Upload GAIC_All_Questions_Answers_Final.json

GAIC_All_Questions_Answers_Final.json

@@ -758,7 +758,7 @@
     {
         "label": "Combinary-16",
         "problem": "Zheng flips an unfair coin 5 times. If the probability of getting exactly 1 head is equal to the probability of getting exactly 2 heads and is nonzero, then the probability of getting exactly 3 heads is \\_\\_\\_\\_.",
-        "answer": "\\frac{40}{243}.",
+        "answer": "\\frac{40}{243}",
         "solution": "Let the probability of getting a head be $p$.\n\nAccording to the given conditions, we have $C_5^1 p(1-p)^4 = C_5^2 p^2(1-p)^3$.\n\nSolving this equation, we find $p = \\frac{1}{3}$.\n\nTherefore, the probability of getting exactly 3 heads is $C_5^3 p^3(1-p)^2 = \\frac{40}{243}$."
     },
     {
@@ -1274,7 +1274,7 @@
     {
         "label": "Alg2-41",
         "problem": "Jasmine invests $\\$ 2,658$ in a retirement account with a fixed annual interest rate of $9 \\%$ compounded continuously. What will the account balance be after 15 years?",
-        "answer": "\\ 9681.72",
+        "answer": "9681.72",
         "solution": "Using the compound interest formula we have $2658\\times (1.09)^{15} = 9681.72$"
     },
     {
@@ -1730,7 +1730,7 @@
     {
         "label": "",
         "problem": "Consider the differential equation $\\frac{dy}{dx} = \\frac{y-1}{x^3}$, where $x\\neq 0$. Find the general solution $y=f(x)$ to the differential equation.",
-        "answer": "y = ce^{-\\frac{1}{x}}+1.",
+        "answer": "y = ce^{-\\frac{1}{x}}+1",
         "solution": "$\\frac{1}{y-1} d y=\\frac{1}{x^{2}} d x, \\ln (y-1)=-x^{-1}+c, y=c e^{-\\frac{1}{x}}+1$."
     },
     {
@@ -1784,7 +1784,7 @@
     {
         "label": "",
         "problem": "[Rank of a matrix]\n Compute the dimension  of the linear subspace generated by the following vectors\n \\[\\left(\\begin{array}{c}\n     1 \\\\\n     1 \\\\\n     1 \\\\\n    1\n   \\end{array}\\right),\\left(\\begin{array}{c}\n     1 \\\\\n     2 \\\\\n     1 \\\\\n    0\n   \\end{array}\\right), \\left(\\begin{array}{c}\n     0 \\\\\n     -1 \\\\\n     3 \\\\\n    4\n   \\end{array}\\right),\\left(\\begin{array}{c}\n     2 \\\\\n     2 \\\\\n     5 \\\\\n    5\n   \\end{array}\\right).\n \\]",
-        "answer": "3.",
+        "answer": "3",
         "solution": "Let\n\\[A=\\left(\n      \\begin{array}{cccc}\n        1 & 1 & 0 & 2 \\\\\n        1 & 2 & -1 & 2 \\\\\n        1 & 1 & 3 & 5 \\\\\n        1 & 0 & 4 & 5 \\\\\n      \\end{array}\n    \\right)\n\\]\nThen\nthe dimension of the linear subspace generated by the column vectors of matrix A is $\\text{rank}(A).$\nBy elementary transformation of matrix, we have\n\\[A\\to \\left(\n      \\begin{array}{cccc}\n        1 & 1 & 0 & 2 \\\\\n        0 & 1 & -1 & 0 \\\\\n        0 & 0 & 3 & 3 \\\\\n        0 & -1 & 4 & 3 \\\\\n      \\end{array}\n    \\right)\\to \\left(\n      \\begin{array}{cccc}\n        1 & 1 & 0 & 2 \\\\\n        0 & 1 & -1 & 0 \\\\\n        0 & 0 & 3 & 3 \\\\\n        0 & 0 & 3 & 3 \\\\\n      \\end{array}\n    \\right)\\to \\left(\n      \\begin{array}{cccc}\n        1 & 1 & 0 & 2 \\\\\n        0 & 1 & -1 & 0 \\\\\n        0 & 0 & 3 & 3 \\\\\n        0 & 0 & 0 & 0 \\\\\n      \\end{array}\n    \\right).\\]\nIt shows that the rank of $A$ is 3. Thus, the dimension is 3."
     },
     {
@@ -1892,7 +1892,7 @@
     {
         "label": "",
         "problem": "Find the limit $$\\lim\\limits_{x\\to 1}\\frac{f(2x^2+x-3)-f(0)}{x-1}$$ given $f'(1)=2$ and $f'(0)=-1$.",
-        "answer": "-5.",
+        "answer": "-5",
         "solution": "Let $g(x)=2x^2+x-3$. Noticing that $g(1)=0$, the desired limit equals $\\lim\\limits_{x\\to 1}\\frac{f(g(x))-f(g(1))}{x-1}$. By the definition of the derivative and the chain rule and noting that $g'(1)=5$, we have\n\\[\n\\lim\\limits_{x\\to 1}\\frac{f(g(x))-f(g(1))}{x-1}=f'(g(1))g'(1)=f'(0)g'(1)=(-1)(5)=-5.\n\\]\\\\"
     },
     {
@@ -1904,7 +1904,7 @@
     {
         "label": "",
         "problem": "Find the values of  $a$ such that the function $f(x)$ is continuous on $\\mathbb{R}$, where $f(x)$ is defined as \n\\[\nf(x)=\\begin{cases} 2x-1, &\\text{if } x\\leq 0,\\\\ \na(x-1)^2-3, & \\text{otherwise.}\n\\end{cases}\n\\]",
-        "answer": "2.",
+        "answer": "2",
         "solution": "By the definition of $f(x)$, we have\n\\begin{align*}\nf(0)&=-1;\\\\\n\\lim\\limits_{x\\to 0^{-}}f(x)&=\\lim\\limits_{x\\to 0^{-}}(2x-1)=2(0)-1=-1;\\\\\n\\lim\\limits_{x\\to 0^{+}}f(x)&=\\lim\\limits_{x\\to 0^{+}}(a(x-1)^2-3)=a(0-1)^2-3=a-3.\n\\end{align*}\n\nTo obtain the continuity of $f(x)$ at $x=0$, we need $-1=a-3$, that is, $a=2$.\n\nSo, the function $f(x)$ is continuous at $x=0$ when $a=2$.\\\\"
     },
     {
@@ -1922,7 +1922,7 @@
     {
         "label": "",
         "problem": "Let $f(3)=-1$, $f'(3)=0$, $g(3)=2$ and $g'(3)=5$. Evaluate $\\left(\\frac{f}{g}\\right)'(3)$.",
-        "answer": "1.25 \\\\",
+        "answer": "1.25 \\",
         "solution": "Use the quotient rule. The quotient rule gives\n\\[ \n\\left(\\frac{f}{g}\\right)' = \\frac{f'g - fg'}{g^2}. \n\\]\n\nNow, using that $f(3) = -1$, $f'(3) = 0$, $g(3) = 2$, and $g'(3) = 5$, we have\n\\[\n \\left(\\frac{f}{g}\\right)'(3) = \\frac{f'(3)g(3) - f(3)g'(3)}{g(3)^2}=  \\frac{0 \\cdot 2 - (-1) \\cdot 5}{2^2} = \\frac{5}{4}.  \\]\\\\"
     },
     {
@@ -1958,19 +1958,19 @@
     {
         "label": "",
         "problem": "Evaluate the limit $\\lim\\limits_{x\\to 0}\\frac{(1+x)^{\\frac{1}{x}}-e}{x}$.",
-        "answer": "-\\frac{ e}{2}.",
+        "answer": "-\\frac{ e}{2}",
         "solution": "We can use L'H\\^{o}pital's Rule to obtain\n\\[\n\\lim\\limits_{x\\to 0}\\frac{\\ln(1+x)}{x}=\\lim\\limits_{x\\to 0}\\frac{\\frac{1}{1+x}}{1}=1.\n\\]\nThen,\n\\[\n\\lim\\limits_{x\\to 0}(1+x)^{\\frac{1}{x}}=\\lim\\limits_{x\\to 0}e^{\\ln{(1+x)^{\\frac{1}{x}}}}=\\lim\\limits_{x\\to 0}e^{\\frac{\\ln(1+x)}{x}}=e^1=e.\n\\]\n\nLet $f(x) (1+x)^{\\frac{1}{x}}$, then $\\lim\\limits_{x\\to 0}f(x)=e$ and the given limit can be written as:\n\\[ \n\\lim_{x\\to 0}\\frac{(1+x)^{\\frac{1}{x}}-e}{x} = \\lim_{x\\to 0}\\frac{f(x) - e}{x}. \n\\]\nNow, find the derivative of \\(f(x)\\) by using the chain rule and the quotient rule:\n\\begin{align*}\n f'(x) = \\frac{d}{dx}(1+x)^{\\frac{1}{x}}= \\frac{d}{dx}e^{\\ln{(1+x)^{\\frac{1}{x}}}}&=\\frac{d}{dx}e^{\\frac{\\ln(1+x)}{x}}\\\\\n &=e^{\\frac{\\ln(1+x)}{x}} \\frac{d}{dx}\\frac{\\ln(1+x)}{x}\\\\\n &=(1+x)^{\\frac{1}{x}}\\cdot\\frac{\\frac{x}{1+x}-\\ln(1+x)}{x^2}.\n \\end{align*}\nUsing L'H\\^{o}pital's Rule again to get\n\\begin{align*}\n\\lim_{x\\to 0}\\frac{f(x) - e}{x}=\\lim\\limits_{x\\to 0}\\frac{f'(x)}{1}&=\\lim\\limits_{x\\to 0}(1+x)^{\\frac{1}{x}}\\cdot \\lim\\limits_{x\\to 0}\\frac{\\frac{x}{1+x}-\\ln(1+x)}{x^2}\\\\\n&=e \\cdot \\lim\\limits_{x\\to 0}\\frac{\\frac{(1+x)-x}{(1+x)^2}-\\frac{1}{1+x}}{2x}\\\\\n&=e \\cdot \\lim\\limits_{x\\to 0}\\frac{-1}{2(1+x)^2}\\\\\n&=-\\frac{e}{2}.\n\\end{align*}\n\nTherefore,\n\\[ \n\\lim_{x\\to 0}\\frac{(1+x)^{\\frac{1}{x}}-e}{x} =-\\frac{e}{2}.\n\\]\\\\"
     },
     {
         "label": "",
         "problem": "Evaluate the series $\\sum\\limits_{n=0}^\\infty \\frac{1}{2n+1}\\left(\\frac12\\right)^{2n+1}$.",
-        "answer": "\\ln\\sqrt{3}.",
+        "answer": "\\ln\\sqrt{3}",
         "solution": "For $x\\in (-1,1)$, we have\n\\[\n\\frac{1}{1-x^2}=\\sum_{n=0}^\\infty x^{2n}.\n\\]\nThe series on the right-hand side converges uniformly on any interval $[-x, x]$ for any $x\\in (0, 1)$.\nTaking the integrals on both sides yields\n\\[\n\\int_0^x\\frac{1}{1-t^2}dt=\\int_0^x\\sum_{n=0}^\\infty t^{2n}dt=\\sum_{n=0}^\\infty\\int_0^x t^{2n}dt=\\sum_{n=0}^\\infty\\frac{1}{2n+1}x^{2n+1}.\n\\]\nNoting that by partial fraction of $\\frac{1}{1-t^2}=\\frac12\\left(\\frac{1}{1+t}+\\frac{1}{1-t}\\right)$, we have, for $x\\in (0,1)$,\n\\[\n\\int_0^x\\frac{1}{1-t^2}dt=\\frac{1}{2}\\int_0^x\\left(\\frac{1}{1+t}+\\frac{1}{1-t}\\right)dt=\\frac12\\ln\\left(\\frac{1+x}{1-x}\\right).\n\\]\nSo, \n\\[\n\\frac12\\ln\\left(\\frac{1+x}{1-x}\\right)=\\sum_{n=0}^\\infty\\frac{1}{2n+1}x^{2n+1}.\n\\]\nTaking $x=\\frac12$ leads to \n\\[\n\\sum_{n=0}^\\infty\\frac{1}{2n+1}\\left(\\frac12\\right)^{2n+1}=\\frac12\\ln 3=\\ln\\sqrt{3}.\n\\]\\\\"
     },
     {
         "label": "",
         "problem": "Evaluate the limit $\\lim\\limits_{n\\to\\infty}\\sum\\limits_{k=0}^{n-1}\\frac{1}{\\sqrt{n^2-k^2}}$.",
-        "answer": "\\frac{\\pi}{2}.",
+        "answer": "\\frac{\\pi}{2}",
         "solution": "To evaluate this limit, we can interpret this sum as a Riemann sum and convert it into an integral.\n\nLet $f(x) = \\frac{1}{\\sqrt{1 - x^2}}$ on the interval $[0, 1)$. Notice that $f(x)$ is integrable on the interval $[0,1)$. \n\nThe given sum can be expressed as:\n\\[ \n\\lim_{n \\to \\infty} \\sum_{k=0}^{n-1} \\frac{1}{\\sqrt{n^2 - k^2}} =\\lim_{n \\to \\infty} \\sum_{k=0}^{n-1} \\frac{1}{n}\\frac{1}{\\sqrt{1 - \\left(\\frac{k}{n}\\right)^2}}= \\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=0}^{n-1} f\\left(\\frac{k}{n}\\right).\n \\]\nBy the definition of definite integral, we have\n\\[ \n\\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=0}^{n-1} f\\left(\\frac{k}{n}\\right) = \\int_{0}^{1} f(x) \\,dx=\\int_0^1\\frac{1}{\\sqrt{1-x^2}}\\,dx .\n\\]\n\nBy a substitution of $x = \\sin(\\theta)$, we have\n\\begin{align*}\n\\int_{0}^{1} \\frac{1}{\\sqrt{1 - x^2}} \\,dx &= \\int_{0}^{\\frac{\\pi}{2}} \\frac{1}{\\sqrt{1 - \\sin^2(\\theta)}} \\cos(\\theta) \\,d\\theta \\\\\n&= \\int_{0}^{\\frac{\\pi}{2}} \\frac{1}{\\cos(\\theta)} \\cos(\\theta) \\,d\\theta\\\\\n& = \\int_{0}^{\\frac{\\pi}{2}} d\\theta = \\frac{\\pi}{2}\n\\end{align*}\nTherefore, we obtain $\\lim_{n \\to \\infty} \\sum_{k=0}^{n-1} \\frac{1}{\\sqrt{n^2 - k^2}}=\\frac{\\pi}{2}$.\\\\\n\n\nAn alternative method to evaluate $\\displaystyle{\\int_0^1\\frac{1}{\\sqrt{1-x^2}}\\,dx}$:\n\\[\n\\int_0^1\\frac{1}{\\sqrt{1-x^2}}\\,dx=\\arcsin{x}\\big|_0^1=\\arcsin(1)-\\arcsin(0)=\\frac{\\pi}{2}-0=\\frac{\\pi}{2}.\n\\] \\\\"
     },
     {
@@ -1982,19 +1982,19 @@
     {
         "label": "",
         "problem": "Find the length of the curve of the entire cardioid $r=1+\\cos{\\theta}$, where the curve is given in polar coordinates.",
-        "answer": "8.",
+        "answer": "8",
         "solution": "We'll use the arc length formula for polar curves:\n\\[\n L = \\int_0^{2\\pi} \\sqrt{r^2 + \\left(\\frac{dr}{d\\theta}\\right)^2} \\,d\\theta.\n  \\]\nFor the cardioid $r = 1 + \\cos{\\theta}$, we have $\\frac{dr}{d\\theta} = -\\sin{\\theta}$.\nNow, substitute $r$ and $\\frac{dr}{d\\theta}$ into the arc length formula and use a change of variables:\n\\begin{align*}\n L &= \\int_0^{2\\pi} \\sqrt{(1 + \\cos{\\theta})^2 + (-\\sin{\\theta})^2} \\,d\\theta \\\\\n &= \\int_0^{2\\pi}\\sqrt{1 + 2\\cos{\\theta} + \\cos^2{\\theta} + \\sin^2{\\theta}} \\,d\\theta\\\\\n &=\\int_0^{2\\pi}\\sqrt{2 + 2\\cos{\\theta}} \\,d\\theta =\\int_0^{2\\pi}2\\left|\\cos\\left(\\frac{\\theta}{2}\\right)\\right| \\,d\\theta\\\\\n  &=\\int_0^{\\pi}4\\left|\\cos\\left(\\alpha\\right)\\right| \\,d\\alpha =8\\int_0^{\\frac{\\pi}{2}}\\cos\\left(\\alpha\\right) \\,d\\alpha =8\\sin \\alpha\\big|_{0}^{\\frac{\\pi}{2}}=8.\n \\end{align*}\nSo, the length of the curve for the entire cardioid \\(r = 1 + \\cos{\\theta}\\) is \\(8\\).\\\\"
     },
     {
         "label": "",
         "problem": "Find the value of the integral $\\displaystyle{\\int_0^1\\frac{1}{(1+x^2)^2}dx}$.",
-        "answer": "\\frac{\\pi}{8} + \\frac{1}{4}.",
+        "answer": "\\frac{\\pi}{8} + \\frac{1}{4}",
         "solution": "Let $x = \\tan \\theta$, then $dx = \\sec^2 \\theta \\,d\\theta$. Substitute these into the integral to obtain\n\\begin{align*}\n  \\int_0^1 \\frac{1}{(1+x^2)^2} \\,dx& = \\int_0^{\\frac{\\pi}{4}} \\frac{1}{(1 + \\tan^2 \\theta)^2} \\sec^2 \\theta \\,d\\theta\\\\\n  &=\\int_0^{\\frac{\\pi}{4}} \\frac{1}{(\\sec^2\\theta)^2} \\sec^2 \\theta \\,d\\theta\\\\\n    &=\\int_0^{\\frac{\\pi}{4}} \\frac{1}{\\sec^2\\theta}  \\,d\\theta\\\\\n    &= \\int_0^{\\frac{\\pi}{4}} \\cos^2 \\theta \\,d\\theta \\\\\n& =  \\frac{1}{2} \\int_0^{\\frac{\\pi}{4}} (1 + \\cos (2\\theta)) \\,d\\theta  = \\frac{1}{2} \\left.\\left[\\theta + \\frac{1}{2}\\sin (2\\theta)\\right]\\right|_0^{\\frac{\\pi}{4}} =  \\frac{\\pi}{8} + \\frac{1}{4}.\n  \\end{align*}\\\\"
     },
     {
         "label": "",
         "problem": "Evaluate the improper integral $\\displaystyle{\\int_0^\\infty \\frac{1}{x^2+2x+2}dx}$.",
-        "answer": "\\frac{\\pi}{4}.",
+        "answer": "\\frac{\\pi}{4}",
         "solution": "We can write\n\\[\n\\int_0^\\infty \\frac{1}{x^2+2x+2}dx=\\int_0^\\infty \\frac{1}{(x + 1)^2 + 1} \\,dx.\n\\]\nNow, making the substitution $u = x + 1$, so $dx = du$, we have\n\\begin{align*}\n\\int_0^\\infty \\frac{1}{x^2+2x+2}dx&=\\int_0^\\infty \\frac{1}{(x + 1)^2 + 1} \\,dx\\\\\n&=\\int_1^\\infty \\frac{1}{u^2 + 1} \\,du\\\\\n&= \\lim_{a \\to \\infty} \\int_1^a \\frac{1}{u^2 + 1} \\,du\\\\\n& = \\lim_{a \\to \\infty} \\arctan(u)\\big|_1^a\\\\\n&=\\lim_{a \\to \\infty} \\left[\\arctan(a) - \\arctan(1)\\right] = \\frac{\\pi}{2} - \\frac{\\pi}{4} = \\frac{\\pi}{4}.\n\\end{align*} \\\\"
     },
     {
@@ -2012,19 +2012,19 @@
     {
         "label": "",
         "problem": "Evaluate the iterated integral $\\displaystyle{\\int_0^1dy\\int_y^1(e^{-x^2}+e^x)dx}$.",
-        "answer": "\\frac{3}{2}-\\frac12 e^{-1}.",
+        "answer": "\\frac{3}{2}-\\frac12 e^{-1}",
         "solution": "Noting that the region of the integration is \n\\[D=\\{(x,y): 0\\leq y\\leq 1, y\\leq x\\leq 1\\}=\\{(x,y): 0\\leq x\\leq 1, 0\\leq y\\leq x\\}\n\\] and the function $f(x,y)=e^{-x^2}+e^x$ is continuous on $D$, we have\n\\begin{align*}\n\\int_0^1dy\\int_y^1(e^{-x^2}+e^x)dx&=\\iint_D(e^{-x^2}+e^x)dx\\\\\n&=\\int_0^1dx\\int_0^x(e^{-x^2}+e^x)dy\\\\\n&=\\int_0^1(e^{-x^2}+e^x) y\\big|_0^x dx\\\\\n&=\\int_0^1(e^{-x^2}+e^x)xdx\\\\\n&=\\int_0^1xe^{-x^2}dx+\\int_0^1xe^xdx.\n\\end{align*}\nBy substitution $t=x^2$, we obtain\n\\[\n\\int_0^1xe^{-x^2}dx=\\frac12\\int_0^1e^{-t}dt=-\\frac12 e^{-t}\\big|_0^1=\\frac12-\\frac12 e^{-1}.\n\\]\nBy integration by parts, we have\n\\[\n\\int_0^1xe^xdx=\\int_0^1xd(e^x)=xe^x\\big|_0^1-\\int_0^1e^xdx=e-e^x\\big|_0^1=e-(e-1)=1.\n\\]\nCombining all the steps, we can obtain\n\\[\n\\int_0^1dy\\int_y^1(e^{-x^2}+e^x)dx=\\left(\\frac12-\\frac12 e^{-1}\\right)+1=\\frac{3}{2}-\\frac12 e^{-1}.\n\\]\\\\"
     },
     {
         "label": "",
         "problem": "Assume that $a_n>0$ for all $n\\in\\mathbb{N}$ and the series $\\displaystyle{\\sum_{n=1}^\\infty a_n}$ converges to $4$. Let $\\displaystyle{R_n=\\sum_{k=n}^\\infty a_k}$ for all $n=1, 2,\\dots$. Evaluate $\\displaystyle{\\sum_{n=1}^\\infty \\frac{a_n}{\\sqrt{R_n}+\\sqrt{R_{n+1}}}}$.",
-        "answer": "2.",
+        "answer": "2",
         "solution": "Noting that $R_n-R_{n+1}=a_n$ for all $n$ and\n\\[ \n\\frac{a_n}{\\sqrt{R_{n}}+\\sqrt{R_{n+1}}}=\\frac{a_n}{\\sqrt{R_{n}} + \\sqrt{R_{n+}}} \\cdot \\frac{\\sqrt{R_{n}} - \\sqrt{R_{n+1}}}{\\sqrt{R_{n}} - \\sqrt{R_{n+1}}}=\\frac{a_n(\\sqrt{R_{n}} - \\sqrt{R_{n+1}})}{R_{n}- R_{n+1}}=\\sqrt{R_{n}} - \\sqrt{R_{n+1}}.\n \\]\n\nTo evaluate the series $\\sum_{n=1}^\\infty \\frac{a_n}{\\sqrt{R_{n+1}} + \\sqrt{R_n}}$, we'll use a telescoping series:\n\\begin{align*}\n\\sum_{n=1}^\\infty \\frac{a_n}{\\sqrt{R_{n}} + \\sqrt{R_{n+1}}}&=\\sum_{n=1}^\\infty (\\sqrt{R_{n}} - \\sqrt{R_{n+1}})\\\\\n&=[\\sqrt{R_{1}} - \\sqrt{R_{2}}]+[\\sqrt{R_{2}} - \\sqrt{R_{3}}]+[\\sqrt{R_{3}} - \\sqrt{R_{4}}]+\\cdots\\\\\n&=\\sqrt{R_{1}}=\\sqrt{\\sum_{n=1}^\\infty a_n}=\\sqrt{4}=2.\n\\end{align*}\n\nTherefore, the series $\\displaystyle{\\sum_{n=1}^\\infty \\frac{a_n}{\\sqrt{R_n}+\\sqrt{R_{n+1}}}}$ converges to $2$.\\\\"
     },
     {
         "label": "",
         "problem": "For any $a>0$ and $b\\in\\mathbb{R}$, use Sterling's formula\n \\[\n \\lim\\limits_{x\\to\\infty}\\frac{\\Gamma(x+1)}{x^x e^{-x}\\sqrt{2\\pi x}}=1\n \\]\n   to evaluate the limit \n \\[\n  \\lim\\limits_{n\\to\\infty}\\frac{\\Gamma(an+b)}{(n!)^a\\: a^{an+b-\\frac12}n^{b-\\frac12-\\frac{a}{2}}(2\\pi)^{\\frac12-\\frac{a}{2}}},\n \\]\n where $\\Gamma(\\alpha)=\\int_0^\\infty t^{\\alpha-1}e^{-t}dt$ is the gamma function defined for any $\\alpha>0$.",
-        "answer": "1.",
+        "answer": "1",
         "solution": "Since $a>0$, we know that $an+b=(an+b-1)+1\\to\\infty$ as $n\\to\\infty$. By Sterling's formula, we have\n\\[\n\\lim\\limits_{n\\to\\infty}\\frac{\\Gamma(an+b)}{(an+b-1)^{an+b-1}e^{-(an+b-1)}\\sqrt{2\\pi(an+b-1)}}=1.\n\\]\nNoting that $\\Gamma(n+1)=n!$, we get\n\\[\n\\lim\\limits_{n\\to\\infty}\\frac{n!}{n^{n}e^{-n}\\sqrt{2\\pi n}}=1.\n\\]\nThus,\n\\[\n\\lim\\limits_{n\\to\\infty}\\frac{(n!)^a}{n^{an}e^{-an}(2\\pi n)^{\\frac{a}{2}}}=1.\n\\]\nThen,\n\\begin{align*}\n &\\lim\\limits_{n\\to\\infty}\\frac{\\Gamma(an+b)}{(n!)^a\\: a^{an+b-\\frac12}n^{b-\\frac12-\\frac{a}{2}}(2\\pi)^{\\frac12-\\frac{a}{2}}}\\\\\n = &\\lim\\limits_{n\\to\\infty}\\frac{\\Gamma(an+b)}{(an+b-1)^{an+b-1}e^{-(an+b-1)}\\sqrt{2\\pi(an+b-1)}}\\cdot\\frac{(an+b-1)^{an+b-1}e^{-(an+b-1)}\\sqrt{2\\pi(an+b-1)}}{(n!)^a\\: a^{an+b-\\frac12}n^{b-\\frac12-\\frac{a}{2}}(2\\pi)^{\\frac12-\\frac{a}{2}}}\\\\\n =&\\lim\\limits_{n\\to\\infty}\\frac{(an+b-1)^{an+b-1}e^{-(an+b-1)}\\sqrt{2\\pi(an+b-1)}}{(n!)^a\\: a^{an+b-\\frac12}n^{b-\\frac12-\\frac{a}{2}}(2\\pi)^{\\frac12-\\frac{a}{2}}}\\\\\n =&\\lim\\limits_{n\\to\\infty}\\frac{(an+b-1)^{an+b-1}e^{-(an+b-1)}\\sqrt{2\\pi(an+b-1)}}{\\: a^{an+b-\\frac12}n^{b-\\frac12-\\frac{a}{2}}(2\\pi)^{\\frac12-\\frac{a}{2}}\\cdot n^{an}e^{-an}(2\\pi n)^{\\frac{a}{2}}}\\cdot\\frac{n^{an}e^{-an}(2\\pi n)^{\\frac{a}{2}}}{(n!)^a}\\\\\n =&\\lim\\limits_{n\\to\\infty}\\frac{(an+b-1)^{an+b-1}e^{-(an+b-1)}\\sqrt{2\\pi(an+b-1)}}{\\: a^{an+b-\\frac12}n^{b-\\frac12-\\frac{a}{2}}(2\\pi)^{\\frac12-\\frac{a}{2}}\\cdot n^{an}e^{-an}(2\\pi n)^{\\frac{a}{2}}}\\\\\n=&\\lim\\limits_{n\\to\\infty}\\frac{(an+b-1)^{an+b-\\frac12}e^{-(b-1)}}{\\: a^{an+b-\\frac12}n^{b-\\frac12} n^{an}}\\\\\n=&\\lim\\limits_{n\\to\\infty}\\frac{(an+b-1)^{an}(an+b-1)^{b-\\frac12}e^{-(b-1)}}{\\: (an)^{an}n^{b-\\frac12} a^{b-\\frac12}}\\\\\n=&\\lim\\limits_{n\\to\\infty}\\left(\\frac{an+b-1}{an}\\right)^{an}\\left(\\frac{an+b-1}{n}\\right)^{b-\\frac12}\\frac{e^{-(b-1)}}{ a^{b-\\frac12}}\\\\\n=&\\lim\\limits_{n\\to\\infty}\\left(1+\\frac{\\frac{b-1}{a}}{n}\\right)^{an}\\left(a+\\frac{b-1}{n}\\right)^{b-\\frac12}\\frac{e^{-(b-1)}}{ a^{b-\\frac12}}.\n\\end{align*}\nNoticing that \n\\[\n\\lim\\limits_{n\\to\\infty}\\left(1+\\frac{x}{n}\\right)^{n}=e^x, \\ \\forall x\\in\\mathbb{R},\n\\]\nwe obtain\n\\[\n\\lim\\limits_{n\\to\\infty}\\left(1+\\frac{\\frac{b-1}{a}}{n}\\right)^{an}=\\lim\\limits_{n\\to\\infty}\\left[\\left(1+\\frac{\\frac{b-1}{a}}{n}\\right)^{n}\\right]^a=\\left[e^{\\frac{b-1}{a}}\\right]^a=e^{b-1}.\n\\]\nNotice also that $\\lim\\limits_{n\\to\\infty}\\left(a+\\frac{b-1}{n}\\right)^{b-\\frac12}=a^{b-\\frac12}$.\n\nTherefore, by putting everything together, we can obtain the limit \n\\[\n \\lim\\limits_{n\\to\\infty}\\frac{\\Gamma(an+b)}{(n!)^a\\: a^{an+b-\\frac12}n^{b-\\frac12-\\frac{a}{2}}(2\\pi)^{\\frac12-\\frac{a}{2}}}=1.\n\\]\\\\"
     },
     {
@@ -2042,25 +2042,25 @@
     {
         "label": "",
         "problem": "Given three vectors $y_1=(1,0,0)^\\top,y_2=(x,0,0)^\\top$ and $y_3=(x^2,0,0)^\\top$. Does there exist a system of three linear homogeneous ODEs such that all of $y_1,y_2,y_3$ are the solution to this homogeneous ODE system?",
-        "answer": "No.",
+        "answer": "No",
         "solution": "Suppose there is such a system. Then, $[y_1,y_2,y_3]$ gives a fundamental matrix of the solution, and $\\det[y_1,y_2,y_3]\\neq 0$. Consider the linear system $C_1y_1+C_2y_2+C_3y_3=\\vec{0}$, which implies $C_1+C_2x+C_3x^2=0$. This quadratic equation cannot hold for all $x\\in\\mathbb{R}$ unless $C_1=C_2=C_3=0$, that is, $y_1,y_2,y_3$ are linearly independent. It implies that the determinant $\\det[y_1,y_2,y_3]=0$, which leads to a contradiction."
     },
     {
         "label": "",
         "problem": "Does the ODE $x^2y''+(3x-1)y'+y=0$ have a nonzero power series solution near $x=0$?",
-        "answer": "No.",
+        "answer": "No",
         "solution": "Assume there exists a power series solution $y=\\sum_{n\\geq 0} c_n x^n$. Plugging it into the equation, we can get the recursive formula $c_{n+1}=(n+1)c_n$ for $n\\geq 0$. Then, $c_n=c_0 n!$. But we know $\\sum_{n\\geq 0} n! x^n$ must be divergent as long as $x\\neq 0$."
     },
     {
         "label": "",
         "problem": "Is $y=0$ a singular solution to $y'=\\sqrt{y}\\ln(\\ln(1+y))$?",
-        "answer": "Yes.",
+        "answer": "Yes",
         "solution": "First, $y=0$ is a singular solution to the ODE. In fact, this is a separable ODE, and the solution is given by $x(y)=\\int_0^y \\frac{dt}{\\sqrt{t}\\ln\\ln(1+t)}$ when $y\\neq 0$. On the other hand, $y=0$ is indeed a solution. So $y=0$ is a singular solution."
     },
     {
         "label": "",
         "problem": "For the ODE system $x'(t)=y+x(x^2+y^2)$ and $y'(t)=-x+y(x^2+y^2)$, is the equilibrium $(x,y)=(0,0)$ stable?",
-        "answer": "No \\\\",
+        "answer": "No \\",
         "solution": "The equilibrium $(x,y)=(0,0)$ for the linear counterpart is a center, as the coefficient matrix has eigenvalues $\\pm i$, purely imaginary. Even if the nonlinearity is locally linear ($o(\\sqrt{x^2+y^2})$ size near $(0,0)$), we cannot tell the type of the equilibrium $(x,y)=(0,0)$ for the nonlinear system. Instead, we can introduce the Lyapunov function $V(x,y)=\\frac{x^2+y^2}{2}$. Along the trajectory, we compute that\n\\[\n\\frac{dV}{dt}=xx'(t)+yy'(t)=xy+2(x^2+y^2)^2-xy>0\\quad\\forall (x,y)\\neq (0,0).\n\\]\nThat is to say, $V(x,y)$ is increasing as $t$ grows. So, any trajectory starting near the origin will penetrate the circles (the trajectories for the linearized system) and leave away from the equilibrium $(x,y)=(0,0)$. Thus, the equilibrium $(x,y)=(0,0)$ for the nonlinear system is unstable."
     },
     {
@@ -2084,13 +2084,13 @@
     {
         "label": "",
         "problem": "Does there exists any nonzero function $f(x)\\in L^2(\\mathbb{R}^n)$ such that $f$ is harmonic in $\\mathbb{R}^n$?",
-        "answer": "No.",
+        "answer": "No",
         "solution": "If there exists such a function $u$, then taking Fourier transform, we get $-|\\xi|^2\\hat{f}(\\xi)=0$. $\\hat{f}\\in L^2(\\mathbb{R}^n)$ and thus it is supported in $\\{\\xi=0\\}$. So, $\\hat{f}=0$ in $L^2$ and by Plancherel theorem $f=0$ in $L^2$. Since harmonic function is smooth, the function $f$ must be identically zero."
     },
     {
         "label": "",
         "problem": "Let $u$ be a harmonic function in $\\mathbb{R}^n$ satisfying $|u(x)|\\leq 100(100+\\ln(100+|x|^{100}))$ for any $x\\in\\mathbb{R}^n$. Can we assert $u$ is a constant?",
-        "answer": "Yes.",
+        "answer": "Yes",
         "solution": "By the gradient estimate for harmonic functions, we have \n\\[\n|\\nabla u(x)|\\leq \\frac{n}{R}\\max\\limits_{\\overline{B(x,R)}}|u(x)|\\leq   \\frac{100n}{R}(100+\\ln(100+R^{100})).\n\\]Let $R\\to\\infty$ and we get $\\nabla u\\equiv 0$. So $u$ must be a constant."
     },
     {
@@ -2108,7 +2108,7 @@
     {
         "label": "",
         "problem": "In how many ways can you arrange the letters in the word ``INTELLIGENCE''?",
-        "answer": "9979200.",
+        "answer": "9979200",
         "solution": "It is given by the multinomial coefficient $\\binom{12}{2, 2, 1, 3, 2, 1, 1}=\\frac{12!}{2!2!1!3!2!1!1!} = 9,979,200$."
     },
     {
@@ -2180,7 +2180,7 @@
     {
         "label": "",
         "problem": "Find the variance of the random variable $X$ if the cumulative distribution function of $X$ is\n    $$F(x) = \\begin{cases} 0, & {\\rm if} \\ x < 1, \\\\ 1 - 2e^{-x}, & {\\rm if} \\ x \\geq 1. \\end{cases}$$",
-        "answer": "0.93\\\\",
+        "answer": "0.93\\",
         "solution": "The random variable $X$ has a point mass at $x=1$. $P(X=1) = 1-2e^{-1}$. \n    $$ \\mathbb{E}[X] = (1)P(X=1) + \\int_1^\\infty xf(x) dx = (1-2e^{-1}) + \\int_1^\\infty 2xe^{-x} dx = 1 + 2e^{-1} $$\n    $$\\mathbb{E}[X^2] = (1^2)P(X=1) + \\int_1^\\infty x^2f(x) dx\n          = (1-2e^{-1}) + \\int_1^\\infty 2x^2e^{-x} dx = 1 + 8e^{-1}.$$\n    $${\\rm Var}[X] = \\mathbb{E}[X^2] - (\\mathbb{E}[X])^2 = 4e^{-1}(1-e^{-1}).$$"
     },
     {
@@ -2198,7 +2198,7 @@
     {
         "label": "",
         "problem": "The joint probability density function for the random variables $X$ and $Y$ is \n      $$f(x, y) = 6e^{-(2x+3y)}, \\ x>0, \\ y>0.$$\n    Calculate the variance of $X$ given that $X>1$ and $Y>2$.",
-        "answer": "0.25.",
+        "answer": "0.25",
         "solution": "The marginal density functions can be found as follows.\n    $$f_X(x) = \\int_0^\\infty f(x, y) dy = 2e^{-2x}, \\ x>0,$$\n    $$f_Y(y) = \\int_0^\\infty f(x, y) dx = 3e^{-3y}, \\ y>0.$$\n    Clearly, $f(x, y) = f_X(x)f_Y(y)$ and this implies that the random variables are independent. Thus, ${\\rm Var}[X|X>1, Y>2] = {\\rm Var}[X|X>1]$. Taking into account $P(X>1) = e^{-2}$, we have\n    $$\\mathbb{E}[X|X>1] = \\int_1^\\infty 2xe^{-2x}\\cdot \\frac{1}{e^{-2}} dx = 1.5,$$\n    $$\\mathbb{E}[X^2|X>1] = \\int_1^\\infty 2x^2e^{-2x}\\cdot \\frac{1}{e^{-2}} dx = 2.5.$$\n    Thus, $${\\rm Var}[X|X>1, Y>2] = {\\rm Var}[X|X>1] = 2.5 - 1.5^2 = 0.25.$$"
     },
     {
@@ -2258,7 +2258,7 @@
     {
         "label": "",
         "problem": "The accompanying data on cube compressive strength (MPa) of concrete specimens are listed as follows:\n\t\t\\[\n\t\t112.3  \\quad   97.0  \\quad  92.7  \\quad  86.0  \\quad  102.0  \\quad  99.2  \\quad  95.8  \\quad  103.5  \\quad  89.0  \\quad  86.7.\n\t\t\\]\n\t\tAssume that the compressive strength for this type of concrete is normally distributed. Suppose the concrete will be used for a particular application unless there is strong evidence that the true average strength is less than 100 MPa. Should the concrete be used under significance level 0.05?",
-        "answer": "Yes. \\\\ \n\t% Yes, the concrete should be used.",
+        "answer": "Yes. \\\n\t% Yes, the concrete should be used.",
         "solution": "We want to test the following hypotheses\n\t\\[\n\tH_0: \\mu=100 \\quad vs. \\quad H_1: \\mu<100.\n\t\\]\n\tThe test statistic is\n\t\\[\n\tt= \\frac{\\bar{x}-\\mu_0}{s/\\sqrt{n}} = \\frac{96.42 - 100}{8.26/\\sqrt{10}} \\approx -1.37.\n\t\\]\n\tThe p-value is \n\t\\[\n\tP(t_{9}\\le -1.37) \\approx 0.102\n\t\\]\n\twhich is greater than 0.05. So, we do not reject $H_0$ and so the concrete should be used."
     },
     {
@@ -2282,7 +2282,7 @@
     {
         "label": "",
         "problem": "The ages of a random sample of five university professors are 39, 54, 61, 72, and 59. Using this\n\t\tinformation, find a 99\\% confidence interval for the population variance of the ages of all professors at the university, assuming that the ages of university professors are normally distributed.",
-        "answer": "(38.90, 2792.41).",
+        "answer": "(38.90, 2792.41)",
         "solution": "We have that $n = 5$ and the sample variance $s^2 = 144.5$. Meanwhile, the critical values for chi-square distribution with degree of freedom 4 are given by $\\chi_{0.995, 4}^2=0.20699$ and $\\chi_{0.005, 4}^2=14.8602$. Thus, the 99\\% confidence interval for the variance is given by \n\t\\[\n\t\\left(\\frac{(n-1)s^2}{\\chi_{0.005, 4}^2}, \\frac{(n-1)s^2}{\\chi_{0.995, 4}^2}\\right) = \\left(\\frac{4\\cdot 144.5}{14.8602}, \\frac{4\\cdot 144.5}{0.20699}\\right) = (38.90, 2792.41).\n\t\\]"
     },
     {
@@ -2312,7 +2312,7 @@
     {
         "label": "",
         "problem": "Let $U_1, U_2, \\ldots$, be i.i.d. ${\\rm Uniform}(0,1)$ random variables and let $X_n=\\left(\\prod_{k=1}^{n} U_k\\right)^{-1/n}$. What is the variance of the asymptotic distribution of $\\frac{\\sqrt{n}(X_n-e)}{e}$ as $n\\to \\infty$?",
-        "answer": "1 \\\\",
+        "answer": "1 \\",
         "solution": "Let $Y_n = \\log X_n = \\frac{1}{n}\\sum_{k=1}^n (-\\log U_k)$. Note that $-\\log U_k$ are i.i.d. with Exponential distribution with parameter 1, having mean $\\mu=1$ and variance $\\sigma^2=1$. By the central limit theorem,\n\\[\n\\frac{\\sqrt{n}(Y_n-\\mu)}{\\sigma} = \\sqrt{n}(Y_n-1) \\xrightarrow{d} N(0,1).\n\\]\nApplying the Delta method with $g(y)=e^y$ such that $g(1)=e$ and $g'(1)=e>0$, we obtain\n\\[\n\\sqrt{n}(g(Y_n)-g(1))\\xrightarrow{d} N(0,[g'(1)]^2),\n\\] \nwhich is equivalent to $\\sqrt{n}(X_n-e) \\xrightarrow{d} N(0,e^2)$, yielding\n\\[\n\\frac{\\sqrt{n}(X_n-e)}{e} \\xrightarrow{d} N(0,1).\n\\]"
     },
     {