WAEC SSCE - Further Mathematics - 2014

Detalhes da Resposta

Since only 3 places are available, the number of ways to choose 3 people out of 9 to fill these seats is given by the combination formula: \[^{9}C_{3} = \frac{9!}{3!(9-3)!} = 84.\] After the selection, the 3 people can be arranged in the available seats in 3! ways, which is 6. Therefore, the total number of ways 9 people can be seated on a bench if only 3 places are available is 84 x 6 = 504. So, the answer is (B) 504.

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To find the length of BC, we need to first find the coordinates of point B and point C. Point B is given as AB = (4, 3). To find point C, we can use the fact that AC + CB = AB. Rearranging, we get CB = AB - AC. Substituting the values we get, CB = \(\begin{pmatrix} 4 \\ 3 \end{pmatrix}\) - \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\) = \(\begin{pmatrix} 2 \\ 6 \end{pmatrix}\) The length of BC is then given by the formula: |BC| = √((x2 - x1)² + (y2 - y1)²), where (x1, y1) and (x2, y2) are the coordinates of points B and C respectively. Substituting the values we get, |BC| = √((2 - 0)² + (6 - 0)²) = √(4 + 36) = √40 = 2√10 Therefore, the length of BC is 2√10, which is option C.

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Expanding the given expression using the identity \((a+b)(a-b)=a^2-b^2\), we get: $$(1-\sin\theta)(1+\sin\theta)=1^2-\sin^2\theta=\cos^2\theta$$ Therefore, the simplified expression is \(\cos^{2} \theta\).

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To find the median class, we first need to calculate the cumulative frequency of each class.

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We can use the equation: distance = initial velocity × time + 0.5 × acceleration × time² We are given the distance as 250m, the time as 6 seconds, and the acceleration as 12\(ms^{-2}\). We need to find the initial velocity. Substituting the values we get, 250 = initial velocity × 6 + 0.5 × 12 × 6² Simplifying, 250 = 6initial velocity + 216 250 - 216 = 6initial velocity 34 = 6initial velocity initial velocity = 34/6 = 5.67 \(ms^{-1}\) (rounded to two decimal places) Therefore, the initial velocity of the particle is 5.67 \(ms^{-1}\), which is option A.

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The nth term of a linear sequence can be expressed as follows: nth term = a + (n-1)d where a is the first term, d is the common difference and n is the number of terms. In this question, the first term is 9 and the common difference is 7, so we have: nth term = 9 + (n-1)7 We are told that the nth term is 380, so we can substitute this into the equation and solve for n: 380 = 9 + (n-1)7 380 - 9 = (n-1)7 371 = (n-1)7 n-1 = 371/7 n-1 = 53 n = 54 Therefore, the value of n is 54.

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To evaluate the limit, we can substitute the value of x into the expression and see what we get. However, in this case, x = 3 would make the denominator 0, and division by 0 is undefined. This suggests that we need to simplify the expression first. We can factor the numerator and denominator, then simplify the expression: \[\lim\limits_{x \to 3} \frac{x^{3} + x^{2} - 12x}{x^{2} - 9} = \lim\limits_{x \to 3} \frac{x(x-3)(x+4)}{(x+3)(x-3)} = \lim\limits_{x \to 3} \frac{x(x+4)}{x+3}\] Now, we can substitute x = 3 into the simplified expression: \[\frac{3(3+4)}{3+3} = \frac{21}{6} = \frac{7}{2}\] Therefore, the answer is \(\frac{7}{2}\).

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To express (14N, 240°) as a column vector, we need to convert the given magnitude and direction into horizontal and vertical components. We can start by drawing a diagram with a vector starting at the origin and pointing in the direction of 240°. We can then use trigonometry to find the horizontal and vertical components. The horizontal component (x-coordinate) is given by the magnitude (14N) multiplied by the cosine of the angle (240°): $$\text{Horizontal component} = 14\cos(240°) = 14\left(-\frac{1}{2}\right) = -7$$ The vertical component (y-coordinate) is given by the magnitude multiplied by the sine of the angle: $$\text{Vertical component} = 14\sin(240°) = 14\left(-\frac{\sqrt{3}}{2}\right) = -7\sqrt{3}$$ Therefore, the column vector representing (14N, 240°) is: $$\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}$$ So the answer is option A: \(\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}\).

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Detalhes da Resposta

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To solve the given equation, we can simplify the left-hand side of the equation as follows: \[\frac{1}{5^{-y}} = 5^y\] Similarly, we can simplify the right-hand side of the equation as follows: \[25(5^{4-2y}) = 25 \times 5^4 \times 5^{-2y} = 25 \times 625 \times 5^{-2y} = 15625 \times 5^{-2y}\] Now we can rewrite the given equation as: \[5^y = 15625 \times 5^{-2y}\] Simplifying further, we get: \[5^{3y} = 15625\] Taking the cube root of both sides, we get: \[5^y = 25\] Thus, y = 2. Therefore, the value of y is 2. Answer: (b) 2.

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The function \(f(x)\) is given as \(f(x) = 3x^3 + 8x^2 + 6x + k\), and we are given that \(f(2) = 1\). To find the value of \(k\), we can substitute \(x = 2\) and \(f(2) = 1\) into the function and solve for \(k\). \begin{align*} f(2) &= 3(2)^3 + 8(2)^2 + 6(2) + k \\ 1 &= 24 + 32 + 12 + k \\ 1 &= 68 + k \\ k &= 1 - 68 \\ k &= -67 \end{align*} Therefore, the value of \(k\) is -67.

Detalhes da Resposta

The surface area, \(A\), of a sphere is given by \(A = 4\pi r^2\), where \(r\) is the radius. We are given that \(\frac{dr}{dt} = 3\text{ cm/s}\) and \(r = 2\text{ cm}\). We want to find \(\frac{dA}{dt}\) at this point. Taking the derivative of the surface area formula with respect to time, we get: \begin{align*} \frac{dA}{dt} &= \frac{d}{dt}(4\pi r^2) \\ &= 8\pi r \frac{dr}{dt} \\ \end{align*} Substituting the given values, we get: \begin{align*} \frac{dA}{dt} &= 8\pi (2\text{ cm})(3\text{ cm/s}) \\ &= 48\pi\text{ cm}^2/\text{s} \\ \end{align*} Therefore, the rate of increase in the surface area when the radius is 2cm is \(48\pi\text{ cm}^2/\text{s}\). So, the answer is (d) \(48\pi cm^{2}s^{-1}\).

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To simplify the given expression, we need to eliminate the square root from the denominator. We can do this by rationalizing the denominator. To do this, we multiply both the numerator and the denominator of the fraction by the conjugate of the denominator. The conjugate of \(3 - \sqrt{2}\) is \(3 + \sqrt{2}\). \begin{align*} \frac{1 + \sqrt{8}}{3 - \sqrt{2}} &= \frac{1 + \sqrt{8}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}}\\ &= \frac{(1 + \sqrt{8})(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})}\\ &= \frac{3 + \sqrt{2} + 4\sqrt{2} + 2\sqrt{2}}{7}\\ &= \frac{5\sqrt{2} + 3 + \sqrt{2}}{7}\\ &= \frac{8\sqrt{2} + 3}{7} \end{align*} Therefore, the simplified form of the expression is \(\frac{8\sqrt{2} + 3}{7}\), which corresponds to option (D).

Detalhes da Resposta

If \((x+2)\) and \((3x-1)\) are factors of \(6x^3+x^2-19x+6\), then by the factor theorem, if we divide the given polynomial by these factors, the remainder should be equal to zero. Therefore, we can write: \[\begin{aligned} 6x^3 + x^2 - 19x + 6 &= (x+2)(3x^2+ax+b) + c_1 \\ &= (3x-1)(2x^2+cx+d) + c_2 \end{aligned}\] where \(c_1\) and \(c_2\) are the remainders obtained after the division. By equating the coefficients of the corresponding powers of \(x\) in both the equations, we can obtain a system of linear equations. Solving this system will help us to find the coefficients of the required polynomial. After solving the system, we get the third factor as \(\boxed{2x-3}\). Therefore, is the correct answer.

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The equation of a circle with center (h,k) and radius r is given by the formula: $$(x-h)^2 + (y-k)^2 = r^2$$ In this case, the center of the circle is (2,-3) and the radius is 2 units. Therefore, we have: $$(x-2)^2 + (y+3)^2 = 2^2$$ Expanding the left-hand side, we get: $$x^2 - 4x + 4 + y^2 + 6y + 9 = 4$$ Simplifying this equation, we get: $$x^2 + y^2 - 4x + 6y + 9 = 0$$ Therefore, the equation of the circle is (A) \(x^{2} + y^{2} - 4x + 6y + 9 = 0\).

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To find the distance between two points in a plane, we can use the distance formula: $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$ where (x1, y1) and (x2, y2) are the coordinates of the two points. In this case, the given points are (2, 5) and (5, 9), so we can substitute the values into the formula: $$d = \sqrt{(5-2)^2 + (9-5)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$$ Therefore, the distance between the two points is 5 units. So, the answer is option (B).

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The mean of a set of numbers is the sum of the numbers divided by the total number of numbers. If we have a set of numbers, the deviations from the mean are obtained by subtracting the mean from each number. In this case, we are given the deviations from the mean of a set of numbers, and we need to find the value of k. Let's start by using the fact that the sum of the deviations from the mean is equal to zero. That is: \[(k+3)^2 + (k+7) - 2 + k + (k+2)^2 = 0\] Expanding the squares, we get: \[k^2 + 6k + 9 + k + 7 - 2 + k + k^2 + 4k + 4 = 0\] Simplifying, we obtain: \[2k^2 + 12k + 18 = 0\] Dividing both sides by 2, we get: \[k^2 + 6k + 9 = 0\] This is a quadratic equation that can be factored as: \[(k + 3)^2 = 0\] Therefore, the only solution is: \[k = -3\] So the answer is -3.

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To solve this problem, we can use the formula: \begin{align*} h &= \frac{v_{0}^{2}}{2g} \end{align*} where h is the maximum height reached, \(v_{0}\) is the initial velocity and g is the acceleration due to gravity. Substituting the given values, we get: \begin{align*} h &= \frac{(15\ ms^{-1})^{2}}{2(10\ ms^{-2})}\\ &= \frac{225\ m^{2}s^{-2}}{20\ m s^{-2}}\\ &= 11.25\ m \end{align*} Therefore, the maximum height reached by the ball is 11.25m. Hence, the correct option is (c) 11.25m.

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To find the upper quartile, we need to first calculate the median, which is the value separating the higher half from the lower half of the data. To do that, we can use the formula: Median = ((n + 1) / 2)th term where n is the total number of data points. In this case, n = 6 + 8 + 14 + 10 + 12 = 50. Median = ((50 + 1) / 2)th term Median = 25.5th term Since we have an even number of data points, the median is the average of the two middle terms, which are the 25th and 26th terms when the data is arranged in order. To find these terms, we can use the cumulative frequency distribution: 10-14: 6 15-19: 6 + 8 = 14 20-24: 14 + 14 = 28 25-29: 28 + 10 = 38 30-34: 38 + 12 = 50 The 25th and 26th terms are in the group 20-24, so the median is: Median = (23 + 24) / 2 = 23.5 Next, we need to find the upper quartile, which is the value separating the upper 25% of the data from the lower 75%. We can use the formula: Upper quartile = ((3 * n) + 1) / 4)th term Upper quartile = ((3 * 50) + 1) / 4)th term Upper quartile = 38th term Looking at the cumulative frequency distribution, we see that the 38th term is in the group 25-29. Therefore, the upper quartile is in the group 25-29. The correct option is (C).

Detalhes da Resposta

Detalhes da Resposta

When we toss a fair coin, there are two possible outcomes: heads or tails. Since the coin is fair, the probability of getting a head on any given toss is 1/2, and the probability of getting a tail is also 1/2. To find the probability of obtaining at least one head in four tosses, we can use the complement rule: the probability of an event happening is equal to one minus the probability of the event not happening. So, the probability of getting at least one head is equal to 1 minus the probability of getting no heads at all. The probability of getting no heads in one toss is 1/2, and since each toss is independent, the probability of getting no heads in four tosses is (1/2) x (1/2) x (1/2) x (1/2) = 1/16. Therefore, the probability of getting at least one head in four tosses is 1 - 1/16 = 15/16. So, the answer is (D) \(\frac{15}{16}\).

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To integrate \((x-\frac{1}{x})^2\) with respect to x, we need to expand the square first: $$(x-\frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2}$$ Now we can integrate term by term: $$\int (x^2 - 2 + \frac{1}{x^2}) dx = \frac{x^3}{3} - 2x - \frac{1}{x} + c$$ where c is the constant of integration. Therefore, the answer is the option (D).

Detalhes da Resposta

Let's first calculate the total number of balls in the box. The box contains 5 red balls and k blue balls, so the total number of balls is 5 + k. The probability of selecting a blue ball can be calculated as follows: P(blue ball) = (number of blue balls) / (total number of balls) We are told that this probability is \(\frac{2}{3}\), so we can set up the following equation: \(\frac{k}{5+k} = \frac{2}{3}\) Cross-multiplying and simplifying, we get: 3k = 10 + 2k k = 10 Therefore, there are 10 blue balls in the box. Answer: 10.

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To find the inverse of a function, you need to switch the roles of x and y, then solve for y. The resulting expression will be the inverse function, denoted as \(h^{-1}\). Starting with the original function: \(h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}\) Switching the roles of x and y: \(x = 2 - \frac{1}{2y - 3}\) Solving for y: \(x - 2 = - \frac{1}{2y - 3}\) \(-2y + 3 = -\frac{1}{x - 2}\) \(-2y = -\frac{1}{x - 2} - 3\) \(y = \frac{1}{2} \cdot \frac{1}{2 - x} + \frac{3}{2}\) \(y = \frac{3x - 7}{2x - 4}, x \neq 2\) Therefore, the inverse function is \(h^{-1} : x \to \frac{3x - 7}{2x - 4}, x \neq 2\). The correct option is (B).

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Detalhes da Resposta

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To find the variance of a set of data, we need to follow these steps: 1. Find the mean of the data set. 2. Subtract the mean from each data point, and then square each difference. 3. Find the sum of the squared differences. 4. Divide the sum by the number of data points. To find the variance of 1, 2, 0, -3, 5, -2, 4, we first need to find the mean: mean = (1 + 2 + 0 - 3 + 5 - 2 + 4) / 7 = 7 / 7 = 1 Next, we subtract the mean from each data point and square the differences: (1 - 1)^2 = 0 (2 - 1)^2 = 1 (0 - 1)^2 = 1 (-3 - 1)^2 = 16 (5 - 1)^2 = 16 (-2 - 1)^2 = 9 (4 - 1)^2 = 9 Then we find the sum of the squared differences: 0 + 1 + 1 + 16 + 16 + 9 + 9 = 52 Finally, we divide the sum by the number of data points: 52 / 7 = 7.43 (rounded to two decimal places) Therefore, the variance of the data set is approximately 7.43. So the correct option is (A) \(\frac{52}{7}\).

Detalhes da Resposta

Since the quadratic equation has equal roots, the discriminant is equal to zero. Therefore, we have: $$(P+1)^2 - 4P^2 = 0$$ Simplifying the equation above, we get: $$P^2 + 2P + 1 - 4P^2 = 0$$ $$-3P^2 + 2P + 1 = 0$$ We can now solve for P using the quadratic formula: $$P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=-3$, $b=2$, and $c=1$. Substituting these values, we get: $$P = \frac{-2 \pm \sqrt{2^2 - 4(-3)(1)}}{2(-3)}$$ Simplifying, we have: $$P = \frac{-2 \pm \sqrt{16}}{-6}$$ Therefore, the solutions are: $$P_1 = \frac{-2 + 4}{-6} = \frac{-1}{3}$$ $$P_2 = \frac{-2 - 4}{-6} = 1$$ Hence, the values of P are $\frac{-1}{3}$ and 1. Therefore, the correct option is $\text{: 1 and }\frac{-1}{3}$.

Detalhes da Resposta

Detalhes da Resposta

Detalhes da Resposta

To differentiate the function \(\sqrt[3]{(3x^3 + 1)}\), we use the chain rule. Let \(u = 3x^3 + 1\), then we have: \[\frac{d}{dx} \sqrt[3]{(3x^3 + 1)} = \frac{d}{du}u^{1/3}\cdot\frac{d}{dx}(3x^3 + 1)\] Simplifying, we get: \[\frac{d}{dx} \sqrt[3]{(3x^3 + 1)} = \frac{1}{3}(3x^3 + 1)^{-2/3}\cdot(9x^2)\] Simplifying further, we get: \[\frac{d}{dx} \sqrt[3]{(3x^3 + 1)} = \frac{3x^2}{\sqrt[3]{(3x^3 + 1)^2}}\] Therefore, the correct option is (b) \(\frac{3x^2}{\sqrt[3]{(3x^3 + 1)^2}}\).

Detalhes da Resposta

Differentiate \(xy^{2}+x^{2}y=4xy\) implicitly with respect to \(x\), treating \(y\) as a function of \(x\) and using the product rule on each term.

\[\underbrace{y^{2}+2xy\frac{dy}{dx}}_{\frac{d}{dx}(xy^{2})}+\underbrace{2xy+x^{2}\frac{dy}{dx}}_{\frac{d}{dx}(x^{2}y)}=\underbrace{4y+4x\frac{dy}{dx}}_{\frac{d}{dx}(4xy)}.\]

Collect the \(\dfrac{dy}{dx}\) terms on one side:

\[(2xy+x^{2}-4x)\frac{dy}{dx}=4y-y^{2}-2xy,\]

\[\frac{dy}{dx}=\frac{4y-y^{2}-2xy}{2xy+x^{2}-4x}.\]

Substitute the point \((1,3)\):

\[\text{Numerator}=4(3)-3^{2}-2(1)(3)=12-9-6=-3,\]

\[\text{Denominator}=2(1)(3)+1^{2}-4(1)=6+1-4=3.\]

\[\frac{dy}{dx}=\frac{-3}{3}=-1.\]

The gradient at \((1,3)\) is \(-1\).

Detalhes da Resposta

(a) With \(x=3i-j\) and \(y=2i+kj\):

\[x\cdot y=(3)(2)+(-1)(k)=6-k,\quad |x|=\sqrt{3^2+(-1)^2}=\sqrt{10},\quad |y|=\sqrt{4+k^2}.\]

Given \(\cos\theta=\dfrac{\sqrt5}{5}=\dfrac{1}{\sqrt5}\):

\[\frac{6-k}{\sqrt{10}\,\sqrt{4+k^2}}=\frac{1}{\sqrt5}.\]

Cross-multiply and square:

\[\sqrt5\,(6-k)=\sqrt{10}\,\sqrt{4+k^2}\;\Rightarrow\;5(6-k)^2=10(4+k^2).\]

\[(6-k)^2=2(4+k^2)\;\Rightarrow\;36-12k+k^2=8+2k^2.\]

\[k^2+12k-28=0\;\Rightarrow\;(k-2)(k+14)=0.\]

So \(k=2\) or \(k=-14\). Both keep \(6-k>0\) (so \(\cos\theta>0\)), hence both are valid: \(\mathbf{k=2}\) or \(\mathbf{k=-14}\).

(b) Take \(A\) as the origin. Then, using the given vectors:

\[\overrightarrow{AB}=\begin{pmatrix}-5\\-1\end{pmatrix}\Rightarrow B=(-5,-1),\qquad \overrightarrow{AC}=\begin{pmatrix}-6\\-9\end{pmatrix}\Rightarrow C=(-6,-9).\]

\[D=B+\overrightarrow{BD}=\begin{pmatrix}-5\\-1\end{pmatrix}+\begin{pmatrix}4\\-7\end{pmatrix}=\begin{pmatrix}-1\\-8\end{pmatrix}\Rightarrow D=(-1,-8).\]

For ABCD (vertices in order) to be a parallelogram, opposite sides must be equal, i.e. \(\overrightarrow{AB}=\overrightarrow{DC}\). Now

\[\overrightarrow{DC}=C-D=\begin{pmatrix}-6-(-1)\\-9-(-8)\end{pmatrix}=\begin{pmatrix}-5\\-1\end{pmatrix}=\overrightarrow{AB}.\]

Since \(\overrightarrow{AB}=\overrightarrow{DC}\), the side AB is equal and parallel to DC. As a check, \(\overrightarrow{AD}=D-A=\begin{pmatrix}-1\\-8\end{pmatrix}\) and \(\overrightarrow{BC}=C-B=\begin{pmatrix}-1\\-8\end{pmatrix}\), so \(\overrightarrow{AD}=\overrightarrow{BC}\) as well. (Equivalently, the diagonals bisect each other: midpoint of \(AC=(-3,-4.5)=\) midpoint of \(BD\).)

Therefore ABCD is a parallelogram.

Detalhes da Resposta

For \(3x^{2}+5x+1=0\) with roots \(\alpha,\beta\), the sum and product of roots are

\[\alpha+\beta=-\frac{5}{3},\qquad \alpha\beta=\frac{1}{3}.\]

Use the identity for the sum of cubes:

\[\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3\alpha\beta(\alpha+\beta).\]

Substitute the values:

\[\alpha^{3}+\beta^{3}=\left(-\frac{5}{3}\right)^{3}-3\left(\frac{1}{3}\right)\left(-\frac{5}{3}\right)=-\frac{125}{27}+\frac{5}{3}.\]

Writing over a common denominator of \(27\):

\[\alpha^{3}+\beta^{3}=-\frac{125}{27}+\frac{45}{27}=-\frac{80}{27}.\]

Therefore

\[27(\alpha^{3}+\beta^{3})=27\times\left(-\frac{80}{27}\right)=-80.\]

Detalhes da Resposta

Let \(u=\log_{2}m\). Using the power law of logarithms, \(\log_{2}m^{3}=3\log_{2}m=3u\). The equation becomes a quadratic in \(u\):

\[u^{2}-3u=10\;\Rightarrow\; u^{2}-3u-10=0.\]

Factorising,

\[(u-5)(u+2)=0\;\Rightarrow\; u=5\quad\text{or}\quad u=-2.\]

Now convert back to \(m\) using \(u=\log_{2}m\Rightarrow m=2^{u}\):

• \(u=5:\quad m=2^{5}=32.\)
• \(u=-2:\quad m=2^{-2}=\dfrac{1}{4}.\)

Both values are positive, so both are valid solutions:

\[m=32\quad\text{or}\quad m=\tfrac{1}{4}.\]

Detalhes da Resposta

(a) Simplify \(\dfrac{5+\sqrt2}{3-\sqrt2}-\dfrac{5-\sqrt2}{3+\sqrt2}\).

Rationalise each fraction. For the first, multiply by \(\dfrac{3+\sqrt2}{3+\sqrt2}\):

\[\frac{(5+\sqrt2)(3+\sqrt2)}{(3-\sqrt2)(3+\sqrt2)}=\frac{15+5\sqrt2+3\sqrt2+2}{9-2}=\frac{17+8\sqrt2}{7}.\]

For the second, multiply by \(\dfrac{3-\sqrt2}{3-\sqrt2}\):

\[\frac{(5-\sqrt2)(3-\sqrt2)}{9-2}=\frac{15-5\sqrt2-3\sqrt2+2}{7}=\frac{17-8\sqrt2}{7}.\]

Subtracting:

\[\frac{17+8\sqrt2}{7}-\frac{17-8\sqrt2}{7}=\frac{16\sqrt2}{7}.\]

So in the form \(a+b\sqrt2\): \(a=0,\ b=\dfrac{16}{7},\) i.e. \(\dfrac{16}{7}\sqrt2.\)

(b) Solve by the determinant (Cramer's) method:

\(3x-y-z=-2,\quad x+5y+2z=5,\quad 2x+3y+z=0.\)

Main determinant:

\[D=\begin{vmatrix}3&-1&-1\\1&5&2\\2&3&1\end{vmatrix}=3(5-6)+1(1-4)-1(3-10)=-3-3+7=1.\]

\[D_x=\begin{vmatrix}-2&-1&-1\\5&5&2\\0&3&1\end{vmatrix}=-2(5-6)+1(5-0)-1(15-0)=2+5-15=-8.\]

\[D_y=\begin{vmatrix}3&-2&-1\\1&5&2\\2&0&1\end{vmatrix}=3(5-0)+2(1-4)-1(0-10)=15-6+10=19.\]

\[D_z=\begin{vmatrix}3&-1&-2\\1&5&5\\2&3&0\end{vmatrix}=3(0-15)+1(0-10)-2(3-10)=-45-10+14=-41.\]

Therefore

\[x=\frac{D_x}{D}=-8,\quad y=\frac{D_y}{D}=19,\quad z=\frac{D_z}{D}=-41.\]

Check in equation (1): \(3(-8)-19-(-41)=-24-19+41=-2.\) Correct.

Detalhes da Resposta

Let the passing probabilities be \(P(K)=\dfrac{9}{10},\ P(W)=\dfrac{4}{5},\ P(A)=x,\) with failing probabilities \(\dfrac{1}{10},\ \dfrac{1}{5},\ (1-x)\) respectively.

(a) Value of x. "Only one passes" means exactly one of the three succeeds while the other two fail:

\[P(\text{only one})=\underbrace{\tfrac{9}{10}\cdot\tfrac{1}{5}\cdot(1-x)}_{K\text{ only}}+\underbrace{\tfrac{1}{10}\cdot\tfrac{4}{5}\cdot(1-x)}_{W\text{ only}}+\underbrace{\tfrac{1}{10}\cdot\tfrac{1}{5}\cdot x}_{A\text{ only}}.\]

\[=\frac{9}{50}(1-x)+\frac{4}{50}(1-x)+\frac{1}{50}x=\frac{13(1-x)+x}{50}=\frac{13-12x}{50}.\]

Set equal to \(\dfrac{9}{50}:\)

\[13-12x=9\Rightarrow 12x=4\Rightarrow x=\frac{1}{3}.\]

(b) Probability that at least one passes. Use the complement (none pass). With \(P(A)=\dfrac13,\) failing \(A=\dfrac23:\)

\[P(\text{none})=\frac{1}{10}\cdot\frac{1}{5}\cdot\frac{2}{3}=\frac{2}{150}=\frac{1}{75}.\]

\[P(\text{at least one})=1-\frac{1}{75}=\frac{74}{75}.\]

Detalhes da Resposta

The panel has \(5\) men and \(3\) women, i.e. \(8\) people, from which a committee of \(3\) is chosen. Order does not matter, so we use combinations.

(a) Number of ways of forming the committee

\[\binom{8}{3}=\frac{8\times 7\times 6}{3\times 2\times 1}=56.\]

There are \(56\) possible committees.

(b) Probability that at least one woman is on the committee

Use the complement "no woman", i.e. an all-men committee chosen from the \(5\) men:

\[\binom{5}{3}=\frac{5\times 4\times 3}{3\times 2\times 1}=10.\]

\[P(\text{no woman})=\frac{10}{56}=\frac{5}{28}.\]

Therefore

\[P(\text{at least one woman})=1-\frac{5}{28}=\frac{23}{28}.\]

Detalhes da Resposta

(a) Differentiate \((x-3)(x^2+5)\).

Expanding first: \((x-3)(x^2+5)=x^3+5x-3x^2-15=x^3-3x^2+5x-15.\)

Differentiating term by term:

\[\frac{dy}{dx}=3x^2-6x+5.\]

(The product rule gives the same result: \((1)(x^2+5)+(x-3)(2x)=x^2+5+2x^2-6x=3x^2-6x+5.\))

(b) \((x+1)^2\) is a factor of \(f(x)=x^3+ax^2+bx+3\).

(i) Values of a and b. If \((x+1)^2\) divides \(f(x)\), then \(x=-1\) is a repeated root, so both \(f(-1)=0\) and \(f'(-1)=0\).

\(f(-1)=-1+a-b+3=0\Rightarrow a-b=-2.\)

\(f'(x)=3x^2+2ax+b,\) so \(f'(-1)=3-2a+b=0\Rightarrow 2a-b=3.\)

Subtracting the first from the second: \(a=5.\) Then \(b=a+2=7.\)

\[a=5,\quad b=7.\]

(ii) Zeros of f(x). Now \(f(x)=x^3+5x^2+7x+3.\) Since \((x+1)^2=x^2+2x+1\) is a factor, divide:

\[x^3+5x^2+7x+3=(x+1)^2(x+3).\]

Therefore the zeros are \(x=-1\) (a double/repeated root) and \(x=-3.\)

Detalhes da Resposta

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Detalhes da Resposta

(a)

\[\begin{aligned} (g\circ f)(x)&=g(f(x))\\ &=\frac{f(x)-1}{f(x)+1}\\ &=\frac{\dfrac{x-3}{2x-1}-1}{\dfrac{x-3}{2x-1}+1}\\ &=\frac{\dfrac{x-3-2x+1}{2x-1}}{\dfrac{x-3+2x-1}{2x-1}}\\ &=\frac{-x-2}{3x-4}. \end{aligned}\]

Hence,

\[\boxed{(g\circ f)(x)=-\frac{x+2}{3x-4}}\]

The restrictions are \(x\ne\frac12\) and \(x\ne\frac43\).

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(b)(i) For \(y=9x-x^3\),

\[y=x(9-x^2)=x(3-x)(3+x).\]

Therefore, the curve cuts the \(x\)-axis at \((-3,0)\), \((0,0)\), and \((3,0)\).

\[\frac{dy}{dx}=9-3x^2.\]

At turning points,

\[9-3x^2=0\implies x=\pm\sqrt3.\]

\[\begin{aligned}y\bigl(\sqrt3\bigr)&=6\sqrt3\approx10.39,\\y\bigl(-\sqrt3\bigr)&=-6\sqrt3\approx-10.39.\end{aligned}\]

Also, \(\dfrac{d^2y}{dx^2}=-6x\); hence \((\sqrt3,6\sqrt3)\) is a maximum point and \((-\sqrt3,-6\sqrt3)\) is a minimum point. The curve is symmetric about the origin.

\(x\)	\(-5\)	\(-4\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)
\(y=9x-x^3\)	\(80\)	\(28\)	\(0\)	\(-10\)	\(-8\)	\(0\)	\(8\)	\(10\)	\(0\)	\(-28\)	\(-80\)

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(b)(ii) The part of the curve from \(0\) to \(3\) lies above the \(x\)-axis. By symmetry, the two bounded areas, from \(-3\) to \(0\) and from \(0\) to \(3\), are equal.

\[\begin{aligned}\text{Total area}&=2\int_0^3(9x-x^3)\,dx\\&=2\left[\frac{9x^2}{2}-\frac{x^4}{4}\right]_0^3\\&=2\left(\frac{81}{2}-\frac{81}{4}\right)\\&=\frac{81}{2}.\end{aligned}\]

\[\boxed{\text{Total area}=\frac{81}{2}\text{ square units}=40.5\text{ square units}.}\]

Detalhes da Resposta

Method. Assumed mean \(A = 1.45\), class width \(c = 0.2\), \(u = \dfrac{x - 1.45}{0.2}\) where \(x\) is the class midpoint.

Mean.

Detalhes da Resposta

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Detalhes da Resposta

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Detalhes da Resposta

The operation is \(m\ast n=\dfrac{m^{2}-n^{2}}{2mn}\).

(a) Evaluate \(-3\ast 2\)

\[-3\ast 2=\frac{(-3)^{2}-2^{2}}{2(-3)(2)}=\frac{9-4}{-12}=\frac{5}{-12}=-\frac{5}{12}.\]

(b) Is \(\ast\) associative?

The operation is associative only if \((m\ast n)\ast p=m\ast(n\ast p)\) for all admissible values. Test the counterexample \(m=1,\;n=2,\;p=3\).

First \(m\ast n=\dfrac{1-4}{2(1)(2)}=-\dfrac{3}{4}\). Then

\[(m\ast n)\ast p=\left(-\tfrac{3}{4}\right)\ast 3=\frac{\left(-\frac{3}{4}\right)^{2}-3^{2}}{2\left(-\frac{3}{4}\right)(3)}=\frac{\frac{9}{16}-9}{-\frac{9}{2}}=\frac{-\frac{135}{16}}{-\frac{9}{2}}=\frac{135}{16}\times\frac{2}{9}=\frac{15}{8}.\]

Next \(n\ast p=\dfrac{4-9}{2(2)(3)}=-\dfrac{5}{12}\). Then

\[m\ast(n\ast p)=1\ast\left(-\tfrac{5}{12}\right)=\frac{1-\frac{25}{144}}{2(1)\left(-\frac{5}{12}\right)}=\frac{\frac{119}{144}}{-\frac{5}{6}}=\frac{119}{144}\times\left(-\frac{6}{5}\right)=-\frac{119}{120}.\]

Since \(\dfrac{15}{8}\neq -\dfrac{119}{120}\), we have \((m\ast n)\ast p\neq m\ast(n\ast p)\). Therefore \(\ast\) is not associative.

Detalhes da Resposta

(a) Frequency distribution table

(b) Cumulative frequency curve

Plot the cumulative frequencies against the corresponding upper class boundaries to obtain the less-than ogive shown below.

(i) Median

\[\frac{N}{2}=\frac{60}{2}=30.\]

From cumulative frequency 30, the score read from the ogive is

\[\boxed{\text{Median}=48.4}\]

(ii) Quartile deviation

\[Q_1=\frac{N}{4}=15\text{th score},\qquad Q_3=\frac{3N}{4}=45\text{th score}.\]

From the ogive, \(Q_1=35.5\) and \(Q_3=57.5\).

\[\text{Quartile deviation}=\frac{Q_3-Q_1}{2}=\frac{57.5-35.5}{2}=\boxed{11.0}.\]