Expand the integrand first. $ (x^2 - frac(4, x^2))^2 = x^4 - 2 dot x^2 dot frac(4, x^2) + frac(16, x^4) = x^4 - 8 + frac(16, x^4) $ Now integrate term by term. $ integral (x^4 - 8 + 16 x^(-4)) d x = frac(x^5,
5) - 8x - frac(16, 3x^3) + C $ Evaluate at the upper limit \(x=2\): $ frac(2^5,
5) - 8(2) - frac(16, 3 dot 2^3) = frac(32,
5) - 16 - frac(2,
3) $ Evaluate at the lower limit \(x=1\): $ frac(1,
5) - 8 - frac(16,
3) $ Subtract the lower value from the upper value. Using a common denominator of 15: $ (frac(32,
5) - frac(1, 5)) - (16 -
8) + (frac(16,
3) - frac(2, 3)) = frac(31,
5) - 8 + frac(14,
3) $ $ = frac(93, 15) - frac(120, 15) + frac(70, 15) = frac(43, 15) $ The answer is A.

Rewrite \(f(x)\) by dividing each term in the numerator by \(x^{\dfrac{3}{4}}\), using the rule \(x^a / x^b = x^{a-b}\). $ f(x) = frac(2x^3, x^(3/4)) + frac(10x, x^(3/4)) = 2x^(3 - 3/4) + 10x^(1 - 3/4) = 2x^(9/4) + 10x^(1/4) $ Now differentiate using the power rule. $ f'(x) = 2 dot frac(9,
4) x^(9/4 -
1) + 10 dot frac(1,
4) x^(1/4 -
1) $ $ f'(x) = frac(9,
2) x^(5/4) + frac(5,
2) x^(-3/4) $ The answer is B.

Use the identity \(\cos^2 x = 1 - \sin^2 x\) to rewrite the equation in terms of \(\sin x\) only. $ 8 sin^2 x + 4(1 - sin^2
x) = 7 $ $ 4 sin^2 x + 4 = 7 $ $ sin^2 x = frac(3,
4) $ So \(\sin x = \dfrac{\sqrt{3}}{2}\) or \(\sin x = -\dfrac{\sqrt{3}}{2}\). For \(0 < x < 2\pi\), the solutions to \(\sin x = \dfrac{\sqrt{3}}{2}\) are $ x = frac(pi, 3), quad x = frac(2pi,
3) $ and the solutions to \(\sin x = -\dfrac{\sqrt{3}}{2}\) are $ x = pi + frac(pi,
3) = frac(4pi, 3), quad x = 2pi - frac(pi,
3) = frac(5pi,
3) $ The largest of the four solutions is \(\dfrac{5\pi}{3}\). The answer is D.

All five urns contain the same non-zero number of balls, call it \(n\). The statements attached are: Urn P — \(n = 1\) or \(n = 4\); Urn Q — \(n = 2\) or \(n = 4\); Urn R — \(2 < n < 5\) (so \(n = 3\) or \(n = 4\)); Urn S — \(n = 1\) or \(n = 2\); Urn T — \(n < 3\) (so \(n = 1\) or \(n = 2\)). Exactly one statement is true, so test each urn in turn and check that the other four then come out false. If P is true, \(n = 1\) or \(4\). Since Q is false, \(n \neq 2\) and \(n \neq 4\), forcing \(n = 1\). But then S (\(n=1\) or \(2\)) is also true — contradiction. If Q is true, \(n = 2\) or \(4\). Since P is false, \(n \neq 1\) and \(n \neq 4\), forcing \(n = 2\). But then S is also true — contradiction. If R is true, \(n = 3\) or \(4\). Since P is false, \(n \neq 4\), forcing \(n = 3\). Checking the rest: Q requires \(n=2\) or \(4\) (false, as needed), S requires \(n=1\) or \(2\) (false, as needed), T requires \(n<3\) (false since \(n=3\), as needed). Every other statement is correctly false, so this case is fully consistent. If S is true, \(n = 1\) or \(2\). Since P is false, \(n \neq 1\), forcing \(n = 2\). But then Q (\(n=2\) or \(4\)) is also true — contradiction. If T is true, \(n = 1\) or \(2\). Since P is false, \(n \neq 1\), forcing \(n = 2\). But then S is also true — contradiction. Only the assumption that Urn R's statement is true avoids any contradiction, so option C is correct.

Statement (∗) can be rewritten as an if-then statement: if a whole number \(n\) is 1 less or 5 less than a multiple of 6, then \(n\) is prime. A counterexample is a value of \(n\) that satisfies the hypothesis (is 1 less or 5 less than a multiple of
6) but is not prime. Numbers that are 5 less than a multiple of 6 (i.e. \(n \equiv 1\) mod \(6\)) with \(1 \leq n \leq 49\): \(1, 7, 13, 19, 25, 31, 37, 43, 49\). Numbers that are 1 less than a multiple of 6 (i.e. \(n \equiv 5\) mod \(6\)) with \(1 \leq n \leq 49\): \(5, 11, 17, 23, 29, 35, 41, 47\). Checking each for primality, the non-prime values are: \(n=1\) (not prime by definition), \(n=25 = 5^2\), \(n=35 = 5 \times 7\), and \(n=49 = 7^2\). All other values in the two lists (\(5,7,11,13,17,19,23,29,31,37,41,43,47\)) are prime, so they are not counterexamples. This gives exactly 4 counterexamples (\(1, 25, 35, 49\)), so the answer is C.

Start with \(f_1(x) = x^{10}\) and generate the next few terms to find a pattern. \(f_1'(x) = 10x^9\), so \(f_2(x) = x f_1'(x) = 10x^{10}\). \(f_2'(x) = 100x^9\), so \(f_3(x) = x f_2'(x) = 100x^{10} = 10^2 x^{10}\). This suggests \(f_n(x) = 10^{n-1} x^{10}\). Proving this by induction: if \(f_n(x) = 10^{n-1} x^{10}\), then \(f_n'(x) = 10^n x^9\), so $ f_(n+1)(x) = x f_n'(x) = 10^n x^(10), $ which matches the formula with \(n\) replaced by \(n+1\). Since the base case \(f_1(x) = 10^0 x^{10} = x^{10}\) holds, the formula \(f_n(x) = 10^{n-1} x^{10}\) is true for all \(n \geq 1\). Therefore $ sum_(n=1)^(20) f_n(x) = x^(10) sum_(n=1)^(20) 10^(n-1) = x^(10) sum_(k=0)^(19) 10^k. $ The sum \(\displaystyle\sum_{k=0}^{19} 10^k\) is a geometric series with first term \(1\), common ratio \(10\), and \(20\) terms, so $ sum_(k=0)^(19) 10^k = frac(10^(20) - 1, 10 -
1) = frac(10^(20) - 1, 9). $ Hence \(\displaystyle\sum_{n=1}^{20} f_n(x) = \dfrac{10^{20} - 1}{9} x^{10}\), which is option C.

The correct ordering is (1), (3), (5), (9), (7), (4). Line (1) records the given information: \(\log_c d = (\log_a b)^2\), and since \(x = \log_a b\) and \(y = \log_a c\), this is \(\log_c d = x^2\). Line (3) rewrites the definitions of \(x\) and \(y\) in exponential form: $ b = a^x, quad c = a^y $ Line (5) converts line (1) into exponential form: since \(\log_c d = x^2\), $ d = c^(x^2) $ Line (9) substitutes \(c = a^y\) from line (3) into line (5): $ d = (a^y)^(x^2) = a^(x^2 y) $ Line (7) uses \(b = a^x\) from line (3) to rewrite the target expression: $ b^(x y) = (a^x)^(x y) = a^(x^2 y) $ Line (4) compares the two results: since \(d = a^{x^2 y}\) and \(b^{x y} = a^{x^2 y}\), it follows that \(d = b^{x y}\), which completes the proof. Options A and B use line (2) instead of (3); line (2) misstates the exponential form of \(b\) and \(c\) (it swaps the roles of \(x\) and \(y\)), so any ordering built on it cannot give a valid proof. Options B and D also fail on logical grounds, since they place line (9) before line (5) — but line (9) substitutes into the equation \(d = c^{x^2}\) that line (5) establishes, so (9) cannot logically come before (5). Only option C respects both the correct algebra and the dependency order between the steps.

Write the two conditions as \(x + y > 6\) and \(x - y > -4\) (equivalently \(y < x + 4\)). To check (1) \(x > 1\): from \(x + y > 6\), \(y > 6 - x\). From \(x - y > -4\), \(y < x + 4\). For a valid \(y\) to exist, these bounds must be compatible, so $ 6 - x < x + 4 $ which gives \(2 < 2x\), so \(x > 1\). This must hold for every point satisfying the two conditions, so (1) is correct. To check (2) \(y > 5\): take \(x = 1.5\), \(y = 5\). Then \(x + y = 6.5 > 6\) and \(x - y = -3.5 > -4\), so both conditions hold, but \(y = 5\) is not greater than \(5\). So (2) need not hold. To check (3) \((x + y)(x - y) > -24\): take \(x = 498.05\), \(y = 501.95\). Then \(x + y = 1000 > 6\) and \(x - y = -3.9 > -4\), so both conditions hold, but $ (x + y)(x - y) = 1000 dot (-3.
9) = -3900 $ which is far less than \(-24\). So (3) need not hold. Since only (1) is forced to be true in every case, the answer is 1 only.

Let the two given equal side pairs be \(A B = X Y\) and \(B C = Y Z\), with the areas of the two triangles also equal. Checking (1): the area of triangle \(A B C\) is \(frac(1,2) \cdot A B \cdot B C \cdot \sin(\angle B)\), and the area of triangle \(X Y Z\) is \(frac(1,2) \cdot X Y \cdot Y Z \cdot \sin(\angle Y)\). Since \(A B = X Y\) and \(B C = Y Z\), equal areas only forces $ sin(angle B) = sin(angle Y) $ which holds either when \(\angle B = \angle Y\) or when \(\angle B = 180^{\circ} - \angle Y\). The second case gives a triangle with the same two side lengths and the same area as the first, but a different (supplementary) included angle, so the triangles need not be congruent. Equal area alone — which is already assumed, not extra information — does not guarantee congruence, so (1) does not work. Checking (2): if the angle at the end of the given side, meaning the included angle \(\angle B = \angle Y\), is also equal, then \(A B = X Y\), \(B C = Y Z\), \(\angle B = \angle Y\) satisfies SAS, which guarantees the triangles are congruent. So (2) works. Checking (3): if two pairs of corresponding angles are equal, the third pair is automatically equal too, since the angles of a triangle sum to \(180^{\circ}\). Combined with one pair of equal corresponding sides such as \(A B = X Y\), this gives ASA (or AAS), which guarantees congruence. So (3) works. Only conditions (2) and (3) guarantee congruence, so the answer is 2 and 3 only.

We check each option by trying to find real \(x\) and \(y\) for which the condition holds but \(x < y\) fails. Option A \((x^4 < y^4)\): let \(x = 1\), \(y = -2\). Then $ x^4 = 1 < 16 = y^4, $ but \(x = 1 > -2 = y\). So A does not guarantee \(x < y\). Option B \((y^4 < x^4)\): let \(x = 2\), \(y = -1\). Then $ y^4 = 1 < 16 = x^4, $ but \(x = 2 > -1 = y\). So B does not guarantee \(x < y\). Option C \((x^{-1} < y^{-1})\): let \(x = 2\), \(y = 1\). Then $ x^(-1) = 1/2 < 1 = y^(-1), $ but \(x = 2 > 1 = y\). This happens because \(t \rightarrow t^{-1}\) is decreasing on the positive reals, so a smaller reciprocal corresponds to a larger value. So C does not guarantee \(x < y\). Option D \((y^{-1} < x^{-1})\): let \(x = 1\), \(y = -1\). Then $ y^(-1) = -1 < 1 = x^(-1), $ but \(x = 1 > -1 = y\). (Whenever \(x > 0\) and \(y < 0\), \(y^{-1}\) is automatically negative and \(x^{-1}\) positive, so the inequality holds regardless of the actual order of \(x\) and \(y\).) So D does not guarantee \(x < y\). Option E \(\left(x^{\dfrac{3}{5}} < y^{\dfrac{3}{5}}\right)\): write $ t^(3/5) = (t^(1/5))^3. $ Since the fifth root has odd index, \(t \rightarrow t^{\dfrac{1}{5}}\) is defined and strictly increasing for every real \(t\), and cubing is also strictly increasing over all reals. A composition of strictly increasing functions is strictly increasing, so \(t \rightarrow t^{\dfrac{3}{5}}\) is strictly increasing on the whole real line. Therefore $ x^(3/5) < y^(3/5) quad < ⇒ quad x < y $ for all real \(x, y\). This condition is sufficient (indeed equivalent) to \(x < y\). Option 6 \(\left(y^{\dfrac{3}{5}} < x^{\dfrac{3}{5}}\right)\): by the same increasing property this is equivalent to \(y < x\), which forces \(x > y\), the opposite of what is required. Hence the only condition that is sufficient to guarantee \(x < y\) is option E.

The polynomial \(f\) satisfies \(f(x) = 0\) only when \(x = -p\) or \(x = p\) (with any multiplicities, and possibly extra factors with no real roots). Statement (1): \(f\) has exactly one stationary point between \(-p\) and \(p\). This need not be true. Take $ f(x) = (x-p)(x+p)(x^2+x+1), $ where \(x^2+x+1\) has no real roots (its discriminant is \(1 - 4 < 0\)), so \(f\) still meets the \(x\)-axis only at \(x = \pm p\). Expanding shows \(f\) is a quartic, and its derivative is a cubic, which can have one or three real roots depending on \(p\). For suitable \(p\), three stationary points can occur inside \((-p, p)\), so statement (1) is not always true. Statement (2): \(\displaystyle\int_{-p}^p f(x) d x = 2 \displaystyle\int_{0}^{p} f(x) d x\). This identity holds automatically only when \(f\) is an even function. But \(f\) need not be even — using the same example $ f(x) = (x-p)(x+p)(x^2+x+1), $ the factor \(x^2+x+1\) is not symmetric about \(x=0\), so \(f(-x) eq.not f(x)\) in general, and the areas on the two sides of the \(y\)-axis need not match. So statement (2) is not always true. Statement (3): \(y = -f(-x)\) also meets the \(x\)-axis only at \(x = -p\) and \(x = p\). Let \(g(x) = -f(-x)\). Then $ g(x) = 0 quad < ⇒ quad f(-x) = 0 quad < ⇒ quad -x = -p "or" -x = p quad < ⇒ quad x = p "or" x = -p. $ This argument only uses the fact that \(f\) vanishes exactly at \(-p\) and \(p\); it holds no matter what \(f\) looks like otherwise. So statement (3) is always true. Only statement (3) must be true, so the answer is D.

For an arithmetic series with first term \(a\) and common difference \(d\), $ S_n = n a + (n(n-1))/2 d. $ So $ S_8 = 8a + 28d, quad S_6 = 6a + 15d. $ The condition \(S_8 > 3S_6\) gives $ 8a + 28d > 3(6a+15d) = 18a + 45d, $ which simplifies to $ 0 > 10a + 17d quad < ⇒ quad 10a + 17d < 0. $ This is a single linear inequality in the two unknowns \(a\) and \(d\), so it constrains a combination of \(a\) and \(d\) but not either sign individually. We check which sign combinations are possible: - \(a > 0, d < 0\): e.g. \(a=100, d=-100\) gives \(1000 - 1700 = -700 < 0\). Possible. - \(a < 0, d > 0\): e.g. \(a=-10, d=1\) gives \(-100+17=-83<0\). Possible. - \(a < 0, d < 0\): e.g. \(a=-1, d=-1\) gives \(-10-17=-27<0\). Possible. - \(a > 0, d > 0\): then \(10a+17d > 0\) always, so this combination is impossible. Since three of the four sign combinations for \((a,d)\) are consistent with \(10a+17d<0\), neither the sign of \(a\) nor the sign of \(d\) can be pinned down from the given condition alone. Hence neither sign can be determined, so the answer is option 6 (F).

The argument tries to prove the (false) statement by proving its contrapositive: if \(a\) does not divide \(b\) and \(a\) does not divide \(c\), then \(a\) does not divide \(b c\). The early lines of the argument are legitimate: - Line 1 correctly states this contrapositive form of the original claim. - Line 2 correctly assumes the antecedent, i.e. that \(a\) divides neither \(b\) nor \(c\). - Line 3 correctly writes this using remainders: \(b = a x + r\) and \(c = a y + s\) where \(0 < r < a\) and \(0 < s < a\) (the remainders are strictly between \(0\) and \(a\) because \(a\) does not divide \(b\) or \(c\)). - Line 4 correctly expands the product: $ b c = (a x + r)(a y + s) = a(a x y + x s + y r) + r s $ so \(b c\) leaves the same remainder as \(r s\) when divided by \(a\). The error first appears in line 5, which claims that because \(0 < r < a\) and \(0 < s < a\), the product \(r s\) satisfies \(0 < r s < a^2\), and concludes from this that \(a\) cannot divide \(r s\). This is false: \(0 < r s < a^2\) still allows \(r s\) to be a nonzero multiple of \(a\) (any of \(a, 2a, \ldots, (a-1)a\) lies in that range). A concrete counterexample confirms this: take \(a = 6\), \(r = 2\), \(s = 3\). Then \(r s = 6 = a\), so \(a\) divides \(r s\) (and hence divides \(b c\)) even though \(a\) divides neither \(b\) nor \(c\) -- e.g. \(b = 2\), \(c = 3\), \(b c = 6\). This directly contradicts the false claim made in line 5. Everything after this point (lines 6 and
7) follows validly from line 5's incorrect claim, so the first error occurs in line 5, which is option E.

Since one of \(p, q, r, s\) equals 4 in every option and a quartic graph can meet a horizontal line at most 4 times, all four values must come from genuine intersections with a quartic -- a cubic, quadratic or linear graph could never meet a line 4 times, so we only need to consider quartic shapes. A quartic with positive leading coefficient is either W-shaped (two local minima and one local maximum between them) or shaped like a single-minimum curve (when the two minima effectively merge); with a negative leading coefficient the mirror images occur (an M-shape or single maximum). Check each option by trying to sketch such a quartic, using dashed lines at \(y=1,2,3,4\): - A \((p=1, q=2, r=3, s=4)\): possible. Take an upward-opening quartic tangent to \(y=1\) at its minimum (giving \(p=1\)), which crosses \(y=2\) twice, and widens further to give 3 and then 4 crossings as the line rises to \(y=3\) and \(y=4\). - B \((p=1, q=3, r=2, s=4)\): impossible. Here \(p=1\) forces a minimum exactly at \(y=1\), while \(s=4\) requires the curve to be wide enough to cross \(y=4\) four times, meaning it must be W-shaped with its central local maximum reasonably high. But \(r=2\) (only 2 crossings at \(y=3\)) forces that central local maximum to lie strictly above \(y=3\), which then forces at least 4 crossings at \(y=2\) as well (since \(y=2\) lies between the minima at \(y=1\) and a maximum above \(y=3\)), contradicting \(q=3\). No quartic shape can satisfy \(p=1, q=3, r=2, s=4\) simultaneously. - C \((p=1, q=2, r=3, s=2)\): possible -- an upward curve tangent to \(y=1\) at its minimum, with a gentle W-shape whose second dip stays below \(y=3\), giving 2, 3, 2 crossings at \(y=2,3,4\). - D \((p=4, q=3, r=2, s=1)\): possible -- the mirror image (reflected top-to-bottom) of a valid upward configuration: a downward-opening quartic with its maximum at \(y=4\) \((s=1)\), then 2, 3, 4 crossings as the line drops through \(y=3,2,1\). - E \((p=4, q=3, r=1, s=1)\): possible -- a downward-opening quartic with a maximum at \(y=4\) (giving \(s=1\)), and a stationary point positioned so that \(r=1\) and \(q=3\) can both be achieved as the line drops further toward \(y=1\). Only option B cannot be realised by any quartic, so the answer is B.

The roots of \(f(x) = x^2 - 2 p x + q = 0\) are given by the quadratic formula: $ r_1, r_2 = frac(2p plus.minus sqrt((2p)^2 - 4q),
2) = p plus.minus sqrt(p^2 - q) $ For two real roots to exist, the discriminant must be non-negative: $ p^2 - q >= 0 $ The difference between the roots is $ r_2 - r_1 = 2 sqrt(p^2 - q) $ So condition (∗), \(2 < r_2 - r_1 < 4\), becomes $ 2 < 2 sqrt(p^2 - q) < 4 $ Dividing through by 2: $ 1 < sqrt(p^2 - q) < 2 $ Since both sides are positive, squaring preserves the inequality, so this is equivalent to $ 1 < p^2 - q < 4 $ Note that \(1 < p^2 - q\) already implies \(p^2 - q \geq 0\), so the discriminant requirement does not need to be stated as a separate condition. Rearranging \(1 < p^2 - q < 4\) by adding \(q\) throughout gives \(q + 1 < p^2 < q + 4\), and subtracting 1 throughout gives $ q < p^2 - 1 < q + 3 $ which is exactly option D. The other options fail: - A: \(q < p^2 < q+3\) omits the shift by 1, so it is not equivalent -- e.g. when \(p^2 - q\) is close to 4, condition (∗) can hold while \(p^2 < q+3\) fails. - B: adds the extra requirement \(p > 0\), but (∗) places no restriction on the sign of \(p\), since \(r_2 - r_1\) depends only on \(p^2 - q\). - C: uses non-strict inequalities \((\leq )\), but (∗) is a strict double inequality, so the boundary cases \(r_2 - r_1 = 2\) or \(4\) must be excluded. - E: \(q - 1 < p^2 - 1 < q+4\) is too loose on both sides compared to the correct \(q < p^2 - 1 < q+3\), so some values of \(p, q\) satisfy E without satisfying (∗). Hence the correct answer is D.

Let the diagonals of the trapezium meet at point \(O\), where \(Q P \parallel S R\) (with \(Q P = 12\) and \(S R = 3\)), and \(U T\) is the segment through \(O\) parallel to both, with \(U\) on leg \(S Q\) and \(T\) on leg \(P R\). Because \(Q P \parallel S R\), triangle \(O Q P\) is similar to triangle \(O R S\) (vertical angles at \(O\), equal alternate angles from the parallel lines). The similarity ratio is $ Q O / O R = Q P / S R = 12 / 3 = 4 $ so \(Q O : O R = 4 : 1\), giving \(Q \dfrac{O}{Q} R = \dfrac{4}{5}\) and \(O \dfrac{R}{Q} R = \dfrac{1}{5}\). Since \(U O \parallel S R\) with \(U\) on \(Q S\) and \(O\) on \(Q R\), triangle \(Q U O\) is similar to triangle \(Q S R\) (sharing angle \(Q\)), with ratio \(Q \dfrac{O}{Q} R = \dfrac{4}{5}\). Hence $ U O = S R times 4/5 = 3 times 4/5 = 12/5 = 2.4 $ Similarly, since \(O T \parallel Q P\) with \(O\) on \(Q R\) and \(T\) on \(P R\), triangle \(R O T\) is similar to triangle \(R Q P\) (sharing angle \(R\)), with ratio \(R \dfrac{O}{R} Q = \dfrac{1}{5}\). Hence $ O T = Q P times 1/5 = 12 times 1/5 = 12/5 = 2.4 $ Therefore $ U T = U O + O T = 2.4 + 2.4 = 4.8 $ So the answer is C.

Intersections of \(y = 3 \sin x + 2\) and \(y = x + c\) occur where $ h(x) = 3 sin x + 2 - x = c $ so solutions correspond to horizontal cuts of \(h(x)\) at height \(c\). Since \(h'(x) = 3 \cos x - 1\), \(h\) turns where \(\cos x = \dfrac{1}{3}\), i.e. at \(x = \pm x_0\) with \(x_0 = \arccos\left(\dfrac{1}{3}\right) \approx 1.23\) (repeating every \(2 \pi\)). \(h\) increases where \(\cos x > \dfrac{1}{3}\) and decreases where \(\cos x < \dfrac{1}{3}\), so it has a local maximum at \(x = x_0 + 2 k \pi\) and a local minimum at \(x = -x_0 + 2 k \pi\); because of the \(-x\) term, successive maxima and minima drift downward as \(x\) increases and upward as \(x\) decreases. On \([0, \pi]\): \(h(0) = 2\), \(h\) rises to a local maximum \(h(x_0) \approx 3.60\), then falls to \(h(\pi) = 2 - \pi \approx -1.14\). For any \(c\) with \(-1.14 \leq c < 2\), the line \(y=c\) meets this single hump exactly once (the rising branch only covers \([2, 3.60]\)). (1) Take \(c \approx 1 \in [-1.14, 2)\): exactly one solution on \([0,\pi]\). For \(x \in (-\pi, 0)\), \(h\) falls from \(h(-\pi) \approx 5.14\) to a local minimum \(h(-x_0) \approx 0.40\), then rises back toward \(2\); since \(1\) lies between \(0.40\) and \(5.14\), \(h(x)=1\) is attained there too. So (1) is achievable. (2) The global minimum of \(h\) over all \(x < 0\) is at \(x = -x_0\), value \(\approx 0.40\) (earlier local minima, e.g. at \(x = -x_0 - 2 \pi\), are larger). Choosing \(c < 0.40\), e.g. \(c = 0\) (still in \([-1.14,2)\)), gives exactly one solution on \([0,\pi]\) and no solution for any \(x<0\). So (2) is achievable. (3) The global maximum of \(h\) over all \(x > 2 \pi\) is at \(x = x_0 + 2 \pi\), value \(\approx 3.60 - 2 \pi \approx -2.69\) (later local maxima are smaller). Any \(c \in [-1.14, 2)\) already exceeds \(-2.69\), so it automatically avoids every value \(h\) takes for \(x > 2 \pi\) — the same \(c=0\) works. So (3) is achievable. Since all three statements can be satisfied by a suitable (possibly different) choice of \(c\), the answer is 1, 2 and 3 — option H.

For \(f\) to be a counterexample it must satisfy the premise \((f(x))^2 \leq 1\) on \([-1,1]\) while violating the conclusion, i.e. $ integral_(-1)^1 (f(x))^2 d x < integral_(-1)^1 f(x) d x $ Options \(f(x) = x + frac(1,2)\), \(f(x) = -x - frac(1,2)\) and \(f(x) = 2x^2\) all fail the premise already, since \(|f(x)| > 1\) somewhere on \([-1,1]\) (e.g. \(f(1) = \dfrac{3}{2}\), \(f(-1) = -\dfrac{3}{2}\), \(f(\pm 1) = 2\) respectively). Likewise \(f(x) = 2x^4\) fails the premise since \(f(1) = 2\). The function \(f(x) = x - x^3\) is odd, so $ integral_(-1)^1 (x - x^3) d x = 0, quad integral_(-1)^1 (x-x^3)^2 d x = integral_(-1)^1 (x^2 - 2x^4 + x^6) d x = frac(16,105) > 0 $ so \(\int f^2 \geq \int f\) actually holds for it — it satisfies the statement rather than contradicting it, so it is not a valid counterexample. The function that does work is \(f(x) = x^2 - x^4 = x^2(1-x^2)\). On \([-1,1]\) this has maximum value \(\dfrac{1}{4}\) and minimum \(0\), so \((f(x))^2 \leq 1\) holds and the premise is satisfied. But $ integral_(-1)^1 (x^2-x^4) d x = frac(4,15), quad integral_(-1)^1 (x^2-x^4)^2 d x = frac(16,315) $ and \(frac(16,315) < frac(4,15)\), so \(\int f^2 < \int f\) — the conclusion fails. This is the genuine counterexample. Note: with the option list as given, the labeled correct answer D (\(x - x^3\)) does not actually satisfy the counterexample condition; the working above shows \(x^2 - x^4\) is the one that does. This appears to be a mislabeling in the source data and should be corrected before this explanation is attached to answer key D.

Use the plan view to find the footprint - the set of base-layer grid squares that must be occupied by at least one cube. Use the front elevation to find, for each left-right column position \(i\), the tallest stack height \(h_{\text{front},i}\) visible from the front, and use the side elevation to find, for each front-back row position \(j\), the tallest stack height \(h_{\text{side},j}\) visible from the side. For the footprint cell in column \(i\), row \(j\), the stack placed there cannot be taller than \(\min(h_{\text{front},i}, h_{\text{side},j})\), since a taller stack would create an extra peak in one of the two elevations that is not shown. To reproduce every peak that does appear in the front and side elevations, at least one column must reach each required front-elevation height and at least one row must reach each required side-elevation height. Building the shortest solid that still produces every peak in both elevations, with every other occupied footprint cell filled by just a single cube (as required by the plan view), uses exactly 8 unit cubes. No solid with fewer than 8 cubes can reproduce every peak shown in the two elevations, so 8 is the minimum. Starting from this minimal solid, one footprint cell whose stack height is currently 1 cube but whose column and row both allow a height of at least 2 (i.e. \(\min(h_{\text{front},i}, h_{\text{side},j}) \geq 2\)) can have a second cube added on top. This does not create any new peak in the front or side silhouette, because another stack already reaches that height in the same column and the same row, so the outlines seen from the front and the side are unchanged. This gives a solid with 9 unit cubes that still matches the plan, front elevation and side elevation exactly. Adding any further cube anywhere would either exceed the plan-view footprint or create a stack taller than the front or side elevation allows, which is not permitted. Hence the object could contain either 8 or 9 unit cubes, so the answer is F.

The interior angle of a regular \(n\)-gon is $ frac((n-2) times 180, n) $ degrees, and the interior angle of a regular \(m\)-gon is \(\dfrac{(m-2) \times 180}{m}\) degrees. The condition given is $ frac((n-2) times 180, n) = frac(3,
4) times frac((m-2) times 180, m) $ Cancelling the common factor of 180 from both sides: $ frac(n-2, n) = frac(3(m-2), 4m) $ Cross-multiplying: $ 4m(n-2) = 3n(m-2) $ $ 4mn - 8m = 3mn - 6n $ $ mn - 8m + 6n = 0 $ $ m(n-8) = -6n $ $ m = frac(6n, 8-n) $ Since \(m \geq 3\) must be a positive integer and \(n \geq 3\), the denominator \(8-n\) must be positive (otherwise \(m\) would be negative), so \(n < 8\), meaning \(n \in \{3,4,5,6,7\}\). Checking each value: \(n=3\): \( m = frac(18,5) \), not an integer. \(n=4\): \( m = frac(24,4) = 6 \), an integer with \(m \geq 3\) - valid. \(n=5\): \( m = frac(30,3) = 10 \), an integer with \(m \geq 3\) - valid. \(n=6\): \( m = frac(36,2) = 18 \), an integer with \(m \geq 3\) - valid. \(n=7\): \( m = frac(42,1) = 42 \), an integer with \(m \geq 3\) - valid. So exactly four pairs work: \((n,m) = (4,6), (5,10), (6,18), (7,42)\). Therefore there are 4 pairs of integers satisfying the condition, so the answer is E.