Mathematics 2014 Jamb Examination

To find the value of x, we first calculate the determinant of the \(2 \times 2\) matrix. The determinant of a matrix \(\begin{vmatrix} a & b \\ c & d \end{vmatrix}\) is given by \(ad - bc\).
Applying this to the given matrix: \[ (-x)(4) - (-1)(12) = -12 \] \[ -4x + 12 = -12 \] Now, we isolate the term with x by subtracting 12 from both sides of the equation: \[ -4x = -12 - 12 \] \[ -4x = -24 \] Finally, divide both sides by -4 to solve for x: \[ x = \frac{-24}{-4} = 6 \] Thus, the value of x that satisfies the determinant equation is 6.

\[ \begin{vmatrix}
0 & 3 & 2 \\
1 & 7 & 8 \\
0 & 5 & 4
\end{vmatrix}\] \[ = 0
\begin{vmatrix}
7 & 8 \\
5 & 4
\end{vmatrix}
- 3
\begin{vmatrix}
1 & 8 \\
0 & 4
\end{vmatrix}
+ 2
\begin{vmatrix}
1 & 7 \\
0 & 5
\end{vmatrix}\] \[ = 0(7 \times 4 - 8 \times 5)
- 3(1 \times 4 - 8 \times 0)
+ 2(1 \times 5 - 7 \times 0)\] \[ = 0(28 - 40) - 3(4 - 0) + 2(5 - 0)\] \[ = 0 - 12 + 10\] \[ = -2\]

For any regular polygon, the sum of the interior angle and the exterior angle at any vertex is 180°.
First, we find the exterior angle: \[ \text{Exterior Angle} = 180^\circ - 135^\circ = 45^\circ \] The sum of all exterior angles of any convex polygon is always 360°. For a regular polygon with \(n\) sides, each exterior angle is \(\dfrac{360^\circ}{n}\).\[ 45^\circ = \frac{360^\circ}{n} \] \[ n = \frac{360^\circ}{45^\circ} \] \[ n = 8 \] A polygon with 8 sides is an octagon. Therefore, a regular octagon has interior angles of 135°.

The capacity of a cylindrical tank refers to its volume \(V\). The formula for the volume of a cylinder is: \[ V = \pi r^2 h \] Where: \(V = 6160\, \text{m}^3\) \(r = 28\, \text{m}\) \(\pi \approx \dfrac{22}{7}\) We need to find the depth (height), \(h\): \[ 6160 = \frac{22}{7} \times (28)^2 \times h \] \[ 6160 = \frac{22}{7} \times 28 \times 28 \times h \] \[ 6160 = 22 \times 4 \times 28 \times h \] \[ 6160 = 2464h \] Divide both sides by 2464: \[ h = \frac{6160}{2464} = 2.5 \, \text{m} \] The depth of the tank is 2.5 metres.

The mid-point \(M\) of a line segment joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is calculated using the mid-point formula: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \] Substitute the coordinates of S(-5, 4) and T(-3, -2): \[ x_{mid} = \frac{-5 + (-3)}{2} = \frac{-8}{2} = -4 \] \[ y_{mid} = \frac{4 + (-2)}{2} = \frac{2}{2} = 1 \] The mid-point is (-4, 1). This point lies exactly halfway between S and T on the Cartesian plane.

The gradient \(m\) of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:\[ m = \dfrac{y_2 - y_1}{x_2 - x_1} \]Substituting the given values \(m = \dfrac{1}{2}\), \((x_1, y_1) = (x, 4)\), and \((x_2, y_2) = (1, 2)\):\[ \frac{1}{2} = \frac{2 - 4}{1 - x} \]\[ \frac{1}{2} = \frac{-2}{1 - x} \]Cross-multiplying to solve for x:\[ 1(1 - x) = 2(-2) \]\[ 1 - x = -4 \]\[ -x = -4 - 1 \]\[ -x = -5 \]\[ x = 5 \]Therefore, the x-coordinate of the first point must be 5 to yield a gradient of 0.5.

To find the segment of the line lying between the axes, we first determine the intercepts.
For the x-intercept, set \(y = 0\):\[ 0 - 4x + 3 = 0 \]\[ 4x = 3 \Rightarrow x = \frac{3}{4} \] So the first point is \(A(\frac{3}{4}, 0)\).
For the y-intercept, set \(x = 0\):\[ y - 4(0) + 3 = 0 \]\[ y + 3 = 0 \Rightarrow y = -3 \] So the second point is \(B(0, -3)\).
The mid-point \(M(x, y)\) is calculated as:\[ x = \dfrac{x_1 + x_2}{2} = \dfrac{\frac{3}{4} + 0}{2} = \frac{3}{8} \]\[ y = \dfrac{y_1 + y_2}{2} = \dfrac{0 + (-3)}{2} = -\frac{3}{2} \] The coordinates of the mid-point are \((\frac{3}{8}, -\frac{3}{2})\).

Using the trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can find \(\cos \theta\):\[ \left(\frac{12}{13}\right)^2 + \cos^2 \theta = 1 \]\[ \frac{144}{169} + \cos^2 \theta = 1 \]\[ \cos^2 \theta = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169} \]\[ \cos \theta = \sqrt{\frac{25}{169}} = \frac{5}{13} \] Now, evaluate \(1 + \cos \theta\):\[ 1 + \frac{5}{13} = \frac{13 + 5}{13} = \frac{18}{13} \]The result is the sum of the unit value and the adjacent ratio of the triangle.

To find the derivative of the function with respect to x, we apply the power rule of differentiation, which states that if \(y = ax^n\), then \(\dfrac{dy}{dx} = nax^{n-1}\).
Differentiating each term of \(y = 4x^3 - 2x^2 + x\) individually:
For \(4x^3\): \(3 \times 4x^{3-1} = 12x^2\)
For \(-2x^2\): \(2 \times -2x^{2-1} = -4x\) For \(x\): \(1 \times x^{1-1} = 1x^0 = 1\)
Combining these results, we get:\[ \frac{dy}{dx} = 12x^2 - 4x + 1 \]This expression represents the instantaneous rate of change or the gradient function of the original curve.

To differentiate \(y = \cos 3x\), we must use the chain rule. The chain rule states that if \(y = f(g(x))\), then \(\dfrac{dy}{dx} = f'(g(x)) \times g'(x)\).
In this case: Let \(u = 3x\), so \(y = \cos u\). Differentiating \(u\) with respect to \(x\):\[ \frac{du}{dx} = 3 \] Differentiating \(y\) with respect to \(u\):\[ \frac{dy}{du} = -\sin u \] Now, multiply them together:\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]\[ \frac{dy}{dx} = (-\sin u) \times 3 \]\[ \frac{dy}{dx} = -3 \sin 3x \]Note that the derivative of cosine is negative sine, and the coefficient of the inner function (3) becomes a multiplier.