Mathematics 2004 Jamb Examination

We are given: \[ \alpha + \beta = 2 \] \[ \text{Distance between } (1, \alpha) \text{ and } (\beta, 1) = 3 \] Using the distance formula: \[ \sqrt{(\beta - 1)^2 + (1 - \alpha)^2} = 3 \] Square both sides: \[ (\beta - 1)^2 + (1 - \alpha)^2 = 9 \] Expand both terms: \[ \beta^2 - 2\beta + 1 + 1 - 2\alpha + \alpha^2 = 9 \] Combine like terms: \[ \alpha^2 + \beta^2 - 2\alpha - 2\beta + 2 = 9 \] Substitute \( \alpha + \beta = 2 \): \[ \alpha^2 + \beta^2 - 2(\alpha + \beta) + 2 = 9 \] \[ \Rightarrow \alpha^2 + \beta^2 - 4 + 2 = 9 \Rightarrow \alpha^2 + \beta^2 - 2 = 9 \] \[ \Rightarrow \boxed{\alpha^2 + \beta^2 = 11} \]

We are given the points \( P(-3, 5) \) and \( Q(5, -3) \)Using the midpoint formula:\[\text{Midpoint} = \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)\]Substitute values:\[= \left( \dfrac{-3 + 5}{2}, \dfrac{5 + (-3)}{2} \right)= \left( \dfrac{2}{2}, \dfrac{2}{2} \right)= \boxed{(1, 1)}\]

We are given a triangle with:\[ \text{One side } = 15\,\text{cm} \text{ (opposite } 60^\circ), \quad x \text{ (opposite } 45^\circ)\]

Using the sine rule:\[ \dfrac{x}{\sin(45^\circ)} = \dfrac{15}{\sin(60^\circ)}\]

Solve for \(x\):\[ x = \dfrac{15 \cdot \sin(45^\circ)}{\sin(60^\circ)}\]

Recall:\[ \sin(45^\circ) = \dfrac{\sqrt{2}}{2}, \quad \sin(60^\circ) = \dfrac{\sqrt{3}}{2}\]

\[ x = \dfrac{15 \cdot \dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{3}}{2}} = 15 \cdot \dfrac{\sqrt{2}}{\sqrt{3}} = 15 \cdot \dfrac{\sqrt{6}}{3} = \boxed{5\sqrt{6}}\]  

We are given:\[ \text{Height of the pole} = 5\sqrt{3}\,\text{m}, \quad \text{Length of shadow} = 5\,\text{m}\] Using the tangent ratio:\[ \tan(\theta) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5\sqrt{3}}{5} = \sqrt{3}\]

Recall:\[\tan(60^\circ) = \sqrt{3}\]

\[\boxed{\theta = 60^\circ}\]  

Let:
\(y = (2 + 3x)(1 - x)\)
Using the product rule:
\(\dfrac{dy}{dx} = \dfrac{d}{dx}(2 + 3x)(1 - x)\)
\(= (2 + 3x)'(1 - x) + (2 + 3x)(1 - x)'\)

Differentiate each part:
\(= 3(1 - x) + (2 + 3x)(-1)\)
\(= 3 - 3x - (2 + 3x)\)
\(= 3 - 3x - 2 - 3x\)
\(= 1 - 6x\)
\(\boxed{\dfrac{dy}{dx} = 1 - 6x}\)

We are given:\[ y = 2x^2(2x - 1)\]

Let:\[ u = 2x^2, \quad v = 2x - 1\]

Then:\[ \dfrac{dy}{dx} = u'v + uv' = 4x(2x - 1) + 2x^2(2)\]\[ = 8x^2 - 4x + 4x^2 = 12x^2 - 4x\]

Now evaluate at \( x = -1 \):\[ \dfrac{dy}{dx} = 12(-1)^2 - 4(-1) = 12 + 4 = \boxed{16}\]  

We are given:\[ y = 3\cos\left(\dfrac{x}{3}\right)\]

Differentiate using the chain rule:\[ \dfrac{dy}{dx} = -3 \cdot \sin\left(\dfrac{x}{3}\right) \cdot \dfrac{1}{3} = -\sin\left(\dfrac{x}{3}\right)\]

Evaluate at \( x = \dfrac{3\pi}{2} \):\[ \dfrac{dy}{dx} = -\sin\left(\dfrac{3\pi}{2} \div 3\right) = -\sin\left(\dfrac{\pi}{2}\right) = \boxed{-1}\]  

The volume of a hemisphere is:\[ V = \dfrac{2}{3} \pi r^3\]

Differentiate with respect to \( r \):\[ \dfrac{dV}{dr} = \dfrac{2}{3} \pi \cdot 3r^2 = 2\pi r^2\]

Evaluate at \( r = 2 \):\[ \dfrac{dV}{dr} = 2\pi (2)^2 = 2\pi \cdot 4 = \boxed{8\pi}\]  

Evaluate:\[ \int\limits_{1}^{3} (x^2 - 1)\,dx\]

Integrate:\[ \int (x^2 - 1)\,dx = \dfrac{x^3}{3} - x\]

Evaluate from 1 to 3:\[ \left[ \dfrac{x^3}{3} - x \right]_1^3\]\[ = \left( \dfrac{27}{3} - 3 \right) - \left( \dfrac{1}{3} - 1 \right)\]\[ = (9 - 3) - \left( \dfrac{1}{3} - 1 \right)\] ​​​​​​​\[ = 6 - (-\dfrac{2}{3}) = \boxed{\dfrac{20}{3}}\]

Given:\[ \text{Millet angle} = 150^\circ, \quad \text{Maize} = 90^\circ, \quad \text{Others} = 60^\circ\]

Then:\[ \text{Beans angle} = 360^\circ - (150^\circ + 90^\circ + 60^\circ) = 60^\circ\]

Millet corresponds to 3000 tonnes, so total harvest is:\[ \dfrac{150}{360} = \dfrac{3000}{\text{Total}} \Rightarrow \text{Total} = \dfrac{3000 \cdot 360}{150} = 7200\,\text{tonnes}\]

Beans is:\[ \dfrac{60}{360} \cdot 7200 = \dfrac{1}{6} \cdot 7200 = \boxed{1200\,\text{tonnes}}\]