WAEC SSCE - Further Mathematics - 2016 (Essay)
Frage 1 Bericht
When \(f(x) = 2x^{3} + mx^{2} + nx + 11\) is divided by \(x^{2} + 5x + 1\), the quotient is \(2x - 5\) and the remainder is \(30x + 16\). Find the values of m and n.
By the division identity, dividend = divisor \(\times\) quotient + remainder:
\[ f(x) = (x^2 + 5x + 1)(2x - 5) + (30x + 16). \]
Expand the product:
\[ (x^2+5x+1)(2x-5) = 2x^3 - 5x^2 + 10x^2 - 25x + 2x - 5 = 2x^3 + 5x^2 - 23x - 5. \]
Add the remainder:
\[ f(x) = 2x^3 + 5x^2 - 23x - 5 + 30x + 16 = 2x^3 + 5x^2 + 7x + 11. \]
Compare with \( f(x) = 2x^3 + mx^2 + nx + 11 \):
\[ m = 5, \qquad n = 7. \]
Antwortdetails
By the division identity, dividend = divisor \(\times\) quotient + remainder:
\[ f(x) = (x^2 + 5x + 1)(2x - 5) + (30x + 16). \]
Expand the product:
\[ (x^2+5x+1)(2x-5) = 2x^3 - 5x^2 + 10x^2 - 25x + 2x - 5 = 2x^3 + 5x^2 - 23x - 5. \]
Add the remainder:
\[ f(x) = 2x^3 + 5x^2 - 23x - 5 + 30x + 16 = 2x^3 + 5x^2 + 7x + 11. \]
Compare with \( f(x) = 2x^3 + mx^2 + nx + 11 \):
\[ m = 5, \qquad n = 7. \]
Frage 2 Bericht
(a) Two Mathematics books, 5 different Physics books and 3 different Chemistry are to be arranged on a shelf. How many arrangements are possible if ;
(i) books on the same subject must stand together? (ii) only the Physics books must stand together?
(b) In a certain community, 13 out of every 20 persons speak English. If 8 persons are selected at random from the community, find, correct to three significant figures, the probability that at least 3 of them speak English.
(a)(i) If books of the same subject must stand together, treat each subject as one block: 3 blocks arranged in \(3!\) ways, with the books inside each block also arranged.
\[3! \times 2! \times 5! \times 3! = 6 \times 2 \times 120 \times 6 = 8640\](ii) If only the Physics books must stand together, treat the 5 Physics books as one block. The items to arrange are: 2 Maths + 1 Physics block + 3 Chemistry \(= 6\) items.
\[6! \times 5! = 720 \times 120 = 86400\](b) \(p = \tfrac{13}{20} = 0.65,\ q = 0.35,\ n = 8\). Find \(P(\text{at least 3})\) via the complement:
\[P(0) = (0.35)^8 = 0.000225\] \[P(1) = \binom{8}{1}(0.65)(0.35)^7 = 0.003346\] \[P(2) = \binom{8}{2}(0.65)^2(0.35)^6 = 0.021747\] \[P(\ge 3) = 1 - (0.000225 + 0.003346 + 0.021747) = 1 - 0.025318 = 0.975\]The probability that at least 3 speak English is \(0.975\) (3 s.f.).
Antwortdetails
(a)(i) If books of the same subject must stand together, treat each subject as one block: 3 blocks arranged in \(3!\) ways, with the books inside each block also arranged.
\[3! \times 2! \times 5! \times 3! = 6 \times 2 \times 120 \times 6 = 8640\](ii) If only the Physics books must stand together, treat the 5 Physics books as one block. The items to arrange are: 2 Maths + 1 Physics block + 3 Chemistry \(= 6\) items.
\[6! \times 5! = 720 \times 120 = 86400\](b) \(p = \tfrac{13}{20} = 0.65,\ q = 0.35,\ n = 8\). Find \(P(\text{at least 3})\) via the complement:
\[P(0) = (0.35)^8 = 0.000225\] \[P(1) = \binom{8}{1}(0.65)(0.35)^7 = 0.003346\] \[P(2) = \binom{8}{2}(0.65)^2(0.35)^6 = 0.021747\] \[P(\ge 3) = 1 - (0.000225 + 0.003346 + 0.021747) = 1 - 0.025318 = 0.975\]The probability that at least 3 speak English is \(0.975\) (3 s.f.).
Frage 3 Bericht
(a) If \(^{k}P_{2} = 72\), find the value of k.
(b) Solve the equation : \(2\cos^{2} \theta - 5\cos \theta = 3; 0° \leq \theta \leq 360°\)
(a) \(^{k}P_2 = k(k - 1) = 72\):
\[k^2 - k - 72 = 0 \ \Rightarrow\ (k - 9)(k + 8) = 0\]Since \(k > 0\), \(k = 9\).
(b) \(2\cos^2\theta - 5\cos\theta = 3\). Let \(c = \cos\theta\):
\[2c^2 - 5c - 3 = 0 \ \Rightarrow\ (2c + 1)(c - 3) = 0\] \[c = -\tfrac{1}{2} \quad\text{or}\quad c = 3\ (\text{rejected, since } |\cos\theta| \le 1)\]So \(\cos\theta = -\tfrac{1}{2}\). In \(0^\circ \le \theta \le 360^\circ\):
\[\theta = 120^\circ \quad\text{and}\quad \theta = 240^\circ\]Antwortdetails
(a) \(^{k}P_2 = k(k - 1) = 72\):
\[k^2 - k - 72 = 0 \ \Rightarrow\ (k - 9)(k + 8) = 0\]Since \(k > 0\), \(k = 9\).
(b) \(2\cos^2\theta - 5\cos\theta = 3\). Let \(c = \cos\theta\):
\[2c^2 - 5c - 3 = 0 \ \Rightarrow\ (2c + 1)(c - 3) = 0\] \[c = -\tfrac{1}{2} \quad\text{or}\quad c = 3\ (\text{rejected, since } |\cos\theta| \le 1)\]So \(\cos\theta = -\tfrac{1}{2}\). In \(0^\circ \le \theta \le 360^\circ\):
\[\theta = 120^\circ \quad\text{and}\quad \theta = 240^\circ\]Frage 4 Bericht
The probabilities that Ago, Sulley and Musa will gain admission to a certain university are \(\frac{4}{5}, \frac{3}{4}\) and \(\frac{2}{3}\) respectively. Find the probability that :
(a) none of them will gain admission ;
(b) only Ago and Sulley will gain admission.
Probabilities of gaining admission: \( P(A)=\tfrac45,\ P(S)=\tfrac34,\ P(M)=\tfrac23 \). The events are independent.
Failure probabilities: \( P(A')=\tfrac15,\ P(S')=\tfrac14,\ P(M')=\tfrac13 \).
(a) None gains admission: \[ P(A')\,P(S')\,P(M') = \frac15 \cdot \frac14 \cdot \frac13 = \frac{1}{60}. \]
(b) Only Ago and Sulley gain admission (Musa fails): \[ P(A)\,P(S)\,P(M') = \frac45 \cdot \frac34 \cdot \frac13 = \frac{12}{60} = \frac{1}{5}. \]
Antwortdetails
Probabilities of gaining admission: \( P(A)=\tfrac45,\ P(S)=\tfrac34,\ P(M)=\tfrac23 \). The events are independent.
Failure probabilities: \( P(A')=\tfrac15,\ P(S')=\tfrac14,\ P(M')=\tfrac13 \).
(a) None gains admission: \[ P(A')\,P(S')\,P(M') = \frac15 \cdot \frac14 \cdot \frac13 = \frac{1}{60}. \]
(b) Only Ago and Sulley gain admission (Musa fails): \[ P(A)\,P(S)\,P(M') = \frac45 \cdot \frac34 \cdot \frac13 = \frac{12}{60} = \frac{1}{5}. \]
Frage 5 Bericht
(a) A body of mass 3kg moves with a velocity of 8ms\(^{-1}\). It collides with a second body moving in the same direction with a velocity of 5ms\(^{-1}\). After collision, the bodies move together with a velocity of 6ms\(^{-1}\). Find the mass of the second body.
(b) If the second body in (a) moves with a velocity of 5ms\(^{-1}\) in the opposite direction as that of the 3kg body with a velocity of 8ms\(^{-1}\), find, correct to two decimal places, the common velocity of the two bodies if they move together after collision.
Momentum is conserved in each collision.
(a) Let the second mass be \(m\). Both bodies move in the same direction, then coalesce at \(6\text{ ms}^{-1}\):
\[ 3(8) + m(5) = (3 + m)(6). \]
\[ 24 + 5m = 18 + 6m \Rightarrow m = 6\text{ kg}. \]
(b) Now the \(6\text{ kg}\) body moves at \(5\text{ ms}^{-1}\) opposite to the \(3\text{ kg}\) body. Take the direction of the \(3\text{ kg}\) body as positive:
\[ 3(8) + 6(-5) = (3 + 6)v. \]
\[ 24 - 30 = 9v \Rightarrow v = \frac{-6}{9} = -0.67\text{ ms}^{-1}\ (\text{2 d.p.}). \]
The common velocity is \(0.67\text{ ms}^{-1}\) in the direction of the \(6\text{ kg}\) body's original motion.
Antwortdetails
Momentum is conserved in each collision.
(a) Let the second mass be \(m\). Both bodies move in the same direction, then coalesce at \(6\text{ ms}^{-1}\):
\[ 3(8) + m(5) = (3 + m)(6). \]
\[ 24 + 5m = 18 + 6m \Rightarrow m = 6\text{ kg}. \]
(b) Now the \(6\text{ kg}\) body moves at \(5\text{ ms}^{-1}\) opposite to the \(3\text{ kg}\) body. Take the direction of the \(3\text{ kg}\) body as positive:
\[ 3(8) + 6(-5) = (3 + 6)v. \]
\[ 24 - 30 = 9v \Rightarrow v = \frac{-6}{9} = -0.67\text{ ms}^{-1}\ (\text{2 d.p.}). \]
The common velocity is \(0.67\text{ ms}^{-1}\) in the direction of the \(6\text{ kg}\) body's original motion.
Frage 6 Bericht
(a) If \(f(x + 2) = 6x^{2} + 5x - 8\), find \(f(5)\).
(b) Express \(\frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}}\) in the form \(p + q\sqrt{r}\), where p, q and r are rational numbers.
(a) To find \(f(5)\) from \( f(x+2) = 6x^2 + 5x - 8 \), set \( x + 2 = 5 \), i.e. \( x = 3 \):
\[ f(5) = 6(3)^2 + 5(3) - 8 = 54 + 15 - 8 = 61. \]
(b) Rationalise by multiplying numerator and denominator by \( 4\sqrt2 + 2\sqrt3 \):
Denominator: \( (4\sqrt2 - 2\sqrt3)(4\sqrt2 + 2\sqrt3) = (4\sqrt2)^2 - (2\sqrt3)^2 = 32 - 12 = 20. \)
Numerator: \( (7\sqrt2 + 3\sqrt3)(4\sqrt2 + 2\sqrt3) = 56 + 14\sqrt6 + 12\sqrt6 + 18 = 74 + 26\sqrt6. \)
\[ \frac{74 + 26\sqrt6}{20} = \frac{37}{10} + \frac{13}{10}\sqrt6. \]
So \( p = \dfrac{37}{10},\ q = \dfrac{13}{10},\ r = 6 \).
Antwortdetails
(a) To find \(f(5)\) from \( f(x+2) = 6x^2 + 5x - 8 \), set \( x + 2 = 5 \), i.e. \( x = 3 \):
\[ f(5) = 6(3)^2 + 5(3) - 8 = 54 + 15 - 8 = 61. \]
(b) Rationalise by multiplying numerator and denominator by \( 4\sqrt2 + 2\sqrt3 \):
Denominator: \( (4\sqrt2 - 2\sqrt3)(4\sqrt2 + 2\sqrt3) = (4\sqrt2)^2 - (2\sqrt3)^2 = 32 - 12 = 20. \)
Numerator: \( (7\sqrt2 + 3\sqrt3)(4\sqrt2 + 2\sqrt3) = 56 + 14\sqrt6 + 12\sqrt6 + 18 = 74 + 26\sqrt6. \)
\[ \frac{74 + 26\sqrt6}{20} = \frac{37}{10} + \frac{13}{10}\sqrt6. \]
So \( p = \dfrac{37}{10},\ q = \dfrac{13}{10},\ r = 6 \).
Frage 7 Bericht
Given that \(p = \begin{pmatrix} 5 \\ 3 \end{pmatrix}, q = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\) and \(r = \begin{pmatrix} 17 \\ 5 \end{pmatrix}\) and \(r = \alpha r + \beta q\), where \(\alpha\) and \(\beta\) are scalars, express q in terms of r and p.
Express \(r\) as a linear combination \(r = \alpha p + \beta q\):
\[\begin{pmatrix} 17 \\ 5 \end{pmatrix} = \alpha\begin{pmatrix} 5 \\ 3 \end{pmatrix} + \beta\begin{pmatrix} -1 \\ 2 \end{pmatrix}\]This gives the equations:
\[5\alpha - \beta = 17,\qquad 3\alpha + 2\beta = 5\]From the first, \(\beta = 5\alpha - 17\). Substitute:
\[3\alpha + 2(5\alpha - 17) = 5 \ \Rightarrow\ 13\alpha = 39 \ \Rightarrow\ \alpha = 3,\quad \beta = -2\]Hence \(r = 3p - 2q\). Make \(q\) the subject:
\[2q = 3p - r \ \Rightarrow\ q = \tfrac{1}{2}(3p - r)\]Check: \(\tfrac{1}{2}\left(3\begin{pmatrix}5\\3\end{pmatrix} - \begin{pmatrix}17\\5\end{pmatrix}\right) = \tfrac{1}{2}\begin{pmatrix}-2\\4\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix} = q\).
Antwortdetails
Express \(r\) as a linear combination \(r = \alpha p + \beta q\):
\[\begin{pmatrix} 17 \\ 5 \end{pmatrix} = \alpha\begin{pmatrix} 5 \\ 3 \end{pmatrix} + \beta\begin{pmatrix} -1 \\ 2 \end{pmatrix}\]This gives the equations:
\[5\alpha - \beta = 17,\qquad 3\alpha + 2\beta = 5\]From the first, \(\beta = 5\alpha - 17\). Substitute:
\[3\alpha + 2(5\alpha - 17) = 5 \ \Rightarrow\ 13\alpha = 39 \ \Rightarrow\ \alpha = 3,\quad \beta = -2\]Hence \(r = 3p - 2q\). Make \(q\) the subject:
\[2q = 3p - r \ \Rightarrow\ q = \tfrac{1}{2}(3p - r)\]Check: \(\tfrac{1}{2}\left(3\begin{pmatrix}5\\3\end{pmatrix} - \begin{pmatrix}17\\5\end{pmatrix}\right) = \tfrac{1}{2}\begin{pmatrix}-2\\4\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix} = q\).
Frage 8 Bericht
Four vectors \(r = \alpha i + \beta j\), where \(\alpha \text{ and } \beta\) are constants, \(s = 2i -j, m = 3i + 2j\) and \(n = i + j\) are such that the magnitude of r is three times as s and is parallel to the vactor (m - n).
(a) Find the values of \(\alpha\) and \(\beta\).
(b) Calculate the magnitude and direction of (r - s).
(a) Since \(r\) is parallel to \(m - n\):
\[m - n = (3i + 2j) - (i + j) = 2i + j\]So \(r = \lambda(2i + j)\), giving \(\alpha = 2\lambda,\ \beta = \lambda\).
The magnitude of \(r\) is three times that of \(s = 2i - j\), where \(|s| = \sqrt{4 + 1} = \sqrt{5}\):
\[|r| = \sqrt{(2\lambda)^2 + \lambda^2} = |\lambda|\sqrt{5} = 3\sqrt{5} \ \Rightarrow\ |\lambda| = 3\]Taking \(\lambda = 3\): \(\alpha = 6,\ \beta = 3\) (the other solution \(\lambda = -3\) gives \(\alpha = -6,\ \beta = -3\)).
(b) Using \(r = 6i + 3j\):
\[r - s = (6 - 2)i + (3 - (-1))j = 4i + 4j\] \[|r - s| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \approx 5.7\]Direction: \(\tan\theta = \dfrac{4}{4} = 1\), so \(\theta = 45^\circ\) above the positive \(x\)-axis (bearing \(045^\circ\)).
Antwortdetails
(a) Since \(r\) is parallel to \(m - n\):
\[m - n = (3i + 2j) - (i + j) = 2i + j\]So \(r = \lambda(2i + j)\), giving \(\alpha = 2\lambda,\ \beta = \lambda\).
The magnitude of \(r\) is three times that of \(s = 2i - j\), where \(|s| = \sqrt{4 + 1} = \sqrt{5}\):
\[|r| = \sqrt{(2\lambda)^2 + \lambda^2} = |\lambda|\sqrt{5} = 3\sqrt{5} \ \Rightarrow\ |\lambda| = 3\]Taking \(\lambda = 3\): \(\alpha = 6,\ \beta = 3\) (the other solution \(\lambda = -3\) gives \(\alpha = -6,\ \beta = -3\)).
(b) Using \(r = 6i + 3j\):
\[r - s = (6 - 2)i + (3 - (-1))j = 4i + 4j\] \[|r - s| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \approx 5.7\]Direction: \(\tan\theta = \dfrac{4}{4} = 1\), so \(\theta = 45^\circ\) above the positive \(x\)-axis (bearing \(045^\circ\)).
Frage 9 Bericht
(a) Without using mathematical tables or calculator, evaluate \(\frac{\frac{3}{2}\log 27 - 3\log 5\sqrt{5}}{\log 0.6}\)
(b) Two linear transformations A and B in the \(O_{xy}\) plane, are defined by :
\(A : (x, y) (x + 2y, -x + y)\)
\(B : (x, y) (2x + 3y, x + 2y)\).
(i) Write down the matrices A and B; (ii) Find the image of the point P(-2, 2) under the linear transformation A followed by B.
(a) Work on the numerator:
\[\tfrac{3}{2}\log 27 = \tfrac{3}{2}\log 3^3 = \tfrac{9}{2}\log 3,\qquad 3\log 5\sqrt{5} = 3\log 5^{3/2} = \tfrac{9}{2}\log 5\] \[\text{Numerator} = \tfrac{9}{2}\log 3 - \tfrac{9}{2}\log 5 = \tfrac{9}{2}\log\tfrac{3}{5} = \tfrac{9}{2}\log 0.6\]Dividing by \(\log 0.6\):
\[\frac{\tfrac{9}{2}\log 0.6}{\log 0.6} = \frac{9}{2} = 4.5\](b)(i) The matrices are:
\[A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix},\qquad B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\](ii) Apply \(A\) to \(P(-2, 2)\):
\[A(P) = (\,-2 + 2(2),\ -(-2) + 2\,) = (2,\ 4)\]Then apply \(B\) to \((2, 4)\):
\[B(2,4) = (2(2) + 3(4),\ 2 + 2(4)) = (16,\ 10)\]The image of \(P\) under \(A\) followed by \(B\) is \((16, 10)\).
Antwortdetails
(a) Work on the numerator:
\[\tfrac{3}{2}\log 27 = \tfrac{3}{2}\log 3^3 = \tfrac{9}{2}\log 3,\qquad 3\log 5\sqrt{5} = 3\log 5^{3/2} = \tfrac{9}{2}\log 5\] \[\text{Numerator} = \tfrac{9}{2}\log 3 - \tfrac{9}{2}\log 5 = \tfrac{9}{2}\log\tfrac{3}{5} = \tfrac{9}{2}\log 0.6\]Dividing by \(\log 0.6\):
\[\frac{\tfrac{9}{2}\log 0.6}{\log 0.6} = \frac{9}{2} = 4.5\](b)(i) The matrices are:
\[A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix},\qquad B = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}\](ii) Apply \(A\) to \(P(-2, 2)\):
\[A(P) = (\,-2 + 2(2),\ -(-2) + 2\,) = (2,\ 4)\]Then apply \(B\) to \((2, 4)\):
\[B(2,4) = (2(2) + 3(4),\ 2 + 2(4)) = (16,\ 10)\]The image of \(P\) under \(A\) followed by \(B\) is \((16, 10)\).
Frage 10 Bericht
A particle is projected vertically upwards from the ground with speed \(30ms^{-1}\). Calculate the :
(a) maximum height reached by the particle;
(b) time taken by the particle to return to the ground;
(c) time(s) taken for the particle to attain a height of 40m above the ground. [Take \(g = 10ms^{-2}\)]
Given \(u = 30\,\text{ms}^{-1},\ g = 10\,\text{ms}^{-2}\), upward taken positive.
(a) At maximum height \(v = 0\). Use \(v^2 = u^2 - 2gh\):
\[0 = 30^2 - 2(10)h \ \Rightarrow\ 20h = 900 \ \Rightarrow\ h = 45\,\text{m}\](b) Time to return to ground:
\[t = \frac{2u}{g} = \frac{2(30)}{10} = 6\,\text{s}\](c) Height \(40\,\text{m}\): use \(h = ut - \tfrac{1}{2}gt^2\):
\[40 = 30t - 5t^2 \ \Rightarrow\ 5t^2 - 30t + 40 = 0 \ \Rightarrow\ t^2 - 6t + 8 = 0\] \[(t - 2)(t - 4) = 0 \ \Rightarrow\ t = 2\,\text{s (going up)},\ t = 4\,\text{s (coming down)}\]Antwortdetails
Given \(u = 30\,\text{ms}^{-1},\ g = 10\,\text{ms}^{-2}\), upward taken positive.
(a) At maximum height \(v = 0\). Use \(v^2 = u^2 - 2gh\):
\[0 = 30^2 - 2(10)h \ \Rightarrow\ 20h = 900 \ \Rightarrow\ h = 45\,\text{m}\](b) Time to return to ground:
\[t = \frac{2u}{g} = \frac{2(30)}{10} = 6\,\text{s}\](c) Height \(40\,\text{m}\): use \(h = ut - \tfrac{1}{2}gt^2\):
\[40 = 30t - 5t^2 \ \Rightarrow\ 5t^2 - 30t + 40 = 0 \ \Rightarrow\ t^2 - 6t + 8 = 0\] \[(t - 2)(t - 4) = 0 \ \Rightarrow\ t = 2\,\text{s (going up)},\ t = 4\,\text{s (coming down)}\]Frage 11 Bericht
A binary operation \(*\) is defined on the set, R, of real numbers by \(m * n = m + n + 2\). Find the :
(a) identity element under the operation ;
(b) inverse of n under the operation .
The operation is \( m * n = m + n + 2 \).
(a) Identity element. Let \(e\) be the identity, so \( m * e = m \) for all \(m\):
\[ m + e + 2 = m \Rightarrow e = -2. \] The identity element is \( e = -2 \).
(b) Inverse of n. Let \(n^{-1}\) be the inverse, so \( n * n^{-1} = e = -2 \):
\[ n + n^{-1} + 2 = -2 \Rightarrow n^{-1} = -4 - n. \] The inverse of \(n\) is \( n^{-1} = -(n+4) \).
Antwortdetails
The operation is \( m * n = m + n + 2 \).
(a) Identity element. Let \(e\) be the identity, so \( m * e = m \) for all \(m\):
\[ m + e + 2 = m \Rightarrow e = -2. \] The identity element is \( e = -2 \).
(b) Inverse of n. Let \(n^{-1}\) be the inverse, so \( n * n^{-1} = e = -2 \):
\[ n + n^{-1} + 2 = -2 \Rightarrow n^{-1} = -4 - n. \] The inverse of \(n\) is \( n^{-1} = -(n+4) \).
Frage 12 Bericht
(a)(i) Write down the expansion of \((1 + x)^{7}\) in ascending powers of x.
(ii) If the coefficients of the fifth, sixth and seventh terms in the expansion in (a)(i) above form a linear sequence(A.P), find the common difference of the A.P.
(b) Using the trapezium rule with ordinates at 1, 2, 3, 4 and 5, calculate, correct to two decimal places,
\(\int_{1}^{5} \sqrt{(2x + 8x^{2})} \mathrm {d} x\).
(a)(i) By the binomial theorem:
\[(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7\](ii) The 5th, 6th and 7th coefficients are \(35,\ 21,\ 7\). They form an A.P. with common difference:
\[d = 21 - 35 = -14 \quad(\text{and } 7 - 21 = -14)\](b) Let \(f(x) = \sqrt{2x + 8x^2}\), with \(h = 1\). Compute the ordinates:
- \(f(1) = \sqrt{10} = 3.1623\)
- \(f(2) = \sqrt{36} = 6.0000\)
- \(f(3) = \sqrt{78} = 8.8318\)
- \(f(4) = \sqrt{136} = 11.6619\)
- \(f(5) = \sqrt{210} = 14.4914\)
Trapezium rule:
\[\int_1^5 f\,dx \approx \tfrac{h}{2}\big[(f_0 + f_4) + 2(f_1 + f_2 + f_3)\big]\] \[= \tfrac{1}{2}\big[(3.1623 + 14.4914) + 2(6 + 8.8318 + 11.6619)\big]\] \[= \tfrac{1}{2}\big[17.6537 + 52.9874\big] = \tfrac{1}{2}(70.6411) = 35.32\]So the integral is approximately \(35.32\) (to 2 decimal places).
Antwortdetails
(a)(i) By the binomial theorem:
\[(1 + x)^7 = 1 + 7x + 21x^2 + 35x^3 + 35x^4 + 21x^5 + 7x^6 + x^7\](ii) The 5th, 6th and 7th coefficients are \(35,\ 21,\ 7\). They form an A.P. with common difference:
\[d = 21 - 35 = -14 \quad(\text{and } 7 - 21 = -14)\](b) Let \(f(x) = \sqrt{2x + 8x^2}\), with \(h = 1\). Compute the ordinates:
- \(f(1) = \sqrt{10} = 3.1623\)
- \(f(2) = \sqrt{36} = 6.0000\)
- \(f(3) = \sqrt{78} = 8.8318\)
- \(f(4) = \sqrt{136} = 11.6619\)
- \(f(5) = \sqrt{210} = 14.4914\)
Trapezium rule:
\[\int_1^5 f\,dx \approx \tfrac{h}{2}\big[(f_0 + f_4) + 2(f_1 + f_2 + f_3)\big]\] \[= \tfrac{1}{2}\big[(3.1623 + 14.4914) + 2(6 + 8.8318 + 11.6619)\big]\] \[= \tfrac{1}{2}\big[17.6537 + 52.9874\big] = \tfrac{1}{2}(70.6411) = 35.32\]So the integral is approximately \(35.32\) (to 2 decimal places).
Frage 13 Bericht
The table shows the distribution of marks scored by some students in a test.
| Marks | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
| No. of students | 3 | 17 | 41 | 85 | 97 | 115 | 101 | 64 | 21 | 6 |
(a)(i) Construct a cumulative frequency table for the distribution ; (ii) Draw a cumulative frequency curve for the distribution.
(b) Use the curve to estimate the :
(i) number of students who scored marks between 32 and 74 ; (ii) pass mark, if 18% of the students failed ; (iii) lowest mark for distinction, if 8% of the students passed with distinction.
(a)(i) Cumulative frequency table
The cumulative frequency is obtained by adding each class frequency to the running total. The curve is plotted against the upper class boundary of each class, so these are tabulated as well. The total number of students is \(N = 550\).
| Marks | Class boundaries | Frequency \(f\) | Cumulative frequency \(\sum f\) |
|---|---|---|---|
| 1 - 10 | 0.5 - 10.5 | 3 | 3 |
| 11 - 20 | 10.5 - 20.5 | 17 | 20 |
| 21 - 30 | 20.5 - 30.5 | 41 | 61 |
| 31 - 40 | 30.5 - 40.5 | 85 | 146 |
| 41 - 50 | 40.5 - 50.5 | 97 | 243 |
| 51 - 60 | 50.5 - 60.5 | 115 | 358 |
| 61 - 70 | 60.5 - 70.5 | 101 | 459 |
| 71 - 80 | 70.5 - 80.5 | 64 | 523 |
| 81 - 90 | 80.5 - 90.5 | 21 | 544 |
| 91 - 100 | 90.5 - 100.5 | 6 | 550 |
(a)(ii) Cumulative frequency curve (ogive)
Plot the cumulative frequency (vertical axis) against the upper class boundary (horizontal axis), starting the curve from \((0.5,\,0)\), and join the points with a smooth S-shaped curve.
(b) Estimates read from the curve
(i) Number of students scoring between 32 and 74. Read the cumulative frequency at each mark from the ogive (dashed guide lines):
\[\text{CF}(32)\approx 61+\frac{32-30.5}{10}\times 85 \approx 74\]\[\text{CF}(74)\approx 459+\frac{74-70.5}{10}\times 64 \approx 481\]Number scoring between 32 and 74:
\[481-74 = 407\ \text{students (approximately)}.\](ii) Pass mark, if 18% failed. The number who failed is
\[18\%\ \text{of}\ 550 = \frac{18}{100}\times 550 = 99.\]The pass mark is the mark below which 99 students lie, i.e. the mark corresponding to \(\text{CF}=99\). Reading horizontally from \(99\) on the vertical axis to the curve and down to the mark axis:
\[\text{pass mark} \approx 30.5+\frac{99-61}{85}\times 10 \approx 35.\](iii) Lowest mark for distinction, if 8% passed with distinction. The distinction group is the top 8%, so the number below the distinction mark is
\[550-\Big(8\%\ \text{of}\ 550\Big)=550-44=506.\]Read the mark corresponding to \(\text{CF}=506\):
\[\text{lowest distinction mark} \approx 70.5+\frac{506-459}{64}\times 10 \approx 78.\]Antwortdetails
(a)(i) Cumulative frequency table
The cumulative frequency is obtained by adding each class frequency to the running total. The curve is plotted against the upper class boundary of each class, so these are tabulated as well. The total number of students is \(N = 550\).
| Marks | Class boundaries | Frequency \(f\) | Cumulative frequency \(\sum f\) |
|---|---|---|---|
| 1 - 10 | 0.5 - 10.5 | 3 | 3 |
| 11 - 20 | 10.5 - 20.5 | 17 | 20 |
| 21 - 30 | 20.5 - 30.5 | 41 | 61 |
| 31 - 40 | 30.5 - 40.5 | 85 | 146 |
| 41 - 50 | 40.5 - 50.5 | 97 | 243 |
| 51 - 60 | 50.5 - 60.5 | 115 | 358 |
| 61 - 70 | 60.5 - 70.5 | 101 | 459 |
| 71 - 80 | 70.5 - 80.5 | 64 | 523 |
| 81 - 90 | 80.5 - 90.5 | 21 | 544 |
| 91 - 100 | 90.5 - 100.5 | 6 | 550 |
(a)(ii) Cumulative frequency curve (ogive)
Plot the cumulative frequency (vertical axis) against the upper class boundary (horizontal axis), starting the curve from \((0.5,\,0)\), and join the points with a smooth S-shaped curve.
(b) Estimates read from the curve
(i) Number of students scoring between 32 and 74. Read the cumulative frequency at each mark from the ogive (dashed guide lines):
\[\text{CF}(32)\approx 61+\frac{32-30.5}{10}\times 85 \approx 74\]\[\text{CF}(74)\approx 459+\frac{74-70.5}{10}\times 64 \approx 481\]Number scoring between 32 and 74:
\[481-74 = 407\ \text{students (approximately)}.\](ii) Pass mark, if 18% failed. The number who failed is
\[18\%\ \text{of}\ 550 = \frac{18}{100}\times 550 = 99.\]The pass mark is the mark below which 99 students lie, i.e. the mark corresponding to \(\text{CF}=99\). Reading horizontally from \(99\) on the vertical axis to the curve and down to the mark axis:
\[\text{pass mark} \approx 30.5+\frac{99-61}{85}\times 10 \approx 35.\](iii) Lowest mark for distinction, if 8% passed with distinction. The distinction group is the top 8%, so the number below the distinction mark is
\[550-\Big(8\%\ \text{of}\ 550\Big)=550-44=506.\]Read the mark corresponding to \(\text{CF}=506\):
\[\text{lowest distinction mark} \approx 70.5+\frac{506-459}{64}\times 10 \approx 78.\]Frage 14 Bericht
Given that (5, 2), (-4, k) and (2, 1) lie on a straight line, find the value of k.
Three points are collinear when the slope between any two pairs is the same.
Using \((5,2)\) and \((2,1)\): \[ m = \frac{2-1}{5-2} = \frac{1}{3}. \]
Using \((2,1)\) and \((-4,k)\) and equating the slope: \[ \frac{k-1}{-4-2} = \frac{1}{3} \Rightarrow \frac{k-1}{-6} = \frac{1}{3}. \]
\[ k - 1 = -2 \Rightarrow k = -1. \]
Antwortdetails
Three points are collinear when the slope between any two pairs is the same.
Using \((5,2)\) and \((2,1)\): \[ m = \frac{2-1}{5-2} = \frac{1}{3}. \]
Using \((2,1)\) and \((-4,k)\) and equating the slope: \[ \frac{k-1}{-4-2} = \frac{1}{3} \Rightarrow \frac{k-1}{-6} = \frac{1}{3}. \]
\[ k - 1 = -2 \Rightarrow k = -1. \]
Frage 15 Bericht
The mean of the numbers 1, 4, k, (k + 4) and 11 is (k + 1). Calculate the :
(a) value of k ;
(b) standard deviation.
(a) Value of k. The mean of \(1, 4, k, (k+4), 11\) is \((k+1)\):
\[ \frac{1 + 4 + k + (k+4) + 11}{5} = k + 1 \Rightarrow \frac{20 + 2k}{5} = k + 1. \]
\[ 20 + 2k = 5k + 5 \Rightarrow 15 = 3k \Rightarrow k = 5. \]
(b) Standard deviation. The numbers become \(1, 4, 5, 9, 11\) with mean \(k+1 = 6\).
Deviations from the mean: \(-5, -2, -1, 3, 5\); their squares: \(25, 4, 1, 9, 25\), summing to \(64\).
\[ \text{Variance} = \frac{64}{5} = 12.8, \qquad \text{S.D.} = \sqrt{12.8} \approx 3.58. \]
Antwortdetails
(a) Value of k. The mean of \(1, 4, k, (k+4), 11\) is \((k+1)\):
\[ \frac{1 + 4 + k + (k+4) + 11}{5} = k + 1 \Rightarrow \frac{20 + 2k}{5} = k + 1. \]
\[ 20 + 2k = 5k + 5 \Rightarrow 15 = 3k \Rightarrow k = 5. \]
(b) Standard deviation. The numbers become \(1, 4, 5, 9, 11\) with mean \(k+1 = 6\).
Deviations from the mean: \(-5, -2, -1, 3, 5\); their squares: \(25, 4, 1, 9, 25\), summing to \(64\).
\[ \text{Variance} = \frac{64}{5} = 12.8, \qquad \text{S.D.} = \sqrt{12.8} \approx 3.58. \]
