Chapter 0 Challenge Problems Solutions

SOLUTION The discriminant of \(g\) is \[\Delta = b^2 - 8 \period \] The function \(g\) has no real zeros if \(\Delta \lt 0. \) Thus, we find the values of \(b\) that satisfies this inequality: \[b^2 - 8 \lt 0 \implies \boxed{-2 \sqrt 2 \lt b \lt 2 \sqrt 2} \]

SOLUTION Let \(x\) be the number of four-point questions and \(y\) be the number of seven-point questions. To solve for both variables, we write a system of two linear equations: \begin{align} x + y &= 16 \label{eq:points-1} \nl 4x + 7y &= 100 \label{eq:points-2} \pd \end{align} Let's solve this system by elimination: Multiplying both sides of \(\eqref{eq:points-1}\) by \(4\) and subtracting it from \(\eqref{eq:points-2}\) gives \[ \ba 4x + 7y - 4(x + y) &= 100 - 4(16) \nl 3y &= 36 \nl y &= 12 \pd \ea \] So by \(\eqref{eq:points-1},\) \(x = 16 - 12\) \(= 4.\) Thus, the exam has 4 four-point questions and 12 seven-point questions.

SOLUTION The function \(f\) is even if and only if \(f(-x) = f(x),\) and it is odd if and only if \(f(-x) = -f(x).\) Replacing \(x\) with \(-x\) gives \[f(-x) = (-x)^n = (-1)^n \, x^n \pd\]

1. For even \(n,\) \((-1)^n = 1.\) Hence, \[f(-x) = (1) \, x^n = x^n = f(x)\] and so \(f\) is even.

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2. If \(n\) is odd, then \((-1)^n = -1.\) Consequently, \[f(-x) = (-1) \, x^n = -x^n = -f(x) \cma\] so \(f\) is odd.

SOLUTION The word collinear means all three points lie on the same straight line. The slope of the line connecting \((-3, 4)\) and \((1, 1)\) is \[m = \frac{1 - 4}{1 - (-3)} = -\frac{3}{4} \pd\] So an equation of the line connecting the two points is, by point-slope form, \[y - 1 = -\tfrac{3}{4} (x - 1) \or y = -\tfrac{3}{4} x + \tfrac{7}{4} \pd\] Substituting \((k, -2)\) into the equation of the line, we attain \[ \ba -2 &= -\tfrac{3}{4} k + \tfrac{7}{4} \nl -8 &= -3k + 7 \nl -3k &= -15 \nl \implies k &= 5 \pd \ea \] Hence, \(\boxed{k = 5}\) ensures that \((k, -2)\) lies on the same line as the other two points.

SOLUTION

Let's call \(y\) the prism's height. Then the prism's volume is \begin{equation} x^2 y = 400 \pd \label{eq:prism-V} \end{equation} The prism's two square faces each have an area of \(x^2,\) and the four rectangular faces each have an area of \(xy.\) So the prism's total surface area is their sum: \begin{equation} S = 2x^2 + 4xy \pd \label{eq:prism-S} \end{equation} To express \(S\) entirely in terms of \(x,\) we express \(y\) in terms of \(x.\) From \(\eqref{eq:prism-V}\) we have \(y = 400/x^2,\) which we substitute into \(\eqref{eq:prism-S}\) to get \[ \ba S(x) &= 2x^2 + 4x \par{\frac{400}{x^2}} \nl &= \boxed{2x^2 + \frac{1600}{x}} \ea \] Because the side length \(x\) can't be negative, the domain of \(S\) in context is \[ \ba \{x \mid x \gt 0\} = \boxed{(0, \infty)} \ea \]

SOLUTION When working with reciprocal trigonometric functions, it is a good idea to rewrite the equation in terms of sines and cosines. Doing so and factoring, we get \[ \ba 3 \sin x - 6 \frac{\sin x}{\cos x} &= 0 \nl (3 \sin x)(1 - 2 \sec x) &= 0 \pd \ea \] Equating each factor to \(0,\) we get \[ \baat{2} 3 \sin x &= 0 \lspace &1 - 2 \sec x &= 0 \nl \sin x &= 0 \lspace &\sec x &= \tfrac{1}{2} \pd \eaat \] But since the range of secant is \((-\infty, -1]\) \(\cup\) \([1, \infty),\) the equation \(\sec x = 1/2\) has no solutions. Thus, the only solutions exist when \(\sin x = 0;\) that is, if \(n\) is any integer, then \[x = n \pi \pd\] But these solutions turn out to be extraneous; the original equation was \[\frac{3}{\csc x} - \frac{6}{\cot x} = 0 \pd\] The domains of cosecant and cotangent both exclude \(x = n \pi,\) so, this equation has no real solutions.

SOLUTION Consider an axis system where \(x = 0\) is particle A's initial position. Then particle B is initially located at \(x = 20.\) Let \(t\) be time in seconds; both particles' positions are functions of \(t.\) The rate of change of each particle is its speed. Since particle A's speed is \(4 \undiv{ft}{sec},\) its position function is \(x_A(t) = 4t.\) Particle B's position is decreasing at a rate given by its speed, \(6 \undiv{ft}{sec},\) and its initial position is \(20.\) Accordingly, its position function is \(x_B(t) = 20 - 6t.\) When both particles collide, their \(x\)-coordinates are equivalent, so \[ \ba x_A(t) &= x_B(t) \nl 4t &= 20 - 6t \nl 10 t &= 20 \nl \implies t &= \boxed{2 \un{sec}} \ea \] Also note that \(x = 4(2)\) \(= 20 - 6(2)\) \(= 8.\) Thus, both particles collide \(8 \un{ft}\) to the right of particle A's initial position.

SOLUTION Observe that the \(50 \degree\) angle is complementary to angles \(\theta_1\) and \(\theta_3.\) Thus, we have \[\theta_1 = \theta_3 = 90 \degree - 50 \degree = 40 \degree \pd\] Also, \(\theta_1\) and \(\theta_2\) are complementary, so \[\theta_2 = 90 \degree - \theta_1 = 50 \degree \pd\] Using trigonometry, it is easiest to first find \(d_3\) and \(d_4,\) as follows: \[ \baat{2} d_3 &= 11 \sin 50 \degree &&\approx 8.426 \cma \nl d_4 &= 11 \cos 50 \degree &&\approx 7.071 \pd \eaat \] Next we find the side lengths of the inner triangle. We see \[ \ba \tan \theta_2 &= \frac{4}{d_1} \nl \implies d_1 &= 4 \cot 50 \degree \approx 3.356 \pd \ea \] Likewise, we have \[ \ba \sin \theta_2 &= \frac{4}{d_2} \nl \implies d_2 &= 4 \csc 50 \degree \approx 5.222 \pd \ea \] In summary, the side lengths are \[ \boxed{ \baat{2} d_1 &\approx 3.356 \qquad &d_2 &\approx 5.222 \nl d_3 &\approx 8.426 &\qquad d_4 &\approx 7.071 \eaat} \] The angles are \[\boxed{\theta_1 = 40 \degree} \qquad \boxed{\theta_2 = 50 \degree} \qquad \boxed{\theta_3 = 40 \degree}\]

SOLUTION The domain is \(\{x \mid x \gt 4\}\) so that neither logarithm has a negative argument. Condensing the logarithms on the left side, we get \[\log_7 \par{x^2 - 4x} = 1 \pd\] Exponentiating both sides with base \(7,\) we see \[ \ba 7^{\log_7 \par{x^2 - 4x}} &= 7^1 \nl x^2 - 4x &= 7 \nl x^2 - 4x - 7 &= 0 \pd \ea \] The Quadratic Formula (see Section 0.7) then gives \[ \ba x &= \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-7)}}{2(1)} \nl &= \frac{4 \pm \sqrt{44}}{2} \nl &= \frac{4 \pm 2 \sqrt{11}}{2} \nl &= 2 \pm \sqrt{11} \pd \ea \] Only \(\boxed{x = 2 + \sqrt{11}}\) \(\approx 5.317\) is in the domain of \(\log_7 x + \log_7(x - 4);\) the other solution is extraneous.

SOLUTION

1. We have \[ \ba (f \circ g)(x) &= f(g(x)) \nl &= f \par{4^x} \nl &= \boxed{\tfrac{1}{2} \par{4^x} - 1} \ea \] The domain is \(\boxed{(-\infty, \infty)}.\) Also, \[ \ba (g \circ f)(x) &= g(f(x)) \nl &= g \par{x/2 - 1} \nl &= \boxed{4^{x/2 - 1}} \ea \] The domain is \(\boxed{(-\infty, \infty)}.\)

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2. The composite function \(f \circ g\) is positive when \[ \ba \tfrac{1}{2} \par{4^x} - 1 &\gt 0 \nl 4^x &\gt 2 \pd \ea \] Since \(2 = 4^{1/2},\) the inequality becomes \(4^x \gt 4^{1/2},\) which is true for \(x \gt 1/2.\) Hence, the solution interval is \[\{x \mid x \gt \tfrac{1}{2}\} = \boxed{\par{\tfrac{1}{2}, \infty}}\]

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3. We have \[ \ba 4^{x/2 - 1} &\gt \par{\tfrac{1}{2} \par{4^x} - 1} + 1 \nl \par{4^{1/2}}^x \, 4^{-1} &\gt 2^{-1} \, \par{2^2}^x \nl 2^x \, 2^{-2} &\gt 2^{-1} \, 2^{2x} \nl 2^{x - 2} &\gt 2^{2x - 1} \pd \ea \] By comparing exponents, we have \[ \ba x - 2 &\gt 2x - 1 \nl x &\lt -1 \pd \ea \] So the solution interval is \[\{x \mid x \lt -1\} = \boxed{(-\infty, -1)}\]

SOLUTION

Let's draw a right triangle in which one angle is \(\theta.\) Think of \(\theta = \acos x\) as \(\theta = \acos(x/1).\) Since cosine is the ratio of the adjacent side length to the hypotenuse length, the side adjacent to \(\theta\) has length \(x\) and the hypotenuse's length is \(1.\) Then the side opposite to \(\theta\) is, by the Pythagorean theorem, \(\sqrt{1 - x^2}.\) Sine is the ratio of the opposite side to the hypotenuse; thus, we get \[\sin \theta = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2} \cma\] as requested. Another method is to let \(\theta = \acos x.\) Then \(\cos x = x\) and \(\theta \in [0, \pi],\) over which \(\sin \theta \geq 0.\) Thus, \[\sin (\acos x) = \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2} \cma\] where the penultimate step is true by the Pythagorean identity. With this method, the relationship \[\sin(\acos x) = \sqrt{1 - x^2}\] holds for \(x \in [-1, 1],\) not just for positive \(x.\)

SOLUTION Exponentiating both sides using the base \(e,\) we get \[ \ba e^{\ln(T_1/T_2)} &= e^{\mu \pi} \nl \frac{T_1}{T_2} &= e^{\mu \pi} \nl \implies T_1 &= T_2 e^{\mu \pi} \pd \ea \] If \(T_1\) \(= 60 \un{lb}\) and \(\mu = 0.6,\) then we have \[ \ba 60 &= T_2 e^{0.6 \pi} \nl T_2 &= \frac{60}{e^{0.6 \pi}} \approx \boxed{9.11 \un{lb}} \ea \]

SOLUTION

Let \(y\) be the length of each horizontal line segment. Each semicircle's radius is \(x/2,\) so its circumference is \(\pi x/2.\) The slot's perimeter of \(50\) is given by \(2y\) plus the two semicircles' circumferences—namely, \begin{align} 2 \par{\frac{\pi x}{2}} + 2y &= 50 \nonum \nl \pi x + 2y &= 50 \label{eq:slot-perimeter} \pd \end{align} Each semicircle's area is \(\tfrac{1}{2} \pi (x/2)^2\) \(= \pi x^2/8,\) and the middle rectangle's area is \(xy.\) Hence, the slot's area is \begin{align} A &= 2 \par{\frac{\pi x^2}{8}} + xy \nonum \nl &= \frac{\pi x^2}{4} + xy \pd \label{eq:slot-area} \end{align} Our goal is to express \(A\) solely in terms of \(x,\) so we want to express \(y\) in terms of \(x.\) From \(\eqref{eq:slot-perimeter},\) solving for \(y\) gives \[ \ba y &= \tfrac{1}{2} (50 - \pi x) \nl &= 25 - \frac{\pi x}{2} \cma \ea \] which we substitute into \(\eqref{eq:slot-area}\) to attain \[ \ba A(x) &= \frac{\pi x^2}{4} + x \par{25 - \frac{\pi x}{2}} \nl &= \boxed{25x - \frac{\pi x^2}{4}} \ea \]

SOLUTION Let \(S_n\) \(= \sum_{i = 1}^n i^3\) \(= [n(n + 1)/2]^2.\) Our first goal is to show that \(S_1\) is true. Then we assume that \(S_k\) (where \(k\) is any positive integer) is true and hold that \(S_{k + 1}\) is true. First observe that \(S_1\) is true because \[\sum_{i = 1}^1 i^3 = 1^3 \equalsCheck \parbr{\frac{1(1 + 1)}{2}}^2 \pd\] Next we assume that \(S_k\) is true—namely, \[\sum_{i = 1}^k i^3 = \parbr{\frac{k(k + 1)}{2}}^2 \pd\] We must show that \(S_{k + 1}\) is consequently true—that is, \[\sum_{i = 1}^{k + 1} i^3 = \parbr{\frac{(k + 1)(k + 2)}{2}}^2 \pd\] We see \[ \ba \sum_{i = 1}^{k + 1} i^3 &= \sum_{i = 1}^k i^3 + (k + 1)^3 \nl &= \parbr{\frac{k(k + 1)}{2}}^2 + (k + 1)^3 \nl &= (k + 1)^2 \par{\frac{k^2}{4} + (k + 1)} \nl &= (k + 1)^2 \par{\frac{k^2 + 4k + 4}{4}} \nl &= (k + 1)^2 \frac{(k + 2)^2}{4} \nl &\equalsCheck \parbr{\frac{(k + 1)(k + 2)}{2}}^2 \pd \ea \] Therefore, by mathematical induction, \(S_n\) is true for all positive integers \(n.\)

SOLUTION Exponentiating both sides of the best-fit equation for the line, we have \[ \ba 10^{\log y} &= 10^{-0.6482 + 0.3633 x} \nl y &= 10^{-0.6482} \, 10^{0.3633 x} \nl y &= 0.2248 \par{2.3083}^x \pd \ea \] Comparing this equation to \(y = ab^x,\) we see \(\boxed{a = 0.2248}\) and \(\boxed{b = 2.3083}.\)

SOLUTION By squaring both equations, we get \[ \ba 25 \sin^2 \alpha + 20 \sin \alpha \cos \beta + 4 \cos^2 \beta &= 36 \cma \nl 25 \cos^2 \alpha + 20 \cos \alpha \sin \beta + 4 \sin^2 \beta &= 1 \pd \ea \] Adding both equations, we get \[ \ba 25(\sin^2 \alpha + \cos^2 \alpha) + 20(\sin \alpha \cos \beta + \cos \alpha \sin \beta) + 4(\cos^2 \beta + \sin^2 \beta) &= 37 \nl 29 + 20(\sin \alpha \cos \beta + \cos \alpha \sin \beta) &= 37 \nl \sin \alpha \cos \beta + \cos \alpha \sin \beta &= \tfrac{8}{20} \nl \sin(\alpha + \beta) &= \boxed{\tfrac{2}{5}} \ea \] Then by the Pythagorean identity, \[\cos(\alpha + \beta) = \sqrt{1 - \sin^2(\alpha + \beta)} = \sqrt{1 - \tfrac{4}{25}} = \boxed{\frac{\sqrt{21}}{5}}\] (We chose the positive solution because cosine is positive for inputs between \(0\) and \(\pi/2\) radians.)