General chapter practice

The governing concept is interpreting strict and inclusive inequalities in a set definition. The condition 2<x excludes 2, while x≤6 includes 6. Because x is natural, the resulting set is U={3,4,5,6}. Thus 3, 4, and 6 are members, but 2 is not. Option D is therefore the only element outside U; confusing < with ≤ would lead to a wrong choice.

A factor of 6 is a natural number that divides 6 exactly, leaving remainder zero. The factor pairs are 1×6 and 2×3, so the positive natural factors are 1, 2, 3, and 6. Zero cannot be a factor because division by zero is undefined, and 12 is not a divisor of 6. Therefore option A is the complete roster form of V.

The governing process is substitution in a set-builder rule. Whole numbers less than 4 are n=0,1,2,3. Substituting into x=2n+1 gives x=1,3,5,7 respectively. Hence C={1,3,5,7}, which is option A. Option B incorrectly starts with n=1; option C ignores the rule, and option D includes 0, which the formula never produces for these values.

The governing concept is counting integers in a half-open interval. Since -3<x, the integer -3 is excluded; since x≤2, the endpoint 2 is included. Listing the permitted integers gives -2,-1,0,1,2. There are five distinct values, so the cardinality of D is 5. Therefore option B is correct; counts of 4, 6, or 7 result from mishandling an endpoint or adding an outside integer.

A roster form lists every element explicitly. The positive multiples of 6 below 30 are obtained as 6×1=6, 6×2=12, 6×3=18, and 6×4=24. The next multiple, 30, is excluded because the condition says less than 30, and 0 is not positive. Thus E={6,12,18,24}, so option A is correct.

Using nonnegative integer exponents, the relevant powers are 10⁰=1, 10¹=10, and 10²=100. The condition x<100 includes 1 and 10 but excludes 100, so G={1,10}. Zero is not the value of 10⁰; it is the exponent in that expression. Therefore option A is correct, while the other choices include the boundary or an invalid value.

The governing concept is membership under inclusive bounds. Because 1≤x≤4 includes both endpoints, the set is H={1,2,3,4}. Consequently, 1∈H, 3∈H, and 4∈H are true statements. The number 5 lies beyond the upper limit, so 5∈H is false. Hence option C is the required answer; treating the question as asking for a true statement would reverse the task.

The governing concept is identifying positive divisors and writing them in roster form. The factor pairs of 9 are 1×9 and 3×3, so its positive factors are 1, 3, and 9. Six is not a factor because 9÷6 is not an integer, and zero cannot be a factor because division by zero is undefined. Therefore option A gives the complete set.

The governing idea is converting strict inequalities into a roster form. Greater than 2 means x>2, so 2 is excluded; less than 9 means x<9, so 9 is also excluded. The natural numbers between these boundaries are 3,4,5,6,7,8. Thus option B contains exactly the required members. The other options include one or both excluded endpoints.

A constant term is a term that contains no variable. In 7a - 4, the term 7a contains the variable a, whereas -4 contains no variable. Therefore, -4 is the constant term. The option 7 is only the coefficient of a in 7a; it is not a separate term in the expression. Exam tip: To identify the constant term, choose the term that has no variable.

The number multiplying a variable is called its coefficient. In 5m, the variable m is multiplied by 5, so the coefficient of m is 5. Option C (1) is incorrect because 1 would be the coefficient if the term were simply m. Exam tip: Identify the number directly multiplying the variable.

3p and 2p are like terms because both contain the variable p to the first power. Add their coefficients: 3p + 2p = (3 + 2)p = 5p. Therefore, option C is correct. Exam tip: when adding or subtracting like terms, operate only on their coefficients; p² is obtained from p × p, not from adding terms.

A monomial is a polynomial containing exactly one term. In \(3x^2\), there is no plus or minus sign separating terms, so it has only one term and is a monomial. \(x+2\) and \(x^2+x\) have two terms each, while \(x^2+x+1\) has three terms; therefore, none of them is a monomial. Exam tip: count the terms separated by plus or minus signs.

Terms in an algebraic expression are generally separated by \(+\) or \(-\) signs. The expression \((4x+9)\) has two terms: \(4x\) and \(9\). Therefore, the correct answer is 2. Choosing 3 is incorrect because the \(+\) sign itself is not a term. Exam tip: Count the parts separated by \(+\) or \(-\) signs, including the first term.

The polynomial \(x^2+3x+2\) contains three distinct terms: \(x^2\), \(3x\), and \(2\). A polynomial with three terms is called a trinomial, so option D is correct. A binomial has exactly two terms, so it is not the correct classification here. Exam tip: count the terms after combining any like terms.

6x and 2x are like terms because both contain x to the first power. Therefore, subtract their coefficients: 6 − 2 = 4. Hence, 6x − 2x = 4x. The option 4x² is incorrect because adding or subtracting like terms does not change the variable’s exponent. Exam tip: identify like terms first, and then operate only on their coefficients.

Like terms must have the same variables raised to the same powers. The variable in \(2a\) is \(a\), whereas the variable in \(3b\) is \(b\); therefore, they are unlike terms. A difference in coefficients alone does not make terms unlike—for example, \(2a\) and \(5a\) are like terms. In an exam, first compare the variables and their powers.

Substituting \(a=2\) gives \(a^2+1=2^2+1=4+1=5\), so option A is correct. Option B is only the value of \(a^2\) and omits the additional 1. In such questions, evaluate the exponent first and then perform addition or subtraction.

Subtract 4 from both sides to isolate \(x\): \(x+4-4=9-4\). Therefore, \(x=5\), so option B is correct. Option C, 9, is the right-hand side of the equation, not the value of \(x\). Exam tip: verify the answer by substituting it back into the original equation: \(5+4=9\).

By the distributive law, 3 must be multiplied by both terms inside the brackets: \(3(x+2)=3\times x+3 imes2=3x+6\). Therefore, option D is correct. Option A is incorrect because it does not multiply 3 by 2. Exam tip: distribute the outside coefficient to every term inside the brackets.