Mathematics 2014 Jamb Examination

To factorise the quadratic expression \(2y^2 - 15xy + 18x^2\), we look for two numbers whose product is equal to the product of the first and last coefficients \(2 \times 18 = 36\) and whose sum is the middle coefficient \(-15\).
The two numbers that satisfy these conditions are \(-3\) and \(-12\), since:\[ -3 \times -12 = 36 \]\[ -3 + (-12) = -15 \]We can now rewrite the middle term and factor by grouping:\[ 2y^2 - 3xy - 12xy + 18x^2 \]\[ y(2y - 3x) - 6x(2y - 3x) \]\[ (2y - 3x)(y - 6x) \]This shows that the expression is decomposed into these two linear factors.

According to the Factor Theorem, if \((y - a)\) is a factor of a polynomial \(f(y)\), then \(f(a) = 0\). In this case, since \(y - 1\) is a factor, we set \(y = 1\) and equate the polynomial to zero.
Let \(f(y) = y^3 + 4y^2 + ky - 6\).
Then:\[ f(1) = (1)^3 + 4(1)^2 + k(1) - 6 = 0 \]\[ 1 + 4 + k - 6 = 0 \]\[ 5 + k - 6 = 0 \]\[ k - 1 = 0 \]\[ k = 1 \]Thus, when \(k = 1\), the expression leaves no remainder when divided by \(y - 1\).

This is a problem of direct variation. The relationship is expressed as:\[ y = kw^2 \]where \(k\) is the constant of variation. First, we find \(k\) using the given values \(y = 8\) and \(w = 2\):\[ 8 = k(2^2) \]\[ 8 = 4k \]\[ k = \frac{8}{4} = 2 \]Now that we have the constant, we find the new value of \(y\) when \(w = 3\):\[ y = 2(3^2) \]\[ y = 2(9) \]\[ y = 18 \]The value of \(y\) increases proportionally with the square of \(w\).

The relationship described is a joint variation, written as:\[ P = \frac{kQ}{R} \]where \(k\) is the constant of proportionality. To find \(k\), we substitute the initial values provided:\[ 27 = \frac{k(36)}{16} \]Simplify the fraction \(36/16\) by dividing both by 4:\[ 27 = \frac{9k}{4} \]Solve for \(k\):\[ 27 \times 4 = 9k \]\[ 108 = 9k \]\[ k = \frac{108}{9} = 12 \]Substituting \(k = 12\) back into the general variation formula gives the specific relation:\[ P = \frac{12Q}{R} \]

To solve the inequality \(\dfrac{x - 5}{x + 3} < -1\), we must not cross-multiply by \(x+3\) because its sign is unknown. Instead, we move all terms to one side:\[ \frac{x - 5}{x + 3} + 1 < 0 \]\[ \frac{x - 5 + 1(x + 3)}{x + 3} < 0 \]\[ \frac{x - 5 + x + 3}{x + 3} < 0 \]\[ \frac{2x - 2}{x + 3} < 0 \]The critical points where the expression can change sign are where the numerator is zero (\(2x - 2 = 0 \Rightarrow x = 1\)) and where the denominator is zero (\(x + 3 = 0 \Rightarrow x = -3\)).
Testing intervals:
1. For \(x < -3\): \(\frac{-}{ -} = +\) (False)
2. For \(-3 < x < 1\): \(\frac{-}{ +} = -\) (True)
3. For \(x > 1\): \(\frac{+}{ +} = +\) (False)
Therefore, the solution is the interval where the expression is negative: \(-3 < x < 1\).

To solve the inequality, we first eliminate the fractions by multiplying every term by the Least Common Multiple (LCM) of the denominators 2, 4, 6, and 12, which is 12.\[ 12\left(\frac{x}{2}\right) + 12\left(\frac{3}{4}\right) \leq 12\left(\frac{5x}{6}\right) - 12\left(\frac{7}{12}\right) \] \[ 6x + 9 \leq 10x - 7 \] Next, we collect like terms. Move \(6x\) to the right side and \(-7\) to the left side: \[ 9 + 7 \leq 10x - 6x \] \[ 16 \leq 4x \] Now, divide both sides by 4: \[ \frac{16}{4} \leq x \] \[ 4 \leq x \] This is equivalent to \(x \geq 4\). The solution set includes all values of x that are greater than or equal to 4.

In an Arithmetic Progression (A.P.), the \(n^{th}\) term is given by the formula \(U_n = a + (n - 1)d\), where \(a\) is the first term and \(d\) is the common difference.
For the 4th term (\(n=4\)): \[ a + 3d = 13 \quad \text{--- (Equation 1)} \] For the 10th term (\(n=10\)): \[ a + 9d = 31 \quad \text{--- (Equation 2)} \] Subtract Equation 1 from Equation 2 to find \(d\): \[ (a + 9d) - (a + 3d) = 31 - 13 \] \[ 6d = 18 \] \[ d = \frac{18}{6} = 3 \] Substitute \(d = 3\) into Equation 1 to find \(a\): \[ a + 3(3) = 13 \] \[ a + 9 = 13 \] \[ a = 4 \] Now, find the 24th term (\(n=24\)): \[ U_{24} = a + 23d \] \[ U_{24} = 4 + 23(3) \] \[ U_{24} = 4 + 69 = 73 \]

In a Geometric Progression (G.P.), the common ratio \(r\) is found by dividing any term by its preceding term: \(r = \dfrac{U_2}{U_1}\).
From the series provided: First term \(U_1 = \sqrt{10} + \sqrt{5}\) Second term \(U_2 = 2\sqrt{10} + 2\sqrt{5}\) Calculating the ratio: \[ r = \frac{2\sqrt{10} + 2\sqrt{5}}{\sqrt{10} + \sqrt{5}} \] Factor out the common factor of 2 in the numerator: \[ r = \frac{2(\sqrt{10} + \sqrt{5})}{\sqrt{10} + \sqrt{5}} \] Cancelling the common binomial factor \((\sqrt{10} + \sqrt{5})\) from the numerator and denominator: \[ r = 2 \] The common ratio is 2, meaning each term is exactly twice the preceding term.

\(x * y = x^y\).
\(x^2 = 12 - x\): \[ x^2 + x - 12 = 0 \] \[ (x + 4)(x - 3) = 0 \] \[ x = -4, 3 \] 

We can solve this matrix equation by expanding it into a system of simultaneous linear equations. Multiplying the matrix by the vector gives:

1. \(5x + 2y = 7\)
2. \(-6x - 7y = -11\)

To find \(y\), we can use the elimination method. Multiply equation (1) by 6 and equation (2) by 5 to eliminate \(x\): \[ 6(5x + 2y) = 6(7) \Rightarrow 30x + 12y = 42 \] \[ 5(-6x - 7y) = 5(-11) \Rightarrow -30x - 35y = -55 \] Add the two resulting equations: \[ (30x - 30x) + (12y - 35y) = 42 - 55 \] \[ -23y = -13 \]