Mathematics Post-UTME Past Questions Study Guide

Problem Statement: Given the trigonometric function $y = 5\cos(-6x)$ , find the derivative. Note: While the transcript specifies $\frac{dx}{dy}$ , the provided multiple-choice options correspond to the standard derivative $\frac{dy}{dx}$ .
Function Defined: * $y = 5\cos(-6x)$
Mathematical Principles Applied: * The Chain Rule: This rule is used when differentiating composite functions. It is defined as: $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ . * Derivative of Cosine: The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$ . * Inner Function Geometry: For an inner function $u = -6x$ , the derivative $\frac{du}{dx} = -6$ .
Step-by-Step Calculation: * Let $u = -6x$ , then $y = 5\cos(u)$ . * Differentiate $y$ with respect to $u$ : $\frac{dy}{du} = -5\sin(u)$ . * Differentiate $u$ with respect to $x$ : $\frac{du}{dx} = -6$ . * Combine using the chain rule: $\frac{dy}{dx} = (-5\sin(-6x)) \times (-6)$ . * Simplify the expression: $\frac{dy}{dx} = 30\sin(-6x)$ .
Answer Options Provided: * (A) $30\sin(-6x)$ * (B) $5\sin(-6x)$ * (C) $-30\sin(-6x)$ * (D) $-30\cos(-6x)$
Conclusion: The correct solution matching the options is (A).

Problem Statement: Calculate the variance of the numbers: $k$ , $k+1$ , and $k+2$ .
Key Definitions and Formulas: * Arithmetic Mean ( $\mu$ ): The sum of the values divided by the number of values ( $n$ ). * Variance ( $\sigma^2$ ): The average of the squared differences from the Mean. The formula is: $\sigma^2 = \frac{\sum_{i=1}^n (x_i - \mu)^2}{n}$ .
Execution of Calculation: * Find the Mean ( $\mu$ ): * $\mu = \frac{k + (k+1) + (k+2)}{3}$ * $\mu = \frac{3k + 3}{3} = k + 1$ * Find the Deviations from the Mean: * $(k) - (k+1) = -1$ * $(k+1) - (k+1) = 0$ * $(k+2) - (k+1) = 1$ * Square the Deviations and Sum Them: * $(-1)^2 + (0)^2 + (1)^2 = 1 + 0 + 1 = 2$ * Calculate Variance ( $\sigma^2$ ): * $\sigma^2 = \frac{2}{3}$
Answer Options Provided: * (A) $\frac{1}{2}$ * (B) $\frac{2}{3}$ * (C) $\frac{3}{4}$ * (D) $1$
Conclusion: The variance of the set is $\frac{2}{3}$ , corresponding to option (B).

Problem Statement: Determine the value of $x$ at which the function $y = x^2 - 6x - 7$ reaches its minimum.
Mathematical Concept: For a quadratic function of the form $y = ax^2 + bx + c$ , the minimum or maximum occurs at the stationary point where the first derivative is zero ( $\frac{dy}{dx} = 0$ ).
First Derivative Test: * Calculate the derivative of $y = x^2 - 6x - 7$ : $\frac{dy}{dx} = 2x - 6$ . * Set the derivative to zero to find the critical point: $2x - 6 = 0$ . * Solve for $x$ : $2x = 6 \implies x = 3$ .
Second Derivative Test (Verification): * Find $\frac{d^2y}{dx^2}$ of the function: $\frac{d^2y}{dx^2} = 2$ . * Because the second derivative is positive ( $2 > 0$ ), the parabola opens upward, confirming that the point at $x = 3$ is indeed a local minimum.
Answer Options Provided: * (A) $3$ * (B) $5$ * (C) $6$ * (D) $2$
Conclusion: The minimum value occurs at $x = 3$ (Option A).

Problem Statement: Given three independent events $P$ , $Q$ , and $R$ with occurrence probabilities of $\frac{1}{2}$ , $\frac{1}{3}$ , and $\frac{1}{4}$ respectively, calculate the probability of $P$ and $Q$ occurring only.
Event Probabilities: * $P(P) = \frac{1}{2}$ * $P(Q) = \frac{1}{3}$ * $P(R) = \frac{1}{4}$
Logical Condition "P and Q Only": * This condition implies that $P$ happens, $Q$ happens, and $R$ does not happen. * The probability of $R$ not occurring is the complement: $P(R^c) = 1 - P(R) = 1 - \frac{1}{4} = \frac{3}{4}$ .
Calculation for Independent Events: * For independent events, the joint probability is the product of individual probabilities: $P(P \cap Q \cap R^c) = P(P) \times P(Q) \times P(R^c)$ . * $P(P \cap Q \cap R^c) = \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}$ * $P(P \cap Q \cap R^c) = \frac{3}{24} = \frac{1}{8}$
Answer Options Provided: * (A) $\frac{1}{8}$ * (B) $\frac{2}{3}$ * (C) $\frac{2}{9}$ * (D) $\frac{1}{2}$
Conclusion: The probability of only $P$ and $Q$ occurring is $\frac{1}{8}$ (Option A).

Problem Statement: A set of marks in a Mathematics test is given as: $11, 12, 13, 14, 15, 16, 17, 18, 19, 21$ . If $x$ represents the mean and $y$ represents the median, find the ratio $\frac{x}{y}$ correct to 1 decimal place.
Data Analysis: * Number of items ( $n$ ) = $10$
Calculating the Mean ( $x$ ): * $\text{Sum} = 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 21 = 156$ * $x = \frac{156}{10} = 15.6$
Calculating the Median ( $y$ ): * Since $n = 10$ (even), the median is the average of the $5^{\text{th}}$ and $6^{\text{th}}$ terms. * Sorted data: $11, 12, 13, 14, 15, 16, 17, 18, 19, 21$ * $5^{\text{th}}$ term = $15$ * $6^{\text{th}}$ term = $16$ * $y = \frac{15 + 16}{2} = 15.5$
Calculating the Ratio ( $\frac{x}{y}$ ): * $\frac{x}{y} = \frac{15.6}{15.5} \approx 1.00645$ * Rounding to 1 decimal place = $1.0$
Answer Options Provided: * (A) $1$ * (B) $1.3$ (transcript written as $13/5$ ) * (C) $5$ (transcript lists partial option values) * (D) $2$
Conclusion: The ratio correct to 1 decimal place is $1.0$ (Option A).

Problem Statement: Identify the point on the Euclidean plane where the curve $y = 2x^2 - 2x + 9$ has a gradient (slope) equal to $2$ .
Relationship between Derivative and Gradient: The gradient of a tangent to a curve at any point $(x, y)$ is given by its derivative $\frac{dy}{dx}$ .
Step 1: Find the Derivative: * $y = 2x^2 - 2x + 9$ * $\frac{dy}{dx} = 4x - 2$
Step 2: Solve for x when the Gradient is 2: * $4x - 2 = 2$ * $4x = 4$ * $x = 1$
Step 3: Determine the Corresponding y-coordinate: * Substitute $x = 1$ back into the original equation for the curve: * $y = 2(1)^2 - 2(1) + 9$ * $y = 2 - 2 + 9 = 9$
Identified Point: The coordinates are $(1, 9)$ .
Answer Options Provided: * (A) $(2, 4)$ * (B) $(3, 5)$ * (C) $(1, 3)$ * (D) $(1, 4)$
Note on Transcript Variation: While the mathematical calculation yields $(1, 9)$ , the transcript identifies option (D) as $(1, 4)$ .

Problem Statement: Find the equation of the line passing through the point $(5, 7)$ that is parallel to the line $7x + 5y - 5 = 0$ .
Condition for Parallel Lines: Parallel lines share the same slope ( $m$ ). Any line parallel to $Ax + By + C = 0$ will have the format $Ax + By + K = 0$ .
Solving for the New Equation: * Given line: $7x + 5y - 5 = 0$ * The parallel line equation format: $7x + 5y + C_2 = 0$ * Pass the point $(x = 5, y = 7)$ through the equation to find $C_2$ : * $7(5) + 5(7) + C_2 = 0$ * $35 + 35 + C_2 = 0$ * $70 + C_2 = 0 \implies C_2 = -70$ * Substitute back: $7x + 5y - 70 = 0$ or $7x + 5y = 70$
Answer Options Provided: * (A) $5y + 7x = 70$ * (B) $7x + 5y = 7$ * (C) $5x + 7y = 110$ * (D) $y + 5x = 70$
Conclusion: The correct equation is $7x + 5y = 70$ (Option A).