Mathematics 2005 Jamb Examination

Using circle theorems: since O is the center, the central angle ∠UOT = 70° subtends arc UT. In triangle UOT (two radii), it is isosceles with base angles\[\angle OUT = \angle UT O = \dfrac{180\deg - 70\deg}{2} = 55\deg.\]

The inscribed angle ∠RST = 100° subtends arc RT not containing S, so the minor central angle \(\angle ROT\) over arc RT is

\[\angle ROT = 2\times \angle RST = 2\times 100\deg = 200\deg,\]

hence the minor arc RT is \(360\deg - 200\deg = 160\deg\), so the inscribed angle at U on chord RT is

\[\angle RUT = \tfrac12\times 160\deg = 80\deg.\]

Finally, at U the angle between UR and UO is the difference between the angle \(\angle RUT\) and the angle between UT and UO (which is 55°):

\[\angle RUO = \angle RUT - \angle OUT = 80\deg - 55\deg = 25\deg.\]

Use the sector area formula

\[A = \dfrac{\theta}{360\deg}\,\pi r^2.\]

Given \(A=55\,\text{cm}^2\), \(r=10\,\text{cm}\) and \(\pi=\dfrac{22}{7}\), solve for \(\theta\):

\[55 = \dfrac{\theta}{360\deg}\cdot \dfrac{22}{7}\cdot 100 \quad \Rightarrow \quad \theta = \dfrac{55\cdot 360\deg}{(22/7)\cdot 100} \]

\[= \dfrac{19800\deg}{\dfrac{2200}{7}} = 198\cdot \dfrac{7}{22}\,\deg = 9\cdot 7\,\deg = 63\deg.\]

Therefore, the angle of the sector is 63°.

Curved surface area (CSA) of a cone is

\[\text{CSA} = \pi r\ell,\]

where \(r\) is the base radius and \(\ell\) is the slant height. Given diameter \(=10\,\text{cm}\Rightarrow r=5\,\text{cm}\) and height \(h=12\,\text{cm}\). Compute \(\ell\) using Pythagoras:

\[\ell = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} \]\[= \sqrt{169} = 13.\]

Hence,

\[\text{CSA} = \pi\cdot 5\cdot 13 = 65\,\pi\,\text{cm}^2.\]

Points equidistant from two intersecting lines lie on the angle bisectors of the pair. When two lines cross (here at 75°), there are two bisectors: one bisects the acute angle (37.5°) and the other bisects the obtuse angle (180° - 75° = 105°, halved to 52.5°).

Thus, the required locus is the set of points on the bisectors of the angles between PQ and ST.

Rewrite the given line in slope–intercept form: \(y = -2x + 5\). Its slope is \(m_1 = -2\). A line perpendicular to it has slope \(m_2 = -\tfrac{1}{m_1} = \tfrac{1}{2}\).

Using point–slope form through \((4,3)\):

\[y - 3 = \tfrac{1}{2}(x - 4) \;\Rightarrow\; y = \tfrac{x}{2} + 1.\]

Multiply through by 2: \(2y = x + 2\), so \(2y - x = 2\).

Formula for chord length corresponding to central angle \(\theta\) (in degrees):

\[\text{chord} = 2r\sin\left(\dfrac{\theta}{2}\right).\]

Here \(r=14\,\text{cm}\) and \(\theta=60°\). Compute:

\[\text{chord} = 2\times 14\times \sin\left(\dfrac{60°}{2}\right) = 28\sin 30° .\]

Since \(\sin 30° = \tfrac{1}{2}\),

\[\text{chord} = 28 \times \tfrac{1}{2} = 14\,\text{cm}.\]

Thus the chord length is 14 cm.

We solve \(\sin\theta = -\tfrac{1}{2}\). The reference angle for \(|\sin\theta|=\tfrac{1}{2}\) is \(30°\).

\(\sin\theta\) is negative in quadrants III and IV, so the solutions in \(0<\theta<360°\) are:

Quadrant III: \(\theta = 180° + 30° = 210°\).

Quadrant IV: \(\theta = 360° - 30° = 330°\).

Therefore \(\theta = 210°,\; 330°.\)

Differentiate using the chain rule. Let \(u=1-2x\), so \(y=u^3\). Then:

\[\dfrac{dy}{dx} \]\[= 3u^2\dfrac{du}{dx} \]\[= 3(1-2x)^2\cdot(-2) \]\[= -6(1-2x)^2.\]

Evaluate at \(x=-1\):

\[1-2(-1)=1+2\]\[=3\quad\Rightarrow\quad \dfrac{dy}{dx}\Big|_{x=-1}\]\[ = -6(3)^2 = -6\times 9 = -54.\]

Thus the derivative at \(x=-1\) is \(-54\).

Use the chain rule. Let \(u = 2x^3 + 3x -4\), so \(y=\sin u\). Then

\[\dfrac{dy}{dx} = \cos u \cdot \dfrac{du}{dx}.\]

Compute \(\dfrac{du}{dx} = 6x^2 + 3\). Therefore

\[\dfrac{dy}{dx} = (6x^2+3)\cos(2x^3+3x-4).\]

This is the required derivative.

Area of a disc: \(A=\pi r^2\). Differentiate with respect to time \(t\):

\[\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}.\]

Given \(\dfrac{dr}{dt}=0.5\ \text{cm/s}\) and \(r=6\ \text{cm}\), substitute:

\[\dfrac{dA}{dt} = 2\pi \times 6 \times 0.5 = 6\pi\ \text{cm}^2/\text{s}.\]

Hence the area is increasing at \(6\pi\ \text{cm}^2/\text{s}\).