CMI Data Science Chapter-wise PYQs

**Given:** - $x, y, z$ are positive real numbers. - $x^2 + y^2 = z^2$. **Required:** Evaluate the expression: **Solution:** *Step 1:* Simplify the given algebraic fraction. The expression is of the form $\frac{A+B}{AB}$, which simplifies to $\frac{A}{AB} + \frac{B}{AB} = \frac{1}{B} + \frac{1}{A}$. Let $A = \log_{y+z} x$ and $B = \log_{z-y} x$. The expression becomes: *Step 2:* Apply the reciprocal property of logarithms, $\frac{1}{\log_b a} = \log_a b$. Applying this property to both terms transforms the expression to: *Step 3:* Apply the product rule for logarithms, $\log_M N + \log_M P = \log_M (NP)$. Combine the two logarithmic terms into a single logarithm: *Step 4:* Simplify the argument of the logarithm using the difference of squares formula, $(a-b)(a+b) = a^2 - b^2$. *Step 5:* Use the given condition $x^2 + y^2 = z^2$ to substitute into the argument. Rearranging the given equation gives $x^2 = z^2 - y^2$. Substituting this into the expression: *Step 6:* Apply the power rule for logarithms, $\log_M (N^p) = p \log_M N$. Since $\log_x x = 1$ for any valid base $x$, the expression evaluates to: **Answer:** $\boxed{2}$ **CORRECT OPTIONS: 4** ["2"]

**Given:** - An integer $n \geq 3$. - Real variables $x_1, x_2, \ldots, x_n$ with the constraint $0 \leq x_i \leq 1$ for all $1 \leq i \leq n$. - $A = x_1 + x_2 + \ldots + x_n = \sum_{i=1}^n x_i$ - $B = x_1 x_2 + x_2 x_3 + \ldots + x_{n-1} x_n + x_n x_1 = \sum_{i=1}^n x_i x_{i+1}$, where indices are taken cyclically ($x_{n+1} = x_1$). **Required:** To determine which of the given statements about the relationship between $A$ and $B$ are true. **Solution:** *Step 1:* Analyze the expression $A-B$. Rearrange the terms by grouping corresponding elements. Factor out $x_i$ from each group. *Step 2:* Determine the sign of $A-B$ using the given constraints. The constraint $0 \leq x_i \leq 1$ implies that for any $i$, both $x_i \geq 0$ and $(1 - x_{i+1}) \geq 0$. Each term $x_i(1 - x_{i+1})$ in the sum is a product of two non-negative numbers, hence each term is non-negative. The sum of non-negative terms is non-negative. This inequality holds for all valid choices of $x_i$. This confirms that statement 1 is true and statement 2 is false. *Step 3:* Analyze the condition for equality, $A=B$. Equality holds if and only if $A-B=0$. A sum of non-negative terms is zero if and only if each term is zero. *Step 4:* Find all solutions to the system of equations from Step 3. The condition $x_i(1 - x_{i+1}) = 0$ implies that for each $i$, either $x_i=0$ or $x_{i+1}=1$. First, assume there exists an index $k$ such that $x_k \in (0, 1)$. From $x_k(1-x_{k+1})=0$, since $x_k \neq 0$, it must be that $x_{k+1}=1$. From $x_{k-1}(1-x_k)=0$, since $1-x_k \neq 0$, it must be that $x_{k-1}=0$. Now, since $x_{k+1}=1$, the condition $x_{k+1}(1-x_{k+2})=0$ implies $1(1-x_{k+2})=0$, so $x_{k+2}=1$. Inductively, $x_j=1$ for all $j > k$ (cyclically). This means $x_{k-2}=1$. The condition for index $k-2$ is $x_{k-2}(1-x_{k-1})=0$. Substituting $x_{k-2}=1$ gives $1(1-x_{k-1})=0$, which implies $x_{k-1}=1$. This contradicts our finding that $x_{k-1}=0$. Therefore, no $x_i$ can be strictly between 0 and 1. All solutions must have $x_i \in \{0, 1\}$. Now, consider solutions where $x_i \in \{0, 1\}$. If the solution is not all zeros or all ones, there must be at least one 0 and at least one 1. In a circular arrangement, this implies there must be an index $k$ such that $x_k=1$ and $x_{k+1}=0$. For this index $k$, the condition $x_k(1 - x_{k+1}) = 0$ becomes $1(1-0) = 1 \neq 0$. This violates the condition. Thus, no solution can contain both 0s and 1s. The only possible solutions are: 1. $x_1 = x_2 = \ldots = x_n = 0$. 2. $x_1 = x_2 = \ldots = x_n = 1$. There are exactly two solutions. This is a finite number. This confirms that statement 3 is true and statement 4 is false. **Answer:** \boxed{1, 3}

**Given:** The real valued function $f(x) = \log_2(x^2 - 5x + 6)$. **Required:** The domain of the function $f(x)$. **Solution:** *Step 1:* State the domain condition for a logarithmic function. The argument of the logarithm must be strictly positive. *Step 2:* Factor the quadratic expression to find its roots. The roots of the quadratic equation $(x-2)(x-3) = 0$ are $x=2$ and $x=3$. *Step 3:* Determine the solution to the inequality. The quadratic expression represents an upward-opening parabola. The value of the expression is positive for values of $x$ that are outside the roots. Therefore, the inequality holds for $x < 2$ or $x > 3$. *Step 4:* Express the domain in interval notation. The domain of $f(x)$ is the union of the intervals where the inequality is satisfied. Answer: $\boxed{(-\infty, 2) \cup (3, \infty)}$

**Given:** A system of two equations: For the expressions to be defined, $x > 0$ and $y > 0$. **Required:** The values of $x$ and $y$ that satisfy both equations. **Solution:** *Step 1:* Introduce substitutions to simplify the equations. Let $a = \sqrt{x}$ and $b = \sqrt{y}$. The system of equations transforms into: *Step 2:* Solve equation (3) for the ratio $\frac{a}{b}$. Let $t = \frac{a}{b}$. Equation (3) becomes $t + \frac{1}{t} = \frac{5}{2}$. Multiplying by $2t$ gives a quadratic equation: Factoring the quadratic yields: The possible values for $t$ are $t = \frac{1}{2}$ or $t = 2$. This implies $\frac{a}{b} = \frac{1}{2}$ or $\frac{a}{b} = 2$. *Step 3:* Simplify equation (4). Combine the terms on the left-hand side: *Step 4:* Substitute the ratios from Step 2 into the simplified equation (4). Case 1: $\frac{a}{b} = 2 \implies a = 2b$. Substitute $a=2b$ into the simplified equation (4): If $b=1$, then $a = 2b = 2$. Case 2: $\frac{a}{b} = \frac{1}{2} \implies b = 2a$. Substitute $b=2a$ into the simplified equation (4): If $a=1$, then $b = 2a = 2$. *Step 5:* Convert the solutions for $(a, b)$ back to $(x, y)$. From Case 1, we have $(a, b) = (2, 1)$. This gives the solution pair $(x, y) = (4, 1)$. From Case 2, we have $(a, b) = (1, 2)$. This gives the solution pair $(x, y) = (1, 4)$. **Answer:** The solutions are $\boxed{(x, y) = (4, 1) \text{ and } (x, y) = (1, 4)}$.

**Given:** - A function $f$ on positive real numbers. - Functional equation: $f(xy) = f(x) + f(y)$ for all $x, y > 0$. - $f(2024) = 2$. **Required:** - The value of $f\left(\frac{1}{2024}\right)$. **Solution:** *Step 1:* Determine the value of $f(1)$ using the functional equation. Let $x=1$ and $y=1$. This implies $f(1) = 0$. *Step 2:* Establish a relationship between $f(x)$ and $f(1/x)$. Let $y = 1/x$ in the functional equation. Using the result from Step 1, $f(1)=0$. *Step 3:* Apply this property for $x = 2024$. *Step 4:* Substitute the given value of $f(2024)$. This corresponds to option 3. **Answer:** $\boxed{-2}$

**Given:** - $f(x)$ is a polynomial of degree 1 with real coefficients. - $g(x)$ is a polynomial of degree 2 with real coefficients. - $h(x)$ is a polynomial of degree 3 with real coefficients. **Required:** To determine which of the given statements about the intersection of the graphs of these polynomials are always true. **Solution:** The intersection points of the graphs of two polynomials, $P(x)$ and $Q(x)$, correspond to the real roots of the equation $P(x) = Q(x)$, or equivalently, the real roots of the difference polynomial $R(x) = P(x) - Q(x)$. **Statement 1: The graphs of $f(x)$ and $g(x)$ intersect at one or more points.** *Step 1:* Form the difference polynomial $R_1(x) = g(x) - f(x)$. Since $\operatorname{deg}(g) = 2$ and $\operatorname{deg}(f) = 1$, the degree of $R_1(x)$ is $\max(2, 1) = 2$. $R_1(x)$ is a quadratic polynomial. *Step 2:* Analyze the roots of $R_1(x)$. A quadratic polynomial with real coefficients does not necessarily have real roots. Its roots can be two non-real complex conjugates. For example, let $g(x) = x^2$ and $f(x) = x - 1$. The discriminant is $\Delta = (-1)^2 - 4(1)(1) = -3 < 0$. Since the discriminant is negative, $R_1(x)$ has no real roots, meaning the graphs of $f(x)$ and $g(x)$ do not intersect. Therefore, statement 1 is not always true. **Statement 2: The graphs of $f(x)$ and $h(x)$ intersect at one or more points.** *Step 1:* Form the difference polynomial $R_2(x) = h(x) - f(x)$. Since $\operatorname{deg}(h) = 3$ and $\operatorname{deg}(f) = 1$, the degree of $R_2(x)$ is $\max(3, 1) = 3$. $R_2(x)$ is a cubic polynomial. *Step 2:* Analyze the roots of $R_2(x)$. A polynomial of odd degree with real coefficients must have at least one real root. This is because its end behavior is opposite as $x \to \infty$ and $x \to -\infty$. Since the polynomial is continuous, the Intermediate Value Theorem guarantees it must cross the x-axis at least once. Therefore, $R_2(x)$ always has at least one real root, and the graphs of $f(x)$ and $h(x)$ always intersect. Statement 2 is always true. **Statement 3: The graphs of $g(x)$ and $h(x)$ intersect at one or more points.** *Step 1:* Form the difference polynomial $R_3(x) = h(x) - g(x)$. Since $\operatorname{deg}(h) = 3$ and $\operatorname{deg}(g) = 2$, the degree of $R_3(x)$ is $\max(3, 2) = 3$. $R_3(x)$ is a cubic polynomial. *Step 2:* Analyze the roots of $R_3(x)$. As established for statement 2, a cubic polynomial with real coefficients always has at least one real root. Therefore, $R_3(x)$ always has at least one real root, and the graphs of $g(x)$ and $h(x)$ always intersect. Statement 3 is always true. Answer: $\boxed{2, 3}$

**Given:** A polynomial $q(x)$ with integer coefficients. $q(1) = 2$ $q(3) = 5$ **Required:** To determine if such a polynomial $q(x)$ exists and provide a justification. **Solution:** 1. State the property for polynomials with integer coefficients. For a polynomial $q(x)$ with integer coefficients, and for any two distinct integers $a$ and $b$, the difference $(a-b)$ must divide the difference $q(a) - q(b)$. 2. Apply this property using the given values. Let $a=3$ and $b=1$. The difference in the inputs is: 3. Calculate the corresponding difference in the polynomial's values. The difference in the outputs is: 4. Check the divisibility condition. The property requires that $(a-b)$ must divide $q(a) - q(b)$. In this case, $2$ must divide $3$. 5. Conclude based on the result. Since $3$ is not an integer multiple of $2$, the condition is not satisfied. Therefore, no such polynomial $q(x)$ with integer coefficients can exist. **Answer:** \boxed{\text{No, such a polynomial does not exist.}} For a polynomial with integer coefficients, $(3-1)$ must divide $(q(3)-q(1))$. However, $2$ does not divide $3$, so no such polynomial can exist.

**Given:** Four statements regarding the existence of roots for polynomials in specified intervals: 1. $f(x) = x^3 + x - 1$ in $[-1,0]$ 2. $f(x) = x^3 + x - 1$ in $[0,1]$ 3. $f(x) = x^3 + x + 1$ in $[0,1]$ 4. $f(x) = x^3 + x + 1$ in $[-1,0]$ **Required:** Identify which of the statements are true. **Solution:** The Intermediate Value Theorem (IVT) is used to determine the existence of a root in an interval. For a continuous function $f(x)$ on $[a,b]$, if $f(a)$ and $f(b)$ have opposite signs, a root must exist in $(a,b)$. Polynomials are continuous for all $x \in \mathbb{R}$. *Step 1:* Analyze statement 1. Let $f(x) = x^3 + x - 1$. Evaluate at the endpoints of $[-1,0]$. Both $f(-1)$ and $f(0)$ are negative. IVT does not guarantee a root. Statement 1 is not necessarily true. *Step 2:* Analyze statement 2. Let $f(x) = x^3 + x - 1$. Evaluate at the endpoints of $[0,1]$. Since $f(0) < 0$ and $f(1) > 0$, the signs are opposite. By IVT, a root exists in $(0,1)$. Statement 2 is true. *Step 3:* Analyze statement 3. Let $f(x) = x^3 + x + 1$. Evaluate at the endpoints of $[0,1]$. Both $f(0)$ and $f(1)$ are positive. IVT does not guarantee a root. Statement 3 is not necessarily true. *Step 4:* Analyze statement 4. Let $f(x) = x^3 + x + 1$. Evaluate at the endpoints of $[-1,0]$. Since $f(-1) < 0$ and $f(0) > 0$, the signs are opposite. By IVT, a root exists in $(-1,0)$. Statement 4 is true. Answer: $\boxed{\text{Statements 2 and 4 are true.}}$