Practice exam 2

(a) The form. At $x=0$ the numerator is $\ln 1=0$ and the denominator is $e^{0}-1=0.$ Both tend to $0$ as $x\to0,$ so the quotient is of the form $\tfrac00$ (Definition 5.18), whatever $a$ is, and l'Hôpital's rule (Fact 5.19) may be used.

(a) Differentiate numerator and denominator separately. By the chain rule, $$ \begin{aligned} \frac{d}{dx}\ln(1+ax)&=\frac{a}{1+ax}\,,\\ \frac{d}{dx}\bigl(e^{2x}-1\bigr)&=2e^{2x}\,. \end{aligned} $$

(a) The limit as an expression in $a.$ The new quotient is continuous at $0$ and its denominator is $2\ne0$ there, so we substitute (Fact 5.17): $$ \begin{aligned} &\lim_{x\to0}\frac{\ln(1+ax)}{e^{2x}-1}\\ &\quad=\lim_{x\to0}\frac{\frac{a}{1+ax}}{2e^{2x}} =\frac{a}{2}\,. \end{aligned} $$

(a) The parameter. The condition $\dfrac{a}{2}=4$ gives $$a=8\,.$$ It is the only value: the equation is linear in $a.$

(b) The conjugate. For $x>2$ both roots are defined. Multiply and divide by the conjugate $\sqrt{x^{2}+6x}+\sqrt{x^{2}-2x}{:}$ $$ \begin{aligned} &\sqrt{x^{2}+6x}-\sqrt{x^{2}-2x}\\ &\quad=\frac{(x^{2}+6x)-(x^{2}-2x)}{\sqrt{x^{2}+6x}+\sqrt{x^{2}-2x}}\\ &\quad=\frac{8x}{\sqrt{x^{2}+6x}+\sqrt{x^{2}-2x}}\,. \end{aligned} $$

(b) Divide by $x>0.$ As $x\to+\infty$ we may take $x>0,$ and then $x=\sqrt{x^{2}}.$ Dividing numerator and denominator by $x$ gives $$ \frac{8}{\sqrt{1+\frac6x}+\sqrt{1-\frac2x}}\,. $$

(b) Let $x\to+\infty.$ $\dfrac6x\to0$ and $\dfrac2x\to0,$ so both roots tend to $1,$ and $$ \begin{aligned} \lim_{x\to+\infty}\Bigl(&\sqrt{x^{2}+6x}\\ &-\sqrt{x^{2}-2x}\Bigr)=4\,. \end{aligned} $$

(a) Choose the parts. The factor $\ln x$ gets simpler when it is differentiated, and $x^{2}$ is easy to integrate (Method 7.22). So $u=\ln x$ (to be differentiated) and $v'=x^{2}$ (to be integrated), which gives $$u'=\frac1x\,,\qquad v=\frac{x^{3}}{3}\,.$$ (This is Example 7.24 of the notes.)

(a) Integrate by parts (Fact 7.21): $$ \begin{aligned} &\int_{1}^{e}x^{2}\ln x\,dx\\ &\quad=\Bigl[\frac{x^{3}}{3}\ln x\Bigr]_{1}^{e}\\ &\qquad-\int_{1}^{e}\frac{x^{3}}{3}\cdot\frac1x\,dx\,. \end{aligned} $$

(a) Evaluate. Since $\ln e=1$ and $\ln 1=0,$ the bracket is $\dfrac{e^{3}}{3}-0,$ and the remaining integrand is $\dfrac{x^{2}}{3}{:}$ $$ \begin{aligned} &=\frac{e^{3}}{3}-0-\Bigl[\frac{x^{3}}{9}\Bigr]_{1}^{e}\\ &=\frac{e^{3}}{3}-\frac{e^{3}}{9}+\frac19\\ &=\frac{2e^{3}+1}{9}\,. \end{aligned} $$

(b) A limit. The integral is improper, so by Definition 8.5 it is the limit as $s\to+\infty$ of the integral over $[0,s].$ (The integrand is continuous on $[0,+\infty),$ since $x^{2}+3x+1\ge1$ there.)

(b) Substitute (Method 8.2) $u=x^{2}+3x+1,$ $du=(2x+3)\,dx.$ When $x=0,$ $u=1;$ when $x=s,$ $u=s^{2}+3s+1.$ Hence $$ \begin{aligned} &\int_{0}^{s}\frac{2x+3}{(x^{2}+3x+1)^{2}}\,dx\\ &\quad=\int_{1}^{s^{2}+3s+1}u^{-2}\,du\\ &\quad=\Bigl[-\frac1u\Bigr]_{1}^{s^{2}+3s+1}\\ &\quad=1-\frac{1}{s^{2}+3s+1}\,. \end{aligned} $$

(b) Let $s\to+\infty.$ Then $s^{2}+3s+1\to+\infty,$ so the last term tends to $0.$ The improper integral converges, and $$ \int_{0}^{+\infty}\frac{2x+3}{(x^{2}+3x+1)^{2}}\,dx=1\,. $$

(a) Put $x=1.$ The equation reads $e^{y}+y+1=2,$ that is $$e^{y}+y=1\,.$$ Clearly $y=0$ is a solution: $e^{0}+0=1.$

(a) The only value. Let $h(y)=e^{y}+y.$ Then $$h'(y)=e^{y}+1>0\,,$$ so $h$ is strictly increasing (Fact 2.23) and takes each value at most once. Hence $y=0$ is the only solution, and $$y(1)=0\,.$$

(b) Differentiate implicitly (Method 4.9) with respect to $x.$ Every $y$ leaves a $y',$ and $x\,e^{y}$ needs the product rule (Fact 3.8) and the chain rule (Fact 4.2): $$ e^{y}+x\,e^{y}y'+y'+3x^{2}=0\,.\tag{1} $$

(b) Insert the point. At $x=1,$ $y=0{:}$ $$1+y'+y'+3=0\,,$$ so $y'(1)=-2.$

(b) The tangent line at $(1,0)$ is $y=y(1)+y'(1)(x-1),$ that is $$y=-2(x-1)=-2x+2\,.$$

(c) Differentiate (1) once more. Term by term, with the product rule on $x\,e^{y}y'{:}$ $$ \begin{aligned} &e^{y}y'+e^{y}y'+x\,e^{y}(y')^{2}\\ &\quad+x\,e^{y}y''+y''+6x=0\,. \end{aligned} $$

(c) Insert $x=1,$ $y=0,$ $y'=-2.$ $$-2-2+4+y''+y''+6=0\,,$$ so $2y''+6=0$ and $y''(1)=-3.$

(c) The Taylor polynomial of degree 2 (Definition 4.19): $$ \begin{aligned} &P_{2}(x)=y(1)+y'(1)(x-1)\\ &\quad+\frac{y''(1)}{2}(x-1)^{2}\\ &\quad=-2(x-1)-\frac32(x-1)^{2}\,. \end{aligned} $$

(c) The shape. $y''(1)=-3<0,$ and $y''$ is continuous, so $y''<0$ on an interval around $x=1.$ Hence $y(x)$ is concave near $x=1$ (Fact 6.20).

(a) The derivative. By the product rule (Fact 3.8), with $\bigl(e^{-x}\bigr)'=-e^{-x},$ $$ \begin{aligned} f'(x)&=(2x+a)e^{-x}\\ &\quad-(x^{2}+ax+b)e^{-x}\\ &=\bigl(-x^{2}+(2-a)x\\ &\qquad+(a-b)\bigr)e^{-x}\,. \end{aligned} $$

(a) Two conditions. Since $e^{-x}\ne0,$ the conditions $f'(1)=0$ and $f'(3)=0$ (Definition 6.12) become $$ \begin{aligned} -1+(2-a)+(a-b)&=0\,,\\ -9+3(2-a)+(a-b)&=0\,, \end{aligned} $$ that is $1-b=0$ and $-3-2a-b=0.$ So $b=1$ and $a=-2{:}$ $$a=-2\,,\qquad b=1\,.$$

(b) Factor $f'.$ Now $f(x)=(x^{2}-2x+1)e^{-x}=(x-1)^{2}e^{-x}$ and $$ \begin{aligned} f'(x)&=\bigl(-x^{2}+4x-3\bigr)e^{-x}\\ &=-(x-1)(x-3)\,e^{-x}\,. \end{aligned} $$

(b) The sign table (Method 2.25). As $e^{-x}>0,$ the sign of $f'$ is the sign of $-(x-1)(x-3)=(1-x)(x-3){:}$

So $f$ goes ↘ on $x<1,$ ↗ on $1<x<3$ and ↘ on $x>3;$ $f'(1)=f'(3)=0,$ with $f(1)=0$ and $f(3)=4e^{-3}.$

(b) The intervals. By Fact 2.23, $f$ is decreasing on $(-\infty,1]$ and on $[3,+\infty),$ and increasing on $[1,3].$

(b) Classify by the first-derivative test (Method 7.10). $f'$ goes from $-$ to $+$ at $x=1{:}$ a local minimum point, with $f(1)=0.$ $f'$ goes from $+$ to $-$ at $x=3{:}$ a local maximum point, with $f(3)=4e^{-3}.$

(c) Global minimum. $f(x)=(x-1)^{2}e^{-x}\ge0$ for all $x,$ and $f(1)=0.$ So $x=1$ is a global minimum point (Definition 6.8), with minimum value $0.$

(c) No global maximum. As $x\to-\infty,$ $(x-1)^{2}\to+\infty$ and $e^{-x}\to+\infty,$ so $f(x)\to+\infty.$ For instance $f(-1)=4e\approx10.9>4e^{-3}.$ Hence $f$ takes arbitrarily large values and has no global maximum. (The local maximum value $4e^{-3}\approx0.2$ is only the largest value on $[1,+\infty),$ where $f\to0$ as $x\to+\infty.$)

(a) $U'_{x}=y$ and $U'_{y}=x,$ so the only stationary point is $(0,0).$ There $A=U''_{xx}=0,$ $B=U''_{xy}=1,$ $C=U''_{yy}=0,$ and $$AC-B^{2}=-1<0\,,$$ so $(0,0)$ is a saddle point (Fact 11.4).

(b) The conditions. The Lagrangian (Definition 11.13) is $$\mathcal{L}(x,y)=xy-\lambda(2x+y-12)\,,$$ and the first-order conditions (Method 11.14) are $$ \begin{aligned} \mathcal{L}'_{x}&=y-2\lambda=0\,,\\ \mathcal{L}'_{y}&=x-\lambda=0\,,\\ 2x+y&=12\,. \end{aligned} $$

(b) Solve. The first two give $y=2\lambda=2x,$ and then $2x+2x=12,$ so $x=3,$ $y=6,$ $\lambda=3.$ The only candidate is $$ \begin{aligned} &(x,y)=(3,\,6)\,,\\ &U(3,6)=18\,. \end{aligned} $$

(b) Why it is the maximum. The budget set $\{(x,y):2x+y=12,\ x\ge0,\ y\ge0\}$ is a segment, closed and bounded, and $U$ is continuous, so a maximum exists (Fact 11.9). It is attained either at an endpoint or at a point satisfying the Lagrange conditions (Watch out 11.15). At the endpoints $$ \begin{aligned} U(6,0)&=0\,,\\ U(0,12)&=0\,, \end{aligned} $$ both less than $18,$ so the maximum is at $(3,6).$ (Alternatively, Method 10.19: substituting $y=12-2x$ gives $U=12x-2x^{2},$ a concave quadratic with its maximum at $x=3.$)

(c) The multiplier. From (b), $\lambda=3.$ The multiplier is the derivative of the optimal value with respect to the constraint level: if $U^{*}(m)$ is the maximal utility with income $m,$ then $$\frac{dU^{*}}{dm}=\lambda$$ (Fact 11.21). So $\lambda=3$ is the marginal utility of income: one more krona raises the maximal utility by approximately $3$ units. It is measured in units of utility per krona.

(c) The estimate. With one more krona, $$ \begin{aligned} U^{*}(13)&\approx U^{*}(12)+\lambda\cdot1\\ &=18+3=21\,. \end{aligned} $$ (The exact value is $U^{*}(13)=\tfrac{13}{4}\cdot\tfrac{13}{2}=21.125.$)

(a) A geometric series. The term with $n=0$ is $1,$ so the series is geometric with first term $1$ and ratio $$k=\frac{2x-1}{5}\,.$$ By Fact 8.7 it converges exactly when $|k|<1.$

(a) Solve $|k|<1.$ $$ \begin{aligned} &\Bigl|\frac{2x-1}{5}\Bigr|<1\\ &\iff-5<2x-1<5\\ &\iff-4<2x<6\\ &\iff-2<x<3\,. \end{aligned} $$ Both endpoints are excluded: at $x=3$ the ratio is $1,$ at $x=-2$ it is $-1,$ and the series diverges. So the series converges exactly for $-2<x<3.$

(a) The sum. For such $x$ the sum is $\dfrac{1}{1-k},$ and $$ \begin{aligned} \frac{1}{1-k}=\frac52&\iff1-k=\frac25\\ &\iff k=\frac35\\ &\iff\frac{2x-1}{5}=\frac35\\ &\iff x=2\,, \end{aligned} $$ which lies in the interval of convergence: $x=2.$

(b) One payment. A payment of $5\,000$ kronor made $j$ years from now is worth $5\,000\cdot1.04^{-j}$ kronor today (that amount grows to $5\,000$ in $j$ years at $4\,\%$).

(b)(i) For ever. The present value of all the payments is the geometric series $$ \begin{aligned} \mathrm{PV}&=\frac{5000}{1.04}+\frac{5000}{1.04^{2}}\\ &\quad+\frac{5000}{1.04^{3}}+\cdots \end{aligned} $$ with first term $5000/1.04$ and ratio $k=1/1.04<1,$ so it converges and $$ \begin{aligned} \mathrm{PV}&=\frac{5000/1.04}{1-1/1.04}\\ &=\frac{5000}{1.04-1}=\frac{5000}{0.04}\\ &=125\,000\text{ kronor}\,. \end{aligned} $$ (This is the perpetuity formula $a/r$ of Example 8.8.)

(b)(ii) For 25 years. With $25$ payments the present value is the finite geometric sum $$ \begin{aligned} \mathrm{PV}_{25}&=\frac{5000}{1.04}+\cdots+\frac{5000}{1.04^{25}}\\ &=\frac{5000}{1.04}\cdot\frac{1-1.04^{-25}}{1-1.04^{-1}}\,, \end{aligned} $$ by the formula $a+ak+\cdots+ak^{n-1}=a\,\dfrac{1-k^{n}}{1-k}$ with $a=5000/1.04,$ $k=1/1.04$ and $n=25$ (Fact 8.9).

(b)(ii) The closed formula. Multiplying numerator and denominator by $1.04$ gives $$ \mathrm{PV}_{25}=5000\cdot\frac{1-1.04^{-25}}{0.04} $$ kronor (approximately $78\,110$ kronor).

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

(a) Factor. Since $a^{2}+a-2=(a-1)(a+2),$ $$\det\mathbf{A}=-(a-1)^{2}(a+2)\,.$$

(a) Exactly one solution precisely when $\det\mathbf{A}\ne0$ (Fact 14.26), that is for $$a\ne1\ \text{ and }\ a\ne-2\,.$$

(b) $a=1.$ All three equations have left-hand side $x+y+z$ but right-hand sides $1,$ $-2,$ $1.$ The first two contradict each other ($R_{2}-R_{1}$ gives the row $(0\;\;0\;\;0\mid-3),$ that is $0=-3$), so there is no solution.

(b) $a=-2.$ A zero determinant says only "not exactly one" (Watch out 15.8); elimination decides. The augmented matrix (Definition 13.15) is reduced as follows: $$ \begin{aligned} &\left(\begin{array}{ccc|c}1&1&-2&1\\1&-2&1&-2\\-2&1&1&1\end{array}\right)\\ &\big\downarrow\ R_{2}-R_{1},\ R_{3}+2R_{1}\\ &\left(\begin{array}{ccc|c}1&1&-2&1\\0&-3&3&-3\\0&3&-3&3\end{array}\right)\\ &\big\downarrow\ R_{3}+R_{2},\ \text{then }R_{2}/(-3)\\ &\left(\begin{array}{ccc|c}1&1&-2&1\\0&1&-1&1\\0&0&0&0\end{array}\right). \end{aligned} $$

(b) All the solutions. The last row is $0=0,$ so the system is consistent with one free variable (Fact 13.20): put $z=t.$ Then $y=1+t$ and $x=1-y+2z=t.$ There are infinitely many solutions: $$ \begin{aligned} &(x,y,z)=(t,\ 1+t,\ t)\,,\\ &t\in\mathbb{R}\,. \end{aligned} $$ Check: $t+(1+t)-2t=1,$ $t-2(1+t)+t=-2,$ $-2t+(1+t)+t=1.$

(c) $a=0.$ The system is $x+y=1,$ $x+z=-2,$ $y+z=1.$ Reducing the augmented matrix (Method 13.16), $$ \begin{aligned} &\left(\begin{array}{ccc|c}1&1&0&1\\1&0&1&-2\\0&1&1&1\end{array}\right)\\ &\big\downarrow\ R_{2}-R_{1}\\ &\left(\begin{array}{ccc|c}1&1&0&1\\0&-1&1&-3\\0&1&1&1\end{array}\right)\\ &\big\downarrow\ R_{3}+R_{2}\\ &\left(\begin{array}{ccc|c}1&1&0&1\\0&-1&1&-3\\0&0&2&-2\end{array}\right). \end{aligned} $$

(c) Back substitution. $2z=-2$ gives $z=-1;$ $-y+z=-3$ gives $y=2;$ $x=1-y=-1.$ So $$ (x,y,z)=(-1,\ 2,\ -1)\,. $$ Check: $-1+2=1,$ $-1-1=-2,$ $2-1=1.$