CMI Data Science Chapter Practice

The problem asks to identify the true statements about the solution set $S$ of the equation $(\ln x)^2 - \ln(x^2) = 3$. **Step 1: Simplify the equation and find the solutions.** The domain of the logarithmic function requires that $x > 0$. Using the power rule for logarithms, $\ln(x^2) = 2\ln x$. The equation becomes: Let $y = \ln x$. Substituting this into the equation gives a quadratic equation in $y$: Factoring the quadratic equation: The solutions for $y$ are $y = 3$ and $y = -1$. Now, we substitute back $\ln x$ for $y$ to find the values of $x$. Case 1: $y = 3$ Case 2: $y = -1$ Both solutions, $e^3$ and $\frac{1}{e}$, are positive, so they are valid. The solution set is $S = \left\{ e^3, \frac{1}{e} \right\}$. **Step 2: Evaluate each statement.** **Statement A: The product of the solutions in $S$ is $e^2$.** The product of the solutions is $e^3 \times \frac{1}{e} = e^{3-1} = e^2$. This statement is **true**. **Statement B: All solutions in $S$ are irrational numbers.** The number $e$ is a transcendental number, which means it is also an irrational number. Any non-zero integer power of $e$ (like $e^3$) and its reciprocal ($1/e$) are also transcendental and therefore irrational. This statement is **true**. **Statement C: The set $S$ contains exactly one solution in the interval $(0, 1)$.** We need to check the values of the solutions. We know that $e \approx 2.718$. - The first solution is $e^3 \approx (2.718)^3 > 2^3 = 8$. So, $e^3 > 1$. - The second solution is $\frac{1}{e} \approx \frac{1}{2.718}$. Since $e > 1$, we have $0 < \frac{1}{e} < 1$. Thus, there is exactly one solution, $\frac{1}{e}$, in the interval $(0, 1)$. This statement is **true**. **Statement D: The sum of the solutions in $S$ is less than 20.** The sum of the solutions is $e^3 + \frac{1}{e}$. Using the approximation $e \approx 2.718$: $e^2 \approx (2.718)^2 \approx 7.389$ $e^3 \approx 7.389 \times 2.718 \approx 20.085$ Since $e^3 > 20$, the sum $e^3 + \frac{1}{e}$ must also be greater than 20. This statement is **false**. Therefore, the correct statements are A, B, and C. Answer: $\boxed{A, B, C}$

The given inequality is $\log_{x} (x^2 - 2x - 2) \ge 1$. **Step 1: Determine the domain of the logarithmic expression.** For the logarithm to be defined, we must satisfy three conditions: 1. The base $x$ must be positive: $x > 0$. 2. The base $x$ must not be equal to 1: $x \neq 1$. 3. The argument $(x^2 - 2x - 2)$ must be positive: $x^2 - 2x - 2 > 0$. To solve $x^2 - 2x - 2 > 0$, we find the roots of the quadratic equation $x^2 - 2x - 2 = 0$. Using the quadratic formula: The parabola opens upwards, so the expression is positive when $x < 1 - \sqrt{3}$ or $x > 1 + \sqrt{3}$. Combining all domain conditions, $x$ must be in $(0, 1) \cup (1, \infty)$ and $(-\infty, 1-\sqrt{3}) \cup (1+\sqrt{3}, \infty)$. The intersection gives the domain: $x \in (1+\sqrt{3}, \infty)$. **Step 2: Solve the inequality by considering cases for the base $x$.** The domain already restricts us to $x > 1+\sqrt{3}$, which means the base $x$ is always greater than 1. For $x > 1$, the logarithmic function $\log_x(u)$ is increasing, so we can solve the inequality without flipping the sign. **Case: $x > 1$** To solve this, we find the roots of $x^2 - 3x - 2 = 0$: The parabola opens upwards, so the inequality holds for $x \le \frac{3 - \sqrt{17}}{2}$ or $x \ge \frac{3 + \sqrt{17}}{2}$. **Step 3: Find the final solution set $S$.** We must find the intersection of the solution from Step 2 with the domain from Step 1. The domain is $x \in (1+\sqrt{3}, \infty)$. The inequality solution is $x \in (-\infty, \frac{3-\sqrt{17}}{2}] \cup [\frac{3+\sqrt{17}}{2}, \infty)$. Let's approximate the values: $1 + \sqrt{3} \approx 1 + 1.732 = 2.732$. $\frac{3 - \sqrt{17}}{2} \approx \frac{3 - 4.123}{2} \approx -0.56$. $\frac{3 + \sqrt{17}}{2} \approx \frac{3 + 4.123}{2} \approx 3.56$. The intersection is between $(2.732, \infty)$ and $(-\infty, -0.56] \cup [3.56, \infty)$. This gives the final solution set $S = [\frac{3+\sqrt{17}}{2}, \infty)$. **Step 4: Evaluate each option.** * **Option 1: The smallest integer belonging to $S$ is 4.** Since $S = [\frac{3+\sqrt{17}}{2}, \infty)$ and we know $\frac{3+\sqrt{17}}{2} \approx 3.56$, the smallest integer in this set is indeed 4. This statement is **true**. * **Option 2: The infimum of $S$ is a rational number.** The infimum (greatest lower bound) of $S$ is $\frac{3+\sqrt{17}}{2}$. Since $\sqrt{17}$ is irrational, the infimum is an irrational number. This statement is **false**. * **Option 3: $S$ is a subset of $(1+\sqrt{3}, \infty)$.** The domain of the original inequality is $(1+\sqrt{3}, \infty)$. The solution set $S$ must be a subset of its domain. Since $\frac{3+\sqrt{17}}{2} \approx 3.56$ and $1+\sqrt{3} \approx 2.732$, we have $\frac{3+\sqrt{17}}{2} > 1+\sqrt{3}$. Therefore, $S = [\frac{3+\sqrt{17}}{2}, \infty)$ is a subset of $(1+\sqrt{3}, \infty)$. This statement is **true**. * **Option 4: $S$ contains the interval $[3, 4]$.** The set $S$ starts at approximately 3.56. The number 3 is not in $S$. Therefore, $S$ does not contain the interval $[3, 4]$. This statement is **false**. Thus, the correct statements are the first and the third. Answer: \boxed{\text{The correct statements are the first and the third.}}

The given equation is **Step 1: Determine the domain of the equation.** The logarithm function $\log_b a$ is defined for $a > 0$, $b > 0$, and $b \neq 1$. For our equation, the argument is $x$. Therefore, the domain is $x > 0$. **Step 2: Simplify the equation using the change of base formula.** We need to express all logarithmic terms with the same base. Let's choose base 2. The term $\log_{1/2} x$ can be rewritten as: **Step 3: Substitute the simplified term back into the equation.** Substituting $-\log_2 x$ for $\log_{1/2} x$ in the original equation gives: **Step 4: Solve the resulting quadratic equation.** Let $y = \log_2 x$. The equation becomes a standard quadratic equation in $y$: Factoring the quadratic expression: This gives two possible values for $y$: $y = -3$ or $y = 2$. **Step 5: Find the corresponding values of $x$.** We substitute back $y = \log_2 x$ to find the solutions for $x$. Case 1: Case 2: Both solutions, $x = 1/8$ and $x = 4$, are positive, so they are valid. The solution set is $S = \{1/8, 4\}$. **Step 6: Evaluate the given options.** Now we check each statement based on the solution set $S$. - **"The equation has exactly two distinct real solutions."** We found two distinct real solutions, $1/8$ and $4$. This statement is **true**. - **"The sum of the solutions is an integer."** The sum of the solutions is This is not an integer. This statement is **false**. - **"The product of the solutions is $\frac{1}{2}$."** The product of the solutions is This statement is **true**. - **"One solution lies in the interval $(0, 1)$ and the other lies in $(1, \infty)$."** The solution $x = 1/8$ satisfies $0 < 1/8 < 1$, so it lies in $(0, 1)$. The solution $x = 4$ satisfies $4 > 1$, so it lies in $(1, \infty)$. This statement is **true**. Therefore, the correct options are the first, third, and fourth statements. Answer: \boxed{The first, third, and fourth statements are true.}

Let $P(x) = a_m x^m + \dots + a_0$ and $Q(x) = b_n x^n + \dots + b_0$ be two non-zero polynomials with real coefficients. This means their leading coefficients, $a_m$ and $b_n$, are non-zero real numbers. **Analysis of Option 1:** This statement claims that $\operatorname{deg}(P(x) + Q(x)) = \max(m, n)$ is always true. Consider the case where $m=n$. The leading term of $P(x) + Q(x)$ would be $(a_m + b_m)x^m$. If $a_m + b_m = 0$ (i.e., $b_m = -a_m$), this leading term vanishes, and the degree of the sum will be less than $m$. **Counterexample:** Let $P(x) = x^2 + 2x$ and $Q(x) = -x^2 + 5$. Here $m=n=2$. The degree of the sum is $1$, which is not equal to $\max(2, 2) = 2$. Thus, the first statement is false. **Analysis of Option 2:** This statement claims that if $m \neq n$, then $\operatorname{deg}(P(x) + Q(x)) = \max(m, n)$. Without loss of generality, assume $m > n$. Then the polynomial sum is: Since $m > n$, the term $a_m x^m$ is the unique term of highest degree in the sum. Its coefficient $a_m$ is non-zero by definition of degree. Therefore, the degree of $P(x) + Q(x)$ is $m$, which is $\max(m, n)$. Thus, the second statement is true. **Analysis of Option 3:** This statement claims that $\operatorname{deg}(P(x) \cdot Q(x)) = m+n$. The product of the polynomials is: The term of highest degree in the product is formed by multiplying the leading terms of $P(x)$ and $Q(x)$. This term is $(a_m b_n) x^{m+n}$. Since the coefficients are real, and $a_m \neq 0$ and $b_n \neq 0$, their product $a_m b_n$ is also a non-zero real number. Therefore, the degree of the product is indeed $m+n$. Thus, the third statement is true. **Analysis of Option 4:** This statement claims that $R(x) = (P(x))^2 + (Q(x))^2$ is always a non-zero polynomial. Let's analyze the degree of $R(x)$. The degree of $(P(x))^2$ is $2m$ and its leading coefficient is $a_m^2$. The degree of $(Q(x))^2$ is $2n$ and its leading coefficient is $b_n^2$. - **Case 1: $m \neq n$.** Assume $m > n$. Then $2m > 2n$. The degree of $R(x)$ is $\max(2m, 2n) = 2m$. Since $m \ge 0$, the degree is non-negative, so $R(x)$ is not the zero polynomial. - **Case 2: $m = n$.** The degree of both $(P(x))^2$ and $(Q(x))^2$ is $2m$. The leading term of $R(x)$ is $(a_m^2 + b_m^2)x^{2m}$. Since $a_m$ and $b_m$ are non-zero real numbers, $a_m^2 > 0$ and $b_m^2 > 0$. Therefore, their sum $a_m^2 + b_m^2$ is strictly positive and cannot be zero. The leading coefficient is non-zero, so the degree of $R(x)$ is $2m$, and $R(x)$ is not the zero polynomial. In both cases, $R(x)$ is a non-zero polynomial. Thus, the fourth statement is true.

The solution involves simplifying the given system of logarithmic equations to find the relationship between $x$, $y$, and $a$. **Step 1: Simplify the system using substitution.** Let $u = \log_a(x)$ and $v = \log_a(y)$. The first equation becomes: Using the change of base formula, $\log_b c = \frac{1}{\log_c b}$, we can rewrite the second equation: So the second equation becomes: **Step 2: Solve the system for $u$ and $v$.** From equation (2), we combine the fractions: Substitute the value of $u+v$ from equation (1): We now have a system for $u$ and $v$: $u+v=4$ and $uv=3$. This means $u$ and $v$ are the roots of the quadratic equation $t^2 - (\text{sum of roots})t + (\text{product of roots}) = 0$. Factoring the quadratic equation: The roots are $t=1$ and $t=3$. Therefore, the set of values for $\{u, v\}$ is $\{1, 3\}$. **Step 3: Relate $u, v$ back to $x, y$.** Since $\{u, v\} = \{\log_a x, \log_a y\} = \{1, 3\}$, we have two possibilities: Case 1: $\log_a x = 1$ and $\log_a y = 3$. This implies $x = a^1 = a$ and $y = a^3$. Case 2: $\log_a x = 3$ and $\log_a y = 1$. This implies $x = a^3$ and $y = a^1 = a$. In both cases, the set of values for $\{x, y\}$ is $\{a, a^3\}$. **Step 4: Evaluate each option.** - **Option 1: $xy = a^4$** In both cases, the product is $xy = a \cdot a^3 = a^4$. This statement is true. - **Option 2: $x+y = a^3+a$** In both cases, the sum is $x+y = a + a^3$. This statement is true. - **Option 3: The possible values for $\log_x y$ are $3$ and $\frac{1}{3}$** In Case 1 ($x=a, y=a^3$): $\log_x y = \log_a (a^3) = 3$. In Case 2 ($x=a^3, y=a$): $\log_x y = \log_{a^3} (a) = \frac{\log_a a}{\log_a a^3} = \frac{1}{3}$. The possible values are indeed $3$ and $1/3$. This statement is true. - **Option 4: $|x-y| = a(a-1)$** The absolute difference is $|a^3 - a| = |a(a^2-1)| = |a(a-1)(a+1)|$. Since $a > 1$, all factors are positive, so $|x-y| = a(a-1)(a+1)$. The statement claims the value is $a(a-1)$, which would only be true if $a+1=1$, implying $a=0$. This contradicts the condition $a > 1$. Therefore, this statement is false. **Conclusion:** The first three statements are necessarily true.

Let's analyze each statement. Let $P(x) = a_m x^m + a_{m-1}x^{m-1} + \dots + a_0$ where $a_m \neq 0$. Let $Q(x) = b_n x^n + b_{n-1}x^{n-1} + \dots + b_0$ where $b_n \neq 0$. **Statement 1: If $m > n$, then $\operatorname{deg}(S) = m$.** **Step 1:** Write the expression for $S(x) = P(x) + Q(x)$. **Step 2:** Identify the term with the highest power of $x$. Since $m > n$, the term with the highest power is $a_m x^m$ from $P(x)$. No term in $Q(x)$ can cancel it. **Step 3:** The leading term of $S(x)$ is $a_m x^m$. Since $a_m \neq 0$, the degree of $S(x)$ is $m$. Thus, this statement is **true**. **Statement 2: The degree of $M(x)$ is always $m+n$.** **Step 1:** Write the expression for $M(x) = P(x) \cdot Q(x)$. **Step 2:** The term with the highest degree in the product is obtained by multiplying the leading terms of $P(x)$ and $Q(x)$. **Step 3:** Since $P(x)$ and $Q(x)$ have real coefficients and are non-zero, their leading coefficients $a_m$ and $b_n$ are non-zero real numbers. The product of two non-zero real numbers is non-zero, so $a_m b_n \neq 0$. Therefore, the degree of $M(x)$ is $m+n$. This statement is **true**. **Statement 3: If a real number $r$ is a root of both $P(x)$ and $Q(x)$, then $r$ is a root of $S(x)$.** **Step 1:** By definition of a root, if $r$ is a root of $P(x)$ and $Q(x)$, then $P(r) = 0$ and $Q(r) = 0$. **Step 2:** Evaluate $S(x)$ at $x=r$. **Step 3:** Substitute the known values. Since $S(r) = 0$, $r$ is a root of $S(x)$. This statement is **true**. **Statement 4: If a real number $r$ is a root of $M(x)$, then $r$ must be a root of both $P(x)$ and $Q(x)$.** **Step 1:** If $r$ is a root of $M(x)$, then $M(r) = 0$. By definition of $M(x)$, this means $P(r) \cdot Q(r) = 0$. **Step 2:** For the product of two real numbers $P(r)$ and $Q(r)$ to be zero, at least one of them must be zero. That is, $P(r) = 0$ or $Q(r) = 0$. **Step 3:** The statement claims that $P(r) = 0$ AND $Q(r) = 0$. This is not necessarily true. We can provide a counterexample. Let $P(x) = x-1$ and $Q(x) = x-2$. Then $M(x) = (x-1)(x-2)$. Let $r=1$. $M(1) = (1-1)(1-2) = 0$, so $r=1$ is a root of $M(x)$. However, while $P(1)=0$, we have $Q(1) = 1-2 = -1 \neq 0$. So $r=1$ is not a root of both polynomials. This statement is **false**.

Let the leading term of $P(x)$ be $a_m x^m$ where $a_m \neq 0$, and the leading term of $Q(x)$ be $b_n x^n$ where $b_n \neq 0$. The degrees are given as $m > 0$ and $n > 0$. We analyze each statement: **1. The degree of the product $P(x) \cdot Q(x)$ is $m+n$.** **Step 1:** Consider the product of the two polynomials. When multiplying, the term with the highest degree is obtained by multiplying the leading terms of $P(x)$ and $Q(x)$. **Step 2:** Identify the leading term of the product. Since $P(x)$ and $Q(x)$ have real coefficients and are non-zero, $a_m \neq 0$ and $b_n \neq 0$. Therefore, their product $a_m b_n \neq 0$. The highest degree term in the product polynomial is $x^{m+n}$ with a non-zero coefficient. Thus, the degree of $P(x) \cdot Q(x)$ is $m+n$. This statement is **always true**. **2. The degree of the sum $P(x) + Q(x)$ is $\max(m, n)$.** This statement is not always true. It fails if the degrees are equal ($m=n$) and the leading coefficients are additive inverses. **Counterexample:** Let $m=n=2$. Consider $P(x) = 3x^2 + 2x$ and $Q(x) = -3x^2 + 5$. The degree of the sum is $1$, but $\max(m, n) = \max(2, 2) = 2$. Since $1 \neq 2$, the statement is **not always true**. **3. The degree of the composition $P(Q(x))$ is $m \cdot n$.** **Step 1:** Consider the composition $P(Q(x))$. Substitute $y = Q(x) = b_n x^n + \dots$. **Step 2:** Find the degree of the leading term. The term with the highest degree in the expansion is $a_m (Q(x))^m$. The highest degree term within $(Q(x))^m$ is $(b_n x^n)^m = b_n^m x^{mn}$. So, the leading term of $P(Q(x))$ is $a_m (b_n^m x^{mn}) = (a_m b_n^m) x^{mn}$. Since $a_m \neq 0$ and $b_n \neq 0$, the coefficient $a_m b_n^m$ is also non-zero. The degree is therefore $m \cdot n$. This statement is **always true**. **4. If $m > n$, the degree of $P(x) + Q(x)$ is $m-n$.** If $m > n$, the degree of $Q(x)$ is strictly less than the degree of $P(x)$. **Step 1:** Write the sum. Since $m > n$, the term $a_m x^m$ is the term with the highest power in the entire sum. Its coefficient $a_m$ is non-zero and is not affected by any term from $Q(x)$. **Step 2:** Determine the degree of the sum. The degree of $P(x) + Q(x)$ is $m$. The statement claims the degree is $m-n$. This is incorrect. **Counterexample:** Let $P(x) = x^3$ ($m=3$) and $Q(x) = x^2$ ($n=2$). Then $m>n$. $P(x) + Q(x) = x^3 + x^2$. The degree is $3$. The statement predicts the degree to be $m-n = 3-2 = 1$. Since $3 \neq 1$, the statement is **not always true**. **Conclusion:** The correct statements are the first and the third.

A key property of polynomials with integer coefficients is that for any two integers $a$ and $b$, the expression $(a-b)$ must divide $(P(a) - P(b))$. We will use this property to evaluate each option. 1. **Analysis of Option 1: "There is no integer $k$ such that $P(k) = 3$."** Let's assume there exists an integer $k$ such that $P(k)=3$. Applying the property: * With $a=k$ and $b=2$: $(k-2)$ must divide $(P(k) - P(2)) = (3 - 5) = -2$. * With $a=k$ and $b=5$: $(k-5)$ must divide $(P(k) - P(5)) = (3 - 2) = 1$. From the second condition, $(k-5)$ must be a divisor of $1$. The only integer divisors of $1$ are $1$ and $-1$. * Case 1: $k-5 = 1 \implies k = 6$. We check this with the first condition: $(k-2) = (6-2) = 4$. Does $4$ divide $-2$? No. * Case 2: $k-5 = -1 \implies k = 4$. We check this with the first condition: $(k-2) = (4-2) = 2$. Does $2$ divide $-2$? Yes. Since $k=4$ is consistent with the necessary divisibility conditions, it is possible for such an integer to exist. For instance, the linear polynomial $P(x) = -x+7$ has integer coefficients and satisfies $P(2)=-2+7=5$ and $P(5)=-5+7=2$. For this polynomial, $P(4) = -4+7=3$. Thus, a polynomial satisfying the conditions can have $P(k)=3$ for an integer $k$. Therefore, this statement is not necessarily true. 2. **Analysis of Option 2: "$P(3)$ is an even integer."** Let's apply the property with $a=3$ and $b=5$: $(3-5)$ must divide $(P(3) - P(5))$. This means $-2$ must divide $(P(3) - 2)$. If an integer is divisible by $-2$, it is an even number. So, $P(3) - 2$ must be an even integer. Since $m$ is an integer, $(m+1)$ is also an integer. This shows that $P(3)$ must be a multiple of $2$, i.e., an even integer. This statement is necessarily true. 3. **Analysis of Option 3: "$P(8) - P(-1)$ is divisible by 9."** Let's apply the property with $a=8$ and $b=-1$: $(a-b)$ must divide $(P(a) - P(b))$. This is a direct consequence of the property. The statement is necessarily true. 4. **Analysis of Option 4: "The degree of $P(x)$ is at least 1."** Let's consider the case where the degree is less than $1$. Since the degree must be a non-negative integer, the only possibility is $\text{deg}(P) = 0$. A polynomial of degree $0$ is a non-zero constant, say $P(x) = c$. From the given conditions: * $P(2) = c = 5$ * $P(5) = c = 2$ This leads to the contradiction $5 = 2$. Therefore, the degree of $P(x)$ cannot be $0$. It must be at least $1$. This statement is necessarily true. Answer: \boxed{\text{Option 1}}