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Form 4 KCSE 2023 Mathematics Paper 2 Mock Examination

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.

Tangent
Calculate the area of the shaded region correct to 2 decimal places  [4𝑚𝑎𝑟𝑘𝑠]

\(OT^{2} =25^{2}-24^{2}\)

\( OT^{2} =49\)

 \( OT =7\) 

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

 \(\frac{4}{\sqrt{5}+\sqrt{2}} - \frac{3}{\sqrt{5}-\sqrt{2}} \) 

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