The exam of 28 September 2026

(a) The form $\tfrac00.$ At $x=0$ the numerator is $e^{0}-1-0=0$ and the denominator is $0,$ for every $a.$ Both numerator and denominator tend to $0,$ so the form is $\tfrac00$ (Definition 5.18) and l'Hôpital's rule (Fact 5.19) applies: $$ \begin{aligned} &\lim_{x\to0}\frac{e^{ax}-1-ax}{3x^{2}}\\ &\quad=\lim_{x\to0}\frac{a e^{ax}-a}{6x}\,. \end{aligned} $$

(a) Again $\tfrac00$: a second application. At $x=0$ the new numerator is $a-a=0$ and the denominator is $0,$ so this is again of the form $\tfrac00.$ A second application gives a quotient that is continuous at $0$ (Fact 5.17): $$ \lim_{x\to0}\frac{a^{2}e^{ax}}{6}=\frac{a^{2}}{6}\,. $$ (Alternatively: the Taylor polynomial $e^{ax}\approx1+ax+\tfrac12a^{2}x^{2}$ gives the same value.)

(a) Every value of $a.$ The condition $\dfrac{a^{2}}{6}=6$ gives $a^{2}=36,$ so $$ a=6\quad\text{or}\quad a=-6\,. $$

(b) The conjugate. Multiply and divide by the conjugate $\sqrt{9x^{2}+2x}+3x{:}$ $$ \begin{aligned} &\sqrt{9x^{2}+2x}-3x\\ &\quad=\frac{(9x^{2}+2x)-9x^{2}}{\sqrt{9x^{2}+2x}+3x}\\ &\quad=\frac{2x}{\sqrt{9x^{2}+2x}+3x}\,. \end{aligned} $$

(b) Divide by $x>0.$ As $x\to+\infty$ we may take $x>0,$ and then $x=\sqrt{x^{2}},$ so $\dfrac{\sqrt{9x^{2}+2x}}{x}=\sqrt{9+2/x}.$ Dividing numerator and denominator by $x$ gives $$ \begin{aligned} &\frac{2x}{\sqrt{9x^{2}+2x}+3x}\\ &\quad=\frac{2}{\sqrt{9+2/x}+3}\,. \end{aligned} $$

(b) Let $x\to+\infty.$ $2/x\to0,$ so the root tends to $\sqrt9=3$ and the expression tends to $2/(3+3).$ So $$ \lim_{x\to+\infty}\Bigl(\sqrt{9x^{2}+2x}-3x\Bigr)=\frac{1}{3}\,. $$

(a) Divide first. The numerator has higher degree than the denominator, so we divide first (Fact 1.22). Since $$ \begin{aligned} &x^{3}-2x^{2}+x\\ &\quad=x(x-1)^{2}\\ &\quad=(x-1)(x^{2}-x)\,, \end{aligned} $$ we get $$ \begin{aligned} &\frac{x^{3}-2x^{2}+x+3}{x-1}\\ &\quad=x^{2}-x+\frac{3}{x-1}\,. \end{aligned} $$

(a) Check by multiplying back. $$ \begin{aligned} &(x-1)(x^{2}-x)+3\\ &\quad=x^{3}-2x^{2}+x+3\,. \end{aligned} $$

(a) Integrate. For $x>1$ we have $x-1>0,$ so an antiderivative of $\dfrac{3}{x-1}$ is $3\ln(x-1)$ without absolute value. Hence, for $x>1,$ $$ \begin{aligned} &\int\frac{x^{3}-2x^{2}+x+3}{x-1}\,dx\\ &\quad=\frac{x^{3}}{3}-\frac{x^{2}}{2}\\ &\qquad+3\ln(x-1)+C\,, \end{aligned} $$ with $C\in\mathbb{R}$ (Fact 7.13).

(b) Substitute $u=x^{3}.$ Then $du=3x^{2}\,dx$ (Method 8.2). When $x=0,$ $u=0;$ when $x=s,$ $u=s^{3}.$ Since $2^{-u}=e^{-u\ln2}$ (Fact 4.4), $$ \begin{aligned} &\int_{0}^{s}x^{2}\,2^{-x^{3}}\,dx\\ &\quad=\frac13\int_{0}^{s^{3}}2^{-u}\,du\\ &\quad=\frac13\Bigl[-\frac{2^{-u}}{\ln2}\Bigr]_{0}^{s^{3}}\\ &\quad=\frac{1}{3\ln2}\bigl(1-2^{-s^{3}}\bigr)\,. \end{aligned} $$

(b) Let $s\to+\infty.$ The improper integral is the limit of these integrals (Definition 8.5). As $s\to+\infty,$ $2^{-s^{3}}\to0,$ so the improper integral converges and $$ \int_{0}^{+\infty}x^{2}\,2^{-x^{3}}\,dx=\frac{1}{3\ln2}\,. $$

(a) A cubic in $y.$ Substituting $x=4$ gives $$ y^{3}+y-2=0\,. $$ Testing small integers (Method 1.19), $y=1$ is a root: $1+1-2=0.$ Dividing out the factor $y-1$ (Fact 1.17), $$ \begin{aligned} &y^{3}+y-2\\ &\quad=(y-1)(y^{2}+y+2)\,. \end{aligned} $$

(a) The only value. The quadratic factor has discriminant $1-8<0,$ so it has no real roots. Hence $$ y(4)=1 $$ is the only possible value.

(b) Differentiate once. Differentiate the equation implicitly with respect to $x$ (Method 4.9); every $y$ is a function of $x,$ and $(x-3)\,y$ is a product: $$ 3y^{2}y'+y+(x-3)\,y'=0\,. \tag{1} $$ At $x=4,$ $y=1{:}$ $3y'+1+y'=0,$ so $y'(4)=-\tfrac{1}{4}.$

(b) Differentiate (1) once more, before any numbers go in. Each $y'$ leaves a $y'',$ and each product gives two terms: $$ \begin{aligned} &6y\,(y')^{2}+3y^{2}y''\\ &\quad+y'+y'+(x-3)\,y''=0\,. \end{aligned} $$

(b) Insert $x=4,$ $y=1,$ $y'=-\tfrac14.$ $$ 6\cdot\tfrac{1}{16}+3y''-\tfrac12+y''=0\,, $$ that is $4y''=\tfrac12-\tfrac38=\tfrac18,$ so $y''(4)=\tfrac{1}{32}.$

(b) The Taylor polynomial (Definition 4.19) of degree $2,$ with $y''(4)$ divided by $2{:}$ $$ \begin{aligned} P_{2}(x)&=y(4)+y'(4)(x-4)\\ &\quad+\frac{y''(4)}{2}(x-4)^{2}\\ &=1-\frac{1}{4}(x-4)\\ &\quad+\frac{1}{64}(x-4)^{2}\,. \end{aligned} $$

(c) Increasing or decreasing. Since $y'(4)=-\tfrac14<0$ and $y'$ is continuous, $y'<0$ on an interval around $x=4,$ so $y$ is decreasing near $x=4$ (Fact 2.23).

(c) Convex or concave. Since $y''(4)=\tfrac{1}{32}>0$ (and $y''$ is continuous: solve the twice-differentiated equation for $y''),$ $y''>0$ near $x=4,$ so $y$ is convex near $x=4$ (Fact 3.22, Fact 6.20).

(a) The derivative. By the quotient rule (Fact 3.11), $f'(x)=\dfrac{N(x)}{(x+2)^{2}}$ with $$ \begin{aligned} N(x)&=(2x+a)(x+2)\\ &\quad-(x^{2}+ax+b)\\ &=x^{2}+4x+2a-b\,. \end{aligned} $$

(a) Two conditions, a linear system. The conditions $f'(1)=0$ and $f(1)=3$ give $$ \begin{aligned} 5+2a-b&=0\,,\\ \frac{1+a+b}{3}&=3\,, \end{aligned} $$ that is $2a-b=-5$ and $a+b=8.$ Adding gives $3a=3,$ so $$ a=1\,,\quad b=7\,. $$

(b) The factored derivative. With $a=1,$ $b=7$ we have $f(x)=\dfrac{x^{2}+x+7}{x+2}$ and $$ \begin{aligned} f'(x)&=\frac{x^{2}+4x-5}{(x+2)^{2}}\\ &=\frac{(x+5)(x-1)}{(x+2)^{2}}\,. \end{aligned} $$

(b) The sign table. The denominator is positive for $x\ne-2,$ so the sign of $f'$ is the sign of $(x+5)(x-1).$ The table is set on its side, in two halves to fit a phone: first the factors, then $f'$ and $f.$ The pole $x=-2$ has a row of its own, where $f'$ and $f$ are not defined (the dash).

Their product gives the sign of $f'{:}$

(b) The intervals (Fact 2.23). $f$ is increasing on $(-\infty,-5]$ and on $[1,+\infty),$ and decreasing on $[-5,-2)$ and on $(-2,1].$ The two intervals of decrease are separate: $f$ is not defined at $-2.$

(b) The local extreme points. By the first-derivative test (Fact 6.18), $x=-5$ is a local maximum point with $$ f(-5)=\frac{25-5+7}{-3}=-9\,, $$ and $x=1$ is a local minimum point with $f(1)=3.$

(c) Divide first. Division with remainder (Fact 1.22) gives $$ f(x)=x-1+\frac{9}{x+2}\,, $$ (check: $(x-1)(x+2)+9=x^{2}+x+7),$ so $$ \begin{aligned} f'(x)&=1-\frac{9}{(x+2)^{2}}\,,\\ f''(x)&=\frac{18}{(x+2)^{3}}\,. \end{aligned} $$

(c) Convex and concave. Hence $f''>0$ for $x>-2$ and $f''<0$ for $x<-2{:}$ $f$ is convex on $(-2,+\infty)$ and concave on $(-\infty,-2)$ (Fact 6.20). (There is no inflection point, since $x=-2$ is not in the domain.)

(c) No global maximum. $f$ has no global maximum: for instance $f(x)=x-1+9/(x+2)\to+\infty$ as $x\to+\infty$ (for $x>-2$ the fraction is positive, so $f(x)>x-1),$ so $f$ takes arbitrarily large values. (The local maximum value $-9$ is even smaller than the local minimum value $3.)$

(a) The stationary points. The first-order conditions (Fact 10.1) are $$ \begin{aligned} \pi'_{x}&=16-4x-2y=0\,,\\ \pi'_{y}&=14-2x-4y=0\,. \end{aligned} $$ From the first equation $y=8-2x;$ the second then gives $14-2x-32+8x=0,$ so $x=3$ and $y=2.$ The only stationary point is $(3,2),$ with $$ \begin{aligned} \pi(3,2)&=48+28-18-12-8\\ &=38\,. \end{aligned} $$

(a) Classification. The second-order partial derivatives are constant: $A=\pi''_{xx}=-4,$ $B=\pi''_{xy}=-2,$ $C=\pi''_{yy}=-4,$ so $$ \begin{aligned} AC-B^{2}&=16-4=12>0\,,\\ A&=-4<0\,. \end{aligned} $$ Hence $(3,2)$ is a local maximum point (Fact 11.4).

(a) Global. Since $A<0$ and $AC-B^{2}>0$ at every point of $\mathbb{R}^{2},$ the function $\pi$ is concave on the whole plane (Fact 10.11), and a stationary point of a concave function is a global maximum point (Fact 10.13). So $\pi$ has the global maximum $$ \pi(3,2)=38\,. $$

(b) One variable on a closed interval. On $S$ we have $y=3-x$ with $0\le x\le3$ (Method 10.19), and $$ \begin{aligned} g(x)&=\pi(x,3-x)\\ &=16x+14(3-x)-2x^{2}\\ &\quad-2x(3-x)-2(3-x)^{2}\\ &=-2x^{2}+8x+24\,. \end{aligned} $$

(b) Existence, and the candidates. $S$ is a closed and bounded set (a segment) and $\pi$ is continuous, so the extreme values exist (Fact 11.9); they are attained at a stationary point of $g$ in $(0,3)$ or at an endpoint. $g'(x)=-4x+8=0$ gives $x=2,$ and $$ \begin{aligned} g(2)&=32\,,\\ g(0)&=24\,,\\ g(3)&=30\,. \end{aligned} $$

(b) The extreme values. Therefore $$ \max_{S}\pi=32\ \text{at}\ (x,y)=(2,1)\,, $$ and $$ \min_{S}\pi=24\ \text{at}\ (0,3)\,. $$

(c) The Lagrange conditions. With the constraint $x+y=3,$ the Lagrangian is $\mathcal{L}=\pi(x,y)-\lambda(x+y-3),$ and the first-order conditions (Method 11.14) are $$ \begin{aligned} 16-4x-2y&=\lambda\,,\\ 14-2x-4y&=\lambda\,,\\ x+y&=3\,. \end{aligned} $$ Subtracting the first two equations gives $2-2x+2y=0,$ that is $x=y+1;$ with $x+y=3$ this gives $y=1,$ $x=2$ (the maximum point found in (b)) and $$ \lambda=16-8-2=6\,. $$

(c) What $\lambda$ means. The multiplier is the derivative of the optimal value with respect to the constraint level (Fact 11.21): $$ \frac{d\pi^{*}}{dc}=\lambda\,, $$ where $\pi^{*}(c)$ is the maximal profit under $x+y=c.$ In words: one more thousand units of capacity raises the maximal profit by about $6$ million kronor.

(c) The estimate. Hence $$ \begin{aligned} \pi^{*}(3.1)&\approx\pi^{*}(3)+\lambda\cdot0.1\\ &=32+0.6=32.6 \end{aligned} $$ million kronor. (The exact value is $\pi^{*}(3.1)=32.585{:}$ the conditions still give $x=y+1,$ so $x=2.05,$ $y=1.05.)$

(a) Geometric, and the domain of the root. The series is geometric with ratio $k=\sqrt{x+3}/2$ (and first term $k,$ since $n$ starts at $1).$ The square root requires $x\ge-3,$ and then $k\ge0.$

(a) Convergence. A geometric series converges exactly when $|k|<1$ (Fact 8.7), that is $$ \begin{aligned} \frac{\sqrt{x+3}}{2}<1&\iff\sqrt{x+3}<2\\ &\iff x+3<4\\ &\iff x<1\,. \end{aligned} $$ So the series converges exactly for $$ -3\le x<1\,. $$ (At $x=-3$ the ratio is $0$ and every term is $0;$ at $x=1$ the ratio is $1$ and the series diverges.)

(a) The sum equal to $1.$ For such $x$ the sum is $\dfrac{k}{1-k}.$ Setting this equal to $1$ gives $k=1-k,$ so $k=\tfrac12,$ i.e. $\sqrt{x+3}=1$ and $$ x=-2\,, $$ which lies in the interval of convergence.

(b) The sum the words describe. The first deposit has earned interest for $n-1$ years, the second for $n-2$ years, …, and the last for $0$ years. Hence $$ \begin{aligned} S_{n}&=1000\bigl(1.05^{n-1}+1.05^{n-2}\\ &\qquad+\cdots+1.05+1\bigr)\,, \end{aligned} $$ a finite geometric sum with first term $1000,$ ratio $1.05$ and $n$ terms.

(b) The closed formula. By the formula for a geometric sum, $$ \begin{aligned} S_{n}&=1000\cdot\frac{1.05^{n}-1}{1.05-1}\\ &=\frac{1000}{0.05}\bigl(1.05^{n}-1\bigr)\\ &=20\,000\bigl(1.05^{n}-1\bigr)\,. \end{aligned} $$

(b) The inequality, and logarithms. The balance exceeds $40\,000$ when $20\,000(1.05^{n}-1)>40\,000,$ i.e. $1.05^{n}>3.$ Taking logarithms (the logarithm is increasing, Method 2.11), $$ \begin{aligned} &n\ln1.05>\ln3\\ &\iff n>\frac{\ln3}{\ln1.05}\\ &\qquad\approx\frac{1.0986}{0.04879}\approx22.5\,. \end{aligned} $$

(b) The smallest $n.$ So the smallest such $n$ is $$ n=23\,. $$ Check: $S_{22}\approx38\,505$ and $S_{23}\approx41\,430.$

(a) The determinant. Expanding along the first row (Method 14.7), $$ \det\mathbf{A}= \begin{vmatrix} 1&1&a\\ 1&a&1\\ 2&1&1 \end{vmatrix} $$ $$ \begin{aligned} &=1\cdot(a-1)-1\cdot(1-2)\\ &\quad+a\cdot(1-2a)\\ &=a-1+1+a-2a^{2}\\ &=2a-2a^{2}=2a(1-a)\,. \end{aligned} $$

(a) Exactly one solution. The system has exactly one solution precisely when $\det\mathbf{A}\ne0$ (Fact 14.26), i.e. for $$ a\ne0\ \text{and}\ a\ne1\,. $$

(b) $a=1.$ The determinant only says not exactly one (Watch out 15.8), so look at the equations. The first two read $x+y+z=1$ and $x+y+z=3,$ which contradict each other $(R_{2}-R_{1}$ gives $0=2).$ The system has no solution.

(b) $a=0.$ The system is $x+y=1,$ $x+z=3,$ $2x+y+z=4.$ The third equation is the sum of the first two $(R_{3}-R_{1}-R_{2}$ gives $0=0),$ so it can be dropped. Put $z=t$ (free); then $x=3-t$ and $y=1-x=t-2.$ There are infinitely many solutions: $$ \begin{aligned} &(x,y,z)=(3-t,\ t-2,\ t)\,,\\ &t\in\mathbb{R}\,. \end{aligned} $$

(c) $a=2$: the staircase. The augmented matrix is reduced as follows (Method 13.16; $R_{i}$ denotes row $i).$ $R_{2}-R_{1}$ and $R_{3}-2R_{1},$ then $R_{3}+R_{2}{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c} 1&1&2&1\\ 1&2&1&3\\ 2&1&1&4 \end{array}\right)\\ &\sim \left(\begin{array}{ccc|c} 1&1&2&1\\ 0&1&-1&2\\ 0&-1&-3&2 \end{array}\right)\\ &\sim \left(\begin{array}{ccc|c} 1&1&2&1\\ 0&1&-1&2\\ 0&0&-4&4 \end{array}\right) \end{aligned} $$

(c) Clear above. $-\tfrac14R_{3};$ then $R_{2}+R_{3};$ then $R_{1}-R_{2}-2R_{3}$ with the new $R_{2}{:}$ $$ \begin{aligned} &\left(\begin{array}{ccc|c} 1&1&2&1\\ 0&1&-1&2\\ 0&0&1&-1 \end{array}\right)\\ &\sim \left(\begin{array}{ccc|c} 1&0&0&2\\ 0&1&0&1\\ 0&0&1&-1 \end{array}\right) \end{aligned} $$

(c) The solution. That is, $$ (x,y,z)=(2,\ 1,\ -1)\,. $$ Check: $2+1-2=1,$ $2+2-1=3,$ $4+1-1=4.$