Mathematics 2019 Jamb Examination

Isolate the term containing \( \displaystyle q \): \[ \frac{mn}{a^2} - 1 = \frac{pq}{b^2} \] Find a common denominator for the left side: \[ \frac{mn - a^2}{a^2} = \frac{pq}{b^2} \] Cross-multiply to solve for \( \displaystyle q \): \[ b^2(mn - a^2) = a^2 pq \implies q = \frac{b^2(mn - a^2)}{a^2 p} \]

Let \( \displaystyle h \) be the height of the tree. Using the tangent ratio: \[ \displaystyle \tan 78^\circ = \frac{h}{60} \] \[ \displaystyle h = 60 \times \tan 78^\circ \approx 60 \times 4.7046 = 282.276 \] To the nearest whole number, the height is 282m.

First, evaluate the expression in the bracket: \[ \displaystyle 2 \otimes 4 = (2 \times 4) + 2 - 4 = 8 + 2 - 4 = 6 \] Now evaluate the full expression using the result: \[ \displaystyle 3 \otimes 6 = (3 \times 6) + 3 - 6 = 18 + 3 - 6 = 15 \]

Total number of pupils \( \displaystyle = 4 + 13 + 30 + 44 + 9 = 100 \).
The angle for a sector is calculated as \( \displaystyle \frac{\text{Frequency}}{\text{Total}} \times 360^\circ \). \[ \displaystyle \text{Angle} = \frac{13}{100} \times 360^\circ = 46.8^\circ \]

Let \( \displaystyle n(P \cap B) = x \). Using the set theory formula: \[ \displaystyle n(P \cup B) = n(P) + n(B) - n(P \cap B) \] \[ \displaystyle 50 = 40 + 30 - x \implies 50 = 70 - x \] \[ \displaystyle x = 70 - 50 = 20 \]

Multiply the numerator and denominator by the conjugate of the denominator, which is \( \displaystyle \sqrt{2} + \sqrt{3} \): \[ \frac{(\sqrt{2} + \sqrt{3})(\sqrt{2} + \sqrt{3})}{(\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3})} \] Numerator: \( \displaystyle 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6} \)
Denominator: \( \displaystyle 2 - 3 = -1 \)
Final result: \( \displaystyle \frac{5 + 2\sqrt{6}}{-1} = -5 - 2\sqrt{6} \)

Length of chord = \(2r \sin (\frac{\theta}{2})\)

Using the complementary angle identity: \( \displaystyle \sin \theta = \cos(90^\circ - \theta) \). \[ \displaystyle p = 90 - 58 = 32^\circ \]

Numerator: \( \displaystyle \frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6} \)
Denominator: Find LCM of 4, 5, 3 which is 60. \[ \displaystyle \frac{15 + 36 - 20}{60} = \frac{31}{60} \] Final expression: \( \displaystyle \frac{5}{6} \div \frac{31}{60} = \frac{5}{6} \times \frac{60}{31} = \frac{50}{31} \)

\[ \begin{array}{r|l} & \underline{6x + 8} \\ x^2 - x - 1 & 6x^3 + 2x^2 - 5x + 1 \\ & 6x^3 - 6x^2 - 6x \\ & \underline{\phantom{6x^3}8x^2 + x + 1} \\ & 8x^2 - 8x - 8 \\ & \underline{\phantom{8x^2}9x + 9} \end{array} \] \[ \text{Quotient} = 6x + 8, \qquad \text{Remainder} = 9x + 9 \]