Mathematics 2023 Jamb Examination

When a coin is tossed 3 times, the total number of outcomes is \(2^3 = 8\). The only outcome with no heads is (T, T, T).
Therefore, the number of outcomes with at least one head is \(8 - 1 = 7\).
Probability = \(\frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{7}{8}\).

First, find the gradient of \(3x + 2y + 4 = 0\): \(2y = -3x - 4 \implies y = -\frac{3}{2}x - 2\). The gradient \(m_1 = -\frac{3}{2}\).
Perpendicular lines have gradients where \(m_1 \times m_2 = -1\), so \(m_2 = \frac{2}{3}\).
Using \(y - y_1 = m(x - x_1)\):\[y - 3 = \frac{2}{3}(x - 2)\]\[3(y - 3) = 2(x - 2) \implies 3y - 9 = 2x - 4\]\[3y = 2x + 5\]

Expand the function first: \(y = x^{2/3}(2x - x^2) = 2x^{5/3} - x^{8/3}\).
Differentiate using the power rule:\[\frac{dy}{dx} = 2\left(\frac{5}{3}x^{2/3}\right) - \left(\frac{8}{3}x^{5/3}\right)\]\[\frac{dy}{dx} = \frac{10x^{2/3}}{3} - \frac{8x^{5/3}}{3}\]

The first five prime numbers are 2, 3, 5, 7, 11.
Mean (\(\bar{x}\)) = \(\frac{2+3+5+7+11}{5} = \frac{28}{5} = 5.6\).
Mean Deviation = \(\frac{\sum |x - \bar{x}|}{n}\):\[|2-5.6| + |3-5.6| + |5-5.6| + |7-5.6| + |11-5.6| = 3.6 + 2.6 + 0.6 + 1.4 + 5.4 = 13.6\]\[\text{M.D.} = \frac{13.6}{5} = 2.72\]

Find intersection points: \(2x^2 + 10 = 4x + 16 \implies 2x^2 - 4x - 6 = 0 \implies x^2 - 2x - 3 = 0\).
Factors: \((x - 3)(x + 1) = 0\), so \(x = 3, -1\).
Area = \(\int_{-1}^{3} [(4x + 16) - (2x^2 + 10)] dx = \int_{-1}^{3} (6 + 4x - 2x^2) dx\)\[[6x + 2x^2 - \frac{2x^3}{3}]_{-1}^{3} = (18 + 18 - 18) - (-6 + 2 + \frac{2}{3}) = 18 - (-3.33) = 21.33 = \frac{64}{3}\]

Expand 402_y in base 10:\[4(y^2) + 0(y^1) + 2(y^0) = 102\]\[4y^2 + 2 = 102 \implies 4y^2 = 100\]\[y^2 = 25 \implies y = 5\]

Apply log laws: \(\log(A) + \log(B) = \log(AB)\) and \(n\log A = \log A^n\):\[\log((y + 8)(y - 8)) = \log(3^2 \times 5^2)\]\[y^2 - 64 = 9 \times 25 = 225\]\[y^2 = 225 + 64 = 289 \implies y = \sqrt{289} = 17\]Since log argument must be positive, y = 17 (though \(\pm 17\) is mathematically from \(y^2\)).

Use the set formula: \(n(E \cup F) = n(E) + n(F) - n(E \cap F)\).\[500 = 350 + 400 - n(E \cap F)\]\[500 = 750 - n(E \cap F) \implies n(E \cap F) = 750 - 500 = 250\]

Use difference of two squares twice:\[(4x^2)^2 - (y^2)^2 = (4x^2 - y^2)(4x^2 + y^2)\]\[= (2x - y)(2x + y)(4x^2 + y^2)\]

A – B = {elements in A not in B} = {1, 3, 5}.
B – A = {elements in B not in A} = {8}.
Union = {1, 3, 5, 8}.