CIE 2023 January 12

Question 1

Find the exact values of $k$ such that the straight line $y=1-k-x$ is a tangent to the curve $y=kx^2+x+2k$.

[4]

Solution

Solution:

Step 1: Set up the problem

We are given the straight line:

$$ y = 1 - k - x $$

and the curve:

$$ y = kx^2 + x + 2k $$

We need to find the values of $k$ such that the line is tangent to the curve. This means the line and the curve should intersect at exactly one point, and their slopes should be equal at that point.

Step 2: Solve for the intersection points

Equating the two equations for $y$, we get:

$$ 1 - k - x = kx^2 + x + 2k $$

Simplify and rearrange:

$$ kx^2 + 2x + 3k - 1 = 0 $$

This is a quadratic equation in $x$. For the line to be tangent to the curve, this quadratic equation must have exactly one solution, which occurs when the discriminant is zero.

Step 3: Find the discriminant

For a quadratic equation of the form $ax^2 + bx + c = 0$, the discriminant is given by:

$$ \Delta = b^2 - 4ac $$

Here, $a = k$, $b = 2$, and $c = 3k - 1$. Substituting these values:

$$ \Delta = 2^2 - 4(k)(3k - 1) $$

Simplify:

$$ \Delta = 4 - 12k^2 + 4k $$ $$ \Delta = -12k^2 + 4k + 4 $$

Step 4: Set the discriminant to zero

For the line to be tangent, $\Delta = 0$. Thus:

$$ -12k^2 + 4k + 4 = 0 $$

Divide through by $-4$ to simplify:

$$ 3k^2 - k - 1 = 0 $$

Step 5: Solve for $k$

Solve the quadratic equation $3k^2 - k - 1 = 0$ using the quadratic formula:

$$ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

where $a = 3$, $b = -1$, and $c = -1$. Substituting these values:

$$ k = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-1)}}{2(3)} $$ $$ k = \frac{1 \pm \sqrt{1 + 12}}{6} $$ $$ k = \frac{1 \pm \sqrt{13}}{6} $$

Step 6: Verify the solutions

For each value of $k$, check that the discriminant is zero and that the line is tangent to the curve.

• For $k = \frac{1 + \sqrt{13}}{6}$, substitute into the original equations to verify tangency.
• For $k = \frac{1 - \sqrt{13}}{6}$, substitute into the original equations to verify tangency.

Both values of $k$ satisfy the tangency condition.

Final Answer

The exact values of $k$ such that the line is tangent to the curve are:

$$ k = \frac{1 + \sqrt{13}}{6} \quad \text{and} \quad k = \frac{1 - \sqrt{13}}{6} $$

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Question 2

A curve has equation $y=(5-x)(x+2)^2$.

• (a) Find the $x$-coordinates of the stationary points on the curve. [4]
• (b) On the axes below, sketch the graph of $y=(5-x)(x+2)^2$, stating the coordinates of the points where the curve meets the axes. [3]

• (c) Find the values of $k$ for which the equation $k=(5-x)(x+2)^2$ has one distinct root only. [3]

Solution

Solution:

Step (a): Finding the $x$-coordinates of the stationary points

The given curve is defined by the equation:

$$ y = (5-x)(x+2)^2 $$

First, expand $y$:

$$ y = (5-x)(x^2 + 4x + 4) $$ $$ y = 5x^2 + 20x + 20 - x^3 - 4x^2 - 4x $$ $$ y = -x^3 + x^2 + 16x + 20 $$

Now, compute the derivative $\frac{dy}{dx}$:

$$ \frac{dy}{dx} = \frac{d}{dx}[-x^3 + x^2 + 16x + 20] $$ $$ \frac{dy}{dx} = -3x^2 + 2x + 16 $$

To find the stationary points, set $\frac{dy}{dx} = 0$:

$$ -3x^2 + 2x + 16 = 0 $$

Solve this quadratic equation using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

where $a = -3$, $b = 2$, and $c = 16$.

$$ x = \frac{-2 \pm \sqrt{2^2 - 4(-3)(16)}}{2(-3)} $$ $$ x = \frac{-2 \pm \sqrt{4 + 192}}{-6} $$ $$ x = \frac{-2 \pm \sqrt{196}}{-6} $$ $$ x = \frac{-2 \pm 14}{-6} $$

This gives:

$$ x = \frac{-2 + 14}{-6} = \frac{12}{-6} = -2 $$ $$ x = \frac{-2 - 14}{-6} = \frac{-16}{-6} = \frac{8}{3} $$

The $x$-coordinates of the stationary points are $x=-2$ and $x=\frac{8}{3}$.

Step (b): Sketching the graph of $y=(5-x)(x+2)^2$

First, find the intercepts:

• $y$-intercept: Substitute $x=0$ into $y$:

$$ y = (5-0)(0+2)^2 = 5 \cdot 4 = 20 $$

The $y$-intercept is $(0,20)$.

• $x$-intercepts: Solve $y=0$:

$$ (5-x)(x+2)^2 = 0 $$ $$ 5-x = 0 \quad \text{or} \quad (x+2)^2 = 0 $$ $$ x=5 \quad \text{or} \quad x=-2 $$

The $x$-intercepts are $(5,0)$ and $(-2,0)$.

Using these points and the stationary points $x=-2$ and $x=\frac{8}{3}$, sketch the curve $y=(5-x)(x+2)^2$. The curve passes through $(0,20)$, $(5,0)$, and $(-2,0)$, with stationary points at $x=-2$ and $x=\frac{8}{3}$.

Step (c): Finding $k$ for which $k=(5-x)(x+2)^2$ has one distinct root

The equation $k=(5-x)(x+2)^2$ can be rewritten as:

$$ y=(5-x)(x+2)^2 \qquad \text{and} \qquad y=k $$

Calculate the corresponding $y$-values:

• At $x=-2$:

$$ y = (5-(-2))((-2)+2)^2 $$ $$ y = 7 \cdot 0 = 0 $$

• At $x=\frac{8}{3}$:

$$ y = \left(5-\frac{8}{3}\right) \left(\frac{8}{3}+2\right)^2 $$ $$ y = \frac{7}{3} \left(\frac{14}{3}\right)^2 $$ $$ y = \frac{7}{3} \cdot \frac{196}{9} $$ $$ y = \frac{1372}{27} $$

From Step (b), the turning points occur at $x=-2$ and $x=\frac{8}{3}$. We can check that $(-2,0)$ is a minimum point and $\left(\frac{8}{3}, \frac{1372}{27}\right)$ is a maximum point.

If we take $0 \le k \le \frac{1372}{27}$, the given equation has at least two roots.

For $k$ to have one distinct real root, we must have:

$$ k > \frac{1372}{27} \quad \text{or} \quad k \lt 0 $$

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Question 3

Find the coefficient of $x^8$ in the expansion of $\left(1-x^2\right)\left(2x - \frac{1}{x}\right)^{10}$. [5]

Solution

Solution

Step 1: Expand the binomial $\left(2x - \frac{1}{x}\right)^{10}$

Using the binomial theorem, we have:

$$ \left(2x - \frac{1}{x}\right)^{10} = \sum_{k=0}^{10} \binom{10}{k} (2x)^k \left(-\frac{1}{x}\right)^{10-k}. $$

Simplify each term in the summation:

$$ (2x)^k \left(-\frac{1}{x}\right)^{10-k} = 2^k x^k \cdot (-1)^{10-k} x^{-(10-k)} = 2^k (-1)^{10-k} x^{2k - 10}. $$

Thus, the expansion becomes:

$$ \left(2x - \frac{1}{x}\right)^{10} = \sum_{k=0}^{10} \binom{10}{k} 2^k (-1)^{10-k} x^{2k - 10}. $$

We need the coefficients of $x^6$ and $x^8$ in the expansion of $\left(2x - \frac{1}{x}\right)^{10}$.

For $x^6$:

$$ 2k - 10 = 6 \implies k = 8. $$

Substitute $k=8$:

$$ \binom{10}{8} 2^8 (-1)^{10-8} = \binom{10}{8} 2^8 (-1)^2 = 45 \cdot 256 = 11520. $$

For $x^8$:

$$ 2k - 10 = 8 \implies k = 9. $$

Substitute $k=9$:

$$ \binom{10}{9} 2^9 (-1)^{10-9} = \binom{10}{9} 2^9 (-1)^1 = -10 \cdot 512 = -5120. $$

Thus,

$$ \left(1 - x^2\right)\left(2x - \frac{1}{x}\right)^{10} = (1-x^2)(\cdots +11520 x^6 -5120 x^8 +\cdots) $$

Step 3: Collect coefficients for $x^6$ and $x^8$

When multiplying by $1 - x^2$, the coefficient of $x^8$ comes from:

$$ (1-x^2) $$ $$ (-1)\cdot (\text{coefficient of } x^6) + (1)\cdot (\text{coefficient of } x^8). $$

Substitute the values:

$$ (-1)(11520) + (1)(-5120) = -11520 - 5120 = -16640. $$

Final Answer

The coefficient of $x^8$ is $\boxed{-16640}$.

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Question 4

(a) Write $3 \lg x-2 \lg y^2-3$ as a single logarithm to base 10. [3]

(b) Solve the equation $\log _3 x+\log _x 3=\frac{5}{2}$. [5]

Solution

Solution

Step 1: Simplify $3 \lg x - 2 \lg y^2 - 3$ as a single logarithm to base 10

We start with the given expression:

$$ 3 \lg x - 2 \lg y^2 - 3. $$

Simplify each term:

1. Recall the logarithmic power rule:

$ a \lg b = \lg b^a $.

$ 3 \lg x = \lg x^3 $

$ 2 \lg y^2 = \lg (y^2)^2 = \lg y^4 $

2. Substitute the simplified terms:

$$ \lg x^3 - \lg y^4 - 3. $$

Combine using logarithmic properties:

1. Recall the logarithmic subtraction rule:

$ \lg a - \lg b = \lg \frac{a}{b} $.

$ \lg x^3 - \lg y^4 = \lg \frac{x^3}{y^4} $.

2. Introduce the constant $-3$:

$ \lg \frac{x^3}{y^4} - 3 = \lg \frac{x^3}{y^4} + \lg 10^{-3} $, because $-3 = \lg 10^{-3}$.

3. Combine using the addition rule:

$ \lg a + \lg b = \lg (ab) $:

$ \lg \frac{x^3}{y^4} + \lg 10^{-3} = \lg \left( \frac{x^3}{y^4} \cdot 10^{-3} \right) $.

Final expression:

$$ \lg \left( \frac{x^3}{10^3 y^4} \right). $$

Step 2: Solve the equation $\log_3 x + \log_x 3 = \frac{5}{2}$

Given the equation:

$$ \log_3 x + \log_x 3 = \frac{5}{2}. $$

Simplify $\log_x 3$:

Using the change of base formula, $ \log_x 3 = \frac{1}{\log_3 x} $. Substitute this into the equation:

$$ \log_3 x + \frac{1}{\log_3 x} = \frac{5}{2}. $$

Let $ y = \log_3 x $. Then the equation becomes:

$$ y + \frac{1}{y} = \frac{5}{2}. $$

Multiply through by $y$ to eliminate the fraction:

$$ y^2 + 1 = \frac{5}{2} y. $$

Rearrange into a standard quadratic form:

$$ 2y^2 - 5y + 2 = 0. $$

Solve the quadratic equation:

Use the quadratic formula $ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $:

$$ y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}. $$

Simplify the solutions:

$$ y = \frac{5 + 3}{4} = 2 \quad \text{or} \quad y = \frac{5 - 3}{4} = \frac{1}{2}. $$

Interpret the solutions:

Recall $ y = \log_3 x $:

1. If $ y = 2 $, then $ \log_3 x = 2 $, so $ x = 3^2 = 9 $.

2. If $ y = \frac{1}{2} $, then $ \log_3 x = \frac{1}{2} $, so $ x = 3^{1/2} = \sqrt{3} $.

Final solutions:

$$ x = 9 \quad \text{or} \quad x = \sqrt{3}. $$

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Question 5

The table shows values of the variables $x$ and $y$, which are related by an equation of the form $y=A b^{x^2}$, where $A$ and $b$ are constants.

$x$	1	1.5	2	2.5
$y$	2.0	11.3	128	2896

(a) Use the data to draw a straight line graph of $\ln y$ against $x^2$. [2]

(b) Use your graph to estimate the values of $A$ and $b$. Give your answers correct to 1 significant figure. [5]

(c) Estimate the value of $y$ when $x=1.75$. [2]

(d) Estimate the positive value of $x$ when $y=20$. [2]

Solution

Solution

Step (a): Plotting the Graph of $\ln y$ Against $x^2$

We are given the equation relating $x$ and $y$:

$$ y = A b^{x^2}. $$

Taking the natural logarithm of both sides:

$$ \ln y = \ln A + x^2 \ln b. $$

This equation is in the form of a straight line $y = mx + c$, where:

• The slope $m = \ln b$
• The intercept $c = \ln A$

We are given the following data:

$$ \begin{array}{|c|c|c|c|c|} \hline x & 1 & 1.5 & 2 & 2.5 \\ \hline y & 2.0 & 11.3 & 128 & 2896 \\ \hline \end{array} $$

To create the straight-line graph:

1. Calculate $x^2$ for each value of $x$.
2. Calculate $\ln y$ for each corresponding value of $y$.

$$ \begin{array}{|c|c|c|c|c|} \hline x & 1 & 1.5 & 2 & 2.5 \\ \hline x^2 & 1 & 2.25 & 4 & 6.25 \\ \hline y & 2.0 & 11.3 & 128 & 2896 \\ \hline \ln y & \ln 2.0 \approx 0.693 & \ln 11.3 \approx 2.431 & \ln 128 \approx 4.855 & \ln 2896 \approx 7.975 \\ \hline \end{array} $$

\begin{tikzpicture}
\draw[step=1cm,black,very thin] (-1,-1) grid (7,8);

\draw[thick,->](-1,0) -- (7,0) node[below]{$x^2$};

\draw[thick,->](0,-1) -- (0,8) node[above]{$\ln y$};

\draw[->]
(0,-.7) -- (1,0.693) -- (2.25,2.431) -- (4,4.855) -- (6.25,7.975);

\end{tikzpicture}

Step (b): Estimating Values of $A$ and $b$

From the graph, we can obtain the slope and intercept.

1. The slope of the line is $\ln b$.
2. The intercept is $\ln A$.

Thus

$$ \ln b=\frac{4.855-0.693}{4-1} =\frac{4.162}{3} =1.387. $$

By inspecting the graph, the intercept is approximately $-0.7$. We can calculate the values of $A$ and $b$ as follows:

$$ b = e^{\text{slope}} \quad \text{and} \quad A = e^{\text{intercept}}. $$

$$ b \approx e^{1.387} \approx 4, \quad A \approx e^{-0.7} \approx 0.5. $$

Thus, the estimated values of $A$ and $b$ are:

$$ A \approx 0.5 \quad \text{(to 1 significant figure)} \quad \text{and} \quad b \approx 4.0 \quad \text{(to 1 significant figure)}. $$

Step (c): Estimating the Value of $y$ When $x = 1.75$

Now, using the equation $y = A b^{x^2}$, we can estimate the value of $y$ for $x = 1.75$.

First, calculate $x^2$:

$$ x^2 = (1.75)^2 = 3.0625. $$

Then, substitute the values of $A \approx 0.5$ and $b \approx 4.0$ into the equation:

$$ y = 0.5 \times 4.0^{3.0625}. $$

Estimate the value of $4.0^{3.0625}$ (using a calculator):

$$ 4.0^{3.0625} \approx 66.6. $$

Thus:

$$ y \approx 0.5 \times 66.6 \approx 33.3. $$

So, the estimated value of $y$ when $x = 1.75$ is approximately $33.3$.

Step (d): Estimating the Positive Value of $x$ When $y = 20$

To estimate the positive value of $x$ when $y = 20$, we use the equation $y = A b^{x^2}$.

First, solve for $x^2$:

$$ y = A b^{x^2} \quad \Rightarrow \quad \frac{y}{A} = b^{x^2} \quad \Rightarrow \quad x^2 = \frac{\ln \left( \frac{y}{A} \right)}{\ln b}. $$

Substitute the known values $A \approx 0.5$, $b \approx 4.0$, and $y = 20$:

$$ x^2 = \frac{\ln \left( \frac{20}{0.5} \right)}{\ln 4.0} = \frac{\ln (40)}{\ln 4.0}. $$

Using a calculator:

$$ \ln 40 \approx 3.6889, \quad \ln 4.0 \approx 1.3863. $$

Thus:

$$ x^2 = \frac{3.6889}{1.3863} \approx 2.66. $$

Finally, take the square root of both sides:

$$ x \approx \sqrt{2.66} \approx 1.63. $$

So, the positive value of $x$ when $y = 20$ is approximately $1.63$.

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Question 6

Given that $ \mathrm{f}^{\prime \prime}(x)=(5 x+2)^{-\frac{2}{5}} $, $ \mathrm{f}^{\prime}(6)=\frac{17}{3} $ and $ \mathrm{f}(6)=\frac{26}{3} $, find an expression for $ \mathrm{f}(x) $. [8]

Solution

Solution

Step 1: Integrate the second derivative to find the first derivative

We are given that the second derivative of the function $f(x)$ is:

$$ f''(x) = (5x + 2)^{-\frac{2}{5}}. $$

To find $f'(x)$, we integrate $f''(x)$ with respect to $x$:

$$ f'(x) = \int (5x + 2)^{-\frac{2}{5}} \, dx. $$

Let $u = 5x + 2$, so that $du = 5 \, dx$, or $dx = \frac{du}{5}$. Now the integral becomes:

$$ f'(x) = \int u^{-\frac{2}{5}} \cdot \frac{du}{5}. $$

Simplifying:

$$ f'(x) = \frac{1}{5} \int u^{-\frac{2}{5}} \, du = \frac{1}{5} \cdot \frac{u^{\frac{3}{5}}}{\frac{3}{5}} = \frac{1}{5} \cdot \frac{5}{3} u^{\frac{3}{5}} = \frac{1}{3} (5x + 2)^{\frac{3}{5}} + C_1. $$

Thus, the first derivative is:

$$ f'(x) = \frac{1}{3} (5x + 2)^{\frac{3}{5}} + C_1. $$

Step 2: Use the initial condition $f'(6)=\frac{17}{3}$ to find $C_1$

We are given that $ f'(6)=\frac{17}{3} $. Substituting $x=6$ into the expression for $f'(x)$:

$$ f'(6) = \frac{1}{3}(5(6)+2)^{\frac{3}{5}} + C_1 = \frac{1}{3}(30+2)^{\frac{3}{5}} + C_1 = \frac{1}{3}\cdot 32^{\frac{3}{5}} + C_1. $$

We also know that:

$$ f'(6)=\frac{17}{3}. $$

So:

$$ \frac{1}{3}\cdot 32^{\frac{3}{5}} + C_1 = \frac{17}{3}. $$

Since $32 = 2^5$:

$$ 32^{\frac{3}{5}} = (2^5)^{\frac{3}{5}} = 2^3 = 8. $$

Therefore:

$$ \frac{8}{3} + C_1 = \frac{17}{3}. $$

Subtracting $\frac{8}{3}$ from both sides:

$$ C_1 = \frac{17}{3} - \frac{8}{3} = \frac{9}{3} = 3. $$

Thus, $C_1 = 3$.

Step 3: Integrate the first derivative to find the original function $f(x)$

Now we have:

$$ f'(x) = \frac{1}{3}(5x + 2)^{\frac{3}{5}} + 3. $$

To find $f(x)$, integrate $f'(x)$:

$$ f(x) = \int \left( \frac{1}{3}(5x + 2)^{\frac{3}{5}} + 3 \right) dx. $$

Split the integral into two parts:

$$ f(x) = \frac{1}{3}\int (5x + 2)^{\frac{3}{5}} \, dx + \int 3 \, dx. $$

For the first integral, let $u = 5x + 2$, so that $du = 5dx$, or $dx = \frac{du}{5}$:

$$ \frac{1}{3}\int (5x + 2)^{\frac{3}{5}} \, dx = \frac{1}{3}\cdot\frac{1}{5}\int u^{\frac{3}{5}} \, du = \frac{1}{15}\cdot\frac{u^{\frac{8}{5}}}{\frac{8}{5}} = \frac{1}{15}\cdot\frac{5}{8}u^{\frac{8}{5}} = \frac{1}{24}(5x + 2)^{\frac{8}{5}}. $$

The second integral is:

$$ \int 3 \, dx = 3x. $$

Thus, the general expression for $f(x)$ is:

$$ f(x) = \frac{1}{24}(5x + 2)^{\frac{8}{5}} + 3x + C_2. $$

Step 4: Use the initial condition $f(6)=\frac{26}{3}$ to find $C_2$

We are given that:

$$ f(6)=\frac{26}{3}. $$

Substitute $x=6$ into the expression for $f(x)$:

$$ f(6) = \frac{1}{24}(5(6)+2)^{\frac{8}{5}} + 3(6) + C_2 = \frac{1}{24}(32)^{\frac{8}{5}} + 18 + C_2. $$

Since:

$$ 32^{\frac{8}{5}} = (2^5)^{\frac{8}{5}} = 2^8 = 256, $$

we get:

$$ f(6) = \frac{1}{24}\cdot256 + 18 + C_2 = \frac{256}{24} + 18 + C_2 = \frac{64}{6} + 18 + C_2. $$

Now use $ f(6)=\frac{26}{3} $:

$$ \frac{64}{6} + 18 + C_2 = \frac{26}{3}. $$

Convert to denominator 3:

$$ \frac{64}{6} = \frac{32}{3}, \qquad 18 = \frac{54}{3}. $$

So:

$$ \frac{32}{3} + \frac{54}{3} + C_2 = \frac{26}{3}. $$

Simplifying:

$$ \frac{86}{3} + C_2 = \frac{26}{3}. $$

Subtract $\frac{86}{3}$ from both sides:

$$ C_2 = \frac{26}{3} - \frac{86}{3} = \frac{-60}{3} = -20. $$

Thus, $C_2 = -20$.

Final Answer

$$ f(x) = \frac{1}{24}(5x + 2)^{\frac{8}{5}} + 3x - 20. $$

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Question 7

(a) A 5-character password is to be formed from the following 13 characters.

Letters	A	B	C	D	E
Numbers	9	8	7	6	5
Symbols	$*$	$\#$	$!$

No character may be used more than once in any password.

(i) Find the number of possible passwords that can be formed. [1]

(ii) Find the number of possible passwords that contain at least one symbol. [2]

(b) Given that $ 16 \times { }^n \mathrm{C}_{12} = (n-10)\times { }^{n+1}\mathrm{C}_{11} $, find the value of $n$. [3]

Solution

Solution

Step (a) - Part (i): Number of possible passwords

We are tasked with forming a 5-character password using 13 available characters, which consist of letters, numbers, and symbols.

These characters are:

$$ \text{Letters: } A,B,C,D,E \quad \text{(5 letters)} $$ $$ \text{Numbers: } 9,8,7,6,5 \quad \text{(5 numbers)} $$ $$ \text{Symbols: } *,\#,! \quad \text{(3 symbols)} $$

The total number of characters available is $13$. Since no character may be used more than once in any password, we use permutations.

The number of possible passwords is:

$$ P(13,5) = \frac{13!}{(13-5)!} = \frac{13!}{8!} $$

Compute:

$$ P(13,5) = 13\times12\times11\times10\times9 = 154440 $$

Thus, the number of possible passwords is $ \boxed{154440} $.

Step (a) - Part (ii): Number of passwords with at least one symbol

Use complementary counting.

Total passwords:

$$ 154440 $$

Now find the number of passwords with no symbols. Only the 10 letters and numbers are used.

$$ P(10,5) = \frac{10!}{(10-5)!} = \frac{10!}{5!} $$

Compute:

$$ P(10,5) = 10\times9\times8\times7\times6 = 30240 $$

Therefore:

$$ 154440-30240=124200 $$

Thus, the number of passwords containing at least one symbol is $ \boxed{124200} $.

Step (b): Solve for $n$

We are given:

$$ 16\times {^n\mathrm{C}_{12}} = (n-10)\times {^{n+1}\mathrm{C}_{11}} $$

Recall:

$$ {^n\mathrm{C}_k} = \frac{n!}{k!(n-k)!} $$

Substitute into the equation:

$$ 16\times\frac{n!}{12!(n-12)!} = (n-10)\times \frac{(n+1)!}{11!(n-10)!} $$

Since $ (n+1)!=(n+1)n! $:

$$ 16\times\frac{n!}{12!(n-12)!} = (n-10)\times \frac{(n+1)(n!)}{11!(n-10)!} $$

Cancel $n!$:

$$ 16\times\frac{1}{12!(n-12)!} = (n-10)\times \frac{(n+1)}{11!(n-10)!} $$

Simplify:

$$ 16\times\frac{1}{12!(n-12)!} = \frac{(n-10)(n+1)}{11!(n-10)!} $$

Multiply both sides by $ 12!(n-12)! $:

$$ 16 = \frac{(n-10)(n+1)\times12!(n-12)!} {11!(n-10)!} $$ $$ 16 = \frac{ (n-10)(n+1)\times12\times11!(n-12)! }{ 11!(n-10)(n-11)(n-12)! } $$ $$ 16 = \frac{12(n+1)}{n-11} $$

Cross multiply:

$$ 16(n-11)=12(n+1) $$ $$ 16n-176=12n+12 $$ $$ 4n=188 $$ $$ \boxed{n=47} $$

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Question 8

The diagram shows part of the curve $ y=2-\frac{3}{x-1} $ and the straight line $ 6y=9-2x. $ The curve intersects the $x$-axis at point $A$ and the line at point $B$. The line intersects the $x$-axis at point $C$. Find the area of the shaded region $ABC$, giving your answer in the form $ p+\ln q, $ where $p$ and $q$ are rational numbers. [11]

Solution

Solution

Step 1: Find the equations of the curve and line

The equation of the curve is:

$$ y=2-\frac{3}{x-1} $$

The equation of the line is:

$$ 6y=9-2x $$

Solving for $y$:

$$ y=\frac{9-2x}{6} $$

Step 2: Find point $A$ where the curve meets the $x$-axis

At the $x$-axis, $ y=0. $

$$ 0=2-\frac{3}{x-1} $$

Solve for $x$:

$$ \frac{3}{x-1}=2 $$ $$ x-1=\frac{3}{2} $$ $$ x=\frac{5}{2} $$

Therefore,

$$ A\left(\frac{5}{2},0\right) $$

Step 3: Find point $B$ where the curve and line intersect

Set the equations equal:

$$ 2-\frac{3}{x-1}=\frac{9-2x}{6} $$

Multiply both sides by $6$:

$$ 6\left(2-\frac{3}{x-1}\right)=9-2x $$ $$ 12-\frac{18}{x-1}=9-2x $$

Rearrange:

$$ 3=-2x+\frac{18}{x-1} $$

Multiply by $(x-1)$:

$$ 3(x-1)=-2x(x-1)+18 $$

Expand:

$$ 3x-3=-2x^2+2x+18 $$ $$ 2x^2+x-21=0 $$

Use the quadratic formula:

$$ x=\frac{-1\pm\sqrt{1^2-4(2)(-21)}}{2(2)} $$ $$ x=\frac{-1\pm\sqrt{169}}{4} $$ $$ x=\frac{-1\pm13}{4} $$

Hence:

$$ x=3 \quad \text{or} \quad x=-3.5 $$

Take the positive value:

$$ x=3 $$

Substitute into the line equation:

$$ y=\frac{9-2(3)}{6} =\frac{3}{6} =\frac{1}{2} $$

Therefore,

$$ B\left(3,\frac{1}{2}\right) $$

Step 4: Find point $C$ where the line meets the $x$-axis

At the $x$-axis, $ y=0. $

$$ 0=\frac{9-2x}{6} $$ $$ 9-2x=0 $$ $$ x=\frac{9}{2} $$

Therefore,

$$ C\left(\frac{9}{2},0\right) $$

Step 5: Find the area of the shaded region

The required area is:

$$ \text{Area} = \int_{2.5}^{3} \left( 2-\frac{3}{x-1} \right)dx + \int_{3}^{4.5} \frac{9-2x}{6}\,dx $$

First Integral

$$ \int_{2.5}^{3} \left( 2-\frac{3}{x-1} \right)dx $$ $$ = \int_{2.5}^{3}2\,dx - \int_{2.5}^{3}\frac{3}{x-1}\,dx $$ $$ = \left[2x\right]_{2.5}^{3} - 3\ln|x-1|\Big|_{2.5}^{3} $$ $$ = \left[2(3)-2(2.5)\right] - 3\ln\left(\frac{3-1}{2.5-1}\right) $$ $$ = 1-3\ln\left(\frac{4}{3}\right) $$

Second Integral

$$ \int_{3}^{4.5}\frac{9-2x}{6}\,dx $$ $$ = \frac{1}{6} \int_{3}^{4.5}(9-2x)\,dx $$ $$ = \frac{1}{6} \left[9x-x^2\right]_{3}^{4.5} $$ $$ = \frac{1}{6} \left[ (9\times4.5-(4.5)^2) - (9\times3-3^2) \right] $$ $$ = \frac{1}{6} \left[ (40.5-20.25) - (27-9) \right] $$ $$ = \frac{1}{6}(20.25-18) $$ $$ = \frac{1}{6}(2.25) =0.375 =\frac{3}{8} $$

Total Area

$$ \text{Area} = \left( 1-3\ln\left(\frac{4}{3}\right) \right) + \frac{3}{8} $$ $$ = \frac{11}{8} - 3\ln\left(\frac{4}{3}\right) $$ $$ = \frac{11}{8} + 3\ln\left(\frac{3}{4}\right) $$ $$ = \frac{11}{8} + \ln\left(\frac{3}{4}\right)^3 $$ $$ = \frac{11}{8} + \ln\left(\frac{27}{64}\right) $$

Thus,

$$ \boxed{ \frac{11}{8} + \ln\frac{27}{64} } $$

*****************************************************

Question 9

In this question, all lengths are in metres.

(a) A particle $P$ has position vector $ \binom{2+12t}{5-5t} $ at time $t$ seconds, $t\geq0$.

(i) Write down the initial position vector of $P$. [1]

(ii) Find the speed of $P$. [2]

(iii) Determine whether $P$ passes through the point with position vector $ \binom{158}{-48}. $ [2]

(b)The diagram shows the triangle $OAC$. The point $B$ lies on $AC$ such that $ AB:AC=1:4. $ Given that $ \overrightarrow{OA}=\mathbf{a}, \quad \overrightarrow{OB}=\mathbf{b}, \quad \overrightarrow{OC}=\mathbf{c}, $ find $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. [3]

Solution

Solution

Step 1: Initial position vector of $P$

The position vector of $P$ is:

$$ \binom{2+12t}{5-5t} $$

At $t=0$:

$$ \binom{2+12(0)}{5-5(0)} = \binom{2}{5} $$

Thus, the initial position vector is:

$$ \boxed{\binom{2}{5}} $$

Step 2: Find the speed of $P$

Differentiate the position vector with respect to $t$:

$$ \mathbf{v}(t) = \binom{\frac{d}{dt}(2+12t)}{\frac{d}{dt}(5-5t)} = \binom{12}{-5} $$

The speed is the magnitude of the velocity vector:

$$ \text{Speed} = \sqrt{12^2+(-5)^2} $$ $$ = \sqrt{144+25} $$ $$ = \sqrt{169} $$ $$ =13 $$

Thus, the speed of $P$ is:

$$ \boxed{13} $$

metres per second.

Step 3: Determine whether $P$ passes through $\binom{158}{-48}$

Set the position vector equal to $ \binom{158}{-48}. $

$$ \binom{2+12t}{5-5t} = \binom{158}{-48} $$

Equating components:

$$ 2+12t=158 $$ $$ 5-5t=-48 $$

From the first equation:

$$ 12t=156 $$ $$ t=13 $$

From the second equation:

$$ -5t=-53 $$ $$ t=\frac{53}{5}=10.6 $$

The values of $t$ are different, so the particle does not pass through the point.

The particle $P$ does not pass through $\boxed{\binom{158}{-48}}$

Step 4: Find $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$

Given:

$$ \overrightarrow{OA}=\mathbf{a}, \quad \overrightarrow{OB}=\mathbf{b}, \quad \overrightarrow{OC}=\mathbf{c} $$

Since $ AB:AC=1:4, $ point $B$ divides $AC$ in the ratio $1:3$.

Now:

$$ \overrightarrow{AC} = \mathbf{c}-\mathbf{a} $$

Therefore:

$$ \overrightarrow{OB} = \overrightarrow{OA} + \frac14\overrightarrow{AC} $$ $$ \mathbf{b} = \mathbf{a} + \frac14(\mathbf{c}-\mathbf{a}) $$

Expand:

$$ \mathbf{b} = \mathbf{a} + \frac14\mathbf{c} - \frac14\mathbf{a} $$ $$ \mathbf{b} = \frac34\mathbf{a} + \frac14\mathbf{c} $$

Multiply by $4$:

$$ 4\mathbf{b} = 3\mathbf{a} + \mathbf{c} $$

Hence:

$$ \boxed{ \mathbf{c} = 4\mathbf{b} - 3\mathbf{a} } $$

Conclusion

(i) Initial position vector: $ \binom{2}{5} $

(ii) Speed: $ 13 $ metres per second

(iii) The particle does not pass through $ \binom{158}{-48} $

(iv) $ \mathbf{c}=4\mathbf{b}-3\mathb

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Question 10

(a) It is given that
2 + cos θ = x for 1 < x < 3 and 2 cosec θ = y for y > 2.
Find y in terms of x. [4]

(b) Solve the equation
3 cos(φ/2) = √3 sin(φ/2)
for -4π < φ < 4π. [5]

Solution

Solution

Step (a): Expressing y in terms of x

We are given:

2 + cos θ = x,   1 < x < 3
2 cosec θ = y,   y > 2

Step 1: Solve for cos θ

cos θ = x - 2

Step 2: Use identity

cosec θ = 1 / sin θ

sin² θ + cos² θ = 1

sin² θ = 1 - (x - 2)²

sin θ = ±√(1 - (x - 2)²)

Step 3: Choose correct sign

Since y > 2, we take sin θ > 0:

sin θ = √(1 - (x - 2)²)

Step 4: Find y

y = 2 cosec θ = 2 / sin θ

y = 2 / √(1 - (x - 2)²)

Thus,

y = 2 / √(1 - (x - 2)²)

Step (b): Solve 3 cos(φ/2) = √3 sin(φ/2)

Step 1: Divide by cos(φ/2)

3 = √3 tan(φ/2)

tan(φ/2) = √3

Step 2: General solution

φ/2 = nπ + π/3

Step 3: Multiply by 2

φ = 2nπ + 2π/3

Step 4: Apply interval -4π < φ < 4π

-4π < 2nπ + 2π/3 < 4π

Divide by 2π:

-2 < n + 1/3 < 2

-7/3 < n < 5/3

n = -2, -1, 0, 1

Step 5: Find values of φ

n = -2: φ = -10π/3
n = -1: φ = -4π/3
n = 0: φ = 2π/3
n = 1: φ = 8π/3

Final answer:

φ = -10π/3, -4π/3, 2π/3, 8π/3

****************************************** CIE Additional Mathematics (Summary)