1\end{array}\right|=⟨-4,-4,4⟩$
2. Check Parallel (Part a): Direction of $L_2$ is ${\mathbf{d}}_2=⟨1,1,-1⟩$. Since ${\mathbf{d}}_1=-4{\mathbf{d}}_2$, the direction vectors are multiples. Yes, they are parallel.
3. Plane Equation (Part b): Normal vector is ${\mathbf{d}}_1=⟨-4,-4,4⟩$ (or scaled $⟨1,1,-1⟩$). $1(x-2)+1(y-3)-1(z-0)=0⟹x+y-z=5$

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🔹 Exercise #123408: Improper Integrals ♾️

Question: Determine convergence/divergence of ${\int }_2^{\infty }\frac{x-\sqrt{x}}{x^2-1}dx$.

✅ Correct Solution:

1. Analyze behavior at infinity: The integrand $\frac{x-\sqrt{x}}{x^2-1}\approx \frac{x}{x^2}=\frac{1}{x}$.
2. Comparison Theorem: For $x\ge 2$: $\frac{x-\sqrt{x}}{x^2-1}\ge \frac{x-\sqrt{x}}{x^2}=\frac{1}{x}-\frac{1}{x^{3/2}}$
3. Test convergence:
   • ${\int }_2^{\infty }\frac{1}{x}dx$ diverges ($p$-integral with $p=1$).
   • ${\int }_2^{\infty }\frac{1}{x^{3/2}}dx$ converges ($p$-integral with $p=1.5$).
   • The difference of a divergent and convergent integral is divergent.
4. Conclusion: The integral diverges.

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