TMUA Mock Tmua-maths-terr-1

• A.The argument is correct: the claim is true for all real $x$ and $y$.
• B.The error is that $(x-y)(x+y) = 0$ does not imply that $x - y = 0$ or $x + y = 0$, since a product of two real numbers can be zero without either factor being zero.
• C.In Case 2, showing that $x = -y$ is consistent with $x^2 = y^2$ does not show that $x = -y$ forces $x = y$; the claim is false whenever $x = -y \neq 0$.
• D.The factorisation $x^2 - y^2 = (x-y)(x+y)$ is not valid for all real $x, y$, so Case 1 and Case 2 do not actually cover all possibilities.
• E.The error occurs earlier: $x^2 - y^2 = 0$ does not follow from $x^2 = y^2$.
• F.The argument correctly shows $x = y$ or $x = -y$, but only Case 1 should have been considered valid, since Case 2 was never actually justified.

Answer: C

• A.The first error is in line (III): $\sin\theta(2\cos\theta-1)=0$ does not imply $\sin\theta=0$ or $\cos\theta=\tfrac12$.
• B.Lines (I) to (III) are correct, but line (IV) draws an unjustified conclusion, since e.g. $\theta = \pi$ satisfies $\sin\theta = 0$ and hence the original equation, but is not a multiple of $2\pi$.
• C.The first error is in line (II): the factorisation of $2\sin\theta\cos\theta - \sin\theta$ is incorrect.
• D.Lines (I) to (III) are correct, but line (IV) draws an unjustified conclusion, since e.g. $\theta = \tfrac{\pi}{3}$ satisfies $\cos\theta = \tfrac12$ and hence the original equation, but is not a multiple of $2\pi$.
• E.The proof is correct as it stands.
• F.The first error is in line (I): the rearrangement is not valid.

Answer: B

• A.The deduction $\frac{\ln x}{x} = \frac{\ln y}{y}$ is valid, but the claim that $f$ is strictly decreasing for all $t > e$ is false; in fact $f$ is strictly increasing there.
• B.The deduction $\frac{\ln x}{x} = \frac{\ln y}{y}$ is valid, and $f$ is indeed strictly decreasing for $t > e$, but this only shows $f$ is injective on $(e, \infty)$; it does not rule out one of $x, y$ being at most $e$ while the other exceeds $e$, so pairs with $x \ne y$ and $x^y = y^x$ can still exist, such as $x = 2, y = 4$.
• C.The deduction $\frac{\ln x}{x} = \frac{\ln y}{y}$ from $x^y = y^x$ is invalid, since taking logarithms of both sides of $x^y = y^x$ does not produce this equation.
• D.The claim is entirely correct: the deduction $\frac{\ln x}{x} = \frac{\ln y}{y}$ is valid, and $f(t) = \frac{\ln t}{t}$ being strictly decreasing on $t > e$ correctly rules out solutions there.
• E.The deduction $\frac{\ln x}{x} = \frac{\ln y}{y}$ is valid and $f$ is strictly decreasing for $t>e$, correctly ruling out solutions with both $x,y>e$, but this reasoning wrongly assumes $x \ne y$ forces both to exceed $e$; there is in fact no valid pair with $x,y>0$, $x\ne y$ and $x^y=y^x$ at all.

Answer: B