Mathematics 2011 Jamb Examination

To find the remainder, we perform polynomial long division of \( X^3 - 2X^2 + 3X - 3 \) by \( X^2 + 1 \).
Step 1: Divide the first term of the dividend (\( X^3 \)) by the first term of the divisor (\( X^2 \)) to get \( X \).
Multiply \( X(X^2 + 1) = X^3 + X \).
Subtract this from the dividend: \[ (X^3 - 2X^2 + 3X - 3) - (X^3 + X) = -2X^2 + 2X - 3 \]Step 2: Divide the new first term (\( -2X^2 \)) by \( X^2 \) to get \( -2 \).
Multiply \( -2(X^2 + 1) = -2X^2 - 2 \).
Subtract this from the current polynomial: \[ (-2X^2 + 2X - 3) - (-2X^2 - 2) = 2X - 1 \]Since the degree of \( 2X - 1 \) is less than the degree of \( X^2 + 1 \), the remainder is \( 2X - 1 \).

This expression is a classic example of the difference of two squares. The general formula for factorising the difference of two squares is: \[ a^2 - b^2 = (a - b)(a + b) \]In the given expression \( 9y^2 - 16x^2 \), we can rewrite the terms as perfect squares: \[ 9y^2 = (3y)^2 \] \[ 16x^2 = (4x)^2 \]Applying the formula where \( a = 3y \) and \( b = 4x \), we get: \[ (3y)^2 - (4x)^2 = (3y - 4x)(3y + 4x) \]This represents the complete factorisation of the quadratic expression.

We can solve this using the substitution method or elimination method. Let's use substitution.
From the second equation: \[ x + 3y = 0 \implies x = -3y \]Now, substitute \( x = -3y \) into the first equation: \[ -2(-3y) - 5y = 3 \] \[ 6y - 5y = 3 \] \[ y = 3 \]Now substitute the value of \( y \) back into the expression for \( x \): \[ x = -3(3) = -9 \]Therefore, the values are \( x = -9 \) and \( y = 3 \).

The relationship is given by direct variation: \[ x = k\sqrt{y} \]First, find the constant of proportionality \( k \) using the values \( x = 81 \) and \( y = 9 \): \[ 81 = k\sqrt{9} \] \[ 81 = 3k \implies k = 27 \]Now, find \( x \) when \( y = 1\frac{7}{9} \). Convert the mixed fraction to an improper fraction: \[ y = \frac{(1 \times 9) + 7}{9} = \frac{16}{9} \]Substitute \( k \) and the new \( y \) into the formula: \[ x = 27 \times \sqrt{\frac{16}{9}} \] \[ x = 27 \times \frac{4}{3} \] \[ x = 36 \]

The relationship for inverse variation with the cube is: \[ T = \frac{k}{R^3} \quad \text{or} \quad k = T \times R^3 \]Find the constant \( k \) using \( R = 3 \) and \( T = \frac{2}{81} \): \[ k = \frac{2}{81} \times 3^3 \] \[ k = \frac{2}{81} \times 27 \] \[ k = \frac{2}{3} \]Now, find \( T \) when \( R = 2 \) using the constant \( k = \frac{2}{3} \): \[ T = \frac{2/3}{2^3} \] \[ T = \frac{2/3}{8} \] \[ T = \frac{2}{3 \times 8} = \frac{2}{24} = \frac{1}{12} \]

To solve the linear inequality, we first expand the terms on both sides of the expression to remove the brackets:\[-6(x + 3) \le 4(x - 2)\]\[-6x - 18 \le 4x - 8\]Next, we rearrange the inequality by collecting terms containing \(x\) on one side and constant terms on the other. By adding \(6x\) to both sides and adding \(8\) to both sides, we obtain:\[-18 + 8 \le 4x + 6x\]\[-10 \le 10x\]Finally, we divide both sides by 10 to isolate \(x\), which gives:\[-1 \le x\]This can be rewritten as \(x \ge -1\). Therefore, any value of \(x\) greater than or equal to \(-1\) satisfies the given inequality.

We begin by rearranging the quadratic inequality to set the right-hand side to zero:\[x^2 + 2x - 15 > 0\]Now, we factorise the quadratic expression. We look for two integers that multiply to \(-15\) and add to \(2\); these are \(5\) and \(-3\):\[(x + 5)(x - 3) > 0\]The critical values where the expression equals zero are \(x = -5\) and \(x = 3\). To satisfy the inequality, the product of the two factors must be positive (greater than zero). This occurs when both factors are positive or both are negative. Testing the intervals, we find that the expression is positive when \(x\) is less than \(-5\) or when \(x\) is greater than \(3\).

This series is an arithmetic progression where the first term \(a = 3\) and the common difference \(d = 6 - 3 = 3\).
The formula for the sum of the first \(n\) terms is:\[S_n = \frac{n}{2} [2a + (n - 1)d]\]Substituting \(n = 18\), \(a = 3\), and \(d = 3\) into the formula:\[S_{18} = \frac{18}{2} [2(3) + (18 - 1)3]\]\[S_{18} = 9 [6 + 17(3)]\]\[S_{18} = 9 [6 + 51]\]\[S_{18} = 9 \times 57\]\[S_{18} = 513\]Thus, the total sum of the first eighteen terms of the series is 513.

In a geometric progression, the \(n\)-th term is \(U_n = ar^{n-1}\).
Given:\[U_2 = ar = 4\]\[U_4 = ar^3 = 16\]We find the common ratio \(r\) by dividing the fourth term by the second term:\[\frac{ar^3}{ar} = \frac{16}{4}\]\[r^2 = 4 \implies r = 2\]Substituting \(r = 2\) into the equation for the second term:\[a(2) = 4 \implies a = 2\]Now, we use the sum formula for a geometric progression where \(r > 1\):\[S_n = \frac{a(r^n - 1)}{r - 1}\]\[S_5 = \frac{2(2^5 - 1)}{2 - 1}\]\[S_5 = \frac{2(32 - 1)}{1} = 2 \times 31 = 62\]

To solve this, we apply the definition of the binary operation \(\oplus\) to the given values \(x = 3\) and \(y = -\frac{2}{3}\):\[x \oplus y = xy + x + y\]\[3 \oplus \left(-\frac{2}{3}\right) = 3\left(-\frac{2}{3}\right) + 3 + \left(-\frac{2}{3}\right)\]First, we calculate the product of the terms:\[3 \times \left(-\frac{2}{3}\right) = -2\]Now, we sum all the resulting parts:\[-2 + 3 - \frac{2}{3}\]\[1 - \frac{2}{3} = \frac{1}{3}\]The evaluation of the binary operation with the specified inputs results in one-third.