Mathematics 2015 Jamb Examination

Convert the binary to base 10 and they convert back to base two
10011₂ + xxxxx₂ + 11100₂ + 101₂ = 1001111₂
(1 × 2⁴ + 0 × 2³ + 1 × 2² + 1 × 2¹ + 1 × 2⁰) + xxxxx₂ +(1 × 2⁴ + 1 × 2³ + 1 × 2² + 0 × 2¹ + 0 × 2⁰) + (1 × 2² + 0 × 2¹ + 1 × 2⁰)
= (16 + 0 + 0 + 2 + 1) + xxxxx₂ + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1)
=(64 + 0 + 0 + 8 + 4 + 2 + 1)
19 + xxxxx₂ + 33 = 79
xxxxx₂ + 52 = 79
xxxxx₂ = 79 − 52
xxxxx₂ = 27₁₀
\[ \begin{array}{c|l}
2 & 27 \\ \hline
2 & 13\ \text{rem}\ 1 \\
2 & 6\ \text{rem}\ 1 \\
2 & 3\ \text{rem}\ 0 \\
2 & 1\ \text{rem}\ 1 \\
  & 0\ \text{rem}\ 1 \quad \uparrow
\end{array}\] 27₁₀ = 11011₂
Therefore xxxxx₂ = 27₁₀ = 11011₂

We use the laws of logarithms: the product law \(\log(MN) = \log M + \log N\), the quotient law \(\log(M/N) = \log M - \log N\), and the power law \(\log M^p = p\log M\).\[ \log_5\left(\dfrac{y^2x^5}{125b^{1/2}}\right) \]\[ = \log_5(y^2x^5) - \log_5(125b^{1/2}) \]\[ = (\log_5 y^2 + \log_5 x^5) - (\log_5 125 + \log_5 b^{1/2}) \]\[ = 2\log_5 y + 5\log_5 x - (\log_5 5^3 + \dfrac{1}{2}\log_5 b) \]\[ = 2\log_5 y + 5\log_5 x - 3 - \dfrac{1}{2}\log_5 b \]\[ = 2\log_5 y + 5\log_5 x - \dfrac{1}{2}\log_5 b - 3 \]

To integrate the polynomial, we apply the power rule for integration: \(\int x^n dx = \dfrac{x^{n+1}}{n+1} + C\) for each term:\[ \int 5x^3 dx = \dfrac{5x^{3+1}}{3+1} = \dfrac{5x^4}{4} \]\[ \int 7x^2 dx = \dfrac{7x^{2+1}}{2+1} = \dfrac{7x^3}{3} \]\[ \int -2x dx = \dfrac{-2x^{1+1}}{1+1} = -x^2 \]\[ \int 5 dx = 5x \]Combining these terms and adding the constant of integration \(C\):\[ \dfrac{5x^4}{4} + \dfrac{7x^3}{3} - x^2 + 5x + C \]

The curved surface area (C.S.A.) of a cone is given by the formula:\[ \text{C.S.A.} = \pi r l \]Where \(r\) is the radius and \(l\) is the slant height. We are given height \(h = 4\) cm and radius \(r = 3\) cm. We find \(l\) using Pythagoras' theorem:\[ l = \sqrt{r^2 + h^2} \]\[ l = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{cm} \]Now calculate the area:\[ \text{C.S.A.} = 3.14 \times 3 \times 5 \]\[ \text{C.S.A.} = 3.14 \times 15 = 47.1 \, \text{cm}^2 \]

To simplify the sum of these fractions, we find the common denominator, which is the product of the two denominators: \((x + 1)(x - 1)\).\[ \dfrac{1}{x+1} + \dfrac{1}{x-1} = \dfrac{(x-1) + (x+1)}{(x+1)(x-1)} \]\[ = \dfrac{x - 1 + x + 1}{(x + 1)(x - 1)} \]\[ = \dfrac{2x}{(x + 1)(x - 1)} \]This result matches option B.

The area \(A\) of a circle is calculated using the formula:\[ A = \pi r^2 \]Given radius \(r = 4\) cm and \(\pi = 3.142\):\[ A = 3.142 \times (4)^2 \]\[ A = 3.142 \times 16 \]\[ A = 50.272 \, \text{cm}^2 \]Correct to one decimal place, the area is 50.3 cm².

The area of an ellipse is given by the formula:\[ \text{Area} = \pi a b \]Where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. The length of the major axis is \(2a = 14\) cm, so \(a = 7\) cm. We take \(\pi\) as \(22/7\).\[ 132 = \dfrac{22}{7} \times 7 \times b \]\[ 132 = 22b \]\[ b = \dfrac{132}{22} = 6 \, \text{cm} \]The minor axis is \(2b = 2 \times 6 = 12\) cm.

The volume \(V\) of a cone is given by the formula:\[ V = \dfrac{1}{3}\pi r^2 h \]Given radius \(r = 5\) cm and height \(h = 6\) cm. Using \(\pi \approx 3.14159\):\[ V = \dfrac{1}{3} \times 3.14159 \times 5^2 \times 6 \]\[ V = 3.14159 \times 25 \times 2 \]\[ V = 3.14159 \times 50 \]\[ V = 157.079 \, \text{cm}^3 \]Rounding to the nearest whole number, we get 157 cm³.

There is a logical contradiction in the question parameters. Usually, \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). However, if \(P(A \cup B) = 1/12\), this is less than either \(P(A)\) or \(P(B)\), which is impossible for a union of sets. Assuming the intended question meant \(P(A \cup B)\) is the unknown and the intersection is 1/12:\[ P(A \cup B) = \dfrac{1}{6} + \dfrac{1}{4} - \dfrac{1}{12} \]\[ P(A \cup B) = \dfrac{2+3-1}{12} = \dfrac{4}{12} = \dfrac{1}{3} \]Correcting the likely error in the prompt's value assignment to solve for the intersection:
If \(P(A \cup B) = 1/3\), then \(P(A \cap B) = 1/6 + 1/4 - 1/3 = 1/12\). Since the options do not provide 1/12, and if we strictly use the formula to find intersection using the prompt's own flawed values:\[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \]\[= \dfrac{1}{6} + \dfrac{1}{4} - \dfrac{1}{12} = \dfrac{4}{12} = \dfrac{1}{3} \]

The power set of a set \(Z\) is the set of all possible subsets of \(Z\). The number of elements in the power set is given by the formula:\[ \text{Power set size} = 2^n \]where \(n\) is the number of elements in set \(Z\).
Set \(Z = \{1, 2, 4, 5\}\) contains \(n = 4\) elements.\[ \text{Power set size} = 2^4 = 16 \]There are 16 possible subsets including the empty set and the set itself.