WAEC SSCE - Further Mathematics - 2017

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The sum of the roots of a quadratic equation of the form \(ax^{2} + bx + c = 0\) is given by \(-\frac{b}{a}\). In this case, the given quadratic equation is \(2x^{2} - 3x + 4 = 0\). Therefore, the sum of the roots, \(\alpha\) and \(\beta\), is given by \(-\frac{-3}{2} = \frac{3}{2}\). Hence, the answer is \(\frac{3}{2}\), which represents the sum of the roots of the given quadratic equation.

Bayanin Amsa

Let the average score for the boys be y. Then the total score for the class is: 10y (boys' total score) + 15x (girls' total score) = 25(90) (total class score) Simplifying, we get: 10y + 15x = 2250 Dividing by 10, we get: y + \(\frac{3}{2}\)x = 225 Subtracting \(\frac{3}{2}\)x from both sides, we get: y = 225 - \(\frac{3}{2}\)x Therefore, the average score for the boys in terms of x is \(225 - \frac{3}{2}x\). So the correct answer is.

Bayanin Amsa

The mode is the value that occurs most frequently in the data set. From the table, we can see that the frequencies for faces 1, 2, 4, 5, and 6 are all different. However, the frequency for face 3 is "y", which is an unknown value. To find the mode, we need to determine the value of y. We know that the die was tossed 150 times, so the sum of all the frequencies should be 150. 12 + 18 + y + 30 + 2y + 45 = 150 Simplifying the equation, we get: 3y + 105 = 150 3y = 45 y = 15 Now we can see that the frequency for face 3 is 15. To find the mode, we need to determine which face has the highest frequency. We can see that face 6 has the highest frequency with 45 occurrences. Therefore, the mode is 6.

Bayanin Amsa

Bayanin Amsa

We can use the definition of logarithms to solve for y. Recall that logarithms are the inverse function of exponential functions. That is, if we have: y = b^x Then the logarithm base b of y is: log_b y = x Using this definition, we can write: log_y (1/8) = 3 As an exponential function: y^3 = 1/8 Simplifying the right side: y^3 = (1/2)^3 Taking the cube root of both sides: y = 1/2 Therefore, the answer is \(\frac{1}{2}\).

Bayanin Amsa

Bayanin Amsa

If the equation \(x^{2} - kx + 9 = 0\) has equal roots, it means that the discriminant is zero. The discriminant is the expression inside the square root in the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a For the equation \(x^{2} - kx + 9 = 0\), the discriminant is: b^2 - 4ac = k^2 - 4(1)(9) = k^2 - 36 Since the roots are equal, the discriminant is zero: k^2 - 36 = 0 k^2 = 36 k = ±6 Therefore, the values of k that make the equation have equal roots are \(\pm6\).

Bayanin Amsa

The line of symmetry of a parabola is a vertical line that passes through the vertex of the parabola. The equation of the line of symmetry for a parabola of the form \(y = a(x - h)^2 + k\) is simply the vertical line \(x = h\). The given curve is a quadratic function of the form \(y = -2x^2 - x + 5\), which is a downward-facing parabola. The line of symmetry for this parabola is the vertical line that passes through its vertex. To find the vertex of the parabola, we can first find the x-coordinate of the vertex using the formula \(x = \frac{-b}{2a}\). In this case, \(a = -2\) and \(b = -1\), so we have: $$x = \frac{-b}{2a} = \frac{-(-1)}{2(-2)} = \frac{1}{4}$$ To find the corresponding y-coordinate, we can substitute this value of x into the equation for the curve: $$y = -2x^2 - x + 5 = -2\left(\frac{1}{4}\right)^2 - \frac{1}{4} + 5 = \frac{41}{8}$$ Therefore, the vertex of the parabola is \(\left(\frac{1}{4}, \frac{41}{8}\right)\), and the equation of its line of symmetry is simply: $$x = \frac{1}{4}$$ So the correct option is \(\boxed{\text{(c) } x = \frac{1}{4}}\).

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To evaluate \(-3 \Delta -1\), we simply need to substitute a = -3 and b = -1 in the definition of the binary operation: \(-3 \Delta -1 = \frac{-3+(-1)}{\sqrt{(-3)(-1)}} = \frac{-4}{\sqrt{3}}\) To simplify this expression, we rationalize the denominator by multiplying both the numerator and denominator by \(\sqrt{3}\): \(\frac{-4}{\sqrt{3}} = \frac{-4}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{-4\sqrt{3}}{3}\) Therefore, the correct answer is.

Bayanin Amsa

We can simplify the given expression as follows: \begin{align*} \frac{1}{(1-\sqrt{3})^2} &= \frac{1}{1-2\sqrt{3}+3} \\ &= \frac{1}{4-2\sqrt{3}} \\ &= \frac{4+2\sqrt{3}}{(4-2\sqrt{3})(4+2\sqrt{3})} \\ &= \frac{4+2\sqrt{3}}{16-12} \\ &= \frac{4+2\sqrt{3}}{4} \\ &= 1 + \frac{1}{2}\sqrt{3} \end{align*} Therefore, the answer is (B) \(1+ \frac{1}{2}\sqrt{3}\).

Bayanin Amsa

To find the equation of a straight line, we need to know either two points that the line passes through or a point and the slope of the line. In this case, we are given one point that the line passes through, (-1,2), and the equation of the line, 2x+3y=6. We can use the point-slope form of the equation of a line to find the equation of the line passing through the given point. The point-slope form is: y - y1 = m(x - x1) where (x1, y1) is the given point and m is the slope of the line. To find the slope of the given line, we can rearrange the equation 2x+3y=6 into the slope-intercept form: y = (-2/3)x + 2 where the slope is -2/3. Substituting the values of the given point and slope into the point-slope form, we get: y - 2 = (-2/3)(x - (-1)) Simplifying this equation, we get: y - 2 = (-2/3)x - (2/3) Multiplying both sides by 3 to eliminate the fraction, we get: 3y - 6 = -2x - 2 Adding 2x to both sides, we get: 2x + 3y = 4 Therefore, the equation of the line passing through the given point is 2x+3y=4. Hence, the correct answer is option (D) 2x+3y=4.

Bayanin Amsa

To find \(\frac{dy}{dx}\) when \(y = \frac{1+x}{1-x}\), we need to use the quotient rule of differentiation, which states that if \(y = \frac{u}{v}\), then $$ \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}. $$ In this case, we have $$ u = 1+x \qquad \text{and} \qquad v = 1-x. $$ Taking the derivatives of these functions, we get $$ \frac{du}{dx} = 1 \qquad \text{and} \qquad \frac{dv}{dx} = -1. $$ Substituting into the quotient rule formula, we get \begin{align*} \frac{dy}{dx} &= \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} \\ &= \frac{1-x+1+x}{(1-x)^2} \\ &= \frac{2}{(1-x)^2}. \end{align*} Therefore, the correct answer is \(\frac{2}{(1-x)^2}\). Option (A) is the correct answer.

Bayanin Amsa

To find the equation of the circle, we need to know its center and radius. We are given that the center is (4,5), so the equation will have the form \((x-4)^2 + (y-5)^2 = r^2\), where \(r\) is the radius. To find the radius, we need to first find the y-intercept of the line. To do this, we can set x = 0 and solve for y: 5x - 2y + 6 = 0 -2y + 6 = 0 -2y = -6 y = 3 So the y-intercept of the line is (0,3). Now we can use the distance formula to find the distance between this point and the center of the circle: \(\sqrt{(0-4)^2 + (3-5)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\) So the radius of the circle is \(2\sqrt{5}\). Now we can substitute the center and radius into the equation of the circle: \((x-4)^2 + (y-5)^2 = (2\sqrt{5})^2\) Simplifying: \((x-4)^2 + (y-5)^2 = 20\) Expanding and rearranging, we get: \(x^2 + y^2 - 8x - 10y + 21 = 0\) Therefore, the equation of the circle is \(x^2 + y^2 - 8x - 10y + 21 = 0\). The answer is (3).

Bayanin Amsa

We can start by simplifying the numerator and the denominator separately. The square root of 128 can be written as the square root of 64 times 2, which simplifies to 8 times the square root of 2. The denominator can be simplified using the difference of squares formula: \begin{align*} \sqrt{32} - 2\sqrt{2} &= \sqrt{16 \times 2} - 2\sqrt{2} \\ &= 4\sqrt{2} - 2\sqrt{2} \\ &= 2\sqrt{2} \end{align*} Now we can substitute these simplifications back into the original expression: \begin{align*} \frac{\sqrt{128}}{\sqrt{32} - 2\sqrt{2}} &= \frac{8\sqrt{2}}{2\sqrt{2}} \\ &= 4 \end{align*} Therefore, the simplified expression is 4.

Bayanin Amsa

The table shows the frequency of each face when a fair die is tossed 150 times. To find the probability of obtaining a 5, we need to look at the frequency of the face with the value 5, which is given as 2y. We are not given the value of y, but we know that the die is fair, which means that each face has an equal probability of appearing on any given toss. Since there are six faces on a die, the probability of obtaining a 5 on any given toss is 1/6. Therefore, the probability of obtaining a 5 in this experiment is the frequency of the face with value 5 (2y) divided by the total number of tosses (150), multiplied by the probability of obtaining a 5 on any given toss (1/6): P(5) = (2y/150) x (1/6) = y/225 We don't know the value of y, but we can simplify the expression by noting that the probabilities of all six faces must add up to 1: P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 We can substitute the expressions for the probabilities in terms of y and simplify: 12/150 + 18/150 + y/150 + 30/150 + 2y/150 + 45/150 = 1 105/150 + 3y/150 = 1 3y/150 = 45/150 y = 15 Substituting y = 15 into the expression for P(5), we get: P(5) = 15/225 = 1/15 Therefore, the probability of obtaining a 5 is 1/15, which corresponds to option C.

Bayanin Amsa

To solve the problem, we need to use the sum formula for cosine: \[\cos(x+y) = \cos x \cos y - \sin x \sin y\] We are given the values of \(\sin x\) and \(\sin y\), so we need to find \(\cos x\) and \(\cos y\). To do this, we use the fact that \(\sin^2 x + \cos^2 x = 1\) and \(\sin^2 y + \cos^2 y = 1\) for any angle x and y. From \(\sin x = \frac{5}{13}\), we can find \(\cos x\) by using the Pythagorean identity: \[\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}\] Similarly, from \(\sin y = \frac{8}{17}\), we can find \(\cos y\): \[\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \frac{15}{17}\] Now we can substitute into the formula for \(\cos(x+y)\): \[\cos(x+y) = \cos x \cos y - \sin x \sin y = \frac{12}{13} \cdot \frac{15}{17} - \frac{5}{13} \cdot \frac{8}{17} = \frac{140}{221}\] Therefore, the answer is \(\frac{140}{221}\).

Bayanin Amsa

The domain of a function is the set of all possible values of x for which the function is defined. In the given function, we have a square root with an expression inside it. For the square root to be defined, the expression inside it must be non-negative. Therefore, we need to solve the inequality: \(4 - 2x \geq 0\) Simplifying this inequality, we get: \(2x \leq 4\) \(x \leq 2\) So the domain of the function f is all real numbers such that \(x \leq 2\). Therefore, the correct answer is.

Bayanin Amsa

Bayanin Amsa

The acceleration, A, of a particle is the rate of change of its velocity, V. So, we need to differentiate the given equation of velocity with respect to time, t to get the equation of acceleration. \[\frac{dV}{dt} = \frac{d}{dt}(3t^2 + 2t - 1)\] \[\frac{dV}{dt} = 6t + 2\] Therefore, the acceleration of the particle after 2 seconds is given by substituting t = 2 into the equation of acceleration as follows: \[A = 6t + 2\] \[A = 6(2) + 2\] \[A = 14 ms^{-2}\] Hence, the correct option is (c) 14\(ms^{-2}\).

Bayanin Amsa

When a fair die is tossed twice, there are 6 possible outcomes for the first toss and 6 possible outcomes for the second toss, giving a total of 6 x 6 = 36 possible outcomes for the two tosses combined. Each possible outcome can be represented by an ordered pair (a,b), where a is the outcome of the first toss and b is the outcome of the second toss. Therefore, the sample space of this experiment has 36 possible outcomes, making the correct answer.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To solve this problem, we need to use the combination formula: nCr = n! / r!(n-r)! where n is the total number of objects, r is the number of objects we want to select, and ! represents factorial, which means the product of all positive integers up to and including that number. In this problem, we want to select 3 girls and 2 boys out of 20 students, where 13 are girls and 7 are boys. The number of ways to select 3 girls out of 13 is: 13C3 = 13! / 3!(13-3)! = 13! / 3!10! = (13x12x11) / (3x2x1) = 286 Similarly, the number of ways to select 2 boys out of 7 is: 7C2 = 7! / 2!(7-2)! = 7! / 2!5! = (7x6) / (2x1) = 21 To find the total number of ways to select 3 girls and 2 boys out of 20, we need to multiply the number of ways to select 3 girls and 2 boys: 286 x 21 = 6006 Therefore, there are 6006 ways to select 3 girls and 2 boys from a class of 20. Hence, the correct answer is option (C) 6006.

Bayanin Amsa

We are given the first four terms of an Arithmetic Progression (A.P.) and we are required to find the 21st term. We can notice that the common difference between any two successive terms is \(d = -1.5 - (-4) = 2.5\). Using the formula for the nth term of an A.P., which is given by: \(a_{n} = a_{1} + (n - 1) d\) where \(a_{n}\) is the nth term, \(a_{1}\) is the first term, \(n\) is the number of terms and \(d\) is the common difference. We can now find the 21st term as follows: \(a_{21} = a_{1} + (21 - 1) d\) Substituting the given values, we get: \(a_{21} = -4 + (21 - 1) \times 2.5 = 46\) Hence, the 21st term of the given A.P. is 46. Therefore, the correct option is (B) 46.

Bayanin Amsa

To evaluate \((2a - \frac{1}{4}b)\), we first need to perform scalar multiplication on vectors \(a\) and \(b\) and then subtract the resulting vectors: \begin{align*} 2a - \frac{1}{4}b &= 2\begin{pmatrix} 2 \\ 3 \end{pmatrix} - \frac{1}{4}\begin{pmatrix} -1 \\ 4 \end{pmatrix}\\ &= \begin{pmatrix} 4 \\ 6 \end{pmatrix} - \begin{pmatrix} -\frac{1}{4} \\ 1 \end{pmatrix}\\ &= \begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix} \end{align*} Therefore, the answer is \(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\).

Bayanin Amsa

To find the value of \(k\), we need to calculate the product of the given matrices and compare it with the given result. Multiplying the first matrix by the second matrix, we get: \(\begin{pmatrix}  2  &  1 \\  4 & 3 \end{pmatrix}\)\(\begin{pmatrix}  5 \\ 4 \end{pmatrix}\) = \(\begin{pmatrix}  2(5)+1(4)  \\  4(5)+3(4) \end{pmatrix}\) = \(\begin{pmatrix}  14  \\  32 \end{pmatrix}\) Now, we can compare this result with the given result: k\(\begin{pmatrix}  17.5 \\ 40.0 \end{pmatrix}\) We see that the first element of the calculated result, 14, is less than the first element of the given result, 17.5. This means that k must be less than 1. Dividing both sides of the given equation by \(\begin{pmatrix}  17.5 \\ 40.0 \end{pmatrix}\), we get: \(\begin{pmatrix}  2  &  1 \\  4 & 3 \end{pmatrix}\)\(\begin{pmatrix}  5 \\ 4 \end{pmatrix}\) / \(\begin{pmatrix}  17.5 \\ 40.0 \end{pmatrix}\) = k Calculating the left-hand side, we get: \(\begin{pmatrix}  2  &  1 \\  4 & 3 \end{pmatrix}\)\(\begin{pmatrix}  5 \\ 4 \end{pmatrix}\) / \(\begin{pmatrix}  17.5 \\ 40.0 \end{pmatrix}\) = \(\begin{pmatrix}  14/35  \\  32/40 \end{pmatrix}\) = \(\begin{pmatrix}  0.4  \\  0.8 \end{pmatrix}\) Therefore, the value of \(k\) is equal to the first element of this result, which is 0.4. Hence, the correct answer is option (C) 0.8.

Bayanin Amsa

To evaluate (3a - 8b), we need to multiply each component of the vectors by their respective scalar, and then add the corresponding components. So, 3a = 3(5i + 4j) = 15i + 12j 8b = 8(3i + 7j) = 24i + 56j Therefore, (3a - 8b) = (15i + 12j) - (24i + 56j) = (-9i - 44j) Hence, the correct answer is option (C) -9i - 44j.

Bayanin Amsa

To express cos150° in surd form, we need to use the trigonometric identity cos(180° - θ) = -cos(θ). Thus, cos150° = cos(180° - 30°) = -cos30°. Using the value of cos30°, which is \(\frac{\sqrt{3}}{2}\), we get: cos150° = -cos30° = -\(\frac{\sqrt{3}}{2}\) Therefore, the correct answer is option B, \(-\frac{\sqrt{3}}{2}\).

Bayanin Amsa

To solve for m, we need to first create a common denominator on the right-hand side of the equation: \[\frac{4}{x+5} - \frac{2}{2x-3} = \frac{4(2x-3) - 2(x+5)}{(x+5)(2x-3)} = \frac{5}{2x-3} - \frac{3}{x+5}\] Now we have: \[\frac{6x+m}{2x^{2}+7x-15} = \frac{5}{2x-3} - \frac{3}{x+5}\] Multiplying both sides by the common denominator of the right-hand side, we get: \[(6x+m)(x+5) = 5(2x^2+7x-15) - 3(2x-3)(2x+5)\] Expanding and simplifying, we get: \[6x^2 + 41x + 5m + 15 = 4x^2 + 7x - 30\] \[2x^2 + 34x + 45 + m = 0\] Now, to find the value of m, we substitute the given roots of the equation (which are -5/2 and 9/2) into the equation and solve for m: \[m = -2x^2 - 34x - 45\] When x = -5/2, \[m = -2\left(\frac{-5}{2}\right)^2 - 34\left(\frac{-5}{2}\right) - 45 = -22\] Therefore, the value of m is -22. Option (d) is correct.

Bayanin Amsa

To find the number of ways to seat 6 students around a circular table, we can use the following formula: N = (n-1)! where N is the number of ways to seat n people around a circular table, and ! means factorial (the product of all positive integers up to a given number). So, for 6 people, we have: N = (6-1)! = 5! Using the definition of factorial, we have: 5! = 5 x 4 x 3 x 2 x 1 = 120 Therefore, there are 120 ways to seat 6 students around a circular table.

Bayanin Amsa

Bayanin Amsa

None

Bayanin Amsa

(a) \(g(x)=x+6\), so the inverse is \(g^{-1}(x)=x-6\).

\[f\circ g^{-1}(x)=f(x-6)=\frac{4-5(x-6)}{2}=\frac{4-5x+30}{2}=\frac{34-5x}{2}\]

(b) \(P\) divides \((7,-5)\) and \((-2,7)\) internally in \(5:4\). Using the section formula \(\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)\) with \(m:n=5:4\):

\[x=\frac{5(-2)+4(7)}{9}=\frac{18}{9}=2,\qquad y=\frac{5(7)+4(-5)}{9}=\frac{15}{9}=\frac{5}{3}\]

\[P=\left(2,\ \tfrac{5}{3}\right)\]

Bayanin Amsa

(a)(i) Magnitudes and directions.

\(m=6i+8j\): \(|m|=\sqrt{6^{2}+8^{2}}=10\). Direction \(=\tan^{-1}\dfrac{8}{6}=53.1^{\circ}\) above the positive \(x\)-axis.

\(n=-8i+\tfrac{7}{3}j\): \(|n|=\sqrt{(-8)^{2}+\left(\tfrac{7}{3}\right)^{2}}=\sqrt{\dfrac{576+49}{9}}=\sqrt{\dfrac{625}{9}}=\dfrac{25}{3}\approx8.33\). It is in the second quadrant, direction \(=180^{\circ}-\tan^{-1}\dfrac{7/3}{8}=180^{\circ}-16.3^{\circ}=163.7^{\circ}\).

(ii) Angle between m and n.

\[m\cdot n=6(-8)+8\left(\tfrac{7}{3}\right)=-48+\tfrac{56}{3}=-\tfrac{88}{3}\]

\[\cos\theta=\frac{m\cdot n}{|m|\,|n|}=\frac{-88/3}{10\times25/3}=\frac{-88}{250}=-0.352\]

\[\theta=\cos^{-1}(-0.352)\approx110.6^{\circ}\]

(b) \(\overrightarrow{PQ}=Q-P=\binom{10}{4}-\binom{-2}{3}=\binom{12}{1}\); \(\overrightarrow{RS}=S-R=\binom{4}{0}-\binom{3}{12}=\binom{1}{-12}\).

\[\overrightarrow{PQ}\cdot\overrightarrow{RS}=(12)(1)+(1)(-12)=0\]

Since the scalar product is zero, \(\overrightarrow{PQ}\perp\overrightarrow{RS}\).

Bayanin Amsa

Resolve each force into components (angles measured anticlockwise from the positive \(x\)-axis).

\(F_{1}=18\,\text{N}\) at \(330^{\circ}\): \(x=18\cos330^{\circ}=15.59\), \(y=18\sin330^{\circ}=-9.00\).

\(F_{2}=10\,\text{N}\) at \(090^{\circ}\): \(x=0\), \(y=10\).

\(F_{3}=25\,\text{N}\) at \(180^{\circ}\): \(x=-25\), \(y=0\).

Sum the components:

\[R_{x}=15.59+0-25=-9.41,\qquad R_{y}=-9+10+0=1.00\]

Magnitude:

\[|R|=\sqrt{(-9.41)^{2}+(1.00)^{2}}=\sqrt{89.55}\approx9.5\ \text{N}\]

Direction: \(R\) lies in the second quadrant. The reference angle is \(\tan^{-1}\dfrac{1}{9.41}=6.1^{\circ}\), so

\[\theta=180^{\circ}-6.1^{\circ}=173.9^{\circ}\ \text{(anticlockwise from the positive } x\text{-axis)}\]

Bayanin Amsa

(a) Pass:fail \(=4:1\), so \(P(\text{pass})=0.8,\ P(\text{fail})=0.2\), with \(n=7\).

(i) At least 3 passed. Let \(X=\) number who pass, \(p=0.8\).

\[P(X\ge3)=1-P(0)-P(1)-P(2)\]

\(P(0)=(0.2)^{7}=0.0000128\), \(P(1)=7(0.8)(0.2)^{6}=0.000358\), \(P(2)=21(0.8)^{2}(0.2)^{5}=0.004301\).

\[P(X\ge3)=1-0.004672=0.995\approx1.00\]

(ii) Between 3 and 6 failed (i.e. \(4\) or \(5\) failed). Let \(Y=\) number who fail, \(p=0.2\).

\(P(Y=4)=\binom{7}{4}(0.2)^{4}(0.8)^{3}=35(0.0016)(0.512)=0.028672\).

\(P(Y=5)=\binom{7}{5}(0.2)^{5}(0.8)^{2}=21(0.00032)(0.64)=0.004301\).

\[P(4\le Y\le5)=0.028672+0.004301=0.032973\approx0.03\]

(b) A fair die thrown 5 times, \(P(\text{six})=\tfrac{1}{6}\). Probability of exactly three sixes:

\[\binom{5}{3}\left(\tfrac{1}{6}\right)^{3}\left(\tfrac{5}{6}\right)^{2}=10\cdot\frac{1}{216}\cdot\frac{25}{36}=\frac{250}{7776}\approx0.03\]

Bayanin Amsa

(a) Add the two fractions over a common denominator:

\[\frac{1}{1-\cos\theta}+\frac{1}{1+\cos\theta}=\frac{(1+\cos\theta)+(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}=\frac{2}{1-\cos^{2}\theta}\]

Since \(1-\cos^{2}\theta=\sin^{2}\theta\):

\[=\frac{2}{\sin^{2}\theta}\]

(b) \(y=x^{2}(x-3)=x^{3}-3x^{2}\). Then \(\dfrac{dy}{dx}=3x^{2}-6x=3x(x-2)\).

Stationary points where \(\dfrac{dy}{dx}=0\): \(x=0\) or \(x=2\).

At \(x=0,\ y=0\Rightarrow(0,0)\). At \(x=2,\ y=8-12=-4\Rightarrow(2,-4)\).

Line joining them: slope \(=\dfrac{-4-0}{2-0}=-2\), through \((0,0)\):

\[y=-2x\quad\text{or}\quad 2x+y=0\]

Distance:

\[\sqrt{(2-0)^{2}+(-4-0)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}\]

Bayanin Amsa

In parallelogram \(MNQR\) (vertices in order), the diagonals \(MQ\) and \(NR\) bisect each other, so they share the same midpoint.

Midpoint of \(MQ\): \(\left(\dfrac{4+8}{2},\dfrac{-6+16}{2}\right)=(6,\ 5)\).

Midpoint of \(NR\): \(\left(\dfrac{10+x}{2},\dfrac{2+y}{2}\right)\).

Equate:

\[\frac{10+x}{2}=6\ \Rightarrow\ x=2,\qquad \frac{2+y}{2}=5\ \Rightarrow\ y=8\]

\[R=(2,\ 8)\]

Bayanin Amsa

(a) Histogram

Since the class intervals have unequal widths, plot frequency density against age.

Age class (years)	Class boundaries	Class width	Frequency	Frequency density \(=\frac{f}{\text{class width}}\)
17 - 19	16.5 - 19.5	3	6	2
20 - 22	19.5 - 22.5	3	9	3
23 - 28	22.5 - 28.5	6	12	2
29 - 34	28.5 - 34.5	6	18	3
35 - 43	34.5 - 43.5	9	18	2

(b) Mean age

Age class	Midpoint, \(x\)	Frequency, \(f\)	\(fx\)
17 - 19	18.0	6	108
20 - 22	21.0	9	189
23 - 28	25.5	12	306
29 - 34	31.5	18	567
35 - 43	39.0	18	702
Total		63	1872

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1872}{63}=29.714\ldots\]

Therefore, the mean age of the patients is \(29.71\text{ years}\), correct to two decimal places.

Bayanin Amsa

Binomial with \(n=10\) bottles, defective probability \(p=0.08\), non-defective \(q=0.92\). Let \(X\) be the number of defective bottles.

\[P(X\ge3)=1-P(0)-P(1)-P(2)\]

\(P(0)=(0.92)^{10}=0.43439\).

\(P(1)=\binom{10}{1}(0.08)(0.92)^{9}=10(0.08)(0.47216)=0.37773\).

\(P(2)=\binom{10}{2}(0.08)^{2}(0.92)^{8}=45(0.0064)(0.51322)=0.14781\).

\[P(X\ge3)=1-0.43439-0.37773-0.14781=0.04007\approx0.04\]

Bayanin Amsa

(a) \(f(x)=\displaystyle\int(4x-x^{2})\,dx=2x^{2}-\dfrac{x^{3}}{3}+C\).

Use \(f(3)=21\): \(2(9)-\dfrac{27}{3}+C=18-9+C=9+C=21\Rightarrow C=12\).

\[f(x)=2x^{2}-\frac{x^{3}}{3}+12\]

(b) Let the AP have first term \(a\) and common difference \(d\). The 2nd, 4th, 8th terms are \(a+d,\ a+3d,\ a+7d\) and form a GP:

\[(a+3d)^{2}=(a+d)(a+7d)\]

\[a^{2}+6ad+9d^{2}=a^{2}+8ad+7d^{2}\Rightarrow 2d^{2}-2ad=0\Rightarrow 2d(d-a)=0\]

Since \(d\neq0\), \(a=d\).

Sum of 3rd and 5th terms is \(20\): \((a+2d)+(a+4d)=2a+6d=20\Rightarrow a+3d=10\).

With \(a=d\): \(4d=10\Rightarrow d=2.5,\ a=2.5\).

(i) First four terms: \(2.5,\ 5,\ 7.5,\ 10\).

(ii) Sum of first ten terms:

\[S_{10}=\frac{10}{2}\big(2a+9d\big)=5\big(5+22.5\big)=5(27.5)=137.5\]

Bayanin Amsa

The continuous class boundaries and cumulative frequencies are:

Height (cm)	Class boundaries (cm)	Frequency	Cumulative frequency
36–40	35.5–40.5	3	3
41–45	40.5–45.5	9	12
46–50	45.5–50.5	21	33
51–55	50.5–55.5	12	45
56–60	55.5–60.5	5	50

(a) The cumulative frequency curve is shown below. The point [35.5, 0 is included before plotting the upper class boundaries against their cumulative frequencies.

(b) Total frequency, \(N=50\).

On the curve, the lower quartile corresponds to cumulative frequency

\[\frac{N}{4}=\frac{50}{4}=12.5.\]

Reading across from \(12.5\) to the curve and down to the height axis gives

\[Q_1\approx 45.6\text{ cm}.\]

The upper quartile corresponds to cumulative frequency

\[\frac{3N}{4}=\frac{3(50)}{4}=37.5.\]

Reading across from \(37.5\) to the curve and down to the height axis gives

\[Q_3\approx 52.0\text{ cm}.\]

Hence,

\[\text{Semi-interquartile range}=\frac{Q_3-Q_1}{2}=\frac{52.0-45.6}{2}=\boxed{3.2\text{ cm}}.\]

Bayanin Amsa

(a)(i) Diagram. The point \(B\) carries the weight \(W=qg\) hanging straight down. String \(BA\) rises to the left at \(30^{\circ}\) to the horizontal (tension \(T_{AB}=48\,\text{N}\)); string \(BD\) rises to the right at \(60^{\circ}\) to the horizontal (tension \(T_{DB}\)). The three forces \(T_{AB},\ T_{DB},\ W\) meet at \(B\).

(ii) Tension in DB. For equilibrium the horizontal components of the two tensions balance:

\[T_{AB}\cos30^{\circ}=T_{DB}\cos60^{\circ}\]

\[48\cos30^{\circ}=T_{DB}\cos60^{\circ}\]

\[T_{DB}=\frac{48\cos30^{\circ}}{\cos60^{\circ}}=\frac{48\left(\frac{\sqrt3}{2}\right)}{\frac12}=48\sqrt3\approx83.1\ \text{N}\]

Bayanin Amsa

Rewrite the integrand by splitting \(x-1=(x+1)-2\):

\[\frac{x-1}{(x+1)^{2}}=\frac{(x+1)-2}{(x+1)^{2}}=\frac{1}{x+1}-\frac{2}{(x+1)^{2}}\]

Integrate term by term:

\[\int\left(\frac{1}{x+1}-\frac{2}{(x+1)^{2}}\right)dx=\ln(x+1)+\frac{2}{x+1}+C\]

Evaluate from \(1\) to \(3\):

\[\left[\ln(x+1)+\frac{2}{x+1}\right]_{1}^{3}=\left(\ln4+\tfrac{1}{2}\right)-\left(\ln2+1\right)\]

\[=\ln4-\ln2-\tfrac{1}{2}=\ln2-\tfrac{1}{2}\approx0.193\]

Bayanin Amsa

By the Factor Theorem, \(g(-1)=0\) and \(g(2)=0\).

\(g(x)=x^{4}+ax^{3}+bx^{2}-16x-12\).

Using \(g(-1)=0\):

\[1-a+b+16-12=0\ \Rightarrow\ -a+b+5=0\ \Rightarrow\ b=a-5\quad(1)\]

Using \(g(2)=0\):

\[16+8a+4b-32-12=0\ \Rightarrow\ 8a+4b-28=0\ \Rightarrow\ 2a+b=7\quad(2)\]

Substitute (1) into (2): \(2a+(a-5)=7\Rightarrow3a=12\Rightarrow a=4\).

Then \(b=4-5=-1\).

\[a=4,\qquad b=-1\]

Bayanin Amsa

(a) Using the laws of logarithms with \(\log_{10}p=a,\ \log_{10}q=b,\ \log_{10}s=c\):

\[\log_{10}\!\left(\frac{p^{\frac{1}{3}}q^{4}}{s^{2}}\right)=\tfrac{1}{3}\log_{10}p+4\log_{10}q-2\log_{10}s=\frac{a}{3}+4b-2c\]

(b) Area of a circle: \(A=\pi r^{2}\). Differentiate with respect to time:

\[\frac{dA}{dt}=2\pi r\,\frac{dr}{dt}\]

With \(\dfrac{dA}{dt}=20\,\text{cm}^2\text{s}^{-1}\) and \(r=6\,\text{cm}\):

\[20=2\pi(6)\frac{dr}{dt}\ \Rightarrow\ \frac{dr}{dt}=\frac{20}{12\pi}=\frac{5}{3\pi}\ \text{cm s}^{-1}\]