TMUA practice questions with worked solutions

Use the identity $(x + y)^2 = x^2 + 2xy + y^2$. Then $x^2 + y^2 = (x + y)^2 - 2xy = 49 - 20 = 29$. A common slip is to subtract $xy$ only once, giving $39$. (Indeed $x = 2$, $y = 5$ works: $4 + 25 = 29$.) So the answer is C.

By the cosine rule, $BC^2 = 5^2 + 8^2 - 2 \times 5 \times 8 \cos 60^\circ = 25 + 64 - 40 = 49$, so $BC = 7$. The area is $\frac{1}{2} \times 5 \times 8 \sin 60^\circ = 10\sqrt{3}$. The area is also $\frac{1}{2} \times BC \times h$ where $h$ is the required distance, so $\frac{7h}{2} = 10\sqrt{3}$ and $h = \frac{20\sqrt{3}}{7}$. Forgetting the factor $\frac{1}{2}$ in the second area formula gives $\frac{10\sqrt{3}}{7}$. So the answer is B.

At $x = 3$, $y = 2$ and $\frac{dy}{dx} = 2x - 4 = 2$, so the normal has gradient $-\frac{1}{2}$. Its equation is $y - 2 = -\frac{1}{2}(x - 3)$, that is $y = \frac{7 - x}{2}$. Setting $x^2 - 4x + 5 = \frac{7 - x}{2}$ gives $2x^2 - 7x + 3 = 0$, so $(2x - 1)(x - 3) = 0$. The root $x = 3$ is the original point, so $Q$ has $x = \frac{1}{2}$. Using gradient $\frac{1}{2}$ for the normal (a sign slip) leads to $\frac{3}{2}$, and $7$ is where the normal meets the $x$-axis. So the answer is D.

The number who study at least one language is $40 - 7 = 33$. So the number who study both is $25 + 18 - 33 = 10$. The student is chosen from the $18$ Spanish students, of whom $10$ also study French, so the probability is $\frac{10}{18} = \frac{5}{9}$. Dividing by $40$ instead gives $\frac{1}{4}$, which is the probability of studying both for a student chosen from the whole group. So the answer is E.

The contrapositive of 'if $P$ then $Q$' is 'if not $Q$ then not $P$'. Here $P$ is $x > 2$ and $Q$ is $x^2 > 4$, so the contrapositive is 'if $x^2 \le 4$, then $x \le 2$'. It is logically equivalent to the original statement. 'If $x^2 > 4$ then $x > 2$' is the converse, and 'if $x \le 2$ then $x^2 \le 4$' is the inverse; both are false (try $x = -3$). '$x > 2$ and $x^2 \le 4$' is the negation. So the answer is D.

Squaring the first equation gives $a^2 + 2ab + b^2 = 36$, so $2ab = 36 - 20 = 16$ and $ab = 8$, which is I. Then $(a - b)^2 = a^2 + b^2 - 2ab = 20 - 16 = 4$, so $|a - b| = 2$, which is II. But $a - b$ could be $2$ or $-2$: the pairs $(a, b) = (4, 2)$ and $(2, 4)$ both satisfy the equations, and the second gives $a - b = -2$. So III need not be true. So the answer is E.

A counterexample needs $a$ to divide $b^2$ but not $b$. For $(4, 6)$: $b^2 = 36 = 4 \times 9$, but $4$ does not divide $6$. For $(3, 6)$, $3$ divides $6$, so the conclusion holds. For $(6, 9)$, $(8, 6)$ and $(12, 8)$, $a$ does not divide $b^2$ ($81$, $36$ and $64$ respectively), so the hypothesis fails and these pairs say nothing about the claim. The tempting trap is to pick a pair where $a$ does not divide $b$ without checking the hypothesis. So the answer is B.

Lines 1 to 3 are correct consequences: any solution of the original equation must satisfy $(x-3)(x+2) = 0$. But squaring is not reversible, so each candidate must be checked. For $x = 3$: $\sqrt{9} = 3$, which works. For $x = -2$: $\sqrt{4} = 2 \ne -2$, which fails, because a square root is never negative. So the only solution is $x = 3$. So the answer is C.