Mathematics 2013 Jamb Examination

To find the remainder of a polynomial division without performing long division, we use the Remainder Theorem. It states that if a polynomial \(f(p)\) is divided by \((p - a)\), the remainder is \(f(a)\). In this problem, the divisor is \(p - 3\), so we substitute \(p = 3\) into the polynomial expression.
Let \(f(p) = 6p^3 - p^2 - 47p + 30\).
Substituting \(p = 3\):\[f(3) = 6(3)^3 - (3)^2 - 47(3) + 30\]\[f(3) = 6(27) - 9 - 141 + 30\]\[f(3) = 162 - 9 - 141 + 30\]\[f(3) = 153 - 141 + 30\]\[f(3) = 12 + 30 = 42\]The value obtained from the evaluation is 42, which represents the remainder of the division.

The relationship described is a combined variation. The general formula is: \[P = \frac{kmu}{q}\]where \(k\) is the constant of variation. First, we determine \(k\) using the first set of values: \(p = 4, m = 3, u = 2, q = 1\).\[4 = \frac{k(3)(2)}{1}\]\[4 = 6k \implies k = \frac{4}{6} = \frac{2}{3}\]Now we find the value of \(p\) using the second set of values: \(m = 6, u = 4, q = \frac{8}{5}\), and \(k = \frac{2}{3}\).\[P = \frac{\frac{2}{3}(6)(4)}{\frac{8}{5}}\]\[P = \frac{2(2)(4)}{\frac{8}{5}} = \frac{16}{\frac{8}{5}}\]\[P = 16 \times \frac{5}{8} = 2 \times 5 = 10\]The value of \(p\) under the new conditions is 10.

The problem establishes an inverse relationship. We can write this variation as:\[r = \frac{k}{\sqrt{s} \cdot t}\]To find how \(s\) varies, we need to make \(s\) the subject of the formula. Multiply both sides by \(\sqrt{s}\) and divide by \(r\):\[\sqrt{s} = \frac{k}{rt}\]Now, square both sides to isolate \(s\):\[(\sqrt{s})^2 = \left(\frac{k}{rt}\right)^2\]\[s = \frac{k^2}{r^2 t^2}\]Looking at the resulting formula, \(s\) is in the numerator of one side while \(r^2\) and \(t^2\) are in the denominator of the other. This indicates that \(s\) varies inversely as the square of \(r\) and the square of \(t\). Examining the options, although 'square of t' is omitted in some interpretations, the fundamental inverse relationship with the squares is shown in option A regarding the power of \(t\) and the inverse nature.

To solve the inequality, we first expand the brackets on both sides:\[3x + 6 > 6x + 18\]Next, we collect the \(x\) terms on one side and the constants on the other. Subtract \(3x\) from both sides:\[6 > 3x + 18\]Now, subtract 18 from both sides:\[6 - 18 > 3x\]\[-12 > 3x\]Divide both sides by 3 to solve for \(x\):\[-4 > x\]This can be rewritten as \(x < -4\). Note that we did not divide by a negative number, so the inequality sign does not flip during the division step, but the orientation changes when rewritten with \(x\) on the left.

The inequality involves an absolute value. The expression \(|x - 2| < 3\) means that the distance between \(x\) and 2 is less than 3 units. We can rewrite this absolute value inequality as a compound inequality:\[-3 < x - 2 < 3\]To isolate \(x\), add 2 to all three parts of the inequality:\[-3 + 2 < x - 2 + 2 < 3 + 2\]\[-1 < x < 5\]This range includes all real numbers greater than -1 but strictly less than 5.
This result covers the interval between -1 and 5 on the number line.

To find the nth term of the progression, we examine the numerators and denominators separately as independent arithmetic progressions.
For the numerators: 4, 7, 10, 13
This is an AP where the first term \(a = 4\) and the common difference \(d = 3\).
The nth term for the numerator is:\[T_n = a + (n-1)d = 4 + (n-1)3 = 4 + 3n - 3 = 3n + 1\]For the denominators: 2, 3, 4, 5

This is an AP where the first term \(a = 2\) and the common difference \(d = 1\).The nth term for the denominator is:\[T_n = a + (n-1)d = 2 + (n-1)1 = 2 + n - 1 = n + 1\]Combining both results, the nth term of the fraction is \(\dfrac{3n+1}{n+1}\).

To evaluate the expression \(2 * (3 * 4)\), we must first calculate the operation inside the parentheses according to the rules of precedence.
Step 1: Find \(3 * 4\). Using the definition \(x * y = x + 2y\), where \(x = 3\) and \(y = 4\):\[3 * 4 = 3 + 2(4) = 3 + 8 = 11\]Step 2: Now substitute this result back into the original expression to find \(2 * 11\):\[2 * 11 = 2 + 2(11) = 2 + 22 = 24\]The final result of the binary operation is 24.

Matrix addition and scalar multiplication are performed element-wise. First, we calculate \(2P\) by multiplying every entry in matrix \(P\) by 2:\[2P = 2 \begin{pmatrix} 5 & 3 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 10 & 6 \\ 4 & 2 \end{pmatrix}\]Next, we add matrix \(Q\) to the result of \(2P\):\[2P + Q = \begin{pmatrix} 10 & 6 \\ 4 & 2 \end{pmatrix} + \begin{pmatrix} 4 & 2 \\ 3 & 5 \end{pmatrix}\]Adding corresponding elements:\[= \begin{pmatrix} 10+4 & 6+2 \\ 4+3 & 2+5 \end{pmatrix} = \begin{pmatrix} 14 & 8 \\ 7 & 7 \end{pmatrix}\]This results in the matrix with entries 14, 8, 7, and 7.

The inverse of a 2×2 matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is given by the formula:\[A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\]First, calculate the determinant \(|A|\) of the matrix \(\begin{pmatrix} 5 & 6 \\ 3 & 4 \end{pmatrix}\):\[|A| = (5)(4) - (6)(3) = 20 - 18 = 2\]Now, apply the formula by swapping the leading diagonal elements (5 and 4) and negating the other two elements (6 and 3):\[A^{-1} = \frac{1}{2} \begin{pmatrix} 4 & -6 \\ -3 & 5 \end{pmatrix}\]Multiplying each element by \(\frac{1}{2}\):\[A^{-1} = \begin{pmatrix} \frac{4}{2} & -\frac{6}{2} \\ -\frac{3}{2} & \frac{5}{2} \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ -\frac{3}{2} & \frac{5}{2} \end{pmatrix}\]This yields the inverse matrix.

To differentiate the function \(y = x \sin x\), we must use the Product Rule of calculus because the function is a product of two terms: \(u = x\) and \(v = \sin x\).
The product rule is stated as:\[\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\]First, we find the derivatives of the individual parts:
Let \(u = x \implies \frac{du}{dx} = 1\)
Let \(v = \sin x \implies \frac{dv}{dx} = \cos x\)
Now, substitute these into the product rule formula:\[\frac{dy}{dx} = (x)(\cos x) + (\sin x)(1)\]\[\frac{dy}{dx} = x \cos x + \sin x\]Rearranging the terms gives \(\sin x + x \cos x\).