Loading....
|
Press & Hold to Drag Around |
|||
|
Click Here to Close |
|||
Question 1 Report
In a science laboratory the mass of a sample of a radioactive material is measured. The mass, \(m\) grams, of the sample at time \(t\) hours after the first measurement is modelled by
\[m = 60e^{-kt},\]
where \(k\) is a positive constant. When \(t = 5\) the mass of the sample is \(45\) grams.
The model \(m = 60e^{-kt}\) is exponential decay. At \(t = 0\) the exponential equals \(1\), so \(60\) grams is the initial mass, and the positive constant \(k\) fixes how fast the sample decays. Every part of the question is solved by the same two-step technique: isolate the exponential on one side, then take natural logarithms, using the fact that \(\ln\left(e^{A}\right) = A\). Natural logarithms are the right choice because the base of the exponential is \(e\).
Substitute the given data point, \(m = 45\) when \(t = 5\), and isolate the exponential by dividing by \(60\):
\[45 = 60e^{-5k} \quad\Rightarrow\quad e^{-5k} = \frac{45}{60} = \frac{3}{4} \qquad \textbf{[M1]}\]The [M1] is for the substitution and the isolation, and it survives an arithmetic slip. Now take natural logarithms of both sides:
\[-5k = \ln\frac{3}{4} \qquad\text{so}\qquad 5k = -\ln\frac{3}{4} = \ln\frac{4}{3} \quad\Rightarrow\quad k = \frac{1}{5}\ln\frac{4}{3} \qquad \textbf{[A1]}\]The [A1] is for reaching the printed form, and the step that makes it work is \(-\ln\dfrac{3}{4} = \ln\dfrac{4}{3}\), because \(-\ln M = \ln M^{-1}\). Since the target is printed, this reciprocal manipulation is the substance of the answer, and a candidate who merely writes the given result earns nothing.
\[k = \frac{1}{5}(0.287682\ldots) = 0.0575364\ldots = 0.0575 \text{ (3 s.f.)} \qquad \textbf{[A1]}\]The final [A1] is for the decimal value. Three significant figures on a number beginning \(0.0\) means the leading zeros do not count, so \(0.0575\) is right and \(0.058\) is only two significant figures. Keep the unrounded value in the calculator for parts (b) and (c).
The [M1] is for the substitution of \(t = 12\) into the model, the [A1] for the value to one decimal place. Use the stored value of \(k\) rather than the rounded \(0.0575\); the rounded version gives \(30.09\), which still rounds to \(30.1\) here but will not always be so forgiving. The frequent error is to forget the minus sign in the index, which gives \(60e^{+0.6904} = 119.7\) grams, a mass larger than the original sample and therefore obviously wrong for a decaying quantity.
Now the mass is known and the time is not, so the same isolation is followed by a logarithm and then a division:
\[20 = 60e^{-kt} \quad\Rightarrow\quad e^{-kt} = \frac{20}{60} = \frac{1}{3} \qquad \textbf{[M1]}\] \[-kt = \ln\frac{1}{3} \quad\Rightarrow\quad kt = \ln 3 \quad\Rightarrow\quad t = \frac{\ln 3}{k} = \frac{1.098612\ldots}{0.0575364\ldots} \qquad \textbf{[M1]}\] \[t = 19.0942\ldots = 19.1 \text{ hours (1 d.p.)} \qquad \textbf{[A1]}\]The first [M1] is for isolating the exponential, the second for taking logarithms and rearranging for \(t\); both are method marks and so survive slips. The [A1] requires the value to one decimal place with the unit. Again \(-\ln\tfrac{1}{3} = \ln 3\).
The mistake this part is built to catch is dividing \(20\) by \(60\) the wrong way round and using \(3\) instead of \(\tfrac{1}{3}\), which gives a negative time. A negative \(t\) is impossible when the mass is falling below its starting value, so the sign of the answer is itself a check on the working.
Check: the results must be consistent with each other and with the physics. The mass falls from \(60\) to \(45\) grams in \(5\) hours, so the half-life is \(\dfrac{\ln 2}{k} = \dfrac{0.6931}{0.05754} \approx 12.0\) hours, which is why the mass at \(t = 12\) is almost exactly half of \(60\), namely \(30.1\) grams, agreeing with part (b). Since \(20\) grams is a further reduction beyond half, the answer to (c) must exceed \(12\) hours, and \(19.1\) does. Substituting back also confirms it: \(60e^{-0.0575364 \times 19.0942} = 60 \times 0.33333 = 20.0\) grams.
The model \(m = 60e^{-kt}\) is exponential decay. At \(t = 0\) the exponential equals \(1\), so \(60\) grams is the initial mass, and the positive constant \(k\) fixes how fast the sample decays. Every part of the question is solved by the same two-step technique: isolate the exponential on one side, then take natural logarithms, using the fact that \(\ln\left(e^{A}\right) = A\). Natural logarithms are the right choice because the base of the exponential is \(e\).
Substitute the given data point, \(m = 45\) when \(t = 5\), and isolate the exponential by dividing by \(60\):
\[45 = 60e^{-5k} \quad\Rightarrow\quad e^{-5k} = \frac{45}{60} = \frac{3}{4} \qquad \textbf{[M1]}\]The [M1] is for the substitution and the isolation, and it survives an arithmetic slip. Now take natural logarithms of both sides:
\[-5k = \ln\frac{3}{4} \qquad\text{so}\qquad 5k = -\ln\frac{3}{4} = \ln\frac{4}{3} \quad\Rightarrow\quad k = \frac{1}{5}\ln\frac{4}{3} \qquad \textbf{[A1]}\]The [A1] is for reaching the printed form, and the step that makes it work is \(-\ln\dfrac{3}{4} = \ln\dfrac{4}{3}\), because \(-\ln M = \ln M^{-1}\). Since the target is printed, this reciprocal manipulation is the substance of the answer, and a candidate who merely writes the given result earns nothing.
\[k = \frac{1}{5}(0.287682\ldots) = 0.0575364\ldots = 0.0575 \text{ (3 s.f.)} \qquad \textbf{[A1]}\]The final [A1] is for the decimal value. Three significant figures on a number beginning \(0.0\) means the leading zeros do not count, so \(0.0575\) is right and \(0.058\) is only two significant figures. Keep the unrounded value in the calculator for parts (b) and (c).
The [M1] is for the substitution of \(t = 12\) into the model, the [A1] for the value to one decimal place. Use the stored value of \(k\) rather than the rounded \(0.0575\); the rounded version gives \(30.09\), which still rounds to \(30.1\) here but will not always be so forgiving. The frequent error is to forget the minus sign in the index, which gives \(60e^{+0.6904} = 119.7\) grams, a mass larger than the original sample and therefore obviously wrong for a decaying quantity.
Now the mass is known and the time is not, so the same isolation is followed by a logarithm and then a division:
\[20 = 60e^{-kt} \quad\Rightarrow\quad e^{-kt} = \frac{20}{60} = \frac{1}{3} \qquad \textbf{[M1]}\] \[-kt = \ln\frac{1}{3} \quad\Rightarrow\quad kt = \ln 3 \quad\Rightarrow\quad t = \frac{\ln 3}{k} = \frac{1.098612\ldots}{0.0575364\ldots} \qquad \textbf{[M1]}\] \[t = 19.0942\ldots = 19.1 \text{ hours (1 d.p.)} \qquad \textbf{[A1]}\]The first [M1] is for isolating the exponential, the second for taking logarithms and rearranging for \(t\); both are method marks and so survive slips. The [A1] requires the value to one decimal place with the unit. Again \(-\ln\tfrac{1}{3} = \ln 3\).
The mistake this part is built to catch is dividing \(20\) by \(60\) the wrong way round and using \(3\) instead of \(\tfrac{1}{3}\), which gives a negative time. A negative \(t\) is impossible when the mass is falling below its starting value, so the sign of the answer is itself a check on the working.
Check: the results must be consistent with each other and with the physics. The mass falls from \(60\) to \(45\) grams in \(5\) hours, so the half-life is \(\dfrac{\ln 2}{k} = \dfrac{0.6931}{0.05754} \approx 12.0\) hours, which is why the mass at \(t = 12\) is almost exactly half of \(60\), namely \(30.1\) grams, agreeing with part (b). Since \(20\) grams is a further reduction beyond half, the answer to (c) must exceed \(12\) hours, and \(19.1\) does. Substituting back also confirms it: \(60e^{-0.0575364 \times 19.0942} = 60 \times 0.33333 = 20.0\) grams.
Question 2 Report
The equation \[3x^2 - 8x + c = 0,\] where \(c\) is a constant, has roots \(\alpha\) and \(\beta\). One root is three times the other, so that \(\beta = 3\alpha\).
The extra piece of information here is a relationship between the roots, \(\beta = 3\alpha\). Combined with the sum of the roots taken from the coefficients, it pins down both roots without any need to solve the quadratic, and the product then delivers the unknown constant \(c\). For \(ax^2 + bx + c = 0\) the standard results are \(\alpha + \beta = -\dfrac{b}{a}\) and \(\alpha\beta = \dfrac{c}{a}\), and the division by \(a\) matters here because the leading coefficient is \(3\), not \(1\).
From \(3x^2 - 8x + c = 0\), so \(a = 3\) and \(b = -8\):
\[\alpha + \beta = -\frac{-8}{3} = \frac{8}{3} \qquad \textbf{[B1]}\]The [B1] is for the correct sum, and the two things to get right are the sign, which turns \(-8\) into \(+\tfrac{8}{3}\), and the division by \(3\). Writing \(\alpha + \beta = 8\) is the standard mistake when the leading coefficient is not \(1\).
Now substitute the given relationship \(\beta = 3\alpha\):
\[\alpha + 3\alpha = 4\alpha = \frac{8}{3} \qquad \textbf{[M1]}\] \[\alpha = \frac{8}{12} = \frac{2}{3}, \qquad \beta = 3\alpha = 2 \qquad \textbf{[A1]}\]The [M1] is for using \(\beta = 3\alpha\) in the sum; the [A1] needs both roots. Note that the question fixes which root is which, so the pair \(\left(\tfrac{2}{3},\, 2\right)\) in that order is what "the value of \(\alpha\) and the value of \(\beta\)" asks for.
Use the product of the roots, which is where \(c\) lives:
\[\alpha\beta = \frac{c}{3} \qquad \textbf{[M1]}\] \[\frac{2}{3} \times 2 = \frac{4}{3} = \frac{c}{3} \quad\Rightarrow\quad c = 4 \qquad \textbf{[A1]}\]The [M1] is for quoting the product relationship, the [A1] for the value. Again the \(a = 3\) is easy to lose: \(\alpha\beta = c\) would give \(c = \tfrac{4}{3}\), which is wrong. An equally acceptable route is substitution: since \(\alpha = \tfrac{2}{3}\) is a root, \(3\left(\tfrac{4}{9}\right) - 8\left(\tfrac{2}{3}\right) + c = 0\), giving \(\tfrac{4}{3} - \tfrac{16}{3} + c = 0\) and \(c = 4\).
The instruction "without further solving" means the answer must come from the symmetric functions rather than from substituting the numbers found in (a). Combine the two fractions over a common denominator:
\[\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} \qquad \textbf{[M1]}\]The [M1] is for that restructuring, which is the whole idea: an expression in reciprocals of squares has been converted into two quantities already known.
\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{8}{3}\right)^2 - 2\left(\frac{4}{3}\right) = \frac{64}{9} - \frac{8}{3} = \frac{64}{9} - \frac{24}{9} = \frac{40}{9} \qquad \textbf{[A1]}\] \[(\alpha\beta)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}, \qquad \text{so } \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{40}{9} \div \frac{16}{9} = \frac{40}{16} = \frac{5}{2} \qquad \textbf{[A1]}\]Two independent accuracy marks, one for \(\alpha^2 + \beta^2\) and one for the final value. Converting \(\tfrac{8}{3}\) to ninths before subtracting is essential; subtracting \(\tfrac{8}{3}\) from \(\tfrac{64}{9}\) as though the denominators matched gives \(\tfrac{56}{9}\) and loses both marks. Also note that dividing by \(\tfrac{16}{9}\) means multiplying by \(\tfrac{9}{16}\), so the ninths cancel neatly.
The wrong turn this part is built to catch is writing \(\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{1}{\alpha^2 + \beta^2}\). Reciprocals do not add like that, as a single numerical case shows: \(\tfrac{1}{1} + \tfrac{1}{4} = 1.25\), whereas \(\tfrac{1}{1 + 4} = 0.2\).
Check: the roots from (a) can be used as verification even though they must not be used as the method. With \(\alpha = \tfrac{2}{3}\) and \(\beta = 2\), \(\dfrac{1}{\alpha^2} = \dfrac{9}{4}\) and \(\dfrac{1}{\beta^2} = \dfrac{1}{4}\), so the sum is \(\dfrac{10}{4} = \dfrac{5}{2}\), matching. And \(3x^2 - 8x + 4 = (3x - 2)(x - 2)\) has roots \(\tfrac{2}{3}\) and \(2\), confirming both (a) and (b).
The extra piece of information here is a relationship between the roots, \(\beta = 3\alpha\). Combined with the sum of the roots taken from the coefficients, it pins down both roots without any need to solve the quadratic, and the product then delivers the unknown constant \(c\). For \(ax^2 + bx + c = 0\) the standard results are \(\alpha + \beta = -\dfrac{b}{a}\) and \(\alpha\beta = \dfrac{c}{a}\), and the division by \(a\) matters here because the leading coefficient is \(3\), not \(1\).
From \(3x^2 - 8x + c = 0\), so \(a = 3\) and \(b = -8\):
\[\alpha + \beta = -\frac{-8}{3} = \frac{8}{3} \qquad \textbf{[B1]}\]The [B1] is for the correct sum, and the two things to get right are the sign, which turns \(-8\) into \(+\tfrac{8}{3}\), and the division by \(3\). Writing \(\alpha + \beta = 8\) is the standard mistake when the leading coefficient is not \(1\).
Now substitute the given relationship \(\beta = 3\alpha\):
\[\alpha + 3\alpha = 4\alpha = \frac{8}{3} \qquad \textbf{[M1]}\] \[\alpha = \frac{8}{12} = \frac{2}{3}, \qquad \beta = 3\alpha = 2 \qquad \textbf{[A1]}\]The [M1] is for using \(\beta = 3\alpha\) in the sum; the [A1] needs both roots. Note that the question fixes which root is which, so the pair \(\left(\tfrac{2}{3},\, 2\right)\) in that order is what "the value of \(\alpha\) and the value of \(\beta\)" asks for.
Use the product of the roots, which is where \(c\) lives:
\[\alpha\beta = \frac{c}{3} \qquad \textbf{[M1]}\] \[\frac{2}{3} \times 2 = \frac{4}{3} = \frac{c}{3} \quad\Rightarrow\quad c = 4 \qquad \textbf{[A1]}\]The [M1] is for quoting the product relationship, the [A1] for the value. Again the \(a = 3\) is easy to lose: \(\alpha\beta = c\) would give \(c = \tfrac{4}{3}\), which is wrong. An equally acceptable route is substitution: since \(\alpha = \tfrac{2}{3}\) is a root, \(3\left(\tfrac{4}{9}\right) - 8\left(\tfrac{2}{3}\right) + c = 0\), giving \(\tfrac{4}{3} - \tfrac{16}{3} + c = 0\) and \(c = 4\).
The instruction "without further solving" means the answer must come from the symmetric functions rather than from substituting the numbers found in (a). Combine the two fractions over a common denominator:
\[\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} \qquad \textbf{[M1]}\]The [M1] is for that restructuring, which is the whole idea: an expression in reciprocals of squares has been converted into two quantities already known.
\[\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{8}{3}\right)^2 - 2\left(\frac{4}{3}\right) = \frac{64}{9} - \frac{8}{3} = \frac{64}{9} - \frac{24}{9} = \frac{40}{9} \qquad \textbf{[A1]}\] \[(\alpha\beta)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}, \qquad \text{so } \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{40}{9} \div \frac{16}{9} = \frac{40}{16} = \frac{5}{2} \qquad \textbf{[A1]}\]Two independent accuracy marks, one for \(\alpha^2 + \beta^2\) and one for the final value. Converting \(\tfrac{8}{3}\) to ninths before subtracting is essential; subtracting \(\tfrac{8}{3}\) from \(\tfrac{64}{9}\) as though the denominators matched gives \(\tfrac{56}{9}\) and loses both marks. Also note that dividing by \(\tfrac{16}{9}\) means multiplying by \(\tfrac{9}{16}\), so the ninths cancel neatly.
The wrong turn this part is built to catch is writing \(\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{1}{\alpha^2 + \beta^2}\). Reciprocals do not add like that, as a single numerical case shows: \(\tfrac{1}{1} + \tfrac{1}{4} = 1.25\), whereas \(\tfrac{1}{1 + 4} = 0.2\).
Check: the roots from (a) can be used as verification even though they must not be used as the method. With \(\alpha = \tfrac{2}{3}\) and \(\beta = 2\), \(\dfrac{1}{\alpha^2} = \dfrac{9}{4}\) and \(\dfrac{1}{\beta^2} = \dfrac{1}{4}\), so the sum is \(\dfrac{10}{4} = \dfrac{5}{2}\), matching. And \(3x^2 - 8x + 4 = (3x - 2)(x - 2)\) has roots \(\tfrac{2}{3}\) and \(2\), confirming both (a) and (b).
Question 3 Report
Figure 1 shows the curve \(C\) with equation \(y = x^2 - 6x + 10\) and the line \(l\) with equation \(y = 2x + k\), where \(k\) is a constant. In Figure 1 the line \(l\) is a tangent to \(C\) at the point \(P\).
This question runs one idea through six parts: the number of times a line meets a curve is the number of real roots of the quadratic you get by eliminating \(y\), and that number is controlled by the discriminant. A tangent touches once, so the discriminant is zero; two distinct crossings need it positive. Once part (c) has produced the quadratic in \(x\), parts (d), (e) and (f) are three different questions about the same discriminant, which is why they are worth so many marks between them.
Half of \(-6\) is \(-3\), so \((x - 3)^2 = x^2 - 6x + 9\) and nine must be given back:
\[x^2 - 6x + 10 = (x - 3)^2 - 9 + 10 = (x - 3)^2 + 1 \qquad \textbf{[M1]}\] \[p = 3, \quad q = 1 \qquad \textbf{[A1]}\]The [M1] is for the completing-the-square structure \((x - 3)^2 - 9 + 10\); the [A1] is for both constants. The requested form has a minus sign inside the bracket, so \(p = 3\) here, positive; watch the sign convention in each question because it changes between papers.
\((x - 3)^2 \ge 0\) always, and is zero only at \(x = 3\), so the smallest value of \(y\) is \(1\), reached at \(x = 3\). The minimum point is \((3,\, 1)\) [B1]. This is written down from the completed square, not found by differentiation, which is what "hence" signals. A very common slip is to give \((-3,\, 1)\): the vertex sits where the bracket vanishes, so the \(x\)-coordinate is \(+3\).
The target is printed, so all the credit is in the elimination. At any common point of \(C\) and \(l\) the two \(y\)-values are equal:
\[x^2 - 6x + 10 = 2x + k \qquad \textbf{[M1]}\] \[x^2 - 6x - 2x + 10 - k = 0 \quad\Rightarrow\quad x^2 - 8x + 10 - k = 0 \qquad \textbf{[A1]}\]The [M1] is for equating the two expressions, the [A1] for the tidy printed form. Subtracting \(2x\) gives \(-8x\), not \(-4x\), and the constant term is \(10 - k\), a single quantity that must be carried as such into the discriminant later.
The line is a tangent, so it meets \(C\) exactly once, so the quadratic from (c) has a repeated root, so its discriminant is zero:
\[b^2 - 4ac = 0 \qquad \textbf{[M1]}\] \[(-8)^2 - 4 \times 1 \times (10 - k) = 0 \qquad \textbf{[M1]}\] \[64 - 40 + 4k = 0 \quad\Rightarrow\quad 4k = -24 \quad\Rightarrow\quad k = -6 \qquad \textbf{[A1]}\]The first [M1] is for knowing that tangency means zero discriminant; the second is for substituting \(a = 1\), \(b = -8\), \(c = 10 - k\) correctly. Expanding \(-4(10 - k)\) as \(-40 - 4k\) is the standard error and loses the [A1], giving \(k = 6\).
For the point of contact, put \(k = -6\) back into the quadratic:
\[x^2 - 8x + 16 = 0 \quad\Rightarrow\quad (x - 4)^2 = 0 \quad\Rightarrow\quad x = 4 \qquad \textbf{[M1]}\] \[y = 2(4) + (-6) = 2, \quad\text{so } P(4,\, 2) \qquad \textbf{[A1]}\]The [M1] here is follow-through: a candidate with the wrong \(k\) can still earn it by solving their own quadratic and finding \(y\) from the line. The [A1] needs the correct pair. The repeated-root shortcut \(x = -\dfrac{b}{2a} = 4\) is quicker and avoids a factorising slip.
Two distinct intersections means two distinct real roots, so the same discriminant must now be strictly positive:
\[64 - 4(10 - k) \gt 0 \qquad \textbf{[M1]}\] \[64 - 40 + 4k \gt 0 \quad\Rightarrow\quad 24 + 4k \gt 0 \qquad \textbf{[M1]}\] \[k \gt -6 \qquad \textbf{[A1]}\]The first [M1] is for using \(b^2 - 4ac \gt 0\), the second for correct expansion, the [A1] for the final set. Two traps: writing \(\ge\) instead of \(\gt\) admits the tangent case, where the two points coincide, so it is wrong; and dividing an inequality by a negative number without reversing it. Notice the structural sense of the answer: \(k = -6\) is the tangent found in (d), and any larger \(k\) lifts the line clear of the touching position so that it cuts twice.
Substituting \(k = 10\) into the quadratic from (c) makes the constant term vanish:
\[x^2 - 8x + 10 - 10 = 0 \quad\Rightarrow\quad x^2 - 8x = 0 \qquad \textbf{[M1]}\] \[x(x - 8) = 0 \quad\Rightarrow\quad x = 0 \text{ or } x = 8 \qquad \textbf{[A1]}\]Using \(y = 2x + 10\) gives the points \((0,\, 10)\) and \((8,\, 26)\) [A1].
The wrong turn this part is designed to catch is dividing through by \(x\) to get \(x = 8\) only. Dividing by a quantity that may be zero destroys the root \(x = 0\), and half the answer with it; always factorise instead. The second [A1] also requires the \(y\)-values, since the question asks for coordinates.
Check: both points must lie on \(C\) as well as on \(l\). At \(x = 0\), \(y = 0 - 0 + 10 = 10\); at \(x = 8\), \(y = 64 - 48 + 10 = 26\). Both agree. Consistency with (e) is a further check: \(k = 10 \gt -6\), so two distinct points were expected, and two were found.
This question runs one idea through six parts: the number of times a line meets a curve is the number of real roots of the quadratic you get by eliminating \(y\), and that number is controlled by the discriminant. A tangent touches once, so the discriminant is zero; two distinct crossings need it positive. Once part (c) has produced the quadratic in \(x\), parts (d), (e) and (f) are three different questions about the same discriminant, which is why they are worth so many marks between them.
Half of \(-6\) is \(-3\), so \((x - 3)^2 = x^2 - 6x + 9\) and nine must be given back:
\[x^2 - 6x + 10 = (x - 3)^2 - 9 + 10 = (x - 3)^2 + 1 \qquad \textbf{[M1]}\] \[p = 3, \quad q = 1 \qquad \textbf{[A1]}\]The [M1] is for the completing-the-square structure \((x - 3)^2 - 9 + 10\); the [A1] is for both constants. The requested form has a minus sign inside the bracket, so \(p = 3\) here, positive; watch the sign convention in each question because it changes between papers.
\((x - 3)^2 \ge 0\) always, and is zero only at \(x = 3\), so the smallest value of \(y\) is \(1\), reached at \(x = 3\). The minimum point is \((3,\, 1)\) [B1]. This is written down from the completed square, not found by differentiation, which is what "hence" signals. A very common slip is to give \((-3,\, 1)\): the vertex sits where the bracket vanishes, so the \(x\)-coordinate is \(+3\).
The target is printed, so all the credit is in the elimination. At any common point of \(C\) and \(l\) the two \(y\)-values are equal:
\[x^2 - 6x + 10 = 2x + k \qquad \textbf{[M1]}\] \[x^2 - 6x - 2x + 10 - k = 0 \quad\Rightarrow\quad x^2 - 8x + 10 - k = 0 \qquad \textbf{[A1]}\]The [M1] is for equating the two expressions, the [A1] for the tidy printed form. Subtracting \(2x\) gives \(-8x\), not \(-4x\), and the constant term is \(10 - k\), a single quantity that must be carried as such into the discriminant later.
The line is a tangent, so it meets \(C\) exactly once, so the quadratic from (c) has a repeated root, so its discriminant is zero:
\[b^2 - 4ac = 0 \qquad \textbf{[M1]}\] \[(-8)^2 - 4 \times 1 \times (10 - k) = 0 \qquad \textbf{[M1]}\] \[64 - 40 + 4k = 0 \quad\Rightarrow\quad 4k = -24 \quad\Rightarrow\quad k = -6 \qquad \textbf{[A1]}\]The first [M1] is for knowing that tangency means zero discriminant; the second is for substituting \(a = 1\), \(b = -8\), \(c = 10 - k\) correctly. Expanding \(-4(10 - k)\) as \(-40 - 4k\) is the standard error and loses the [A1], giving \(k = 6\).
For the point of contact, put \(k = -6\) back into the quadratic:
\[x^2 - 8x + 16 = 0 \quad\Rightarrow\quad (x - 4)^2 = 0 \quad\Rightarrow\quad x = 4 \qquad \textbf{[M1]}\] \[y = 2(4) + (-6) = 2, \quad\text{so } P(4,\, 2) \qquad \textbf{[A1]}\]The [M1] here is follow-through: a candidate with the wrong \(k\) can still earn it by solving their own quadratic and finding \(y\) from the line. The [A1] needs the correct pair. The repeated-root shortcut \(x = -\dfrac{b}{2a} = 4\) is quicker and avoids a factorising slip.
Two distinct intersections means two distinct real roots, so the same discriminant must now be strictly positive:
\[64 - 4(10 - k) \gt 0 \qquad \textbf{[M1]}\] \[64 - 40 + 4k \gt 0 \quad\Rightarrow\quad 24 + 4k \gt 0 \qquad \textbf{[M1]}\] \[k \gt -6 \qquad \textbf{[A1]}\]The first [M1] is for using \(b^2 - 4ac \gt 0\), the second for correct expansion, the [A1] for the final set. Two traps: writing \(\ge\) instead of \(\gt\) admits the tangent case, where the two points coincide, so it is wrong; and dividing an inequality by a negative number without reversing it. Notice the structural sense of the answer: \(k = -6\) is the tangent found in (d), and any larger \(k\) lifts the line clear of the touching position so that it cuts twice.
Substituting \(k = 10\) into the quadratic from (c) makes the constant term vanish:
\[x^2 - 8x + 10 - 10 = 0 \quad\Rightarrow\quad x^2 - 8x = 0 \qquad \textbf{[M1]}\] \[x(x - 8) = 0 \quad\Rightarrow\quad x = 0 \text{ or } x = 8 \qquad \textbf{[A1]}\]Using \(y = 2x + 10\) gives the points \((0,\, 10)\) and \((8,\, 26)\) [A1].
The wrong turn this part is designed to catch is dividing through by \(x\) to get \(x = 8\) only. Dividing by a quantity that may be zero destroys the root \(x = 0\), and half the answer with it; always factorise instead. The second [A1] also requires the \(y\)-values, since the question asks for coordinates.
Check: both points must lie on \(C\) as well as on \(l\). At \(x = 0\), \(y = 0 - 0 + 10 = 10\); at \(x = 8\), \(y = 64 - 48 + 10 = 26\). Both agree. Consistency with (e) is a further check: \(k = 10 \gt -6\), so two distinct points were expected, and two were found.
Question 4 Report
\[\mathrm{f}(x) = (1 + 2x)^6 (1 - x)^4\]
Eleven marks on a product of two binomial expansions, ending with a numerical estimate. The strategy is to expand each bracket separately as far as \(x^2\), then multiply the two series keeping only the terms up to \(x^2\). Part (d) then exploits the fact that a truncated series is an excellent approximation when \(x\) is small.
So \((1 + 2x)^6 = 1 + 12x + 60x^2 + \ldots\). Three separate B marks, one for the structure and one for each of the \(x\) and \(x^2\) terms, so a slip in one does not cost the others. The point to watch is \((2x)^2 = 4x^2\): the 2 is squared as well, which is why the coefficient is \(15 \times 4 = 60\) and not 15.
The alternating signs come from the powers of \(-x\): odd powers are negative and even powers positive, so \((-x)^2 = +x^2\) and the \(x^2\) coefficient is \(+6\), not \(-6\).
The result is printed, so the marks are for the multiplication. Collect the products whose indices add to each required power: \[\mathrm{f}(x) = \bigl(1 + 12x + 60x^2 + \ldots\bigr)\bigl(1 - 4x + 6x^2 + \ldots\bigr)\] \[\text{Constant: } 1 \times 1 = 1 \quad \textbf{B1}\] \[\text{Term in } x: \ 1(-4) + 12(1) = -4 + 12 = 8 \quad \textbf{B1}\] \[\text{Term in } x^2: \ 1(6) + 12(-4) + 60(1) = 6 - 48 + 60 = 18 \quad \textbf{B1}\] There are exactly three ways to produce an \(x^2\) term, and all three must appear: constant times \(x^2\), \(x\) times \(x\), and \(x^2\) times constant. Terms in \(x^3\) and beyond are ignored because only the first three are wanted, so there is no need to expand either bracket further. Note the near-cancellation in the \(x^2\) coefficient: \(6 - 48 + 60\) is only 18, so dropping any one contribution changes the answer dramatically.
Match the numerical bases to the algebraic ones. With \(x = 0.01\), \[1 + 2x = 1 + 0.02 = 1.02 \quad \text{and} \quad 1 - x = 1 - 0.01 = 0.99 \quad \textbf{B1}\] so \(\mathrm{f}(0.01) = (1.02)^6 \times (0.99)^4\) exactly. Substituting into the truncated expansion: \[\text{Estimate} = 1 + 8(0.01) + 18(0.01)^2 \quad \textbf{M1}\] \[= 1 + 0.08 + 0.0018 = 1.0818 \ \text{(4 d.p.)} \quad \textbf{A1}\] The B1 is for identifying that \(x = 0.01\) reproduces both bases, which is the insight the part is testing. Note \((0.01)^2 = 0.0001\), so the \(x^2\) term contributes \(0.0018\); the successive terms shrink by roughly a factor of 100, which is exactly why three terms suffice for four decimal places.
Evaluate the target directly: \((1.02)^6 = 1.126162\ldots\) and \((0.99)^4 = 0.960596\ldots\), whose product is \(1.081787\ldots\), which rounds to \(1.0818\). The three-term estimate therefore agrees to all four decimal places, confirming both the expansion and the substitution. A structural check on part (c) is also worth doing: the constant term of \(\mathrm{f}(x)\) must be \(1^6 \times 1^4 = 1\), and the \(x\) coefficient must be \(6(2) + 4(-1) = 8\), matching the answer.
Eleven marks on a product of two binomial expansions, ending with a numerical estimate. The strategy is to expand each bracket separately as far as \(x^2\), then multiply the two series keeping only the terms up to \(x^2\). Part (d) then exploits the fact that a truncated series is an excellent approximation when \(x\) is small.
So \((1 + 2x)^6 = 1 + 12x + 60x^2 + \ldots\). Three separate B marks, one for the structure and one for each of the \(x\) and \(x^2\) terms, so a slip in one does not cost the others. The point to watch is \((2x)^2 = 4x^2\): the 2 is squared as well, which is why the coefficient is \(15 \times 4 = 60\) and not 15.
The alternating signs come from the powers of \(-x\): odd powers are negative and even powers positive, so \((-x)^2 = +x^2\) and the \(x^2\) coefficient is \(+6\), not \(-6\).
The result is printed, so the marks are for the multiplication. Collect the products whose indices add to each required power: \[\mathrm{f}(x) = \bigl(1 + 12x + 60x^2 + \ldots\bigr)\bigl(1 - 4x + 6x^2 + \ldots\bigr)\] \[\text{Constant: } 1 \times 1 = 1 \quad \textbf{B1}\] \[\text{Term in } x: \ 1(-4) + 12(1) = -4 + 12 = 8 \quad \textbf{B1}\] \[\text{Term in } x^2: \ 1(6) + 12(-4) + 60(1) = 6 - 48 + 60 = 18 \quad \textbf{B1}\] There are exactly three ways to produce an \(x^2\) term, and all three must appear: constant times \(x^2\), \(x\) times \(x\), and \(x^2\) times constant. Terms in \(x^3\) and beyond are ignored because only the first three are wanted, so there is no need to expand either bracket further. Note the near-cancellation in the \(x^2\) coefficient: \(6 - 48 + 60\) is only 18, so dropping any one contribution changes the answer dramatically.
Match the numerical bases to the algebraic ones. With \(x = 0.01\), \[1 + 2x = 1 + 0.02 = 1.02 \quad \text{and} \quad 1 - x = 1 - 0.01 = 0.99 \quad \textbf{B1}\] so \(\mathrm{f}(0.01) = (1.02)^6 \times (0.99)^4\) exactly. Substituting into the truncated expansion: \[\text{Estimate} = 1 + 8(0.01) + 18(0.01)^2 \quad \textbf{M1}\] \[= 1 + 0.08 + 0.0018 = 1.0818 \ \text{(4 d.p.)} \quad \textbf{A1}\] The B1 is for identifying that \(x = 0.01\) reproduces both bases, which is the insight the part is testing. Note \((0.01)^2 = 0.0001\), so the \(x^2\) term contributes \(0.0018\); the successive terms shrink by roughly a factor of 100, which is exactly why three terms suffice for four decimal places.
Evaluate the target directly: \((1.02)^6 = 1.126162\ldots\) and \((0.99)^4 = 0.960596\ldots\), whose product is \(1.081787\ldots\), which rounds to \(1.0818\). The three-term estimate therefore agrees to all four decimal places, confirming both the expansion and the substitution. A structural check on part (c) is also worth doing: the constant term of \(\mathrm{f}(x)\) must be \(1^6 \times 1^4 = 1\), and the \(x\) coefficient must be \(6(2) + 4(-1) = 8\), matching the answer.
Question 5 Report
The curve \(C\) has equation \[y = \frac{2x + 3}{x - 1}, \qquad x \ne 1.\]
Figure 5 shows part of \(C\) together with its two asymptotes, drawn as broken lines. The curve crosses the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).
Ten marks on the rational curve \(y = \dfrac{2x + 3}{x - 1}\). The figure shows part of \(C\) with its two asymptotes drawn as broken lines, and the curve crossing the axes at \(A\) and \(B\). Everything here follows from two habits: look at where the denominator vanishes, and look at what happens for numerically large \(x\).
A vertical asymptote occurs where the denominator is zero but the numerator is not, because the fraction then grows without bound: \[x - 1 = 0 \ \Rightarrow\ x = 1 \quad \textbf{B1}\] For the horizontal asymptote, divide numerator and denominator by \(x\): \[y = \frac{2 + \dfrac{3}{x}}{1 - \dfrac{1}{x}} \to \frac{2}{1} = 2 \quad \text{as} \ x \to \pm\infty, \quad \text{so} \ y = 2 \quad \textbf{B1}\] Both are B marks, awarded for the correct statements with no method credit available. Each answer must be an equation: "\(x = 1\)", not "1", and "\(y = 2\)", not "2". A shortcut worth knowing is that for \(\dfrac{ax+b}{cx+d}\) the horizontal asymptote is \(y = \dfrac{a}{c}\), here \(y = \dfrac{2}{1}\).
A curve crosses the \(x\)-axis where \(y = 0\). A fraction is zero only when its numerator is zero: \[2x + 3 = 0 \ \Rightarrow\ x = -\tfrac{3}{2}, \qquad A\left(-\tfrac{3}{2},\, 0\right) \quad \textbf{M1 A1}\] It crosses the \(y\)-axis where \(x = 0\): \[y = \frac{2(0) + 3}{0 - 1} = \frac{3}{-1} = -3, \qquad B(0,\, -3) \quad \textbf{A1}\] The M1 is for setting the numerator to zero, which is the idea being tested; the two A marks are for the two points. Watch the sign in \(B\): the denominator at \(x = 0\) is \(-1\), so \(y\) is negative, which agrees with the figure.
This part explains why \(y = 2\) is an asymptote rather than an ordinary value of the function. Suppose such an \(x\) existed and derive a contradiction: \[\frac{2x + 3}{x - 1} = 2 \ \Rightarrow\ 2x + 3 = 2(x - 1) = 2x - 2 \quad \textbf{M1}\] \[3 = -2, \ \text{which is impossible; hence no such } x \text{ exists.} \quad \textbf{A1}\] The M1 is for clearing the fraction, the A1 for reaching an impossible statement and saying so. Because the target is given, the marks are entirely for the argument; asserting "the curve never reaches its asymptote" without algebra earns nothing. Note that the \(2x\) terms cancel, which is precisely the algebraic signature of a horizontal asymptote at \(y = 2\).
Do not multiply by \((x - 1)\), whose sign is unknown. Record \(x \ne 1\), then multiply by \((x - 1)^2\), which is positive for all \(x \ne 1\) and so preserves the inequality: \[(2x + 3)(x - 1) \ge 3(x - 1)^2 \quad \textbf{M1}\] \[2x^2 + x - 3 \ge 3x^2 - 6x + 3 \ \Rightarrow\ 0 \ge x^2 - 7x + 6 \quad \textbf{A1}\] \[(x - 1)(x - 6) \le 0 \ \Rightarrow\ 1 \le x \le 6, \quad \text{and excluding } x = 1: \ 1 \lt x \le 6 \quad \textbf{A1}\] A positive quadratic is at or below the axis between its roots, which is why this part gives a single bounded interval rather than two pieces. The final A1 depends on the strict inequality at \(x = 1\), the point where the curve is undefined.
Check the intercepts against the figure: \(A\) lies to the left of the origin and \(B\) below it, which matches \(\left(-\tfrac{3}{2}, 0\right)\) and \((0, -3)\). For part (d), test the ends and the middle: at \(x = 2\), \(y = \dfrac{7}{1} = 7 \ge 3\), so the interval is right; at \(x = 6\), \(y = \dfrac{15}{5} = 3\) exactly, confirming the inclusive bound; at \(x = 7\), \(y = \dfrac{17}{6} = 2.83\ldots\), which is less than 3, confirming the interval stops at 6. As \(x\) increases beyond 6 the curve is descending towards \(y = 2\) from above, which is consistent with part (c).
Ten marks on the rational curve \(y = \dfrac{2x + 3}{x - 1}\). The figure shows part of \(C\) with its two asymptotes drawn as broken lines, and the curve crossing the axes at \(A\) and \(B\). Everything here follows from two habits: look at where the denominator vanishes, and look at what happens for numerically large \(x\).
A vertical asymptote occurs where the denominator is zero but the numerator is not, because the fraction then grows without bound: \[x - 1 = 0 \ \Rightarrow\ x = 1 \quad \textbf{B1}\] For the horizontal asymptote, divide numerator and denominator by \(x\): \[y = \frac{2 + \dfrac{3}{x}}{1 - \dfrac{1}{x}} \to \frac{2}{1} = 2 \quad \text{as} \ x \to \pm\infty, \quad \text{so} \ y = 2 \quad \textbf{B1}\] Both are B marks, awarded for the correct statements with no method credit available. Each answer must be an equation: "\(x = 1\)", not "1", and "\(y = 2\)", not "2". A shortcut worth knowing is that for \(\dfrac{ax+b}{cx+d}\) the horizontal asymptote is \(y = \dfrac{a}{c}\), here \(y = \dfrac{2}{1}\).
A curve crosses the \(x\)-axis where \(y = 0\). A fraction is zero only when its numerator is zero: \[2x + 3 = 0 \ \Rightarrow\ x = -\tfrac{3}{2}, \qquad A\left(-\tfrac{3}{2},\, 0\right) \quad \textbf{M1 A1}\] It crosses the \(y\)-axis where \(x = 0\): \[y = \frac{2(0) + 3}{0 - 1} = \frac{3}{-1} = -3, \qquad B(0,\, -3) \quad \textbf{A1}\] The M1 is for setting the numerator to zero, which is the idea being tested; the two A marks are for the two points. Watch the sign in \(B\): the denominator at \(x = 0\) is \(-1\), so \(y\) is negative, which agrees with the figure.
This part explains why \(y = 2\) is an asymptote rather than an ordinary value of the function. Suppose such an \(x\) existed and derive a contradiction: \[\frac{2x + 3}{x - 1} = 2 \ \Rightarrow\ 2x + 3 = 2(x - 1) = 2x - 2 \quad \textbf{M1}\] \[3 = -2, \ \text{which is impossible; hence no such } x \text{ exists.} \quad \textbf{A1}\] The M1 is for clearing the fraction, the A1 for reaching an impossible statement and saying so. Because the target is given, the marks are entirely for the argument; asserting "the curve never reaches its asymptote" without algebra earns nothing. Note that the \(2x\) terms cancel, which is precisely the algebraic signature of a horizontal asymptote at \(y = 2\).
Do not multiply by \((x - 1)\), whose sign is unknown. Record \(x \ne 1\), then multiply by \((x - 1)^2\), which is positive for all \(x \ne 1\) and so preserves the inequality: \[(2x + 3)(x - 1) \ge 3(x - 1)^2 \quad \textbf{M1}\] \[2x^2 + x - 3 \ge 3x^2 - 6x + 3 \ \Rightarrow\ 0 \ge x^2 - 7x + 6 \quad \textbf{A1}\] \[(x - 1)(x - 6) \le 0 \ \Rightarrow\ 1 \le x \le 6, \quad \text{and excluding } x = 1: \ 1 \lt x \le 6 \quad \textbf{A1}\] A positive quadratic is at or below the axis between its roots, which is why this part gives a single bounded interval rather than two pieces. The final A1 depends on the strict inequality at \(x = 1\), the point where the curve is undefined.
Check the intercepts against the figure: \(A\) lies to the left of the origin and \(B\) below it, which matches \(\left(-\tfrac{3}{2}, 0\right)\) and \((0, -3)\). For part (d), test the ends and the middle: at \(x = 2\), \(y = \dfrac{7}{1} = 7 \ge 3\), so the interval is right; at \(x = 6\), \(y = \dfrac{15}{5} = 3\) exactly, confirming the inclusive bound; at \(x = 7\), \(y = \dfrac{17}{6} = 2.83\ldots\), which is less than 3, confirming the interval stops at 6. As \(x\) increases beyond 6 the curve is descending towards \(y = 2\) from above, which is consistent with part (c).
Question 6 Report
Figure 2 shows the parallelogram \(OACB\), in which \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). The point \(M\) is the midpoint of \(BC\). The line \(OM\) meets the diagonal \(AB\) at the point \(X\).
Twelve marks of vector geometry in a parallelogram. Figure 2 shows \(OACB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), \(M\) the midpoint of \(BC\), and \(X\) the point where \(OM\) crosses the diagonal \(AB\). The whole question turns on one powerful idea: if \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, then any point of the plane has exactly one expression as \(\lambda\mathbf{a} + \mu\mathbf{b}\). Writing the same point two ways therefore forces the coefficients to match, and that gives simultaneous equations.
In the parallelogram \(OACB\), the side \(AC\) is equal and parallel to \(OB\), so travelling \(O\) to \(A\) to \(C\) is \(\mathbf{a}\) followed by \(\mathbf{b}\): \[\overrightarrow{OC} = \mathbf{a} + \mathbf{b} \quad \textbf{B1}\]
Since the result is given, the marks are for the route. First find the side \(BC\): \[\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\mathbf{a} + \mathbf{b}) - \mathbf{b} = \mathbf{a}, \ \text{so} \ \overrightarrow{BM} = \tfrac{1}{2}\mathbf{a} \quad \textbf{M1}\] Then travel \(O\) to \(B\) to \(M\): \[\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM} = \mathbf{b} + \tfrac{1}{2}\mathbf{a} = \tfrac{1}{2}\mathbf{a} + \mathbf{b} \quad \textbf{A1}\]
Along \(OM\), \(X\) is a fraction \(t\) of the way: \[\overrightarrow{OX} = t\left(\tfrac{1}{2}\mathbf{a} + \mathbf{b}\right) = \tfrac{1}{2}t\,\mathbf{a} + t\,\mathbf{b} \quad \textbf{B1}\] Along \(AB\), start at \(A\) and move a fraction \(s\) of the way towards \(B\). The direction is \[\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a} \quad \textbf{B1}\] so \[\overrightarrow{OX} = \overrightarrow{OA} + s\,\overrightarrow{AB} = \mathbf{a} + s(\mathbf{b} - \mathbf{a}) = (1 - s)\mathbf{a} + s\,\mathbf{b} \quad \textbf{B1}\] Three independent B marks. Both expressions must be written purely in terms of \(\mathbf{a}\), \(\mathbf{b}\) and the single parameter, with the brackets expanded ready for comparison.
The two expressions describe the same point, and \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel, so the coefficients may be equated: \[\text{Coefficients may be compared because } \mathbf{a} \ \text{and} \ \mathbf{b} \ \text{are not parallel.} \quad \textbf{M1}\] \[\mathbf{a}: \ \tfrac{1}{2}t = 1 - s; \qquad \mathbf{b}: \ t = s \quad \textbf{A1}\] Substituting \(t = s\) into the first equation: \[\tfrac{1}{2}s = 1 - s \ \Rightarrow\ \tfrac{3}{2}s = 1 \ \Rightarrow\ s = \tfrac{2}{3}, \qquad t = \tfrac{2}{3} \quad \textbf{A1}\] The M1 rewards the justification as well as the act of comparing; it is the mathematical content of the part. The final A1 needs both values.
Since \(\overrightarrow{AX} = s\,\overrightarrow{AB} = \tfrac{2}{3}\overrightarrow{AB}\), the point \(X\) is two thirds of the way from \(A\) to \(B\), leaving one third: \[AX : XB = 2 : 1 \quad \textbf{B1}\] Read from \(s\) directly as \(s : (1 - s)\). Quoting \(1 : 2\) reverses the segments.
Use either expression with \(s = t = \tfrac{2}{3}\). Taking the second gives \(\overrightarrow{OX} = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\): \[\overrightarrow{OX} = \tfrac{1}{3}(4\mathbf{i} + \mathbf{j}) + \tfrac{2}{3}(-2\mathbf{i} + 5\mathbf{j}) = \left(\tfrac{4}{3} - \tfrac{4}{3}\right)\mathbf{i} + \left(\tfrac{1}{3} + \tfrac{10}{3}\right)\mathbf{j} = \tfrac{11}{3}\mathbf{j} \quad \textbf{M1}\] The \(\mathbf{i}\) components cancel exactly, so \(\overrightarrow{OX}\) points straight up the \(\mathbf{j}\) direction and its modulus is simply the size of that one component: \[\left|\overrightarrow{OX}\right| = \tfrac{11}{3} \quad \textbf{A1}\] Using the first expression is an equally good check: \(\tfrac{2}{3}\left(\tfrac{1}{2}\mathbf{a} + \mathbf{b}\right) = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\), the same vector.
With the given components, \(\overrightarrow{OM} = \tfrac{1}{2}(4\mathbf{i} + \mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) = 0\mathbf{i} + \tfrac{11}{2}\mathbf{j}\), so \(OM\) itself lies along the \(\mathbf{j}\)-axis and \(\tfrac{2}{3}\) of it is \(\tfrac{11}{3}\mathbf{j}\), confirming part (f) independently. The ratio in part (e) is also verifiable numerically: \(A\) is at \((4, 1)\), \(B\) at \((-2, 5)\), and the point two thirds of the way along is \(\left(4 - \tfrac{2}{3}(6),\ 1 + \tfrac{2}{3}(4)\right) = \left(0,\ \tfrac{11}{3}\right)\), exactly the \(X\) found.
Twelve marks of vector geometry in a parallelogram. Figure 2 shows \(OACB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), \(M\) the midpoint of \(BC\), and \(X\) the point where \(OM\) crosses the diagonal \(AB\). The whole question turns on one powerful idea: if \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, then any point of the plane has exactly one expression as \(\lambda\mathbf{a} + \mu\mathbf{b}\). Writing the same point two ways therefore forces the coefficients to match, and that gives simultaneous equations.
In the parallelogram \(OACB\), the side \(AC\) is equal and parallel to \(OB\), so travelling \(O\) to \(A\) to \(C\) is \(\mathbf{a}\) followed by \(\mathbf{b}\): \[\overrightarrow{OC} = \mathbf{a} + \mathbf{b} \quad \textbf{B1}\]
Since the result is given, the marks are for the route. First find the side \(BC\): \[\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\mathbf{a} + \mathbf{b}) - \mathbf{b} = \mathbf{a}, \ \text{so} \ \overrightarrow{BM} = \tfrac{1}{2}\mathbf{a} \quad \textbf{M1}\] Then travel \(O\) to \(B\) to \(M\): \[\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM} = \mathbf{b} + \tfrac{1}{2}\mathbf{a} = \tfrac{1}{2}\mathbf{a} + \mathbf{b} \quad \textbf{A1}\]
Along \(OM\), \(X\) is a fraction \(t\) of the way: \[\overrightarrow{OX} = t\left(\tfrac{1}{2}\mathbf{a} + \mathbf{b}\right) = \tfrac{1}{2}t\,\mathbf{a} + t\,\mathbf{b} \quad \textbf{B1}\] Along \(AB\), start at \(A\) and move a fraction \(s\) of the way towards \(B\). The direction is \[\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a} \quad \textbf{B1}\] so \[\overrightarrow{OX} = \overrightarrow{OA} + s\,\overrightarrow{AB} = \mathbf{a} + s(\mathbf{b} - \mathbf{a}) = (1 - s)\mathbf{a} + s\,\mathbf{b} \quad \textbf{B1}\] Three independent B marks. Both expressions must be written purely in terms of \(\mathbf{a}\), \(\mathbf{b}\) and the single parameter, with the brackets expanded ready for comparison.
The two expressions describe the same point, and \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel, so the coefficients may be equated: \[\text{Coefficients may be compared because } \mathbf{a} \ \text{and} \ \mathbf{b} \ \text{are not parallel.} \quad \textbf{M1}\] \[\mathbf{a}: \ \tfrac{1}{2}t = 1 - s; \qquad \mathbf{b}: \ t = s \quad \textbf{A1}\] Substituting \(t = s\) into the first equation: \[\tfrac{1}{2}s = 1 - s \ \Rightarrow\ \tfrac{3}{2}s = 1 \ \Rightarrow\ s = \tfrac{2}{3}, \qquad t = \tfrac{2}{3} \quad \textbf{A1}\] The M1 rewards the justification as well as the act of comparing; it is the mathematical content of the part. The final A1 needs both values.
Since \(\overrightarrow{AX} = s\,\overrightarrow{AB} = \tfrac{2}{3}\overrightarrow{AB}\), the point \(X\) is two thirds of the way from \(A\) to \(B\), leaving one third: \[AX : XB = 2 : 1 \quad \textbf{B1}\] Read from \(s\) directly as \(s : (1 - s)\). Quoting \(1 : 2\) reverses the segments.
Use either expression with \(s = t = \tfrac{2}{3}\). Taking the second gives \(\overrightarrow{OX} = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\): \[\overrightarrow{OX} = \tfrac{1}{3}(4\mathbf{i} + \mathbf{j}) + \tfrac{2}{3}(-2\mathbf{i} + 5\mathbf{j}) = \left(\tfrac{4}{3} - \tfrac{4}{3}\right)\mathbf{i} + \left(\tfrac{1}{3} + \tfrac{10}{3}\right)\mathbf{j} = \tfrac{11}{3}\mathbf{j} \quad \textbf{M1}\] The \(\mathbf{i}\) components cancel exactly, so \(\overrightarrow{OX}\) points straight up the \(\mathbf{j}\) direction and its modulus is simply the size of that one component: \[\left|\overrightarrow{OX}\right| = \tfrac{11}{3} \quad \textbf{A1}\] Using the first expression is an equally good check: \(\tfrac{2}{3}\left(\tfrac{1}{2}\mathbf{a} + \mathbf{b}\right) = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\), the same vector.
With the given components, \(\overrightarrow{OM} = \tfrac{1}{2}(4\mathbf{i} + \mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) = 0\mathbf{i} + \tfrac{11}{2}\mathbf{j}\), so \(OM\) itself lies along the \(\mathbf{j}\)-axis and \(\tfrac{2}{3}\) of it is \(\tfrac{11}{3}\mathbf{j}\), confirming part (f) independently. The ratio in part (e) is also verifiable numerically: \(A\) is at \((4, 1)\), \(B\) at \((-2, 5)\), and the point two thirds of the way along is \(\left(4 - \tfrac{2}{3}(6),\ 1 + \tfrac{2}{3}(4)\right) = \left(0,\ \tfrac{11}{3}\right)\), exactly the \(X\) found.
Question 7 Report
\(\mathrm{f}(x) = 5 + 8x - 2x^2\)
Here \(\mathrm{f}(x) = 5 + 8x - 2x^2\) has a negative coefficient of \(x^2\), so the graph is a downward parabola with a maximum, not a minimum. The requested form \(A - B(x - C)^2\) is built for exactly that shape: a positive constant \(A\) with a non-negative quantity \(B(x - C)^2\) subtracted from it. Every part of the question then follows from that single completed square, which is why the later parts all say "hence".
First reorder into descending powers and take out the factor \(-2\) from the \(x^2\) and \(x\) terms only:
\[5 + 8x - 2x^2 = -2\left(x^2 - 4x\right) + 5 \qquad \textbf{[M1]}\]The [M1] is for extracting \(-2\) correctly. Dividing \(8x\) by \(-2\) gives \(-4x\), and that sign change is where most marks are lost: \(-2(x^2 + 4x) + 5\) is a different function.
\[x^2 - 4x = (x - 2)^2 - 4 \qquad \textbf{[M1]}\]The second [M1] is for completing the square inside the bracket: half of \(-4\) is \(-2\), and \(2^2 = 4\) is subtracted back.
\[\mathrm{f}(x) = -2\left[(x - 2)^2 - 4\right] + 5 = -2(x - 2)^2 + 8 + 5 = 13 - 2(x - 2)^2 \qquad \textbf{[A1]}\] \[A = 13, \quad B = 2, \quad C = 2 \qquad \textbf{[A1]}\]The first [A1] is for the correct completed square, the second for identifying the three constants against the requested form. Multiplying \(-2\) by \(-4\) gives \(+8\), so the constant becomes \(8 + 5 = 13\); a candidate who writes \(-8 + 5 = -3\) loses both accuracy marks while keeping both method marks. The question says \(A\), \(B\) and \(C\) are positive, which is a built-in check: if any of your three values comes out negative, the sign handling has gone wrong somewhere.
Since \((x - 2)^2 \ge 0\) for all real \(x\), the term \(2(x - 2)^2\) is never negative, so subtracting it can only reduce \(13\). Therefore \(\mathrm{f}(x) \le 13\), with equality exactly when \((x - 2)^2 = 0\).
The maximum value of \(\mathrm{f}(x)\) is \(13\) [B1], occurring at \(x = 2\) [B1]. These are two independent B marks, so one can be earned without the other. No differentiation is required, and none is expected; the completed square already contains the answer.
"Hence" and "exact values in simplest surd form" together rule out the quadratic formula as the intended route and rule out a decimal answer entirely. Use the completed square:
\[13 - 2(x - 2)^2 = 0 \quad\Rightarrow\quad (x - 2)^2 = \frac{13}{2} \qquad \textbf{[M1]}\] \[x - 2 = \pm\sqrt{\frac{13}{2}} = \pm\frac{\sqrt{13}}{\sqrt{2}} = \pm\frac{\sqrt{26}}{2} \qquad \textbf{[M1]}\] \[x = 2 \pm \frac{\sqrt{26}}{2} = \frac{4 + \sqrt{26}}{2} \quad\text{or}\quad \frac{4 - \sqrt{26}}{2} \qquad \textbf{[A1]}\]The first [M1] is for isolating the squared bracket, the second for taking the square root with \(\pm\) attached. That \(\pm\) is the trap this part is built to catch: taking only the positive root gives one solution instead of two and the accuracy mark is lost. The rationalising step \(\sqrt{13/2} = \sqrt{26}/2\) is what "simplest surd form" is asking for; a surd left in a denominator is not simplest form.
This is where the shape of the graph does the work. The coefficient of \(x^2\) is negative, so the parabola opens downwards, is above the axis between its two roots and below the axis outside them [M1]. Hence
\[\frac{4 - \sqrt{26}}{2} \lt x \lt \frac{4 + \sqrt{26}}{2} \qquad \textbf{[A1]}\]The [M1] is for the correct region ("between the roots"), and it is a follow-through mark: a candidate whose roots in (c) were wrong can still earn it by placing their own roots correctly. The [A1] needs the exact roots from (c) in a single correct inequality. The classic error is to give \(x \lt \frac{4 - \sqrt{26}}{2}\) or \(x \gt \frac{4 + \sqrt{26}}{2}\), which is the answer for a positive coefficient of \(x^2\), and would be right for \(\mathrm{f}(x) \lt 0\) here.
Check: \(\sqrt{26} \approx 5.10\), so the roots are about \(-0.55\) and \(4.55\). Test the midpoint \(x = 2\): \(\mathrm{f}(2) = 5 + 16 - 8 = 13 \gt 0\), which sits inside the stated interval, and \(\mathrm{f}(5) = 5 + 40 - 50 = -5\), which is outside it and negative. Both agree with the answer.
Here \(\mathrm{f}(x) = 5 + 8x - 2x^2\) has a negative coefficient of \(x^2\), so the graph is a downward parabola with a maximum, not a minimum. The requested form \(A - B(x - C)^2\) is built for exactly that shape: a positive constant \(A\) with a non-negative quantity \(B(x - C)^2\) subtracted from it. Every part of the question then follows from that single completed square, which is why the later parts all say "hence".
First reorder into descending powers and take out the factor \(-2\) from the \(x^2\) and \(x\) terms only:
\[5 + 8x - 2x^2 = -2\left(x^2 - 4x\right) + 5 \qquad \textbf{[M1]}\]The [M1] is for extracting \(-2\) correctly. Dividing \(8x\) by \(-2\) gives \(-4x\), and that sign change is where most marks are lost: \(-2(x^2 + 4x) + 5\) is a different function.
\[x^2 - 4x = (x - 2)^2 - 4 \qquad \textbf{[M1]}\]The second [M1] is for completing the square inside the bracket: half of \(-4\) is \(-2\), and \(2^2 = 4\) is subtracted back.
\[\mathrm{f}(x) = -2\left[(x - 2)^2 - 4\right] + 5 = -2(x - 2)^2 + 8 + 5 = 13 - 2(x - 2)^2 \qquad \textbf{[A1]}\] \[A = 13, \quad B = 2, \quad C = 2 \qquad \textbf{[A1]}\]The first [A1] is for the correct completed square, the second for identifying the three constants against the requested form. Multiplying \(-2\) by \(-4\) gives \(+8\), so the constant becomes \(8 + 5 = 13\); a candidate who writes \(-8 + 5 = -3\) loses both accuracy marks while keeping both method marks. The question says \(A\), \(B\) and \(C\) are positive, which is a built-in check: if any of your three values comes out negative, the sign handling has gone wrong somewhere.
Since \((x - 2)^2 \ge 0\) for all real \(x\), the term \(2(x - 2)^2\) is never negative, so subtracting it can only reduce \(13\). Therefore \(\mathrm{f}(x) \le 13\), with equality exactly when \((x - 2)^2 = 0\).
The maximum value of \(\mathrm{f}(x)\) is \(13\) [B1], occurring at \(x = 2\) [B1]. These are two independent B marks, so one can be earned without the other. No differentiation is required, and none is expected; the completed square already contains the answer.
"Hence" and "exact values in simplest surd form" together rule out the quadratic formula as the intended route and rule out a decimal answer entirely. Use the completed square:
\[13 - 2(x - 2)^2 = 0 \quad\Rightarrow\quad (x - 2)^2 = \frac{13}{2} \qquad \textbf{[M1]}\] \[x - 2 = \pm\sqrt{\frac{13}{2}} = \pm\frac{\sqrt{13}}{\sqrt{2}} = \pm\frac{\sqrt{26}}{2} \qquad \textbf{[M1]}\] \[x = 2 \pm \frac{\sqrt{26}}{2} = \frac{4 + \sqrt{26}}{2} \quad\text{or}\quad \frac{4 - \sqrt{26}}{2} \qquad \textbf{[A1]}\]The first [M1] is for isolating the squared bracket, the second for taking the square root with \(\pm\) attached. That \(\pm\) is the trap this part is built to catch: taking only the positive root gives one solution instead of two and the accuracy mark is lost. The rationalising step \(\sqrt{13/2} = \sqrt{26}/2\) is what "simplest surd form" is asking for; a surd left in a denominator is not simplest form.
This is where the shape of the graph does the work. The coefficient of \(x^2\) is negative, so the parabola opens downwards, is above the axis between its two roots and below the axis outside them [M1]. Hence
\[\frac{4 - \sqrt{26}}{2} \lt x \lt \frac{4 + \sqrt{26}}{2} \qquad \textbf{[A1]}\]The [M1] is for the correct region ("between the roots"), and it is a follow-through mark: a candidate whose roots in (c) were wrong can still earn it by placing their own roots correctly. The [A1] needs the exact roots from (c) in a single correct inequality. The classic error is to give \(x \lt \frac{4 - \sqrt{26}}{2}\) or \(x \gt \frac{4 + \sqrt{26}}{2}\), which is the answer for a positive coefficient of \(x^2\), and would be right for \(\mathrm{f}(x) \lt 0\) here.
Check: \(\sqrt{26} \approx 5.10\), so the roots are about \(-0.55\) and \(4.55\). Test the midpoint \(x = 2\): \(\mathrm{f}(2) = 5 + 16 - 8 = 13 \gt 0\), which sits inside the stated interval, and \(\mathrm{f}(5) = 5 + 40 - 50 = -5\), which is outside it and negative. Both agree with the answer.
Question 8 Report
The set of values of \(x\) for which \[x^2 + bx + c \lt 0\] is \(-2 \lt x \lt 7\), where \(b\) and \(c\) are constants.
Nine marks built on one idea: a quadratic inequality and the roots of the quadratic are two views of the same object. If \(x^2 + bx + c \lt 0\) exactly on \(-2 \lt x \lt 7\), then the endpoints of that interval must be the roots, because a positive quadratic is negative precisely between its roots. That single observation makes part (a) immediate.
The roots are \(-2\) and \(7\), and the coefficient of \(x^2\) is 1, so the quadratic factorises as \[x^2 + bx + c = (x + 2)(x - 7) \quad \textbf{M1}\] Expanding, \[= x^2 - 7x + 2x - 14 = x^2 - 5x - 14 \quad \textbf{A1}\] \[b = -5, \qquad c = -14 \quad \textbf{A1}\] The M1 is for using the interval endpoints as roots. An equally valid route uses sum and product of roots: \(-b = (-2) + 7 = 5\) so \(b = -5\), and \(c = (-2)(7) = -14\). Either way the final A1 needs both constants, with the signs right. The commonest slip is writing \((x - 2)(x + 7)\), which corresponds to the interval \(-7 \lt x \lt 2\) instead.
Substitute the constants and collect everything on one side. The \(-14\) appears on both sides and cancels, which is the point of the numbers chosen: \[x^2 - 5x - 14 \ge 4x - 14 \ \Rightarrow\ x^2 - 9x \ge 0 \quad \textbf{M1}\] \[x(x - 9) \ge 0 \quad \textbf{A1}\] The critical values are \(0\) and \(9\), and since the parabola opens upwards it is at or above the axis outside its roots: \[x \le 0 \quad \text{or} \quad x \ge 9 \quad \textbf{A1}\] Note that \(x = 0\) is a genuine solution here, so the inequality is inclusive at both ends. Do not divide through by \(x\); that would lose the root \(x = 0\) and is invalid because \(x\) may be negative or zero.
Use the factorised numerator from part (a). A rational expression changes sign at a zero of the numerator and at a zero of the denominator, so all three matter: \[\frac{(x + 2)(x - 7)}{x + 1} \le 0, \qquad \text{critical values } x = -2, \ -1, \ 7 \quad \textbf{M1}\] Build a sign table over the four regions. To the left of \(-2\) the two numerator factors are negative and the denominator is negative, giving an overall negative value; each subsequent critical value flips the sign: \[\lt 0 \ \text{for} \ x \lt -2; \quad \gt 0 \ \text{for} \ -2 \lt x \lt -1; \quad \lt 0 \ \text{for} \ -1 \lt x \lt 7; \quad \gt 0 \ \text{for} \ x \gt 7 \quad \textbf{A1}\] The two numerator roots may be included, because the fraction is then exactly zero; the denominator root must be excluded, because the expression is undefined there: \[x \le -2 \quad \text{or} \quad -1 \lt x \le 7 \quad \textbf{A1}\] Notice the asymmetry in the final answer: an inclusive bound at \(-2\) and at \(7\), but a strict one at \(-1\). Getting that right is exactly what the last A1 is testing.
Part (a) is self-checking: \((-2)^2 - 5(-2) - 14 = 4 + 10 - 14 = 0\) and \(7^2 - 35 - 14 = 0\), so both endpoints are roots as required. For part (c), test one value per region: at \(x = -3\) the value is \(\dfrac{(-1)(-10)}{-2} = -5 \le 0\), so that region is in; at \(x = -1.5\) it is \(\dfrac{(0.5)(-8.5)}{-0.5} = 8.5\), so that region is out; at \(x = 0\) it is \(-14 \le 0\), in; at \(x = 8\) it is \(\dfrac{(10)(1)}{9} \gt 0\), out. That matches the answer exactly.
Nine marks built on one idea: a quadratic inequality and the roots of the quadratic are two views of the same object. If \(x^2 + bx + c \lt 0\) exactly on \(-2 \lt x \lt 7\), then the endpoints of that interval must be the roots, because a positive quadratic is negative precisely between its roots. That single observation makes part (a) immediate.
The roots are \(-2\) and \(7\), and the coefficient of \(x^2\) is 1, so the quadratic factorises as \[x^2 + bx + c = (x + 2)(x - 7) \quad \textbf{M1}\] Expanding, \[= x^2 - 7x + 2x - 14 = x^2 - 5x - 14 \quad \textbf{A1}\] \[b = -5, \qquad c = -14 \quad \textbf{A1}\] The M1 is for using the interval endpoints as roots. An equally valid route uses sum and product of roots: \(-b = (-2) + 7 = 5\) so \(b = -5\), and \(c = (-2)(7) = -14\). Either way the final A1 needs both constants, with the signs right. The commonest slip is writing \((x - 2)(x + 7)\), which corresponds to the interval \(-7 \lt x \lt 2\) instead.
Substitute the constants and collect everything on one side. The \(-14\) appears on both sides and cancels, which is the point of the numbers chosen: \[x^2 - 5x - 14 \ge 4x - 14 \ \Rightarrow\ x^2 - 9x \ge 0 \quad \textbf{M1}\] \[x(x - 9) \ge 0 \quad \textbf{A1}\] The critical values are \(0\) and \(9\), and since the parabola opens upwards it is at or above the axis outside its roots: \[x \le 0 \quad \text{or} \quad x \ge 9 \quad \textbf{A1}\] Note that \(x = 0\) is a genuine solution here, so the inequality is inclusive at both ends. Do not divide through by \(x\); that would lose the root \(x = 0\) and is invalid because \(x\) may be negative or zero.
Use the factorised numerator from part (a). A rational expression changes sign at a zero of the numerator and at a zero of the denominator, so all three matter: \[\frac{(x + 2)(x - 7)}{x + 1} \le 0, \qquad \text{critical values } x = -2, \ -1, \ 7 \quad \textbf{M1}\] Build a sign table over the four regions. To the left of \(-2\) the two numerator factors are negative and the denominator is negative, giving an overall negative value; each subsequent critical value flips the sign: \[\lt 0 \ \text{for} \ x \lt -2; \quad \gt 0 \ \text{for} \ -2 \lt x \lt -1; \quad \lt 0 \ \text{for} \ -1 \lt x \lt 7; \quad \gt 0 \ \text{for} \ x \gt 7 \quad \textbf{A1}\] The two numerator roots may be included, because the fraction is then exactly zero; the denominator root must be excluded, because the expression is undefined there: \[x \le -2 \quad \text{or} \quad -1 \lt x \le 7 \quad \textbf{A1}\] Notice the asymmetry in the final answer: an inclusive bound at \(-2\) and at \(7\), but a strict one at \(-1\). Getting that right is exactly what the last A1 is testing.
Part (a) is self-checking: \((-2)^2 - 5(-2) - 14 = 4 + 10 - 14 = 0\) and \(7^2 - 35 - 14 = 0\), so both endpoints are roots as required. For part (c), test one value per region: at \(x = -3\) the value is \(\dfrac{(-1)(-10)}{-2} = -5 \le 0\), so that region is in; at \(x = -1.5\) it is \(\dfrac{(0.5)(-8.5)}{-0.5} = 8.5\), so that region is out; at \(x = 0\) it is \(-14 \le 0\), in; at \(x = 8\) it is \(\dfrac{(10)(1)}{9} \gt 0\), out. That matches the answer exactly.
Question 9 Report
Two quadratic inequalities, 8 marks in total. The method for both is the same: get everything on one side, factorise, find the critical values, and then decide the sign of the expression in each region. A sketch of the parabola or a sign table is what makes the last step reliable. The second part carries the extra trap of a variable denominator.
Factorise first: \[2x^2 - 5x - 3 = (2x + 1)(x - 3) \quad \textbf{M1}\] The M1 is for an attempt to factorise or to solve the quadratic; the formula or completing the square would earn it equally. The critical values are where the expression is zero: \[x = -\tfrac{1}{2} \quad \text{and} \quad x = 3 \quad \textbf{A1}\] Now the sign. The coefficient of \(x^2\) is positive, so the parabola opens upwards and lies above the axis outside its two roots and below between them. Since we want the expression to be positive, \[x \lt -\tfrac{1}{2} \quad \text{or} \quad x \gt 3 \quad \textbf{A1}\] The final A1 needs both regions and the word "or". Writing \(-\tfrac{1}{2} \gt x \gt 3\) is meaningless, because no number is simultaneously less than \(-\tfrac{1}{2}\) and greater than 3.
The one thing you must not do is multiply both sides by \((x - 2)\), because that expression changes sign at \(x = 2\) and multiplying an inequality by a negative quantity reverses it. The safe manoeuvre is to multiply by \((x - 2)^2\), which is positive for every \(x \ne 2\) and so leaves the inequality sign alone. Record the excluded value first: \(x \ne 2\), since the original expression is undefined there.
\[(x + 4)(x - 2) \le 3(x - 2)^2 \quad \textbf{M1}\] \[x^2 + 2x - 8 \le 3x^2 - 12x + 12 \quad \textbf{A1}\] \[0 \le 2x^2 - 14x + 20 \ \Rightarrow\ x^2 - 7x + 10 \ge 0 \quad \textbf{A1}\] \[(x - 2)(x - 5) \ge 0 \ \Rightarrow\ x \le 2 \ \text{or} \ x \ge 5 \quad \textbf{M1}\]Finally impose the exclusion. The value \(x = 2\) satisfies \((x-2)(x-5) \ge 0\) but is not in the domain of the original inequality, so it must be removed: \[x \lt 2 \quad \text{or} \quad x \ge 5 \quad \textbf{A1}\] The second M1 is for solving the new quadratic inequality correctly in structure; the last A1 is reserved for the fully correct final set, including the strict inequality at \(x = 2\) and the inclusive one at \(x = 5\). The A marks are not follow-through, so an expansion slip earlier costs them.
An alternative route that also earns full marks is to subtract 3 and combine into a single fraction: \[\frac{x+4}{x-2} - 3 = \frac{x + 4 - 3(x-2)}{x-2} = \frac{10 - 2x}{x - 2} \le 0,\] then use a sign table on the critical values \(x = 2\) and \(x = 5\). Both routes must end with \(x = 2\) excluded.
Test one value from each region. In part (a), \(x = 0\) gives \(-3\), which is not positive, correctly excluding the middle region; \(x = 4\) gives \(9 \gt 0\). In part (b), \(x = 0\) gives \(\dfrac{4}{-2} = -2 \le 3\), so \(x \lt 2\) belongs; \(x = 3\) gives \(\dfrac{7}{1} = 7\), which is not at most 3, so the middle region is correctly excluded; \(x = 5\) gives exactly \(\dfrac{9}{3} = 3\), so the boundary is correctly inclusive. Three substitutions settle a whole inequality.
Two quadratic inequalities, 8 marks in total. The method for both is the same: get everything on one side, factorise, find the critical values, and then decide the sign of the expression in each region. A sketch of the parabola or a sign table is what makes the last step reliable. The second part carries the extra trap of a variable denominator.
Factorise first: \[2x^2 - 5x - 3 = (2x + 1)(x - 3) \quad \textbf{M1}\] The M1 is for an attempt to factorise or to solve the quadratic; the formula or completing the square would earn it equally. The critical values are where the expression is zero: \[x = -\tfrac{1}{2} \quad \text{and} \quad x = 3 \quad \textbf{A1}\] Now the sign. The coefficient of \(x^2\) is positive, so the parabola opens upwards and lies above the axis outside its two roots and below between them. Since we want the expression to be positive, \[x \lt -\tfrac{1}{2} \quad \text{or} \quad x \gt 3 \quad \textbf{A1}\] The final A1 needs both regions and the word "or". Writing \(-\tfrac{1}{2} \gt x \gt 3\) is meaningless, because no number is simultaneously less than \(-\tfrac{1}{2}\) and greater than 3.
The one thing you must not do is multiply both sides by \((x - 2)\), because that expression changes sign at \(x = 2\) and multiplying an inequality by a negative quantity reverses it. The safe manoeuvre is to multiply by \((x - 2)^2\), which is positive for every \(x \ne 2\) and so leaves the inequality sign alone. Record the excluded value first: \(x \ne 2\), since the original expression is undefined there.
\[(x + 4)(x - 2) \le 3(x - 2)^2 \quad \textbf{M1}\] \[x^2 + 2x - 8 \le 3x^2 - 12x + 12 \quad \textbf{A1}\] \[0 \le 2x^2 - 14x + 20 \ \Rightarrow\ x^2 - 7x + 10 \ge 0 \quad \textbf{A1}\] \[(x - 2)(x - 5) \ge 0 \ \Rightarrow\ x \le 2 \ \text{or} \ x \ge 5 \quad \textbf{M1}\]Finally impose the exclusion. The value \(x = 2\) satisfies \((x-2)(x-5) \ge 0\) but is not in the domain of the original inequality, so it must be removed: \[x \lt 2 \quad \text{or} \quad x \ge 5 \quad \textbf{A1}\] The second M1 is for solving the new quadratic inequality correctly in structure; the last A1 is reserved for the fully correct final set, including the strict inequality at \(x = 2\) and the inclusive one at \(x = 5\). The A marks are not follow-through, so an expansion slip earlier costs them.
An alternative route that also earns full marks is to subtract 3 and combine into a single fraction: \[\frac{x+4}{x-2} - 3 = \frac{x + 4 - 3(x-2)}{x-2} = \frac{10 - 2x}{x - 2} \le 0,\] then use a sign table on the critical values \(x = 2\) and \(x = 5\). Both routes must end with \(x = 2\) excluded.
Test one value from each region. In part (a), \(x = 0\) gives \(-3\), which is not positive, correctly excluding the middle region; \(x = 4\) gives \(9 \gt 0\). In part (b), \(x = 0\) gives \(\dfrac{4}{-2} = -2 \le 3\), so \(x \lt 2\) belongs; \(x = 3\) gives \(\dfrac{7}{1} = 7\), which is not at most 3, so the middle region is correctly excluded; \(x = 5\) gives exactly \(\dfrac{9}{3} = 3\), so the boundary is correctly inclusive. Three substitutions settle a whole inequality.
Question 10 Report
The figure shows the curve \(C\) with equation \(y = 4x - x^2\). The curve meets the \(x\)-axis at the origin \(A\) and at the point \(B\). The shaded region \(R\) is bounded by the curve \(C\) and the \(x\)-axis.
Seven marks combining curve sketching with integration. The figure shows the curve \(C\) with equation \(y = 4x - x^2\) meeting the \(x\)-axis at the origin \(A\) and at \(B\), with the region \(R\) shown shaded between the curve and the \(x\)-axis. The three parts use three different techniques on the same curve: factorising to find intercepts, differentiating to find the maximum, and integrating to find an area. Recognising which tool answers which question is most of the skill here.
A curve meets the \(x\)-axis where \(y = 0\), so set the expression to zero and factorise rather than expand or use the formula: \[4x - x^2 = 0 \ \Rightarrow\ x(4 - x) = 0 \quad \textbf{M1}\] \[x = 0 \ \text{(the point } A) \quad \text{or} \quad x = 4, \ \text{so } B \ \text{is } (4,\, 0) \quad \textbf{A1}\] The result is printed, so the marks are for the working. The M1 is for setting \(y = 0\) and factorising; the A1 needs the root \(x = 4\) identified as \(B\) rather than merely listed. Do not divide by \(x\) to "simplify": that loses the root \(x = 0\), which the figure identifies as \(A\).
At a maximum the gradient is zero, so differentiate and solve: \[\frac{\mathrm{d}y}{\mathrm{d}x} = 4 - 2x = 0 \ \Rightarrow\ x = 2 \quad \textbf{M1}\] \[y = 4(2) - 2^2 = 8 - 4 = 4, \ \text{so the maximum point is } (2,\, 4) \quad \textbf{A1}\] Both coordinates are needed. Because the coefficient of \(x^2\) is negative, the parabola opens downwards and this stationary point is a maximum, which agrees with the figure and means no second-derivative test is required here. Symmetry offers a check with no calculus at all: the vertex sits midway between the roots \(0\) and \(4\), so \(x = 2\).
The area between a curve and the \(x\)-axis is the definite integral of \(y\) between the two points where the curve meets that axis, and those limits are exactly the intercepts found in part (a): \[\text{Area} = \int_{0}^{4} \bigl(4x - x^2\bigr)\,\mathrm{d}x \quad \textbf{M1}\] Integrate by raising each index by one and dividing by the new index: \[= \left[2x^2 - \frac{x^3}{3}\right]_{0}^{4} \quad \textbf{A1}\] Evaluate at the upper limit and subtract the value at the lower limit, which is zero here: \[= \left(2(16) - \frac{64}{3}\right) - 0 = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3} \quad \textbf{A1}\] So the area of \(R\) is \(\dfrac{32}{3}\) square units. "Exact" means the fraction must be kept: \(10.7\) or \(10.67\) loses the final accuracy mark. The M1 is for a definite integral with the correct limits, the first A1 for correct integration, and the second for the evaluated exact value. Since this is a definite integral, no constant of integration is needed; it would cancel between the two limits.
An area can always be sanity-checked against a rectangle that contains it. The region \(R\) sits inside the rectangle of width 4 and height 4, of area 16, and it is a rounded arch filling rather more than half of that box, so a value of \(\dfrac{32}{3} = 10.67\) is the right order of magnitude. In fact for any parabolic arch the area is exactly two thirds of the enclosing rectangle, and \(\tfrac{2}{3} \times 16 = \dfrac{32}{3}\), which confirms the answer exactly. A further check on part (b): the maximum value 4 is the height of that rectangle, consistent with the vertex touching its top edge.
Seven marks combining curve sketching with integration. The figure shows the curve \(C\) with equation \(y = 4x - x^2\) meeting the \(x\)-axis at the origin \(A\) and at \(B\), with the region \(R\) shown shaded between the curve and the \(x\)-axis. The three parts use three different techniques on the same curve: factorising to find intercepts, differentiating to find the maximum, and integrating to find an area. Recognising which tool answers which question is most of the skill here.
A curve meets the \(x\)-axis where \(y = 0\), so set the expression to zero and factorise rather than expand or use the formula: \[4x - x^2 = 0 \ \Rightarrow\ x(4 - x) = 0 \quad \textbf{M1}\] \[x = 0 \ \text{(the point } A) \quad \text{or} \quad x = 4, \ \text{so } B \ \text{is } (4,\, 0) \quad \textbf{A1}\] The result is printed, so the marks are for the working. The M1 is for setting \(y = 0\) and factorising; the A1 needs the root \(x = 4\) identified as \(B\) rather than merely listed. Do not divide by \(x\) to "simplify": that loses the root \(x = 0\), which the figure identifies as \(A\).
At a maximum the gradient is zero, so differentiate and solve: \[\frac{\mathrm{d}y}{\mathrm{d}x} = 4 - 2x = 0 \ \Rightarrow\ x = 2 \quad \textbf{M1}\] \[y = 4(2) - 2^2 = 8 - 4 = 4, \ \text{so the maximum point is } (2,\, 4) \quad \textbf{A1}\] Both coordinates are needed. Because the coefficient of \(x^2\) is negative, the parabola opens downwards and this stationary point is a maximum, which agrees with the figure and means no second-derivative test is required here. Symmetry offers a check with no calculus at all: the vertex sits midway between the roots \(0\) and \(4\), so \(x = 2\).
The area between a curve and the \(x\)-axis is the definite integral of \(y\) between the two points where the curve meets that axis, and those limits are exactly the intercepts found in part (a): \[\text{Area} = \int_{0}^{4} \bigl(4x - x^2\bigr)\,\mathrm{d}x \quad \textbf{M1}\] Integrate by raising each index by one and dividing by the new index: \[= \left[2x^2 - \frac{x^3}{3}\right]_{0}^{4} \quad \textbf{A1}\] Evaluate at the upper limit and subtract the value at the lower limit, which is zero here: \[= \left(2(16) - \frac{64}{3}\right) - 0 = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3} \quad \textbf{A1}\] So the area of \(R\) is \(\dfrac{32}{3}\) square units. "Exact" means the fraction must be kept: \(10.7\) or \(10.67\) loses the final accuracy mark. The M1 is for a definite integral with the correct limits, the first A1 for correct integration, and the second for the evaluated exact value. Since this is a definite integral, no constant of integration is needed; it would cancel between the two limits.
An area can always be sanity-checked against a rectangle that contains it. The region \(R\) sits inside the rectangle of width 4 and height 4, of area 16, and it is a rounded arch filling rather more than half of that box, so a value of \(\dfrac{32}{3} = 10.67\) is the right order of magnitude. In fact for any parabolic arch the area is exactly two thirds of the enclosing rectangle, and \(\tfrac{2}{3} \times 16 = \dfrac{32}{3}\), which confirms the answer exactly. A further check on part (b): the maximum value 4 is the height of that rectangle, consistent with the vertex touching its top edge.
Would you like to proceed with this action?