From the linear equation \(x - 3y + 1 = 0\), we get \(y = \dfrac{x + 1}{3}\). Substituting into \(3x^2 - 7xy = 5\) gives $ 3x^2 - 7x dot frac(x+1,
3) = 5 $ Multiplying both sides by 3: $ 9x^2 - 7x(x+1) = 15 $ $ 9x^2 - 7x^2 - 7x - 15 = 0 $ $ 2x^2 - 7x - 15 = 0 $ For a quadratic \(a x^2 + b x + c = 0\), the sum of the roots equals \(-\dfrac{b}{a}\). Here that sum is $ frac(-(-7),
2) = frac(7,
2) = 3.5 $ So the sum of the two values of \(x\) is \(3.5\). The answer is D.

Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\), rewrite \(\sin^2 \theta + 3\cos \theta = 3\) as $ 1 - cos^2 theta + 3cos theta = 3 $ $ cos^2 theta - 3cos theta + 2 = 0 $ This factors as $ (cos theta - 1)(cos theta -
2) = 0 $ Since \(\cos \theta \leq 1\) always, the factor \(\cos \theta = 2\) has no solutions, so we need \(\cos \theta = 1\). On the interval \(0 \leq \theta \leq 4\pi\), \(\cos \theta = 1\) occurs at $ theta = 0, space 2pi, space 4pi $ This gives 3 solutions. The answer is D.

The midpoint of the segment joining \((2,-6)\) and \((5,4)\) is $ (frac(2+5,2), frac(-6+4,2)) = (frac(7,2), -1) $ The slope of the segment is $ frac(4-(-6), 5-2) = frac(10,
3) $ The perpendicular bisector has slope \(-frac(3,10)\) (negative reciprocal) and passes through \((frac(7,2), -1)\), so its equation is $ y + 1 = -frac(3,10)(x - frac(7,2)) $ Setting \(y = 0\) to find the \(x\)-intercept: $ 1 = -frac(3,10)(x - frac(7,2)) $ $ -frac(10,
3) = x - frac(7,
2) $ $ x = frac(7,
2) - frac(10,
3) = frac(21,
6) - frac(20,
6) = frac(1,
6) $ The answer is B.

Factor the left-hand side: \((x^2-1)(x-2) = (x-1)(x+1)(x-2)\). The critical points are \(x=-1\), \(x=1\), and \(x=2\), splitting the number line into four intervals. Test the sign of the product in each. For \(x\leftarrow 1\) (e.g. \(x=-2\)): \((x-1)<0\), \((x+1)<0\), \((x-2)<0\), product negative. For \(-10\), \((x-2)<0\), product positive. For \(10\), \((x+1)>0\), \((x-2)<0\), product negative. For \(x>2\) (e.g. \(x=3\)): all three factors positive, product positive. So \((x-1)(x+1)(x-2) > 0\) holds exactly on \(-1 < x < 1\) and \(x > 2\), giving answer E.

Start from \(y = -\log_10 (1 - x)\). Multiply both sides by \(-1\): $ -y = log_10 (1 -
x) $ Convert from logarithmic to exponential form (base 10): $ 10^(-y) = 1 - x $ Solve for \(x\): $ x = 1 - 10^(-y) $ This matches answer D.

Since \(x + 2\) is a factor of \(P(x) = x^3 + 4c x^2 + x(c+1)^2 - 6\), the factor theorem gives \(P(-2) = 0\). Substitute \(x = -2\): $ P(-2) = (-2)^3 + 4c(-2)^2 + (-2)(c+1)^2 - 6 = -8 + 16c - 2(c+1)^2 - 6 $ Expand \((c+1)^2 = c^2 + 2c + 1\): $ P(-2) = -8 + 16c - 2c^2 - 4c - 2 - 6 = -2c^2 + 12c - 16 $ Set \(P(-2) = 0\) and divide by \(-2\): $ c^2 - 6c + 8 = 0 $ Factor: $ (c - 2)(c -
4) = 0 $ So \(c = 2\) or \(c = 4\), and the sum of the possible values of \(c\) is \(2 + 4 = 6\), matching answer D.

The bag has \(3n\) balls in total. It is easier to find the probability that the two balls picked ARE the same colour, then subtract from \(1\). For any one colour, the probability of drawing that colour first and then the same colour again (without replacement) is $ frac(n, 3n) times frac(n - 1, 3n -
1) $ There are three colours, and these events are mutually exclusive, so $ P("same colour") = 3 times frac(n, 3n) times frac(n - 1, 3n -
1) = frac(n - 1, 3n -
1) $ Therefore the probability that the two balls are not the same colour is $ P("different colours") = 1 - frac(n - 1, 3n -
1) = frac((3n -
1) - (n - 1), 3n -
1) = frac(2n, 3n -
1) $ This matches option C.

Using the laws of indices, the left-hand side can be written as a single power: $ a^x b^(2x) c^(3x) = (a b^2 c^3)^x $ So the equation becomes $ (a b^2 c^3)^x = 2 $ Taking \(\log_10\) of both sides and using \(\log_10(m^x) = x \log_10 m\): $ x log_10 (a b^2 c^3) = log_10 2 $ Dividing both sides by \(\log_10 (a b^2 c^3)\): $ x = frac(log_10 2, log_10 (a b^2 c^3)) $ This matches the option \(\dfrac{\log_10 2}{\log_10 (a b^2 c^3)}\), which is answer F.

For \(2x^2 - 11x + c = 0\), the sum and product of the roots are $ "sum" = frac(11, 2), quad "product" = frac(c,
2) $ Let the smaller root be \(\alpha\), so the larger root is \(\alpha + 2\) (since the roots differ by \(2\)). Using the sum of roots: $ alpha + (alpha +
2) = frac(11,
2) $ $ 2 alpha = frac(11,
2) - 2 = frac(7,
2) $ $ alpha = frac(7,
4) $ So the two roots are $ alpha = frac(7, 4), quad alpha + 2 = frac(15,
4) $ Using the product of roots: $ frac(c,
2) = frac(7,
4) times frac(15,
4) = frac(105, 16) $ Solving for \(c\): $ c = 2 times frac(105, 16) = frac(105,
8) $ This matches option A.

Reflecting the curve \(y = \cos x\) in the horizontal line \(y = 1\) maps each point \((x, y)\) to \((x, 2 - y)\), since the point and its image are equidistant from \(y=1\) on opposite sides. Applying this to \(y = \cos x\) gives the reflected curve $ y = 2 - cos x $ Translating a curve by \(\dfrac{\pi}{4}\) units in the positive \(x\)-direction replaces \(x\) with \(x - \dfrac{\pi}{4}\) in the equation. So the final curve is $ y = 2 - cos(x - frac(pi, 4)) $ This matches option D.

Let \(t = 2^x\). The equation \(2^{2x} - 8 \cdot 2^x + 15 = 0\) becomes $ t^2 - 8t + 15 = 0 $ Factoring, $ (t - 3)(t -
5) = 0 $ so \(t = 3\) or \(t = 5\), giving \(2^x = 3\) or \(2^x = 5\). Taking logarithms, $ x = log_2 3 quad "or" quad x = log_2 5 $ The sum of the roots is $ log_2 3 + log_2 5 = log_2 15 $ Converting to base 10 using the change of base formula, $ log_2 15 = frac(log_10 15, log_10
2) $ This matches option E.

The cross-section is an equilateral triangle with side \(2x\), so its area is $ A = frac(sqrt(3),
4) (2x)^2 = sqrt(3) x^2 $ The volume of the prism is this area times the length \(d\): $ V = sqrt(3) x^2 d $ The total surface area consists of two triangular ends plus three rectangular faces, each of width \(2x\) and length \(d\): $ T = 2 sqrt(3) x^2 + 3(2x)d = 2 sqrt(3) x^2 + 6xd $ Setting \(V = T\) as given, $ sqrt(3) x^2 d = 2 sqrt(3) x^2 + 6xd $ Collecting terms with \(d\) on one side, $ sqrt(3) x^2 d - 6xd = 2 sqrt(3) x^2 $ $ d(sqrt(3) x^2 - 6x) = 2 sqrt(3) x^2 $ Dividing both sides by \(x\), $ d(sqrt(3) x -
6) = 2 sqrt(3) x $ So $ d = frac(2 sqrt(3) x, sqrt(3) x -
6) $ Dividing numerator and denominator by \(\sqrt{3}\), and using \(\dfrac{6}{\sqrt{3}} = 2 \sqrt{3}\), $ d = frac(2x, x - 2 sqrt(3)) $ This matches option D.

Write the quartic as $ x^4 - 4x^3 + 4x^2 - 10 = x^2(x^2 - 4x +
4) - 10 = x^2(x-2)^2 - 10. $ Let \(u = x^2 - 2x = (x-1)^2 - 1\). Then \(x^2(x-2)^2 = (x^2-2x)^2 = u^2\), so the equation becomes \(u^2 = 10\), giving \(u = \sqrt{10}\) or \(u = -\sqrt{10}\). Since \(u = (x-1)^2 - 1\), the minimum value of \(u\) is \(-1\), so \(u \geq -1\) for all real \(x\). \(\sqrt{10} \approx 3.16 \geq -1\), so this branch is valid. Solving \((x-1)^2 - 1 = \sqrt{10}\) gives \((x-1)^2 = 1 + \sqrt{10} > 0\), which has two real solutions for \(x\). \(-\sqrt{10} \approx -3.16 < -1\), so this branch is impossible: \((x-1)^2 - 1 = -\sqrt{10}\) would require \((x-1)^2 = 1 - \sqrt{10} < 0\), which has no real solutions. So the equation has exactly \(2\) real roots. The answer is C.

For a graph of \(\log y\) against \(\log x\) to be a straight line, \(\log y\) must be a linear function of \(\log x\), that is $ log y = m dot log x + c $ for constants \(m\) and \(c\). Check option D: \(y = a x^b\). Taking logarithms of both sides, $ log y = log a + b log x, $ which is linear in \(\log x\) with slope \(b\) and intercept \(\log a\) (both constants since \(a\) and \(b\) are constants). This gives a straight line. The other options fail to produce a linear relationship between \(\log y\) and \(\log x\): - A: \(y^b = a^x\) gives \(b \log y = x \log a\), so \(\log y\) is linear in \(x\), not in \(\log x\). - B: \(y = a b^x\) gives \(\log y = \log a + x \log b\), again linear in \(x\), not \(\log x\). - C: \(y^2 = a + x^b\) gives \(2 \log y = \log(a + x^b)\), which is not linear in \(\log x\). - E: \(y^x = a^b\) gives \(x \log y = b \log a\), so \(\log y = \dfrac{b \log a}{x}\), not linear in \(\log x\). Only option D produces a straight line. The answer is D.

Expand the integrand and integrate term by term: $ integral_0^1 (x-a)^2 d x = integral_0^1 (x^2 - 2 a x + a^2) d x = [frac(x^3,
3) - a x^2 + a^2 x]_0^1 = frac(1,
3) - a + a^2. $ This is a quadratic function of \(a\): $ f(a) = a^2 - a + frac(1, 3). $ Since the coefficient of \(a^2\) is positive, \(f(a)\) has a minimum where \(f'(a) = 0\): $ f'(a) = 2a - 1 = 0 arrow.r.double a = frac(1, 2). $ Substituting \(a = \dfrac{1}{2}\): $ f(frac(1, 2)) = (frac(1, 2))^2 - frac(1,
2) + frac(1,
3) = frac(1,
4) - frac(1,
2) + frac(1,
3) = frac(3, 12) - frac(6, 12) + frac(4, 12) = frac(1, 12). $ So the smallest possible value of the integral is \(\dfrac{1}{12}\). The answer is A.

Write every base as a product of primes 2 and 5: \(10 = 2 \times 5\), \(20 = 2^2 \times 5\), \(8 = 2^3\), \(125 = 5^3\). Numerator: $ 10^(c - 2d) times 20^(2c+d) = 2^(c-2d) 5^(c-2d) times 2^(2(2c+d)) 5^(2c+d) = 2^(5c) 5^(3c-d) $ Denominator: $ 8^c times 125^(c+d) = 2^(3c) 5^(3(c+d)) = 2^(3c) 5^(3c+3d) $ Dividing: $ frac(2^(5c) 5^(3c-d), 2^(3c) 5^(3c+3d)) = 2^(5c-3c) 5^((3c-d)-(3c+3d)) = 2^(2c) 5^(-4d) = frac(2^(2c), 5^(4d)) $ Since 2 and 5 are different primes, there is no cancellation between the two factors. For this to be an integer: - \(2^{2c}\) must not sit in a denominator, so \(c \geq 0\) - \(5^{4d}\) (the denominator power) must not exceed the numerator, so \(d \leq 0\) Because \(c\) and \(d\) are non-zero integers, this forces \(c > 0\) and \(d < 0\). The correct answer is E.

The equation is \(a x^2 + (a-2)x = 2\), which rearranges to $ a x^2 + (a-2)x - 2 = 0 $ This quadratic factorises neatly: $ (a x - 2)(x +
1) = a x^2 + a x - 2x - 2 = a x^2 + (a-2)x - 2 $ So the equation is \((a x - 2)(x+1) = 0\), giving roots $ x = -1 quad "or" quad x = frac(2,
a) $ (the second root exists since \(a \neq 0\) is given). These two roots are real for every non-zero \(a\), so the only way they fail to be distinct is if \(\dfrac{2}{a} = -1\), i.e. \(a = -2\). Check with the discriminant as well: for \(a x^2 + (a-2)x - 2 = 0\), $ Delta = (a-2)^2 - 4a(-2) = a^2 - 4a + 4 + 8a = a^2 + 4a + 4 = (a+2)^2 $ Since \((a+2)^2 \geq 0\) always, and equals zero only when \(a = -2\), the roots are real and distinct for every \(a \neq -2\). The correct answer is D.

Work on \(0 \leq x \leq \pi\), excluding \(x = \dfrac{\pi}{2}\) where \(\tan x\) is undefined. Step 1: solve \(-1 \leq \tan x \leq 1\). On \([0, \dfrac{\pi}{2})\), \(\tan x\) increases from \(0\) to \(\infty\), and is non-negative throughout, so \(\tan x \leq 1\) is the binding condition: $ x in [0, frac(pi, 4)] $ On \((\dfrac{\pi}{2}, \pi]\), \(\tan x\) increases from \(-\infty\) up to \(\tan \pi = 0\), and is non-positive throughout, so \(\tan x \geq -1\) is the binding condition: $ x in [frac(3pi, 4), pi] $ So \(-1 \leq \tan x \leq 1\) holds on $ [0, frac(pi, 4)] union [frac(3pi, 4), pi] $ Step 2: solve \(\sin 2x \geq \dfrac{1}{2}\). As \(x\) ranges over \([0, \pi]\), \(2x\) ranges over \([0, 2\pi]\). Since \(\sin \theta \leq 0\) for \(\theta \in [\pi, 2\pi]\), the inequality \(\sin \theta \geq frac(1,2)\) only holds for $ theta in [frac(pi, 6), frac(5pi, 6)] $ so, dividing by 2, $ x in [frac(pi, 12), frac(5pi, 12)] $ Step 3: intersect the two regions. Compare with \(\dfrac{\pi}{4} = \dfrac{3\pi}{12}\): since \(\dfrac{5\pi}{12} > \dfrac{3\pi}{12}\), $ [frac(pi, 12), frac(5pi, 12)] inter [0, frac(pi, 4)] = [frac(pi, 12), frac(pi, 4)] $ which has length \(\dfrac{\pi}{4} - \dfrac{\pi}{12} = \dfrac{3\pi - \pi}{12} = \dfrac{2\pi}{12} = \dfrac{\pi}{6}\). For the other piece, \([\dfrac{\pi}{12}, \dfrac{5\pi}{12}] \cap [\dfrac{3\pi}{4}, \pi] = \emptyset\), since \(\dfrac{5\pi}{12} < \dfrac{9\pi}{12} = \dfrac{3\pi}{4}\). So the total length of intervals satisfying both conditions is \(\dfrac{\pi}{6}\). The correct answer is B.

The geometric series has terms \(T_n = 4r^{n-1}\), so $ T_1 = 4, quad T_2 = 4r, quad T_4 = 4r^3. $ For \(T_1, T_2, T_4\) to form three successive terms of an arithmetic series, the middle term equals the average of the outer two: $ 2 T_2 = T_1 + T_4. $ Substituting, $ 2 (4r) = 4 + 4r^3 \ 8r = 4 + 4r^3 \ 2r = 1 + r^3 \ r^3 - 2r + 1 = 0. $ Since \(r = 1\) is clearly a root, factor it out: $ r^3 - 2r + 1 = (r - 1)(r^2 + r -
1) = 0. $ So either \(r = 1\) or \(r^2 + r - 1 = 0\). Since the sum to infinity requires \(0 < r < 1\), discard \(r=1\) (it also would make the series non-convergent) and solve the quadratic: $ r = frac(-1 plus.minus sqrt(5), 2). $ Only the positive root lies in \((0,1)\): $ r = frac(sqrt(5) - 1, 2). $ The sum to infinity of the geometric series is $ S_infinity = frac(T_1, 1 - r) = frac(4, 1 - frac(sqrt(5)-1, 2)) = frac(4, frac(3 - sqrt(5), 2)) = frac(8, 3 - sqrt(5)). $ Rationalising the denominator, $ S_infinity = frac(8, 3-sqrt(5)) dot frac(3+sqrt(5), 3+sqrt(5)) = frac(8(3+sqrt(5)), 9 -
5) = frac(8(3+sqrt(5)),
4) = 2(3+sqrt(5)). $ This is option D.

We need the coefficient of \(x^2\) in $ (4 - x^2)[(1+2x+3x^2)^6 - (1+4x^3)^5]. $ Writing \(f(x) = (1+2x+3x^2)^6\) and \(g(x) = (1+4x^3)^5\), expand the bracket: $ (4-x^2)(f(x) - g(x)) = 4f(x) - 4g(x) - x^2 f(x) + x^2 g(x). $ Only low-order terms of \(f\) and \(g\) are needed. Since \(g(x) = (1+4x^3)^5\) only contains powers of \(x\) that are multiples of \(3\) (i.e. \(1, x^3, x^6, \ldots\)), its coefficient of \(x^0\) is \(1\) and its coefficient of \(x^2\) is \(0\). For \(f(x) = (1+2x+3x^2)^6\), write \(u = 2x + 3x^2\) so that \(f = (1+u)^6 = 1 + 6u + 15u^2 + \ldots\). The coefficient of \(x^0\) in \(f\) is \(1\). For the coefficient of \(x^2\), only \(6u\) and \(15u^2\) can contribute an \(x^2\) term. From \(u = 2x+3x^2\), the \(x^2\) term of \(6u\) has coefficient \(6 \cdot 3 = 18\). From $ u^2 = (2x+3x^2)^2 = 4x^2 + 12x^3 + 9x^4, $ the \(x^2\) term of \(15u^2\) has coefficient \(15 \cdot 4 = 60\). So the coefficient of \(x^2\) in \(f(x)\) is \(18 + 60 = 78\). Now collect the coefficient of \(x^2\) in the full expression: - from \(4f(x)\): \(4 \cdot 78 = 312\) - from \(-4g(x)\): \(-4 \cdot 0 = 0\) - from \(-x^2 f(x)\): this contributes minus the coefficient of \(x^0\) in \(f\), i.e. \(-1\) - from \(x^2 g(x)\): this contributes the coefficient of \(x^0\) in \(g\), i.e. \(+1\) Adding these, $ 312 + 0 - 1 + 1 = 312. $ This is option G (\(312\)).