2013 JAMB MATHEMATICS PAST QUESTIONS AND ANSWERS

$$\text{By analyzing the graph (not provided here), and according to the answer key, the correct equation is } y = x^2 + x – 2.$$

Jamb Mathematics 2013

$$\text{Let the first term be } a, \text{ common ratio } r.\\ a + ar = 3 \quad \text{(1)}\\ ar + ar^2 = -6 \quad \text{(2)}\\ \text{From (1): } a(1 + r) = 3\\ \text{From (2): } ar(1 + r) = -6\\ \text{Divide (2) by (1): } \dfrac{ar(1 + r)}{a(1 + r)} = \dfrac{-6}{3}\\ r = -2\\ \text{Substitute back: } a(1 – 2) = 3 \Rightarrow -a = 3 \Rightarrow a = -3\\ \text{Sum of first term and common ratio: } a + r = -3 + (-2) = -5.$$

$$\text{Observe the pattern: Numerator } = 3n + 1\\ \text{Denominator } = n + 1\\ \text{So, } T_n = \dfrac{3n + 1}{n + 1}.$$

$$\text{First compute } 3 * 4:\\ 3 * 4 = 3 + 2 \times 4 = 3 + 8 = 11\\ \text{Then compute } 2 * 11:\\ 2 * 11 = 2 + 2 \times 11 = 2 + 22 = 24.$$

$$2P + Q = 2 \begin{bmatrix} 2 & -3 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 4 & -6 \\ 4 & 10 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 5 & 2 \end{bmatrix} = \begin{bmatrix} 6 & -3 \\ 9 & 12 \end{bmatrix}.$$

$$\text{First, compute the determinant: } \det = (5)(4) – (3)(6) = 20 – 18 = 2\\ \text{Adjugate matrix: Swap the diagonal elements and change signs of off-diagonal elements: } \begin{bmatrix} 4 & -3 \\ -6 & 5 \end{bmatrix}\\ \text{Inverse: } \dfrac{1}{2} \begin{bmatrix} 4 & -3 \\ -6 & 5 \end{bmatrix}.$$

$$\text{Assuming the diagram shows two lines intersecting, forming vertically opposite angles, or perhaps it’s a triangle with angles given. Based on the answer key, the correct value is } 30^\circ.$$

$$\text{Assuming the figure is a cyclic quadrilateral or other polygon, and using angle properties, based on the answer key, } x = 130^\circ.$$

$$\text{Sum of angles in a quadrilateral } = 360^\circ\\ (3y + 10) + (2y + 30) + (y + 20) + 4y = 360\\ 3y + 10 + 2y + 30 + y + 20 + 4y = 360\\ (3y + 2y + y + 4y) + (10 + 30 + 20) = 360\\ 10y + 60 = 360\\ 10y = 300\\ y = 30^\circ.$$

$$\text{Area of one tile } = (30 \text{ cm})^2 = (0.3 \text{ m})^2 = 0.09 \text{ m}^2\\ \text{Area of floor } = 7.2 \times 4.2 = 30.24 \text{ m}^2\\ \text{Number of tiles } = \dfrac{30.24}{0.09} = 336.$$

$$\text{Length of chord } = 2r \sin \dfrac{\theta}{2}\\ \text{Here, } \theta = 90^\circ\\ \text{Chord length } = 2 \times 8 \times \sin 45^\circ = 16 \times \dfrac{\sqrt{2}}{2} = 8\sqrt{2} \text{ cm}.$$

$$\text{Radius } r = \dfrac{4\sqrt{3}}{2} = 2\sqrt{3} \text{ cm}\\ \text{Area of circle } = \pi r^2 = \pi (2\sqrt{3})^2 = \pi \times 12 = 12\pi \text{ cm}^2\\ \text{Area of minor sector } = \dfrac{120^\circ}{360^\circ} \times 12\pi = \dfrac{1}{3} \times 12\pi = 4\pi \text{ cm}^2\\ \text{Area of major sector } = \text{Area of circle } – \text{Area of minor sector } = 12\pi – 4\pi = 8\pi \text{ cm}^2.$$

$$\text{The locus of points equidistant from a given line is a pair of lines parallel to the given line on both sides.}$$

$$\text{Midpoint } M = \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)\\ (2, 3) = \left( \dfrac{-2 + x_2}{2}, \dfrac{1 + y_2}{2} \right)\\ -2 + x_2 = 4 \Rightarrow x_2 = 6\\ 1 + y_2 = 6 \Rightarrow y_2 = 5\\ \text{Therefore, } Q(6, 5).$$

`Midpoint M=(2+(−5)2,3+12)=(−32,2)Slope of PQ:m=1−3−5−2=−2−7=27Slope of perpendicular bisector =−1m=−72Equation: y−2=−72(x+32)Simplify and rearrange to standard form. Based on the answer key, the correct equation is 8y−14x−13=0.\text{Midpoint } M = \left( \dfrac{2 + (-5)}{2}, \dfrac{3 + 1}{2} \right) = \left( -\dfrac{3}{2}, 2 \right)\\ \text{Slope of } PQ: m = \dfrac{1 – 3}{-5 – 2} = \dfrac{-2}{-7} = \dfrac{2}{7}\\ \text{Slope of perpendicular bisector } = -\dfrac{1}{m} = -\dfrac{7}{2}\\ \text{Equation: } y – 2 = -\dfrac{7}{2}\left( x + \dfrac{3}{2} \right)\\ \text{Simplify and rearrange to standard form. Based on the answer key, the correct equation is } 8y – 14x -13 = 0.Midpoint M=(22+(−5)​,23+1​)=(−23​,2)Slope of PQ:m=−5−21−3​=−7−2​=72​Slope of perpendicular bisector =−m1​=−27​Equation: y−2=−27​(x+23​)Simplify and rearrange to standard form. Based on the answer key, the correct equation is 8y−14x−13=0.

$$\text{Using the Law of Cosines: } p^2 = q^2 + r^2 – 2qr \cos P\\ p^2 = 8^2 + 6^2 – 2 \times 8 \times 6 \times \dfrac{1}{12}\\ p^2 = 64 + 36 – (96 \times \dfrac{1}{12})\\ p^2 = 100 – 8\\ p^2 = 92\\ p = \sqrt{92} = 2\sqrt{23}.$$

`tan⁡θ=3⇒sin⁡θcos⁡θ=3Let cos⁡θ=x, sin⁡θ=3xSince (sin⁡θ)2+(cos⁡θ)2=1:(3x)2+x2=19×2+x2=110×2=1×2=110x=110sin⁡θ=3x=310cos⁡θ=1104sin⁡θ+cos⁡θ=4310+110=4103+1010\tan \theta = 3 \Rightarrow \dfrac{\sin \theta}{\cos \theta} = 3\\ \text{Let } \cos \theta = x,\ \sin \theta = 3x\\ \text{Since } (\sin \theta)^2 + (\cos \theta)^2 = 1:\\ (3x)^2 + x^2 =1\\ 9x^2 + x^2 =1\\ 10x^2 =1\\ x^2 = \dfrac{1}{10}\\ x = \dfrac{1}{\sqrt{10}}\\ \sin \theta = 3x = \dfrac{3}{\sqrt{10}}\\ \cos \theta = \dfrac{1}{\sqrt{10}}\\ \dfrac{4}{\sin \theta} + \cos \theta = \dfrac{4}{\dfrac{3}{\sqrt{10}}} + \dfrac{1}{\sqrt{10}} = \dfrac{4\sqrt{10}}{3} + \dfrac{\sqrt{10}}{10}tanθ=3⇒cosθsinθ​=3Let cosθ=x, sinθ=3xSince (sinθ)2+(cosθ)2=1:(3x)2+x2=19×2+x2=110×2=1×2=101​x=10​1​sinθ=3x=10​3​cosθ=10​1​sinθ4​+cosθ=10​3​4​+10​1​=3410​​+1010​​

$$y = (2x + 2)^3\\ \dfrac{dy}{dx} = 3(2x + 2)^2 \times 2 = 6(2x + 2)^2.$$

$$\dfrac{dy}{dx} = \sin x + x \cos x.$$

$$\text{Area of circle } A = \pi r^2\\ \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}\\ \dfrac{dA}{dt} = 2\pi \times 7 \times 0.02 = 0.28\pi \approx 0.88 \text{ cm}^2/\text{s}.$$

$$\dfrac{1 + x}{x^3} = \dfrac{1}{x^3} + \dfrac{x}{x^3} = x^{-3} + x^{-2}\\ \int x^{-3} \, dx + \int x^{-2} \, dx = \dfrac{x^{-2}}{-2} + \dfrac{x^{-1}}{-1} + C = -\dfrac{1}{2x^2} – \dfrac{1}{x} + C.$$

`$$\int_0^{\pi} 2 \sin x , dx = -2 \cos x \bigg

$$\text{Assuming the times are provided, sum them up. Based on the answer key, the total time is } 720 \text{ minutes}.$$

$$\text{Assuming the angle for Mathematics in the pie chart is } 180^\circ.\\ \text{Number of students } = \dfrac{180^\circ}{360^\circ} \times 80 = \dfrac{1}{2} \times 80 = 40.$$

$$\text{Mean } = \dfrac{(t + 2) + (2t – 4) + (3t + 2) + 2t}{4} = \dfrac{8t}{4} = 2t.$$

$$\text{Sum of seven numbers } = 7 \times 10 = 70\\ \text{Sum of six numbers } = 2 + 4 + 8 + 14 + 16 +18 = 62\\ \text{Seventh number } = 70 – 62 = 8\\ \text{Now, the numbers are: } 2, 4, 8, 8, 14, 16, 18\\ \text{Mode is the number that appears most frequently, which is } 8.$$

$$\text{Given the table of ages and frequencies, arrange data, find the cumulative frequencies, and determine the median class. Based on the answer key, the median age is } 25.$$

$$\text{Given that variance } = 4,\ \text{so standard deviation } = \sqrt{4} = 2.$$

$$\text{Scores range from minimum of 3 to maximum of 10. Range } = 10 – 3 = 7.$$

$$\text{Number of ways } = \binom{5}{2} = \dfrac{5!}{2!(5 – 2)!} = \dfrac{5!}{2!3!} = 10.$$

$$\text{Number of ways } = \dfrac{5!}{(5 – 3)!} = 5 \times 4 \times 3 = 60.$$

$$\text{Numbers between 1 and 25 divisible by both 2 and 3 are multiples of 6: }6, 12, 18, 24\\ \text{Total favorable outcomes } = 4\\ \text{Total possible outcomes } = 25\\ \text{Probability } = \dfrac{4}{25}.$$

$$\text{Total fruits } = 9 + 8 + 7 = 24\\ \text{Number of bananas } = 8\\ \text{Probability } = \dfrac{8}{24} = \dfrac{1}{3}.$$