Mathematics 2008 Jamb Examination

Rationalise the denominator:\[ \dfrac{1 + \sqrt{2}}{1 - \sqrt{2}} \times \dfrac{1 + \sqrt{2}}{1 + \sqrt{2}} = \dfrac{(1 + \sqrt{2})^2}{1 - 2} \]Expand the numerator:\[ (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2} \]Thus:\[ \dfrac{3 + 2\sqrt{2}}{-1} = -3 - 2\sqrt{2} \]This corresponds to \( x = -3 \) and \( y = -2 \).

Compute elements of each set:
For X:
\( n = 0 \rightarrow 1 \)
\( n = 2 \rightarrow 5 \)
\( n = 3 \rightarrow 10 \)
Thus X = {1, 5, 10}.
For Y:
\( n = 2 \rightarrow 3 \)
\( n = 3 \rightarrow 4 \)
\( n = 5 \rightarrow 6 \)
Thus Y = {3, 4, 6}.
No common elements occur between X and Y, so the intersection is the empty set.

Use the inclusion–exclusion principle:
Total customers = Mathematics + English − Both\[ 30 + 20 - 10 = 40 \]This represents the total unique customers.

Start with:\[ L = \dfrac{4}{3}M\sqrt{PQ} \]Multiply both sides by \( \dfrac{3}{4M} \):\[ \dfrac{3L}{4M} = \sqrt{PQ} \]Square both sides:\[ \dfrac{9L^2}{16M^2} = PQ \]Divide by P:\[ Q = \dfrac{9L^2}{16M^2 P} \]This matches option A.

A polynomial divisible by \( x - a \) satisfies f(a) = 0.
Substitute x = 4:\[ 2(4^2) - k(4) - 12 = 0 \]\[ 32 - 4k - 12 = 0 \]\[ 20 = 4k \]\[ k = 5 \].

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

Direct proportionality gives:\[ x - 3 = ky^2 \]Given x = 5 and y = 2:\[ 5 - 3 = 4k \Rightarrow k = \dfrac{2}{4} = \dfrac{1}{2} \]For y = 6:\[ x - 3 = \dfrac{1}{2}(36) = 18 \]\[ x = 21 \].

Factorise the quadratic:\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]The product is ≥ 0 when x is outside the roots.
Thus the solution set is:
\( x \le 2 \) or \( x \ge 3 \).

Let first term = a and common difference = d.
\( T_5 = a + 4d = 24 \)
\( T_{11} = a + 10d = 96 \)
Subtract equations:\[ 6d = 72 \Rightarrow d = 12 \]Substitute back:\[ a + 48 = 24 \Rightarrow a = -24 \].

To check commutativity, compare \(x * y = 2x - 3y + 2\) with \(y * x = 2y - 3x + 2\). These are not equal for all positive integers, so the operation is not commutative.
To test closure, consider \(x = 1, y = 2\): \[1 * 2 = 2(1) - 3(2) + 2 = 2 - 6 + 2 = -2\] Since \(-2\) is not a positive integer, the operation is not closed on the set of positive integers.
It is therefore neither commutative nor closed.