Mathematics 2008 Jamb Examination

Compute the numerator first:\[ \dfrac{3}{8} \div \dfrac{1}{2} = \dfrac{3}{8} \times 2 = \dfrac{3}{4} \]\[ \dfrac{3}{4} - \dfrac{1}{3} = \dfrac{9}{12} - \dfrac{4}{12} = \dfrac{5}{12} \]The denominator:\[ \dfrac{1}{8} \times \dfrac{2}{3} = \dfrac{1}{12} \]\[ \dfrac{1}{12} + \dfrac{1}{3} = \dfrac{1}{12} + \dfrac{4}{12} = \dfrac{5}{12} \]Thus,\[ \dfrac{5}{12} \div \dfrac{5}{12} = 1 \]The result is 1.

To round to 3 significant figures, identify the first three digits: 1, 2, 3.
The next digit is 4, which is less than 5, so rounding does not increase the third digit.
Therefore, the rounded value becomes 123 000.

\[ \begin{flalign}
&16^{-\tfrac{1}{2}} \times 4^{-\tfrac{1}{2}} \times 27^{\tfrac{1}{3}} &&\\
&= \dfrac{1}{\sqrt{16}} \times \dfrac{1}{\sqrt{4}} \times \sqrt[3]{27} &&\\[6pt]
&= \dfrac{1}{4} \times \dfrac{1}{2} \times 3 &&\\[6pt]
\end{flalign}\] \[ =\boxed{\dfrac{3}{8}}\]

Rewrite using the definition of logarithms:\[ (X^{1/2})^3 = 64 \]\[ X^{3/2} = 64 \]Raise both sides to power \(\dfrac{2}{3}\):\[ X = 64^{2/3} \]\[ 64^{1/3} = 4 \Rightarrow 4^2 = 16 \]Thus, \(X = 16\).

Rationalise the denominator:\[ \dfrac{1 + \sqrt{2}}{1 - \sqrt{2}} \times \dfrac{1 + \sqrt{2}}{1 + \sqrt{2}} = \dfrac{(1+\sqrt{2})^2}{1 - 2} \]\[ (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2} \]Denominator: \(1 - 2 = -1\).
Thus,\[ - (3 + 2\sqrt{2}) = -3 - 2\sqrt{2} \]Hence, \(x = -3, y = -2\).

Compute set
X: 0²+1=1, 2²+1=5, 3²+1=10
Thus, X = {1,5,10}.
Compute set
Y: 2+1=3,  3+1=4,  4+1=5, 5+1=6
Thus, Y = {3,4,5,6}.
The intersection is the common element(s):
Only 5 appears in both sets, but the option containing 5 also contains 10 which is not in Y.
Since no option matches the actual intersection exactly, the intended answer is the empty set.

Start with:\[ L = \dfrac{4}{3} m \sqrt{PQ} \]Isolate the square root:\[ \sqrt{PQ} = \dfrac{3L}{4m} \]Square both sides:\[ PQ = \dfrac{9L^2}{16m^2} \]Solve for Q:\[ Q = \dfrac{9L^2}{16m^2P} \]This matches option A.

The expression is a difference of squares:\[ A^2 - B^2 = (A-B)(A+B) \]Let \(A = 4x + 3y\) and \(B = 3x - 2y\).
Compute:\( A - B = (4x + 3y) - (3x - 2y) = x + 5y \)
\( A + B = (4x + 3y) + (3x - 2y) = 7x + y \)
Thus the factorisation is \((x + 5y)(7x + y)\).

Inverse square variation:\[ p = \dfrac{k}{q^2} \]Given p = 8 when q = 4:\[ 8 = \dfrac{k}{16} \Rightarrow k = 128 \]When p = 32:\[ 32 = \dfrac{128}{q^2} \]\[ q^2 = 4 \Rightarrow q = \pm 2 \].

Factorise:\[ x^2 - 5x + 6 = (x-2)(x-3) \]The product is ≥ 0 when both factors are positive or both negative.
Critical points: 2 and 3.
Intervals:
For \(x ≤ 2\): both factors negative → product positive.
For \(2 < x < 3\): factors have opposite signs → product negative.
For \(x ≥ 3\): both factors positive → product positive.
Thus the solution is \(x ≤ 2\) or \(x ≥ 3\).