1. A carpenter has \(3\) metres long wood to cut into \(7\) equal pieces. He approximates each piece as \(42.8 \, \text{cm}\) to the nearest millimetre. Calculate the percentage error in the approximation. (3 marks)
\[ \text{Error} = 42 \tfrac{6}{7} \, \text{cm} - 42 \tfrac{4}{5} = 0.05714 \, \text{cm} \] \[ \% \, \text{Error} = \left( \frac{0.05714}{42 \tfrac{6}{7}} \right) \times 100 \] \[ = 0.1333 \% \]3. Find the number of terms of the geometrical series \(\frac{1}{16} + \frac{1}{8} + \frac{1}{4} + \dots\) that are needed to make a sum of \(2^{16} - \frac{1}{16}\). (3 marks)
\[ a = \frac{1}{16}, \quad r = 2 \] \[ S_n = a \frac{r^n - 1}{r - 1} \] \[ 2^{16} - \frac{1}{16} = \frac{1}{16} (2^n - 1) \] \[ 2^{16} - \frac{1}{16} = \frac{1}{16} (2^n - 1) \] \[ (2^{16})(16) = \frac{2^n}{16} \times 16 - \frac{1}{16} \times 16 \] \[ 2^n = 2^{16} \cdot 2^4 \] \[ 2^n = 2^{20} \quad \Rightarrow \quad n = 20 \]4. A point \(P'(-4, -6)\) is the image of point \(P\) under a transformation \(R\) followed by \(T\). Given \( T = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} \) and \(R= \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix} \), find the coordinates of \(P\).
The first step is to get the get the combined matrix of transformation i.e \(TR\)
\[TR= \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix} \] \[= \begin{pmatrix} 6 & 2 \\ 3 & 0 \end{pmatrix} \] \[ \begin{pmatrix} 6 & 2 \\ 3 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -4 \\ -6 \end{pmatrix} \] \[ \begin{cases} 6x + 2y = -4 \\ 3x = -6 \end{cases} \] From the second equation: \[ 3x = -6 \quad \Rightarrow \quad x = -2 \] Substitute into the first equation: \[ 6(-2) + 2y = -4 \] \[ -12 + 2y = -4 \] \[ 2y = 8 \quad \Rightarrow \quad y = 4 \] \[ \therefore \; P(-2, 4) \]Find the possible values of \(c\) for which the equation \(x^2 + cx + (2c+5) = 0\) has a repeated root.
For a quadratic equation in the form \(ax^2 + bx + c = 0\), a repeated root occurs when the discriminant, \(b^2 - 4ac\), is equal to zero.
Given the equation \(x^2 + cx + (2c+5) = 0\), we have:
Substitute these values into the discriminant formula and set it to zero:
\(b^2 - 4ac = 0\)
\(\;c^2 - 4(1)(2c+5) = 0\)
\(\;c^2 - 8c - 20 = 0\)
Now, solve this quadratic equation for \(c\). Factorization gives:
\((c - 10)(c + 2) = 0\)
This gives two possible values for \(c\):
\(c - 10 = 0 \;\;\Rightarrow\;\; c = 10\)
\(c + 2 = 0 \;\;\Rightarrow\;\; c = -2\)
Thus, the possible values of \(c\) are \(\;10\;\) and \(\;-2\).
Solve the equation \(2\cos^2 x + \sin x = 1\) for \(-180^{\circ} \leq x \leq 180^{\circ}\).
\(2(1-\sin^2 x) + \sin x = 1\)
\(2-2 \sin^2 x + \sin x = 1\)
Let \(\sin x = y\)
\(2-2y^2 + y = 1\)
\(-2y^2 + y+1 = 0\)
\(2y^2 -y-1 = 0\)
\(2y^2 -2y+y-1 = 0\)
\(2y(y-1)+1(y-1) = 0\)
\((2y+1)(y-1) = 0\)
\(y=-\frac{1}{2} or y=1\)
\(\Rightarrow \sin x =1 \: \text{or} \: sin x = -\frac{1}{2}\)
\(\Rightarrow x = \sin^{-1}(1) \: or \: x = \sin^{-1}(-\frac{1}{2})\)
\(\Rightarrow x = 90^{\circ} \: or \: x = 210^{\circ} , 330^{\circ} \: (ignore)\)
Note that the value of \(x\) must be in the range \(-180^{\circ} \leq x \leq 180^{\circ}\) hence \(210^{\circ} \: and \: 330^{\circ}\) are outside the range
\(\therefore x = 90^{\circ} \)
The position vectors of \(P\) and \(Q\) are \( \vec{OP} = -\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \quad \text{and} \quad \vec{OQ} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} \) respectively. A point \(R\) divides \(PQ\) externally in the ratio \(2:1\). Calculate the distance of \(R\) from the origin correct to \(1\) decimal place.
\[ \overrightarrow{OR} = 2\begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} + (-1)\begin{pmatrix} -1 \\ 2 \\ -3 \end{pmatrix} \] \[ = \begin{pmatrix} 6 \\ -4 \\ 2 \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix} \] \[ = \begin{pmatrix} 7 \\ -6 \\ -1 \end{pmatrix} \] \[ |\overrightarrow{OR}| = \sqrt{7^2 + (-6)^2 + (-1)^2} \] \[ = \sqrt{49 + 36 + 1} = \sqrt{86} \approx 9.3 \,\text{units} \]8. The cash price of a T.V set is Ksh \(70,000\). Mueni bought the set on hire purchase terms by paying 12 equal monthly instalments of Ksh \(7,000\) each, with the first instalment treated as the deposit. Calculate the monthly rate of compound interest charged on the balance correct to 2 decimal places. (3 marks)
\[ P = 70000 - 7000 \] \[ 63000 \left(1 + \frac{r}{100}\right)^{11} = 77000 \] \[ \left(1 + 0.01r\right)^{11} = \frac{11}{9} \] \[ 1 + 0.01r = 1.01841 \] \[ 0.01r = 0.01841 \] \[ r = \frac{0.01841}{0.01} \] \[ r = 1.841\% \]\(\textbf{9}\) Find the equation of the tangent to the circle \(x^2 + y^2 - 4x - 6y + 3 = 0\) at point \((5, 4)\). (4 marks)
\[ x^2 - 4x + y^2 - 8y = -3 \]15. Solve\ for \( x\) in the equation
\(\log_{4} x - \log_{8} 2x = \frac{1}{2}\) $$\log_{4} x - \frac{\log_{4} 2x}{\log_{4} 8} = \frac{1}{2}$$ $$\log_{4} x - \frac{\log_{4} 2 + \log_{4} x}{\log_{4} 8} = \frac{1}{2}$$ $$\log_{4} x - \left( \frac{\log_{4} 2}{\log_{4} 8} + \frac{\log_{4} x}{\log_{4} 8} \right) = \frac{1}{2}$$ You are right. My apologies for the oversight. Here is the correct LaTeX for the equation and the subsequent steps, starting from the exact line you pointed out. $$y - \left( \frac{1}{3} + \frac{y}{3/2} \right) = \frac{1}{2}$$ $$y - \left( \frac{1}{3} + \frac{2y}{3} \right) = \frac{1}{2}$$ $$y - \left( \frac{1+2y}{3} \right) = \frac{1}{2}$$ $$6y - 2(1+2y) = 3$$ $$6y - 2 - 4y = 3$$ $$2y - 2 = 3$$ $$2y = 5$$ $$y = \frac{5}{2} = 2.5$$ $$\log_{4} x = y = 2.5$$ $$4^{2.5} = x \implies x = 32$$A hot water tap can fill a bath in 6 minutes, while a cold water tap can fill the same bath in 4 minutes. The drain pipe can empty the bath in 8 minutes. The two taps and the drain pipe are fully opened for \(1\frac{1}{2}\) minutes, after which the drain pipe is closed. Calculate the total time taken to fill the bath.
\[ \text{Hot} = 6 \quad\quad \text{Cold} = 4 \quad\quad \text{Drain} = 8 \] \[ \text{In one minute:} \quad = \frac{1}{6} + \frac{1}{4} - \frac{1}{8} \quad = \frac{7}{24} \] \[ 1 \text{ min} = \frac{7}{24} \] \[ 1.5 \text{ min} \times \frac{7}{24} = \frac{7}{16} \] \[ \text{Rem fraction} = 1 - \frac{7}{16} = \frac{9}{16} \] \[ \text{In 1 min:} \quad \frac{1}{6} + \frac{1}{4} = \frac{5}{12} \] \[ \text{If 1 min} = \frac{5}{12} \] \[ \times \frac{9}{16} \] \[ = 1 \times \frac{9}{16} \div \frac{5}{12} \] \[ = 1.35 \, \text{mins} \]5 (a) Expand \( \left( 2 - \tfrac{1}{3}x \right)^5 \) in ascending powers of \(x\) leaving the coefficients as fractions in their simplest form. (1 mark)
(b) Use the first four terms of the expansion in (a) to estimate the value of \( (1.9)^5 \) (2 marks)
Solve the equation \(6\cos^2x + \sin x = 4\) for \(0^\circ \leq x \leq 360^\circ\).