0606/12 May June 2026: Problems and Solutions

0606/12 May June 2026

1. Draw the graph of $y = 4\cos x - 2$ for $-360^\circ \leq x \leq 360^\circ$
   ✒ Solution
   1. Write the amplitude
      ✒ Solution

      $$ \begin{aligned} \text{Amplitude} = 4 \end{aligned} $$
   2. Write the period
      ✒ Solution

      $$ \begin{aligned} \text{Period} = 360^\circ \end{aligned} $$
   3. Find the values of $k$ where $4\cos x - 2 = k$ has at least two solutions.
      ✒ Solution

      The maximum value of the graph is $2$ and the minimum value is $-6$.

      For the equation to have $2$ or more solutions in the given domain, the horizontal line $y=k$ must intersect the graph at least twice.

      Therefore, $k$ must lie within the range of the function.

      $$ \begin{aligned} -6 \leq k \leq 2 \end{aligned} $$
2. $C$ lies on the perpendicular bisector of the line $AB$. $A$ is $(-2, 3)$ and $B$ is $(8, -3)$. Given that the point $C$ is $(0, c)$, find the value of $c$.
   ✒ Solution

   $$ \begin{aligned} \text{Midpoint of } AB &= \left( \frac{-2 + 8}{2}, \frac{3 + (-3)}{2} \right) \\ &= (3, 0) \end{aligned} $$ $$ \begin{aligned} \text{Gradient of } AB &= \frac{-3 - 3}{8 - (-2)} \\ &= \frac{-6}{10} \\ &= -\frac{3}{5} \end{aligned} $$ $$ \begin{aligned} \text{Gradient of perpendicular bisector} &= \frac{5}{3} \end{aligned} $$ $$ \begin{aligned} \text{Equation of perpendicular bisector: } y - 0 &= \frac{5}{3}(x - 3) \\ y &= \frac{5}{3}x - 5 \end{aligned} $$

   Since $C(0, c)$ lies on this line, substitute $x = 0$ and $y = c$:

   $$ \begin{aligned} c &= \frac{5}{3}(0) - 5 \\ c &= -5 \end{aligned} $$
   1. The point $D$ is $(2, -1)$ and the value of $BD$ is $\displaystyle\frac{\sqrt{n}}{2}AD$. Find the value of $n$.
      ✒ Solution

      $$ \begin{aligned} BD &= \sqrt{(2 - 8)^2 + (-1 - (-3))^2} \\ &= \sqrt{(-6)^2 + 2^2} \\ &= \sqrt{36 + 4} \\ &= \sqrt{40} \\ &= 2\sqrt{10} \end{aligned} $$ $$ \begin{aligned} AD &= \sqrt{(2 - (-2))^2 + (-1 - 3)^2} \\ &= \sqrt{4^2 + (-4)^2} \\ &= \sqrt{16 + 16} \\ &= \sqrt{32} \\ &= 4\sqrt{2} \end{aligned} $$

      Substitute the lengths into the given equation:

      $$ \begin{aligned} BD &= \left(\frac{\sqrt{n}}{2}\right) AD \\ 2\sqrt{10} &= \left(\frac{\sqrt{n}}{2}\right) (4\sqrt{2}) \\ 2\sqrt{10} &= 2\sqrt{2n} \\ \sqrt{10} &= \sqrt{2n} \\ 10 &= 2n \\ n &= 5 \end{aligned} $$
3. The polynomial $\mathrm{p}(x)$ is $6x^3 - x^2 - 5x + 2$.
   1. Show that $(x+2)$ is not a factor of $\mathrm{p}(x)$ and find the remainder when $\mathrm{p}(x)$ is divided by $(x+2)$
      ✒ Solution

      $$ \begin{aligned} \mathrm{p}(-2) &= 6(-2)^3 - (-2)^2 - 5(-2) + 2 \\ &= 6(-8) - 4 + 10 + 2 \\ &= -48 - 4 + 12 \\ &= -40 \end{aligned} $$

      Since $\mathrm{p}(-2) \neq 0$, $(x+2)$ is not a factor of $\mathrm{p}(x)$.

      The remainder is $-40$.

   2. Express $\mathrm{p}(x)$ as a product of its linear factors.
      ✒ Solution

      Test for factors using the factor theorem:

      $$ \begin{aligned} \mathrm{p}(1) &= 6(1)^3 - (1)^2 - 5(1) + 2 = 2 \neq 0 \\ \mathrm{p}(-1) &= 6(-1)^3 - (-1)^2 - 5(-1) + 2 = -6 - 1 + 5 + 2 = 0 \end{aligned} $$

      Therefore, $(x+1)$ is a factor.

      By inspection or algebraic division:

      $$ \begin{aligned} 6x^3 - x^2 - 5x + 2 &= (x + 1)(6x^2 - 7x + 2) \end{aligned} $$

      Factorizing the quadratic expression:

      $$ \begin{aligned} 6x^2 - 7x + 2 &= 6x^2 - 3x - 4x + 2 \\ &= 3x(2x - 1) - 2(2x - 1) \\ &= (3x - 2)(2x - 1) \end{aligned} $$

      Hence, as a product of linear factors:

      $$ \begin{aligned} \mathrm{p}(x) &= (x + 1)(3x - 2)(2x - 1) \end{aligned} $$
   3. Given that $6u^6 - u^4 - 5u^2 + 2 = 0$, find the values of $u$.
      ✒ Solution

      Let $x = u^2$. Then the equation becomes:

      $$ \begin{aligned} 6x^3 - x^2 - 5x + 2 &= 0 \end{aligned} $$

      From part (b), we have the roots for $x$:

      $$ \begin{aligned} (x + 1)(3x - 2)(2x - 1) &= 0 \\ x &= -1, \quad x = \frac{2}{3}, \quad x = \frac{1}{2} \end{aligned} $$

      Substitute $x = u^2$ back:

      $$ \begin{aligned} u^2 &= -1 \quad \text{(No real solutions)} \\ u^2 &= \frac{2}{3} \implies u = \pm \sqrt{\frac{2}{3}} = \pm \frac{\sqrt{6}}{3} \\ u^2 &= \frac{1}{2} \implies u = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} \end{aligned} $$

      The values of $u$ are $\pm \frac{\sqrt{6}}{3}$ and $\pm \frac{\sqrt{2}}{2}$.

4. The diagram below shows a circle with a shaded sector $AOB$.
   1. Given that the perimeter of the shaded region is equal to the length of the major arc $ACB$, find the exact value of angle $\alpha$ in radians.
      ✒ Solution

      $$ \begin{aligned} \text{Perimeter of the shaded region} &= r + r + r\alpha \\ &= 2r + r\alpha \end{aligned} $$ $$ \begin{aligned} \text{Angle of the major arc } ACB &= 2\pi - \alpha \end{aligned} $$ $$ \begin{aligned} \text{Length of the major arc } ACB &= r(2\pi - \alpha) \end{aligned} $$

      Equating the two perimeters:

      $$ \begin{aligned} 2r + r\alpha &= r(2\pi - \alpha) \\ 2 + \alpha &= 2\pi - \alpha \\ 2\alpha &= 2\pi - 2 \\ \alpha &= \pi - 1 \end{aligned} $$
   2. Given that the exact area of the shaded region is $18(\pi^2 - 1)$, find the value of $r$.
      ✒ Solution

      $$ \begin{aligned} \text{Area of the shaded region} &= \frac{1}{2}r^2\alpha \end{aligned} $$

      Substitute $\alpha = \pi - 1$ and equate to the given area:

      $$ \begin{aligned} \frac{1}{2}r^2(\pi - 1) &= 18(\pi^2 - 1) \\ \frac{1}{2}r^2(\pi - 1) &= 18(\pi - 1)(\pi + 1) \\ \frac{1}{2}r^2 &= 18(\pi + 1) \\ r^2 &= 36(\pi + 1) \\ r &= \sqrt{36(\pi + 1)} \\ r &= 6\sqrt{\pi + 1} \end{aligned} $$
5. A curve is written as $y = \ln((2x-4)^5)$. Find the equation of the tangent to the curve at $x=3$.
   ✒ Solution

   First, find the $y$-coordinate when $x=3$:

   $$ \begin{aligned} y &= \ln((2(3) - 4)^5) \\ y &= \ln(2^5) \\ y &= 5\ln 2 \end{aligned} $$

   Simplify the equation of the curve using logarithm properties before differentiating:

   $$ \begin{aligned} y &= 5\ln(2x - 4) \end{aligned} $$

   Differentiate with respect to $x$:

   $$ \begin{aligned} \frac{dy}{dx} &= 5 \left( \frac{1}{2x - 4} \right) (2) \\ &= \frac{10}{2x - 4} \end{aligned} $$

   Find the gradient $m$ at $x=3$:

   $$ \begin{aligned} m &= \frac{10}{2(3) - 4} \\ &= \frac{10}{2} \\ &= 5 \end{aligned} $$

   Equation of the tangent line:

   $$ \begin{aligned} y - y_1 &= m(x - x_1) \\ y - 5\ln 2 &= 5(x - 3) \\ y &= 5x - 15 + 5\ln 2 \end{aligned} $$
6. $y =\displaystyle \frac{5x+1}{x-2}$, Find the approximate change in $x$ as $y$ increases from $4$ by the small value $h$.
   ✒ Solution

   First, find the initial value of $x$ when $y = 4$:

   $$ \begin{aligned} 4 &= \frac{5x + 1}{x - 2} \\ 4(x - 2) &= 5x + 1 \\ 4x - 8 &= 5x + 1 \\ -9 &= x \end{aligned} $$

   Differentiate $y$ with respect to $x$ using the quotient rule:

   $$ \begin{aligned} \frac{dy}{dx} &= \frac{(x - 2)(5) - (5x + 1)(1)}{(x - 2)^2} \\ &= \frac{5x - 10 - 5x - 1}{(x - 2)^2} \\ &= \frac{-11}{(x - 2)^2 \end{aligned} $$

   Find the value of $\frac{dy}{dx}$ at $x = -9$:

   $$ \begin{aligned} \frac{dy}{dx} &= \frac{-11}{(-9 - 2)^2} \\ &= \frac{-11}{(-11)^2} \\ &= \frac{-11}{121} \\ &= -\frac{1}{11} \end{aligned} $$

   Using the approximation formula $\displaystyle\delta y \approx \frac{dy}{dx} \delta x$:

   $$ \begin{aligned} h &\approx -\frac{1}{11} \delta x \\ \delta x &\approx -11h \end{aligned} $$

   The approximate change in $x$ is $-11h$.

7. Given that the first term in an arithmetic sequence is $400$, and the common difference is $-2$,
   1. Find the 81st term and the 100th term.
      ✒ Solution

      Given $a = 400$ and $d = -2$.

      $$ \begin{aligned} u_{81} &= a + 80d \\ &= 400 + 80(-2) \\ &= 400 - 160 \\ &= 240 \end{aligned} $$ $$ \begin{aligned} u_{100} &= a + 99d \\ &= 400 + 99(-2) \\ &= 400 - 198 \\ &= 202 \end{aligned} $$
   2. Find the sum of $u_{81}+u_{82}+\dots +u_{100}$.
      ✒ Solution

      The number of terms $n$ in this series is $100 - 81 + 1 = 20$.

      $$ \begin{aligned} S &= \frac{n}{2}(u_{81} + u_{100}) \\ &= \frac{20}{2}(240 + 202) \\ &= 10(442) \\ &= 4420 \end{aligned} $$
8. The following terms are from a geometric sequence for $x>0$.
   
   2nd term: $11x - 1$,
   3rd term: $x + 5$
   4th term: $x - 1$
   1. Find the common ratio and the first term of the sequence.
      ✒ Solution

      For a geometric sequence, the ratio between consecutive terms is constant.

      $$ \begin{aligned} \frac{x + 5}{11x - 1} &= \frac{x - 1}{x + 5} \\ (x + 5)^2 &= (11x - 1)(x - 1) \\ x^2 + 10x + 25 &= 11x^2 - 11x - x + 1 \\ x^2 + 10x + 25 &= 11x^2 - 12x + 1 \\ 0 &= 10x^2 - 22x - 24 \\ 0 &= 5x^2 - 11x - 12 \\ 0 &= (5x + 4)(x - 3) \end{aligned} $$

      Since $x > 0$, we choose $x = 3$.

      Calculate the terms:

      $$ \begin{aligned} \text{2nd term} &= 11(3) - 1 = 32 \\ \text{3rd term} &= 3 + 5 = 8 \\ \text{4th term} &= 3 - 1 = 2 \end{aligned} $$

      Find the common ratio $r$:

      $$ \begin{aligned} r &= \frac{8}{32} \\ r &= \frac{1}{4} \end{aligned} $$

      Find the first term $a$:

      $$ \begin{aligned} ar &= 32 \\ a\left(\frac{1}{4}\right) &= 32 \\ a &= 128 \end{aligned} $$
   2. given that the sum to infinity is $\displaystyle\frac{k}{3}$, find the value of $k$.
      ✒ Solution

      $$ \begin{aligned} S_\infty &= \frac{a}{1 - r} \\ &= \frac{128}{1 - \frac{1}{4}} \\ &= \frac{128}{\frac{3}{4}} \\ &= \frac{512}{3} \end{aligned} $$

      Equating to the given expression:

      $$ \begin{aligned} \frac{k}{3} &= \frac{512}{3} \\ k &= 512 \end{aligned} $$
9. Simplify the expression and express in Log base $e$. $$ \frac{1}{\log_x(e)} - \ln(x - 1) $$
   ✒ Solution

   Using the change of base rule, $\displaystyle\frac{1}{\log_x(e)} = \log_e(x) = \ln(x)$.

   $$ \begin{aligned} \frac{1}{\log_x(e)} - \ln(x - 1) &= \ln(x) - \ln(x - 1) \\ &= \ln\left(\frac{x}{x - 1}\right) \end{aligned} $$
   1. Solve for $x$, where $x > 0$. $$ 2 \log_5(2x) = 1 + \log_5(9x - 10) $$
      ✒ Solution

      $$ \begin{aligned} \log_5((2x)^2) &= \log_5(5) + \log_5(9x - 10) \\ \log_5(4x^2) &= \log_5(5(9x - 10)) \\ 4x^2 &= 45x - 50 \\ 4x^2 - 45x + 50 &= 0 \\ 4x^2 - 40x - 5x + 50 &= 0 \\ 4x(x - 10) - 5(x - 10) &= 0 \\ (4x - 5)(x - 10) &= 0 \end{aligned} $$ $$ \begin{aligned} x &= \frac{5}{4} \quad \text{or} \quad x = 10 \end{aligned} $$

      Both values satisfy $x > 0$ and $9x - 10 > 0$.

10. The position vectors of $A$ and $B$ relative to origin are given as $\overrightarrow{OA}\ (4, -2)$ and $\overrightarrow{OB}$ as $(-4, 6)$. The point $C$ lies on $AB$ such that $\overrightarrow{AC}$ to $\overrightarrow{CB}$ is $3:1$. The point $D$ is such that $\overrightarrow{OD}$ is $\lambda \overrightarrow{OA}$. Given that $\overrightarrow{CD}$ is $(5, -5.5)$
    1. Find the value of $\overrightarrow{OD}$
       ✒ Solution

       First, find the vector $\overrightarrow{AB}$:

       $$ \begin{aligned} \overrightarrow{AB} &= \overrightarrow{OB} - \overrightarrow{OA} \\ &= \begin{pmatrix} -4 \\ 6 \end{pmatrix} - \begin{pmatrix} 4 \\ -2 \end{pmatrix} \\ &= \begin{pmatrix} -8 \\ 8 \end{pmatrix} \end{aligned} $$

       Since $\overrightarrow{AC} : \overrightarrow{CB} = 3:1$, the vector $\overrightarrow{AC}$ is $\displaystyle\frac{3}{4}$ of $\overrightarrow{AB}$:

       $$ \begin{aligned} \overrightarrow{AC} &= \frac{3}{4}\overrightarrow{AB} \\ &= \frac{3}{4} \begin{pmatrix} -8 \\ 8 \end{pmatrix} \\ &= \begin{pmatrix} -6 \\ 6 \end{pmatrix} \end{aligned} $$

       Using the triangle rule, find $\overrightarrow{OC}$:

       $$ \begin{aligned} \overrightarrow{OC} &= \overrightarrow{OA} + \overrightarrow{AC} \\ &= \begin{pmatrix} 4 \\ -2 \end{pmatrix} + \begin{pmatrix} -6 \\ 6 \end{pmatrix} \\ &= \begin{pmatrix} -2 \\ 4 \end{pmatrix} \end{aligned} $$

       We are given $\overrightarrow{CD} = \begin{pmatrix} 5 \\ -5.5 \end{pmatrix}$. Using the triangle rule again to find $\overrightarrow{OD}$:

       $$ \begin{aligned} \overrightarrow{OD} &= \overrightarrow{OC} + \overrightarrow{CD} \\ &= \begin{pmatrix} -2 \\ 4 \end{pmatrix} + \begin{pmatrix} 5 \\ -5.5 \end{pmatrix} \\ &= \begin{pmatrix} 3 \\ -1.5 \end{pmatrix} \end{aligned} $$
    2. Find the value of $\lambda$
       ✒ Solution

       $$ \begin{aligned} \overrightarrow{OD} &= \lambda \overrightarrow{OA} \\ \begin{pmatrix} 3 \\ -1.5 \end{pmatrix} &= \lambda \begin{pmatrix} 4 \\ -2 \end{pmatrix} \end{aligned} $$

       Comparing the first components:

       $$ \begin{aligned} 3 &= 4\lambda \\ \lambda &= \frac{3}{4} = 0.75 \end{aligned} $$
11. The line $L$ passes through $(0,6)$ and $(1.5,0)$ and intersects the curve $y=\displaystyle\frac{10}{4x+1}$ at points $P$ and $Q$. Find the area of the shaded region.
    ✒ Solution

    First, find the equation of line $L$:

    $$ \begin{aligned} \text{Gradient } m &= \frac{0 - 6}{\frac{3}{2} - 0} = -4 \\ y\text{-intercept } c &= 6 \\ y &= -4x + 6 \end{aligned} $$

    Find the points of intersection by equating the line and the curve:

    $$ \begin{aligned} -4x + 6 &= \frac{10}{4x + 1} \\ (-4x + 6)(4x + 1) &= 10 \\ -16x^2 - 4x + 24x + 6 &= 10 \\ -16x^2 + 20x - 4 &= 0 \\ 4x^2 - 5x + 1 &= 0 \\ (4x - 1)(x - 1) &= 0 \\ x &= \frac{1}{4} \quad \text{or} \quad x = 1 \end{aligned} $$

    The area of the shaded region is the definite integral of the upper curve (the line) minus the lower curve between $x=\frac{1}{4}$ and $x=1$.

    $$ \begin{aligned} \text{Area} &= \int_{\frac{1}{4}}^{1} \left[ (-4x + 6) - \frac{10}{4x + 1} \right] \, dx \\ &= \left[ -2x^2 + 6x - \frac{10}{4}\ln|4x + 1| \right]_{\frac{1}{4}}^{1} \\ &= \left[ -2x^2 + 6x - \frac{5}{2}\ln(4x + 1) \right]_{\frac{1}{4}}^{1} \end{aligned} $$

    Evaluate at the upper limit $x = 1$:

    $$ \begin{aligned} \text{Upper value} &= -2(1)^2 + 6(1) - \frac{5}{2}\ln(4(1) + 1) \\ &= 4 - \frac{5}{2}\ln(5) \end{aligned} $$

    Evaluate at the lower limit $x = \displaystyle\frac{1}{4}$:

    $$ \begin{aligned} \text{Lower value} &= -2\left(\frac{1}{4}\right)^2 + 6\left(\frac{1}{4}\right) - \frac{5}{2}\ln\left(4\left(\frac{1}{4}\right) + 1\right) \\ &= -2\left(\frac{1}{16}\right) + \frac{3}{2} - \frac{5}{2}\ln(2) \\ &= -\frac{1}{8} + \frac{12}{8} - \frac{5}{2}\ln(2) \\ &= \frac{11}{8} - \frac{5}{2}\ln(2) \end{aligned} $$

    Calculate the total area:

    $$ \begin{aligned} \text{Area} &= \left(4 - \frac{5}{2}\ln(5)\right) - \left(\frac{11}{8} - \frac{5}{2}\ln(2)\right) \\ &= \frac{32}{8} - \frac{11}{8} - \frac{5}{2}\ln(5) + \frac{5}{2}\ln(2) \\ &= \frac{21}{8} - \frac{5}{2} (\ln(5) - \ln(2)) \\ &= \frac{21}{8} - \frac{5}{2}\ln\left(\frac{5}{2}\right) \end{aligned} $$
12. show that values $n \geq 3$ satisfy the equation ${}^{n+2}C_3 - {}^nC_3 = n^2$
    ✒ Solution

    Expand the combinations formulas:

    $$ \begin{aligned} {}^{n+2}C_3 &= \frac{(n + 2)!}{3! (n + 2 - 3)!} \\ &= \frac{(n + 2)(n + 1)n(n - 1)!}{6(n - 1)!} \\ &= \frac{n(n + 1)(n + 2)}{6} \end{aligned} $$ $$ \begin{aligned} {}^nC_3 &= \frac{n!}{3! (n - 3)!} \\ &= \frac{n(n - 1)(n - 2)(n - 3)!}{6(n - 3)!} \\ &= \frac{n(n - 1)(n - 2)}{6} \end{aligned} $$

    Subtract the two expressions:

    $$ \begin{aligned} {}^{n+2}C_3 - {}^nC_3 &= \frac{n(n + 1)(n + 2)}{6} - \frac{n(n - 1)(n - 2)}{6} \\ &= \frac{n}{6} \left[ (n + 1)(n + 2) - (n - 1)(n - 2) \right] \\ &= \frac{n}{6} \left[ (n^2 + 3n + 2) - (n^2 - 3n + 2) \right] \\ &= \frac{n}{6} \left[ 6n \right] \\ &= n^2 \end{aligned} $$

    Since the definition of ${}^nC_3$ requires $n \geq 3$ to be valid (we cannot choose $3$ items from fewer than $3$ items), the equation holds true for all integers $n \geq 3$.

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