Numbers and expressions, sets and logic (Mathematics I)

By the product of a sum and a difference, (a+b)(a−b)=a²−b², the answer is x²−9. The option x²+9 comes from a sign error, and x²−6x+9 comes from confusing this with (x−3)². You can also check term by term: x²−3x+3x−9.

By (a+b)²=a²+2ab+b², we get x²+2·x·2+2² = x²+4x+4. The option x²+4 is the classic slip of forgetting the middle term 2ab. Expanding (x+2)(x+2) with the distributive law gives x²+2x+2x+4, which confirms the answer.

2x·x=2x², 2x·(−4)+3·x=−8x+3x=−5x, and 3·(−4)=−12, so the answer is 2x²−5x−12. The −11x version comes from computing −8−3 instead of adding −8x and 3x, and 2x²+5x−12 comes from a sign error.

The two numbers with product 6 and sum 5 are 2 and 3, so the answer is (x+2)(x+3). For (x+1)(x+6) the product is 6 but the sum is 7, and (x−2)(x−3) expands to x²−5x+6, whose signs do not match.

With the cross (grouping) method, split 2=2×1 and 3=1×3; the cross sum 2×3+1×1=7 matches, so the answer is (2x+1)(x+3). Expanding (2x+3)(x+1) gives 2x²+5x+3, where the coefficient of x does not match. Getting into the habit of checking by expanding matters.

Split 3=3×1 and 8=(−4)×(−2); the cross sum 3×(−2)+(−4)×1=−10 matches, giving (3x−4)(x−2). Expanding (3x−2)(x−4) gives 3x²−14x+8, and (3x+4)(x+2) gives 3x²+10x+8, whose signs are reversed.

Split 6=2×3 and −2=(−1)×2; the cross sum 2×2+(−1)×3=1 matches, giving (2x−1)(3x+2). Expanding (2x+1)(3x−2) gives 6x²−x−2, in which the sign of the coefficient of x is reversed.

Group the last three terms as −(y²−4y+4)=−(y−2)², giving x²−(y−2)². Applying A²−B²=(A+B)(A−B) with A=x and B=y−2 gives (x+y−2)(x−y+2). Distributing the sign of B wrongly leads to the other forms. Expanding to see whether you return to the original is a good check.

Putting x²=t gives t²−5t+4=(t−1)(t−4). Returning to x² gives (x²−1)(x²−4), and applying A²−B² to each gives (x+1)(x−1)(x+2)(x−2). Factorising should be carried on until nothing can be split further.

Putting x²+x=t gives t²−8t+12=(t−2)(t−6). Substituting back gives (x²+x−2)(x²+x−6)=(x+2)(x−1)(x+3)(x−2). The pair with product 12 and sum −8 is −2 and −6; −3 and −4 give a sum of −7, which does not match.

√12=√(4×3)=√4×√3=2√3. The square factor 4 comes outside the radical. The option 3√2 comes from wrongly splitting 12 as 9×2 (that product is 18, so it is wrong). Numerically √12≒3.46 and 2√3≒3.46, which agree.

First take the square factors outside: √27=3√3 and √12=2√3. Then collect like terms: 3√3+2√3=5√3. The option √39 comes from wrongly adding the numbers under the radicals. Combining what is under radicals is only allowed for multiplication and division.

By (a+b)(a−b)=a²−b², we get (√5)²−(√2)²=5−2=3. The option 7 comes from adding as 5+2, and √3 comes from leaving the radical in place.

Multiply numerator and denominator by √3: √3/(√3×√3)=√3/3. The denominator becomes (√3)²=3. The option √3/9 comes from wrongly taking the denominator to be 3².

Multiply numerator and denominator by the conjugate √5+√3; the denominator becomes 5−3=2. Hence 2(√5+√3)/2=√5+√3. The option 2(√5+√3) comes from forgetting to cancel the final 2, and (√5+√3)/4 is the form that appears when the denominator is wrongly taken to be 5+3=8.

By (a+b)²=a²+2ab+b², we get 3+2√3·√2+2=5+2√6. The option 5 comes from forgetting the middle term 2ab, and 5+√6 comes from forgetting to multiply by 2.

Since π>3, the inside 3−π is negative. For a negative value |a|=−a, so |3−π|=−(3−π)=π−3. Removing the bars and leaving 3−π would give a negative number, and an absolute value (a distance) cannot be negative, which is a useful check.

From 4<7<9 we get 2<√7<3, so the integer part is a=2. The fractional part is the number minus its integer part, b=√7−2, which satisfies 0≦b<1. The value √7−3 is negative, so it cannot be a fractional part.

Rationalising gives x=2+√3 and 1/x=2−√3. Since x+1/x=4, we get x²+1/x²=(x+1/x)²−2=16−2=14. The option 16 comes from leaving (x+1/x)² and forgetting the −2. Indeed x²=7+4√3 and 1/x²=7−4√3, whose sum is 14.

x³+1/x³=(x+1/x)³−3(x+1/x)=27−9=18. The option 27 comes from forgetting the correction term −3(x+1/x), and 24 from subtracting only 3. The value 7 is x²+1/x², a different quantity.

Moving terms gives 3x>9, and dividing by the positive number 3 gives x>3. Dividing by a positive number does not reverse the inequality sign. The option x>9 comes from forgetting to divide by 3.

Moving terms gives −2x≦6. Dividing by the negative number −2 reverses the inequality sign, so x≧−3. The option x≦−3 is the classic slip of forgetting to reverse it. Substituting x=0 gives 3≦9, which holds, so the correct answer must contain 0, and x≧−3 does.

Move 2x to the left and −1 to the right to get 2x<6, hence x<3. The option x<2 comes from wrongly computing 5−1=4 and writing 2x<4. Checking with x=0 gives −1<5, which holds, so 0 must be a solution, and it is.

Multiply both sides by 6 to get 3x−6>2x, then move terms to get x>6. The option x>1 comes from forgetting to multiply the constant −1 by 6 and writing 3x−1>2x. When clearing denominators, multiply the constant terms by the same number too.

The first gives x>2 (2 is not included), and the second gives 2x≦8, so x≦4 (4 is included). The common part is 2<x≦4. Whether an endpoint is included carries over directly from the original signs (> and ≦).

There are two points on the number line at distance 5 from the origin, so x=5 and x=−5, written together as x=±5. Giving only the positive one loses a solution.

The form |X|<c means −c<X<c. From −3<x−2<3, adding 2 to each part gives −1<x<5. The option −3<x<3 comes from ignoring the shift of −2. Substituting x=0 gives |−2|=2<3, which holds, matching the fact that 0 lies in the correct range.

The form |X|≧c gives the outside: X≦−c or X≧c. From x+1≦−4 or x+1≧4 we get x≦−5 or x≧3. The sandwiched form belongs to |X|≦c, and mixing the two up is the classic slip. At x=0 we have |1|=1, which fails, matching an answer that excludes 0.

Rearranging gives 3x<a+3, so x<(a+3)/3. For this to agree with x<4 we need (a+3)/3=4, hence a=9. The option a=12 comes from dropping the constant 3 and setting a/3=4, and a=1 from setting a+3=4. With a=9 both sides are equal at x=4, which confirms it.

We need x>−2 and x≦3, that is −2<x≦3. The integers in this range are −1, 0, 1, 2 and 3, so there are 5 of them. Since −2 is not included (x>−2 has no equality), 6 is wrong. Writing out the endpoints and counting is the reliable way.

A∩B is the set of elements belonging to both. Of the elements 2, 4, 6 and 8 of B, those also in A are 2, 4 and 6. The set {1, 2, 3, 4, 5, 6, 8} is A∪B, the union, and confusing ∩ with ∪ is the classic slip.

A∪B is every element belonging to at least one of them. Adding 8, which is only in B, to the elements of A gives {1, 2, 3, 4, 5, 6, 8}. A repeated element is written only once. The set {2, 4, 6} is the intersection A∩B.

The complement consists of the elements of U that do not belong to A. Removing the even numbers leaves the odd numbers {1, 3, 5, 7, 9}. A complement is always taken after the universal set U has been fixed.

By De Morgan's laws, taking complements swaps ∪ and ∩, so the complement of A∪B is Ā∩B̄. A Venn diagram confirms that being in neither A nor B is the same as being outside A and outside B. The expression Ā∪B̄ is instead the complement of A∩B.

The implication x=3 ⇒ x²=9 is true. The converse x²=9 ⇒ x=3 is false, since x=−3 is a counterexample. Being the hypothesis of a true implication makes x=3 a sufficient but not necessary condition.

Since x>2 ⇒ x>0 is true, x>0 is a necessary condition. On the other hand x>0 ⇒ x>2 is false, with x=1 as a counterexample, so it is not sufficient. It helps to remember through inclusion of ranges that the wider condition (x>0) is the necessary one.

In proof by contradiction you assume the negation of what you want to prove. The negation of "is irrational" is "is rational". Writing it as a fraction q/p in lowest terms and proceeding, both p and q turn out to be even, contradicting lowest terms, which shows the assumption was wrong.

The converse swaps hypothesis and conclusion, giving "If x²=1 then x=1". This converse is in fact false, since x=−1 is a counterexample, so it is an example of a true statement whose converse need not be true. Negating both parts gives the inverse, and doing both gives the contrapositive.

The contrapositive is the negation of the conclusion implying the negation of the hypothesis. The negation of "n is even" is "n is odd", and the negation of "n² is even" is "n² is odd", so the contrapositive is "If n is odd then n² is odd". That form is easy to prove directly and can be used to prove the original statement.

Since a=b ⇒ |a|=|b| is true, |a|=|b| is a necessary condition. The converse |a|=|b| ⇒ a=b is false, with a=1, b=−1 as a counterexample (|1|=|−1| but 1≠−1), so it is not sufficient. Hence it is necessary but not sufficient.