1. Problem 1: Match the functions in Column A with their types in Column B and their graphs in Column C. 1.1. Given functions: $f(x)=3\ln(x)-2$, $f(x)=10-4x$, $f(x)=2^x+7$, $f(x)=x^2-4x+4$. 1.2. Rule: logarithmic functions have form $\ln(\cdot)$ and rise then flatten, exponential functions have form $a^x$ and grow/decay rapidly, linear functions are of form $mx+b$, and quadratic functions are parabolas of form $ax^2+bx+c$. 1.3. Identify types: $f(x)=3\ln(x)-2$ is a logarithmic function. 1.4. Identify types: $f(x)=10-4x$ is a linear function. 1.5. Identify types: $f(x)=2^x+7$ is an exponential function. 1.6. Identify types: $f(x)=x^2-4x+4$ is a quadratic function. 1.7. Match to graphs by shape and key points: the parabola with vertex near $(2,0)$ is the quadratic, so $x^2-4x+4$ matches Graph 1. 1.8. The curve that rises and flattens (logarithmic shape) matches $3\ln(x)-2$, so $3\ln(x)-2$ matches Graph 2. 1.9. The exponential growth curve passing through $(0,8)$ fits $2^x+7$, so $2^x+7$ matches Graph 3. 1.10. The straight line matches $10-4x$, so $10-4x$ matches Graph 4. 1.11. Final answers for Problem 1: i) $3\ln(x)-2$ = logarithmic = Graph 2. 1.12. ii) $10-4x$ = linear = Graph 4. 1.13. iii) $2^x+7$ = exponential = Graph 3. 1.14. iv) $x^2-4x+4$ = quadratic = Graph 1. 2. Problem 2: Intersection properties when $f$ is strictly increasing and $g$ is strictly decreasing and $f(x_0)=g(x_0)$. 2.1. Rule: for $x>x_0$, $f(x)>f(x_0)$ and $g(x)$ for $x>x_0$. 2.3. There cannot be another intersection point $x_1>x_0$ because $f(x)-g(x)$ is strictly increasing for $x>x_0$. 2.4. For $x\le x_0$ we have $g(x)\ge f(x)$ by the monotonicity argument reversed. 2.5. Correct options: "$f(x)\ge g(x)$ for all $x\ge x_0$" is true. 2.6. Correct options: "$g(x)\ge f(x)$ for all $x_0\ge x$" (i.e. for all $x\le x_0$) is true. 2.7. All other listed statements are false. 3. Problem 3: $a_n=\dfrac{12n^2}{3n+27}-\dfrac{4n^2+7}{n+15}$, compute $\lim_{n\to\infty}a_n$. 3.1. Use algebraic simplification to combine fractions. 3.2. Compute numerator: $12n^2(n+15)-(4n^2+7)(3n+27)$. 3.3. Expand: $12n^3+180n^2-(12n^3+108n^2+21n+189)=72n^2-21n-189$. 3.4. Denominator: $(3n+27)(n+15)=3n^2+72n+405$. 3.5. So $a_n=\dfrac{72n^2-21n-189}{3n^2+72n+405}$. 3.6. Divide numerator and denominator by $n^2$ and take limit: leading coefficients give $\dfrac{72}{3}=24$. 3.7. Final answer: $\displaystyle \lim_{n\to\infty}a_n=24$. 4. Problem 4: How many curves have a unique tangent at the origin out of the four described curves. 4.1. Rule: a unique tangent exists at a point if the left and right derivatives coincide and are finite. 4.2. Curve 1 is a "V" (like $y=|x|$) and has different left and right slopes, so no unique tangent. 4.3. Curve 2 is a parabola with vertex at the origin (smooth), so derivative at $0$ exists and tangent is unique. 4.4. Curve 3 is a cubic-like smooth curve through the origin (example $y=x^3$), derivative at $0$ exists, so tangent is unique. 4.5. Curve 4 is a smooth cubic-like curve with an inflection at the origin, derivative exists, so tangent is unique. 4.6. Count: Curves 2, 3, and 4 have a unique tangent, so the number is $3$. 5. Problem 5: Evaluate $\displaystyle \lim_{n\to\infty} e^{\sqrt{n}}\Big[\log\Big(1+\dfrac{28}{n}\Big)-\Big(\dfrac{e^{21/n}-1}{\sqrt{2\pi n}}\Big)^{1/n}\Big]$. 5.1. For large $n$, $\log\Big(1+\dfrac{28}{n}\Big)=\dfrac{28}{n}+O\Big(\dfrac{1}{n^2}\Big)$. 5.2. Also $e^{21/n}-1=\dfrac{21}{n}+O\Big(\dfrac{1}{n^2}\Big)$, so $\dfrac{e^{21/n}-1}{\sqrt{2\pi n}}=O\Big(n^{-3/2}\Big)$. 5.3. Then $\Big(\dfrac{e^{21/n}-1}{\sqrt{2\pi n}}\Big)^{1/n}=\exp\Big(\dfrac{1}{n}\ln\big(O(n^{-3/2})\big)\Big)=1+O\Big(\dfrac{\ln n}{n}\Big)$. 5.4. Therefore the bracket behaves like $\dfrac{28}{n}-1+o(1)$, which tends to $-1$ as $n\to\infty$ after multiplying by $e^{\sqrt{n}}$. 5.5. Since $e^{\sqrt{n}}\to\infty$, the whole expression tends to $-\infty$. 5.6. Final answer: the limit is $-\infty$. 6. Problem 6: $a_n=\dfrac{6+10+14+\ldots+(4n-2)}{n^2}$, find $\lim_{n\to\infty}a_n$. 6.1. The numerator is an arithmetic series with common difference $4$ and first term $6$ and last term $4n-2$. 6.2. Sum formula: $S_n=\dfrac{\text{number of terms}\times(\text{first} + \text{last})}{2}$. 6.3. Using $n$ terms, $S_n=n\cdot\dfrac{6+(4n-2)}{2}=n\cdot\dfrac{4n+4}{2}=n(2n+2)=2n^2+2n$. 6.4. Divide by $n^2$: $\dfrac{2n^2+2n}{n^2}=2+\dfrac{2}{n}$. 6.5. Take limit $n\to\infty$: $\displaystyle \lim_{n\to\infty}a_n=2$. 7. Problem 7: Compute $5\lim_{x\to21^{+}}\lfloor x\rfloor-3\lim_{x\to4^{-}}\lfloor x\rfloor$. 7.1. For $x\to21^{+}$ we have $x>21$ close to 21, so $\lfloor x\rfloor=21$ and the limit is $21$. 7.2. For $x\to4^{-}$ we have $x<4$ close to 4, so $\lfloor x\rfloor=3$ and the limit is $3$. 7.3. Evaluate expression: $5\cdot21-3\cdot3=105-9=96$. 7.4. Final answer: $96$. 8. Problem 8: Algorithms with limits as $n\to\infty$: $a_n=\dfrac{n^2+5n}{6n^2+1}$, $b_n=\dfrac{1}{8}+(-1)^n\dfrac{1}{n}$, $c_n=\dfrac{e^n+4}{7e^n}$. 8.1. Compute limits: $\lim a_n=\dfrac{1}{6}\approx0.166\ldots$. 8.2. $\lim b_n=\dfrac{1}{8}=0.125$. 8.3. $\lim c_n=\dfrac{1}{7}\approx0.142857\ldots$. 8.4. Ordering from smallest to largest: $b_n$ (0.125) $<$ $c_n$ (0.142857) $<$ $a_n$ (0.1667). 8.5. Evaluate given statements: "Error in estimation by Algorithm 2 will be 0.500" is false. 8.6. "Error in estimation by Algorithm 2 will give the minimum error" is true. 8.7. "Error in estimation by Algorithm 2 will give the maximum error" is false. 8.8. "Both Algorithm 1 and Algorithm 2 will give the same error and that will be the maximum" is false. 8.9. "Error in estimation by Algorithm 1 will be 0.166 approximately" is true. 9. Problem 9: New algorithm error given by $a_n-b_n$ with $a_n,b_n$ as above; choose correct options. 9.1. Limits: $\lim a_n=1/6\approx0.1667$, $\lim b_n=1/8=0.125$, so $\lim(a_n-b_n)=1/24\approx0.0416667>0$. 9.2. Compare errors: for large $n$, $a_n-b_n0$, so "The error in estimation using the new algorithm is less than the error in estimation using Algorithm 1" is true. 9.3. Check "The error in estimation using Algorithm 2 is less than the error in estimation using the new algorithm": this would require $b_n\frac{1}{6}$, so the statement is false. 9.4. Final choices: only the first option is correct for Problem 9. Final summary of numeric answers and selections: - Problem 1 matches: i) log -> Graph 2; ii) linear -> Graph 4; iii) exponential -> Graph 3; iv) quadratic -> Graph 1. - Problem 2: correct options are statements 1 and 4. - Problem 3: limit = $24$. - Problem 4: number of curves with a unique tangent at the origin = $3$. - Problem 5: limit = $-\infty$. - Problem 6: limit = $2$. - Problem 7: value = $96$. - Problem 8: correct statements are that Algorithm 2 gives the minimum error and Algorithm 1 has error approximately $0.166$. - Problem 9: only the statement "The error in estimation using the new algorithm is less than the error in estimation using Algorithm 1" is correct.