The question presents one or more propositions and asks the student to determine whether each is true or false, providing a justified proof or counterexample for each.
The two diagonals of the quadrilateral $Q$ are perpendicular. Consider the following statements: I One of the diagonals of $Q$ is a line of symmetry of $Q$. II The midpoints of the sides of $Q$ are the vertices of a square. Which of these statements is/are necessarily true for the quadrilateral $Q$ ?
Which one of the following functions provides a counterexample to the statement: if $\mathrm { f } ^ { \prime } ( x ) > 0$ for all real $x$, then $\mathrm { f } ( x ) > 0$ for all real $x$.
The diagram shows an example of a mountain profile. [Figure] This consists of upstrokes which go upwards from left to right, and downstrokes which go downwards from left to right. The example shown has six upstrokes and six downstrokes. The horizontal line at the bottom is known as sea level. A mountain profile of order $n$ consists of $n$ upstrokes and $n$ downstrokes, with the condition that the profile begins and ends at sea level and never goes below sea level (although it might reach sea level at any point). So the example shown is a mountain profile of order 6. Mountain profiles can be coded by using U to indicate an upstroke and D to indicate a downstroke. The example shown has the code UDUUUDUDDUDD. A sequence of U's and D's obtained from a mountain profile in this way is known as a valid code. Which of the following statements is/are true? I If a valid code is written in reverse order, the result is always a valid code. II If each $U$ in a valid code is replaced by $D$ and each $D$ by $U$, the result is always a valid code. III If U is added at the beginning of a valid code and D is added at the end of the code, the result is always a valid code.
Consider the following attempt to solve the equation $4 x \sqrt { 2 x - 1 } = 10 x - 5$ : $$\begin{aligned}
4 x \sqrt { 2 x - 1 } & = 10 x - 5 \\
4 x \sqrt { 2 x - 1 } & = 5 ( 2 x - 1 ) \\
16 x ^ { 2 } ( 2 x - 1 ) & = 25 ( 2 x - 1 ) ^ { 2 } \\
16 x ^ { 2 } & = 25 ( 2 x - 1 ) \\
16 x ^ { 2 } - 50 x + 25 & = 0 \\
( 8 x - 5 ) ( 2 x - 5 ) & = 0
\end{aligned}$$ The solutions of the original equation are $x = \frac { 5 } { 8 }$ and $x = \frac { 5 } { 2 }$. Which one of the following is true?
$a , b$ and $c$ are real numbers. Given that $a b = a c$, which of the following statements must be true? I $\quad a = 0$ II $b = 0$ or $c = 0$ III $b = c$
Consider the following conjecture: If $N$ is a positive integer that consists of the digit 1 followed by an odd number of 0 digits and then a final digit 1 , then $N$ is a prime number. Here are three numbers: I $\quad N = 101$ (which is a prime number) II $\quad N = 1001$ (which equals $7 \times 11 \times 13$ ) III $N = 10001$ (which equals $73 \times 137$ ) Which of these provide(s) a counterexample to the conjecture?
A student attempts to solve the equation $$\cos x + \sin x \tan x = 2 \sin x - 1$$ in the range $0 \leq x \leq 2 \pi$. The student's attempt is as follows: $\cos x + \sin x \tan x = 2 \sin x - 1$ So $\quad \cos x - \sin x + \sin x \tan x - \sin x = - 1$ So $\quad ( \sin x - \cos x ) ( \tan x - 1 ) = - 1$ So $\quad \sin x - \cos x = - 1$ or $\tan x - 1 = - 1$ So $\quad ( \sin x - \cos x ) ^ { 2 } = 1$ or $\tan x = 0$ So $\quad 2 \sin x \cos x = 0$ or tan $x = 0$ So $x = 0 , \frac { \pi } { 2 } , \pi , \frac { 3 \pi } { 2 } , 2 \pi$ Which of the following best describes this attempt?
For which one of the following statements can the fact that $12 ^ { 2 } + 16 ^ { 2 } = 20 ^ { 2 }$ be used to produce a counterexample? A If $a , b$ and $c$ are positive integers which satisfy the equation $a ^ { 2 } + b ^ { 2 } = c ^ { 2 }$, and the three numbers have no common divisor, then two of them are odd and the other is even. B The equation $a ^ { 4 } + b ^ { 2 } = c ^ { 2 }$ has no solutions for which $a , b$ and $c$ are positive integers. C The equation $a ^ { 4 } + b ^ { 4 } = c ^ { 4 }$ has no solutions for which $a , b$ and $c$ are positive integers. D If $a , b$ and $c$ are positive integers which satisfy the equation $a ^ { 2 } + b ^ { 2 } = c ^ { 2 }$, then one is the arithmetic mean of the other two.
A student makes the following claim: For all integers $n$, the expression $4 \left( \frac { 9 n + 1 } { 2 } - \frac { 3 n - 1 } { 2 } \right)$ is divisible by 3 . Here is the student's argument: $$\begin{aligned}
4 \left( \frac { 9 n + 1 } { 2 } - \frac { 3 n - 1 } { 2 } \right) & = 2 \left( 2 \left( \frac { 9 n + 1 } { 2 } - \frac { 3 n - 1 } { 2 } \right) \right) \\
& = 2 ( 9 n + 1 - 3 n - 1 ) \\
& = 2 ( 6 n ) \\
& = 12 n \\
& = 3 ( 4 n )
\end{aligned}$$ which is always a multiple of 3 . So the expression $4 \left( \frac { 9 n + 1 } { 2 } - \frac { 3 n - 1 } { 2 } \right)$ is always divisible by 3 . Which one of the following is true? A The argument is correct. B The argument is incorrect, and the first error occurs on line (I). C The argument is incorrect, and the first error occurs on line (II). D The argument is incorrect, and the first error occurs on line (III). E The argument is incorrect, and the first error occurs on line (IV). F The argument is incorrect, and the first error occurs on line (V). G The argument is incorrect, and the first error occurs on line (VI).
Consider the following statement: Every positive integer $N$ that is greater than 6 can be written as the sum of two non-prime integers that are greater than 1 . Which of the following is/are counterexample(s) to this statement? I $\quad N = 5$ II $\quad N = 7$ III $N = 9$ A none of them B I only C II only D III only E I and II only F I and III only G II and III only H I, II and III
A student is asked to prove whether the following statement (*) is true or false: (*) For all real numbers $a$ and $b , | a + b | < | a | + | b |$ The student's proof is as follows: Statement (*) is false. A counterexample is $a = 3 , b = 4$, as $| 3 + 4 | = 7$ and $| 3 | + | 4 | = 7$, but $7 < 7$ is false. Which of the following best describes the student's proof? A The statement ( $*$ ) is true, and the student's proof is not correct. B The statement (*) is false, but the student's proof is not correct: the counterexample is not valid. C The statement (*) is false, but the student's proof is not correct: the student needs to give all the values of $a$ and $b$ where $| a + b | < | a | + | b |$ is false. D The statement (*) is false, but the student's proof is not correct: the student should have instead stated that for all real numbers $a$ and $b , | a + b | \leq | a | + | b |$. E The statement (*) is false, and the student's proof is fully correct.
Consider the following claim about positive integers $a , b$ and $c$ : if $a$ is a factor of $b c$, then $a$ is a factor of $b$ or $a$ is a factor of $c$ Which of the following provide(s) a counterexample to this claim? I $a = 5 , b = 10 , c = 20$ II $\quad a = 8 , b = 4 , c = 4$ III $a = 6 , b = 7 , c = 12$ A none of them B I only C II only D III only E I and II only F I and III only G II and III only H I, II and III
On which line is the first error in the following argument? A $\sin ^ { 2 } x + \cos ^ { 2 } x = 1$ for all values of $x$. B Therefore $\cos x = \sqrt { 1 - \sin ^ { 2 } x }$ for all values of $x$. C Hence $1 + \cos x = 1 + \sqrt { 1 - \sin ^ { 2 } x }$ for all values of $x$. D Thus $( 1 + \cos x ) ^ { 2 } = \left( 1 + \sqrt { 1 - \sin ^ { 2 } x } \right) ^ { 2 }$ for all values of $x$. E Substituting $x = \pi$ gives $0 = 4$.
A student attempts to solve the following problem, where $a$ and $b$ are non-zero real numbers: Show that if $a ^ { 2 } - 4 b ^ { 3 } \geq 0$ then there exist real numbers $x$ and $y$ such that $a = x y ( x + y )$ and $b = x y$. Consider the following attempt: $$\begin{aligned}
& ( x - y ) ^ { 2 } \geq 0 \\
& \text { so } \quad x ^ { 2 } + y ^ { 2 } - 2 x y \geq 0 \\
& \text { so } \quad ( x + y ) ^ { 2 } - 4 x y \geq 0 \\
& \text { so } \quad x ^ { 2 } y ^ { 2 } ( x + y ) ^ { 2 } - 4 x ^ { 3 } y ^ { 3 } \geq 0 \\
& \text { so } \quad a ^ { 2 } - 4 b ^ { 3 } \geq 0
\end{aligned}$$ Which of the following best describes this attempt? A It is completely correct. B It is incorrect, but it would be correct if written in the reverse order. C It is incorrect, but the student has correctly proved the converse. D It is incorrect because there is an error in line (II). $\mathbf { E }$ It is incorrect because there is an error in line (III). F It is incorrect because there is an error in line (IV).
Consider the following statement about the positive integer $n$ if $n$ is prime, then $n ^ { 2 } + 2$ is not prime Which of the following is a counterexample to this statement? I $n = 2$ II $n = 3$ III $n = 4$ A none of them B I only C II only D III only E I and II only F I and III only G II and III only H I, II and III
Consider the following claim: The difference between two consecutive positive cube numbers is always prime. Here is an attempted proof of this claim: $$\text { I } \quad ( x + 1 ) ^ { 3 } = x ^ { 3 } + 3 x ^ { 2 } + 3 x + 1$$ II Taking $x$ to be a positive integer, the difference between two consecutive cube numbers can be expressed as $( x + 1 ) ^ { 3 } - x ^ { 3 } = 3 x ^ { 2 } + 3 x + 1$ III It is impossible to factorise $3 x ^ { 2 } + 3 x + 1$ into two linear factors with integer coefficients because its discriminant is negative. IV Therefore for every positive integer value of $x$ the integer $3 x ^ { 2 } + 3 x + 1$ cannot be factorised. V Hence, the difference between two consecutive cube numbers will always be prime. Which of the following best describes this proof? A The proof is completely correct, and the claim is true. B The proof is completely correct, but there are counterexamples to the claim. C The proof is wrong, and the first error occurs on line I. D The proof is wrong, and the first error occurs on line II. E The proof is wrong, and the first error occurs on line III. F The proof is wrong, and the first error occurs on line IV. G The proof is wrong, and the first error occurs on line V.
Which of the following statements is/are true? I For all real numbers $x$ and for all positive integers $n , x < n$ II For all real numbers $x$, there exists a positive integer $n$ such that $x < n$ III There exists a real number $x$ such that for all positive integers $n , x < n$ A none of them B I only C II only D III only E I and II only F I and III only G II and III only H I, II and III
In this question, $a _ { 1 } , \ldots , a _ { 100 }$ and $b _ { 1 } , \ldots , b _ { 100 }$ and $c _ { 1 } , \ldots , c _ { 100 }$ are three sequences of integers such that $$a _ { n } \leq b _ { n } + c _ { n }$$ for each $n$. Which of the following statements must be true? I (minimum of $\left. a _ { 1 } , \ldots , a _ { 100 } \right) \leq$ (minimum of $\left. b _ { 1 } , \ldots , b _ { 100 } \right) + \left( \right.$ minimum of $\left. c _ { 1 } , \ldots , c _ { 100 } \right)$ II (minimum of $\left. a _ { 1 } , \ldots , a _ { 100 } \right) \geq$ (minimum of $\left. b _ { 1 } , \ldots , b _ { 100 } \right) +$ (minimum of $c _ { 1 } , \ldots , c _ { 100 }$ ) III (maximum of $\left. a _ { 1 } , \ldots , a _ { 100 } \right) \leq$ (maximum of $\left. b _ { 1 } , \ldots , b _ { 100 } \right) +$ (maximum of $c _ { 1 } , \ldots , c _ { 100 }$ ) A none of them B I only C II only D III only E I and II only F I and III only G II and III only H I, II and III
A student answered the following question: $a$ and $b$ are non-zero real numbers. Prove that the equation $x ^ { 3 } + a x ^ { 2 } + b = 0$ has three distinct real roots if $27 b \left( b + \frac { 4 a ^ { 3 } } { 27 } \right) < 0$ Here is the student's solution: I We differentiate $y = x ^ { 3 } + a x ^ { 2 } + b$ to get $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } + 2 a x = x ( 3 x + 2 a )$ Solving $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ shows that the stationary points are at $( 0 , b )$ and $\left( - \frac { 2 a } { 3 } , b + \frac { 4 a ^ { 3 } } { 27 } \right)$ II If $27 b \left( b + \frac { 4 a ^ { 3 } } { 27 } \right) < 0$, then $b$ and $b + \frac { 4 a ^ { 3 } } { 27 }$ must have opposite signs, and so one of the stationary points is above the $x$-axis and one is below. III If the cubic has three distinct real roots, then one of the stationary points is above the $x$-axis and one is below. IV Hence if $27 b \left( b + \frac { 4 a ^ { 3 } } { 27 } \right) < 0$, then the equation has three distinct real roots. Which one of the following options best describes the student's solution? A It is a completely correct solution. B The student has instead proved the converse of the statement in the question. C The solution is wrong, because the student should have stated step II after step III. D The solution is wrong, because the student should have shown the converse of the result in step II. E The solution is wrong, because the student should have shown the converse of the result in step III.
A student attempts to answer the following question. What is the largest number of consecutive odd integers that are all prime? The student's attempt is as follows: I There are two consecutive odd integers that are prime (for example: 17, 19). II Any three consecutive odd integers can be written in the form $n - 2 , n , n + 2$ for some $n$. III If $n$ is one more than a multiple of 3 , then $n + 2$ is a multiple of 3 . IV If $n$ is two more than a multiple of 3 , then $n - 2$ is a multiple of 3 . V The only other possibility is that $n$ is a multiple of 3 . VI In each case, one of the integers is a multiple of 3 , so not prime. VII Therefore the largest number of consecutive odd integers that are all prime is two. Which of the following best describes this attempt? A It is completely correct. B It is incorrect, and the first error is on line I. C It is incorrect, and the first error is on line II. D It is incorrect, and the first error is on line III. E It is incorrect, and the first error is on line IV. F It is incorrect, and the first error is on line V. G It is incorrect, and the first error is on line VI. H It is incorrect, and the first error is on line VII.
A student draws a triangle that is acute-angled or obtuse-angled but not right-angled. The student counts the number of straight lines that divide the triangle into two triangles, at least one of which is right-angled. Which of the following statements is/are true? I The student can draw a triangle for which there is exactly 1 such straight line. II The student can draw a triangle for which there are exactly 2 such straight lines. III The student can draw a triangle for which there are exactly 3 such straight lines. A none of them B I only C II only D III only E I and II only F I and III only G II and III only H I, II and III
A student made an error while proving the following claim that he thought was true. Claim: For any sets $A$, $B$, $C$, we have $A \backslash ( B \cap C ) \subseteq ( A \backslash B ) \cap ( A \backslash C )$. The student's proof: If I show that every element of the set $A \backslash ( B \cap C )$ is in the set $( A \backslash B ) \cap ( A \backslash C )$, the proof is complete. Now, let $x \in A \backslash ( B \cap C )$. (I) From this, $x \in A$ and $x \notin ( B \cap C )$. (II) From this, $x \in A$ and $( x \notin B$ and $x \notin C )$. (III) From this, $( x \in A$ and $x \notin B )$ and $( x \in A$ and $x \notin C )$. (IV) From this, $x \in A \backslash B$ and $x \in A \backslash C$. (V) From this, $x \in [ ( A \backslash B ) \cap ( A \backslash C ) ]$. In which of the numbered steps did this student make an error? A) I B) II C) III D) IV E) V
Let A, B and C be sets. I. If $A \cup B = A \cup C$ then $B = C ^ { \prime }$. II. If $\mathrm { A } \cap \mathrm { B } = \varnothing$ then $\mathrm { A } \backslash \mathrm { B } = \mathrm { A } ^ { \prime }$. III. If $A \cup B = A$ then $B \backslash A = \varnothing$. Which of these propositions are always true? A) Only I B) Only II C) Only III D) I and II E) II and III
A student made an error while proving the following claim that he thought was true. Claim: Let $f : X \rightarrow Y$ be a function, and let $A$ and $B$ be subsets of $X$. Then $f ( A \cap B ) = f ( A ) \cap f ( B )$. The student's proof: If I show that the sets $f ( A \cap B )$ and $f ( A ) \cap f ( B )$ are subsets of each other, the proof is complete. Now let $c \in f ( A \cap B )$. I. There exists a $d \in A \cap B$ such that $c = f ( d )$. II. Since $d \in A$ and $d \in B$, we have $f ( d ) \in f ( A )$ and $f ( d ) \in f ( B )$. Thus $c = f ( d ) \in f ( A ) \cap f ( B )$. On the other hand, let $c \in f ( A ) \cap f ( B )$. III. We have $c \in f ( A )$ and $c \in f ( B )$. From this, there exists an $a \in A$ such that $c = f ( a )$ and a $\mathrm { b } \in \mathrm { B }$ such that $c = f ( b )$. IV. Since $c = f ( a )$ and $c = f ( b )$, we have $a = b$. V. Since $a \in A , b \in B$ and $a = b$, we have $a \in A \cap B$ and thus $c = f ( a ) \in f ( A \cap B )$. In which of the numbered steps did this student make an error? A) I B) II C) III D) IV E) V