Published on November 10th 2023 | 1 min , 174 words
Section I (50 Marks)
1. Solve for \(x\) (4mks)
\( ( \log_{3}{x})^2 - \frac{1}{2} \log_{3}{x} =\frac{3}{2}\)
\(solution\)
\(let \log_{3}{x}\: be \: y\)
\(y^2-\frac{1}{2}y=\frac{3}{2}\)
\(2y^2-y-3=0\)
\((2y-3)(y+1)=0\)
\(y=\frac{3}{2} \: or \: y=-1\)
\(\log_{3}{x}=\frac{3}{2}\)
\(or\)
\(\log_{3}{x}=-1\)
\(x=(3)^{\frac{3}{2}}\)
\(or \)
\(x=(3)^{-1}\)
\(x=5.196\) or \(x=\frac{1}{3}\)
2. In the figure below PT is a tangent to the circle from an external point P. 𝑃𝑇 = 24 𝑐𝑚 and 𝑂𝑃 = 25 𝑐𝑚.
\(OT^{2} =25^{2}-24^{2}\)
\( OT^{2} =49\)
\( OT =7\)
let \(\angle TOP \: be\: x\)
tan \(x^{\circ}\) =\(\frac{24}{7}\)
x= \(73.74^{ \circ}\)
Area of \(\triangle PTO=\frac{1}{2}\times24\times7\)
\(=84\)
Area of sector = \(\frac{73.74}{360}\times \frac{22}{7} \times 7^{2}\)
\(=31.54\)
Area of the shaded region \(=84-31.54\)
\(=52.46cm^{2}\)
3. Find the value of 𝑤 in the expression \(wx^{2}-\frac{3}{2}x+\frac{1}{16}\) is a perfect square, given that 𝑤 is a constant [2𝑚𝑎𝑟𝑘𝑠]
\(b^{2}=4ac\)
\(\Rightarrow (\frac{3}{2})^{2}=4\times\frac{1}{16}w\)
\(\frac{9}{4}=\frac{1}{4}w\)
\(w=9\)
4. Simplify
\(\frac{4}{\sqrt{5}+\sqrt{2}} - \frac{3}{\sqrt{5}-\sqrt{2}} \)