Set 01 · Paper 2

The contrapositive of if $P$ then $Q$'' is if $\lnot Q$ then $\lnot P$''. Here $P$ is $x>3$ and $Q$ is . So is and is . The contrapositive is ``if then ''.

By the base-angles theorem and its converse, a triangle has two equal sides if and only if the two angles opposite them are equal. So $A$ holds exactly when $B$ holds: $A \implies B$ (two equal angles force the opposite sides equal) and $B \implies A$ (two equal sides force the opposite angles equal). Each implies the other, so is both necessary and sufficient for .

Since $2$ and $3$ are coprime, an integer is divisible by both exactly when it is divisible by their product $2 \times 3 = 6$ (equivalently $\mathrm{lcm}(2,3)=6$). So divisible by ''. For example satisfy both descriptions, and any number missing a factor of or of fails both.

The statement is $\forall x,\ f(x) > 0$. Its negation is $\exists x,\ \lnot\big(f(x)>0\big)$, i.e. . The complement of '' (which includes ), so the negation is ``there is a real number with ''.

If $(a_n)$ is geometric with ratio $r$ then $a_{n+1}=r a_n$, so : the squares form a geometric sequence with ratio . Thus is true. The converse claims ``if is geometric then is geometric''. This fails: take . The ratios are not constant, so is not geometric, yet has constant ratio . Hence the converse is false.

The equation has two distinct real roots exactly when the discriminant is positive: $36 - 4c > 0 \iff c < 9$. So $B \iff (c<9)$. A condition is sufficient for if it implies , and not necessary if some value with fails the condition. Test : it implies (sufficient), but satisfies yet not (so not necessary).

This is Euclid's lemma: if $a$ divides a product $bc$ and $a$ is coprime to $b$, then $a$ must divide $c$. (Since $$, the prime factors of cannot be supplied by , so they must all come from .) Hence follows necessarily. The others can fail: with we have and , yet , , , and .

The general area formula is $\tfrac12\, AB \cdot BC \sin(\angle B)$. The student's calculation $$ equals this only when , i.e. when , so that and are perpendicular and serve as base and height. That right-angle assumption is exactly what is used but never stated.

``$A \subseteq B$'' says every element of $A$ lies in $B$, i.e. $x \in A \implies x \in B$. Taking the contrapositive: , that is , which is exactly . Complementation reverses inclusion, so this is equivalent to .

The inequality $x + 1/x \geq 2$ holds for all positive $x$ (it rearranges to $(x-1)^2 \geq 0$ after multiplying by the positive ). But for negative the multiplication reverses the inequality, and the claim fails. Test : , which is not . So is a counter-example. The positive test values all satisfy the inequality.

I must hold: putting $x=3$ in $f(x)=f(-x)$ gives $f(3)=f(-3)$ directly. II need not hold: an even function need not have a derivative at ; for example is even but has no stationary point at (the derivative does not exist there). III need not hold: is even with exactly one root (an odd count), since the root at is unpaired. So only I must be true.

The $\boldsymbol{(\Leftarrow)}$ part is correct: if $n=2m$ then $n^2=4m^2$, clearly even. The direction starts correctly in steps (1) and (2) (if then ), but step (3) is the flaw: there is no general rule that is even implies is even'' as a rule is unjustified and not a standard result. The correct proof of uses the contrapositive: if is odd, write , then , which is odd.

The map $u \mapsto 2u-3$ has fixed point $L$ where $L = 2L-3$, giving $$. So if then every term equals , and the statement holds. Writing shows the gap from doubles each step: for , grows without bound, so the sequence is unbounded, B fails. Whether the sequence is increasing or decreasing depends on the sign of : for , the terms are decreasing, A fails. The recurrence is affine, not a pure ratio, so it is not geometric, D fails. For , , E fails. Only the fixed-point conditional is forced for every .

Corresponding angles formed by a transversal are equal \emph{if and only if} the two lines it crosses are parallel. The student asserts the two corresponding angles are both $50^\circ$; this equality holds precisely when $\ell_1 \parallel \ell_2$. That parallelism is exactly the fact used but never stated. Without it, the angle at could be anything.

For $f(x)=e^{mx}$ and $g(x)=e^{nx}$ with , both functions are strictly increasing and everywhere, so the hypotheses hold. The quotient is , which is strictly increasing when but strictly \emph{decreasing} when . Option B has , giving , strictly decreasing, a genuine counter-example. The claim fails because dividing two increasing functions can produce a decreasing one.

By inclusion–exclusion, $|A\cap B| = |A| + |B| - |A\cup B| \geq 6 + 6 - 10 = 2$, since . So II must hold, and II forces I ( has at least elements, hence is non-empty). III need not hold: take ; then . Therefore exactly I and II must be true.

On a closed interval, the minimum of a polynomial is attained either at a stationary point or at an endpoint; the student only compares stationary values. Checking the endpoints: $f(-3) = (-27) - (-9) = -18$ and $f(3) = 27 - 9 = 18$. Since , the true minimum is at , not . Every computation in lines (1)--(3) is correct; the invalid inference is line (4), which equates minimum on the interval''.

From step (3), $g(f(a))=g(f(b))$. Since $g$ is one-to-one (equal outputs imply equal inputs), $f(a)=f(b)$. Since is one-to-one, implies . This contradicts from step (2), so must be one-to-one. The step uses both hypotheses in the correct order and reaches the required contradiction.

From step (3), $S_n = \frac{a}{1-r} - \frac{ar^n}{1-r}$. To show it suffices to show . Since , gives , and (step 2), the fraction is positive. Subtracting it from gives something strictly less, completing the proof. Option E gives exactly this argument.