9709_s24_qp

Question 1

Topic – ALV: 2.3

A curve has equation $y=2\tan x-5\sin x$ for $0≤x<\frac{1}{2}\pi$.

Find the x-coordinate of the stationary point of the curve. Give your answer correct to 3 significant figures.

▶️Answer/Explanation

Step 1: Find the derivative
The function is $ y = 2\tan x – 5\sin x $. Differentiate with respect to $ x $:
The derivative of $ 2\tan x $ is $ 2 \cdot \sec^2 x = 2\sec^2 x $ (since $ \frac{d}{dx} \tan x = \sec^2 x $),
The derivative of $ -5\sin x $ is $ -5 \cdot \cos x = -5\cos x $ (since $ \frac{d}{dx} \sin x = \cos x $).

Thus, the derivative is:
\[ \frac{dy}{dx} = 2\sec^2 x – 5\cos x \].

Step 2: Set the derivative equal to zero
For a stationary point, the slope is zero:
\[ 2\sec^2 x – 5\cos x = 0 \]
\[ 2\sec^2 x = 5\cos x \].

Step 3: Simplify using trigonometric identities
Recall that $ \sec x = \frac{1}{\cos x} $, so $ \sec^2 x = \frac{1}{\cos^2 x} $. Substitute this into the equation:
\[ 2 \cdot \frac{1}{\cos^2 x} = 5\cos x \]
\[ \frac{2}{\cos^2 x} = 5\cos x \].

Step 4: Solve for $ \cos x $
Multiply both sides by $ \cos^2 x $ (noting that $ \cos x \neq 0 $ in $ 0 \leq x < \frac{\pi}{2} $, where $ \cos x > 0 $):
\[ 2 = 5\cos^3 x \]
\[ \cos^3 x = \frac{2}{5} \]
\[ \cos^3 x = 0.4 \]
\[ \cos x = (0.4)^{1/3} \].

Step 5: Calculate $ \cos x $
We need the cube root of 0.4:
$ 0.4 = \frac{2}{5} $,
$ \cos x = \sqrt[3]{0.4} $

Using a calculator:
\[ (0.4)^{1/3} \approx 0.7368063 \].
So, $ \cos x \approx 0.7368063 $.

Step 6: Find $ x $
Since $ \cos x = 0.7368063 $, take the inverse cosine:
\[ x = \cos^{-1}(0.7368063) \].

Using a calculator (in radians):
\[ x \approx 0.74195 \text{ radians} \].

Step 7: Verify the interval
The domain is $ 0 \leq x < \frac{\pi}{2} $, and $ \frac{\pi}{2} \approx 1.5708 $. Since $ 0.74195 $ is between 0 and 1.5708, it satisfies the given interval.

Step 8: Round to 3 significant figures
The value $ 0.74195 $ to 3 significant figures is:
First three digits: 0.7, 4, 1,
Next digit is 9 (which is ≥ 5), so round up 1 to 2,
Thus, $ x \approx 0.742 $.

Final Answer
The x-coordinate of the stationary point is $ x = 0.742 $ (correct to 3 significant figures).

Question 2

Topic – ALV: 3.4

A curve has equation $x^2 \ln y + y^2 + 4x = 9$.

Find the gradient of the curve at the point $(2, 1)$.

▶️Answer/Explanation

Start with the equation:

$ x^2 \ln y + y^2 + 4x = 9 $.

Differentiate both sides with respect to $ x $:
For $ x^2 \ln y $, use the product rule: $ 2x \ln y + x^2 \cdot \frac{1}{y} \cdot \frac{dy}{dx} $.
For $ y^2 $, use the chain rule: $ 2y \frac{dy}{dx} $.
For $ 4x $, the derivative is $ 4 $.
The right side, $ 9 $, has a derivative of $ 0 $.

So, the equation becomes:
$ 2x \ln y + x^2 \cdot \frac{1}{y} \cdot \frac{dy}{dx} + 2y \frac{dy}{dx} + 4 = 0 $.

Solve for $ \frac{dy}{dx} $:
$ x^2 \cdot \frac{1}{y} \cdot \frac{dy}{dx} + 2y \frac{dy}{dx} = -4 – 2x \ln y $,
$ \frac{dy}{dx} \left( \frac{x^2}{y} + 2y \right) = -4 – 2x \ln y $,
$ \frac{dy}{dx} = \frac{-4 – 2x \ln y}{\frac{x^2}{y} + 2y} $.

Now, substitute $ x = 2 $, $ y = 1 $:
Numerator: $ -4 – 2(2) \ln 1 = -4 – 4 \cdot 0 = -4 $.
Denominator: $ \frac{2^2}{1} + 2 \cdot 1 = 4 + 2 = 6 $.

Thus:
$ \frac{dy}{dx} = \frac{-4}{6} = -\frac{2}{3} $.

The gradient at (2, 1) is $ -\frac{2}{3} $.

———-Markscheme————-

Attempt use of product rule for differentiating $x^2 \ln y$

Obtain $2x \ln y + \frac{x^2}{y} \frac{dy}{dx}$

Obtain $+2y \frac{dy}{dx}$

Substitute $x=2, y=1$ in equation equal to 0 involving at least one $\frac{dy}{dx}$ and solve for $\frac{dy}{dx}$

Obtain $-\frac{11}{6}$

Question 3

Topic – ALV: 3.4

(a) Sketch on the same diagram the graphs of $y=|3x-8|$ and $y=5-x$.

(b) Solve the inequality $|3x-8|<5-x$.

▶️Answer/Explanation

(a) Sketching the graphs
$y = |3x – 8|$: This is a V-shaped graph. Find the vertex where $3x – 8 = 0$, so $x = \frac{8}{3} \approx 2.67$. At $x = 0$, $y = |0 – 8| = 8$. At $x = 3$, $y = |9 – 8| = 1$. The graph drops to 0 at $x = 2.67$ and rises symmetrically.
$y = 5 – x$: A straight line. At $x = 0$, $y = 5$. At $x = 5$, $y = 0$. It slopes downward with a gradient of $-1$.
On the same diagram, plot $y = |3x – 8|$ (V-shape, vertex at $(2.67, 0)$, points like $(0, 8)$, $(3, 1)$) and $y = 5 – x$ (line from $(0, 5)$ to $(5, 0)$). They intersect where $|3x – 8| = 5 – x$.

(b) Solve $|3x – 8| < 5 – x$
Since $|3x – 8|$ is absolute value, consider two cases based on the expression inside:

Case 1: $3x – 8 \geq 0$ (i.e., $x \geq \frac{8}{3}$):
Inequality becomes $3x – 8 < 5 – x$.
Solve: $3x + x < 5 + 8 \Rightarrow 4x < 13 \Rightarrow x < \frac{13}{4} = 3.25$.
Combine with $x \geq \frac{8}{3} \approx 2.67$. So, $2.67 \leq x < 3.25$.

Case 2: $3x – 8 < 0$ (i.e., $x < \frac{8}{3}$):
Inequality becomes $-(3x – 8) < 5 – x$.
Simplify: $-3x + 8 < 5 – x \Rightarrow -3x + x < 5 – 8 \Rightarrow -2x < -3 \Rightarrow x > \frac{3}{2} = 1.5$.
Combine with $x < \frac{8}{3} \approx 2.67$. So, $1.5 < x < 2.67$.
Solution: Combine the intervals: $1.5 < x < 3.25$.

(c) Largest integer $N$ for $|3e^{0.1N} – 8| < 5 – e^{0.1N}$
This resembles part (b) with $e^{0.1N}$ replacing $x$. From (b), the inequality $|3x – 8| < 5 – x$ holds for $1.5 < x < 3.25$. So, $1.5 < e^{0.1N} < 3.25$.

Solve $e^{0.1N} > 1.5$: $0.1N > \ln 1.5 \approx 0.405 \Rightarrow N > 4.05$.
Solve $e^{0.1N} < 3.25$: $0.1N < \ln 3.25 \approx 1.179 \Rightarrow N < 11.79$.
$N$ is an integer, so $N$ ranges from 5 to 11.

Check boundaries:
$N = 5$: $e^{0.5} \approx 1.649$, $|3 \cdot 1.649 – 8| = |4.947 – 8| = 3.053$, $5 – 1.649 = 3.351$. $3.053 < 3.351$, true.
$N = 11$: $e^{1.1} \approx 3.004$, $|3 \cdot 3.004 – 8| = |9.012 – 8| = 1.012$, $5 – 3.004 = 1.996$. $1.012 < 1.996$, true.
$N = 12$: $e^{1.2} \approx 3.320$, $|3 \cdot 3.320 – 8| = |9.96 – 8| = 1.96$, $5 – 3.320 = 1.68$. $1.96 < 1.68$, false.

Largest integer $N$ is 11.

Question 4

Topic – ALV: 1.5

(a) Show that $3 \tan 2\theta + \tan(\theta + 45^{\circ}) = \frac{\tan^2 \theta + 8 \tan \theta + 1}{1 – \tan^2 \theta}$.

(b) Hence solve the equation $3 \tan 2\theta + \tan(\theta + 45^{\circ}) = 4$ for $0^{\circ} < \theta < 180^{\circ}$.

▶️Answer/Explanation

1. Express $\tan 2\theta$:
\[
\tan 2\theta = \frac{2 \tan \theta}{1 – \tan^2 \theta}.
\]
So,
\[
3 \tan 2\theta = 3 \cdot \frac{2 \tan \theta}{1 – \tan^2 \theta} = \frac{6 \tan \theta}{1 – \tan^2 \theta}.
\]

2. Express $\tan(\theta + 45^\circ)$:
Use the angle addition formula:
\[
\tan(\theta + 45^\circ) = \frac{\tan \theta + \tan 45^\circ}{1 – \tan \theta \cdot \tan 45^\circ} = \frac{\tan \theta + 1}{1 – \tan \theta \cdot 1} = \frac{\tan \theta + 1}{1 – \tan \theta}.
\]

3. Combine the left side:
\[
3 \tan 2\theta + \tan(\theta + 45^\circ) = \frac{6 \tan \theta}{1 – \tan^2 \theta} + \frac{\tan \theta + 1}{1 – \tan \theta}.
\]
To combine, we need a common denominator. Let $t = \tan \theta$, and note that $1 – \tan^2 \theta = (1 – t^2)$.

Rewrite the second term with the denominator $1 – t^2$:
\[
1 – t = -(t – 1), \quad \text{so} \quad 1 – t^2 = (1 – t)(1 + t) = -(t – 1)(1 + t).
\]
However, it’s easier to express both terms over $1 – t^2$:
\[
\frac{\tan \theta + 1}{1 – \tan \theta} = \frac{t + 1}{1 – t}.
\]
Now, $1 – t = -\frac{1 – t^2}{1 + t}$ (since $1 – t^2 = (1 – t)(1 + t)$):
\[
\frac{t + 1}{1 – t} = \frac{t + 1}{-(t – 1)} = -\frac{t + 1}{t – 1}.
\]
But let’s try:
\[
\frac{t + 1}{1 – t} = \frac{t + 1}{(1 – t)} \cdot \frac{(1 + t)}{(1 + t)} = \frac{(t + 1)(1 + t)}{1 – t^2} = \frac{t + t^2 + 1 + t}{1 – t^2} = \frac{t^2 + 2t + 1}{1 – t^2}.
\]
So,
\[
\tan(\theta + 45^\circ) = \frac{t^2 + 2t + 1}{1 – t^2}.
\]

Now add:
\[
\frac{6t}{1 – t^2} + \frac{t^2 + 2t + 1}{1 – t^2} = \frac{6t + t^2 + 2t + 1}{1 – t^2} = \frac{t^2 + 8t + 1}{1 – t^2}.
\]
This matches the right side, $\frac{\tan^2 \theta + 8 \tan \theta + 1}{1 – \tan^2 \theta}$.

Thus, the equation is proven.

(b) Solve $3 \tan 2\theta + \tan(\theta + 45^\circ) = 4$ for $0^\circ < \theta < 180^\circ$

From (a), we have:
\[
3 \tan 2\theta + \tan(\theta + 45^\circ) = \frac{\tan^2 \theta + 8 \tan \theta + 1}{1 – \tan^2 \theta}.
\]
Set this equal to 4:
\[
\frac{\tan^2 \theta + 8 \tan \theta + 1}{1 – \tan^2 \theta} = 4.
\]
Let $t = \tan \theta$, so:
\[
\frac{t^2 + 8t + 1}{1 – t^2} = 4.
\]
Solve for the numerator:
\[
t^2 + 8t + 1 = 4(1 – t^2).
\]
\[
t^2 + 8t + 1 = 4 – 4t^2.
\]
\[
t^2 + 8t + 1 – 4 + 4t^2 = 0.
\]
\[
5t^2 + 8t – 3 = 0.
\]
This is a quadratic equation. Use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}, \quad \text{where } a = 5, b = 8, c = -3.
\]
\[
t = \frac{-8 \pm \sqrt{8^2 – 4(5)(-3)}}{2(5)} = \frac{-8 \pm \sqrt{64 + 60}}{10} = \frac{-8 \pm \sqrt{124}}{10} = \frac{-8 \pm 2\sqrt{31}}{10} = \frac{-4 \pm \sqrt{31}}{5}.
\]
So,
\[
t = \frac{-4 + \sqrt{31}}{5} \quad \text{or} \quad t = \frac{-4 – \sqrt{31}}{5}.
\]
Now, $\sqrt{31} \approx 5.568$, so:
\[
-4 + \sqrt{31} \approx -4 + 5.568 = 1.568, \quad \frac{1.568}{5} \approx 0.3136.
\]
\[
-4 – \sqrt{31} \approx -4 – 5.568 = -9.568, \quad \frac{-9.568}{5} \approx -1.9136.
\]
$t = \tan \theta$, so $\theta = \tan^{-1}(0.3136)$ or $\theta = \tan^{-1}(-1.9136)$.
Since $0^\circ < \theta < 180^\circ$, $\tan \theta$ can be positive (first quadrant) or negative (second quadrant).

1. For $t = \frac{-4 + \sqrt{31}}{5} \approx 0.3136$ (positive):
\[
\theta = \tan^{-1}(0.3136) \approx 17.4^\circ \quad \text{(first quadrant)}.
\]
In the second quadrant (where $\tan \theta$ is negative but the value is positive here), we don’t need to adjust as it’s already positive.

2. For $t = \frac{-4 – \sqrt{31}}{5} \approx -1.9136$ (negative):
\[
\theta = \tan^{-1}(-1.9136) \approx -62.3^\circ.
\]
Since $\theta$ must be in $0^\circ$ to $180^\circ$, convert the negative angle:
\[
\theta = 180^\circ – 62.3^\circ = 117.7^\circ \quad \text{(second quadrant)}.
\]

Now, check if these $\theta$ values satisfy the original equation, considering $\tan 2\theta$ and $\tan(\theta + 45^\circ)$ are defined (i.e., denominators $1 – \tan^2 \theta \neq 0$ and $1 – \tan \theta \neq 0$):
For $\theta \approx 17.4^\circ$, $\tan 17.4^\circ \approx 0.3136$, $1 – (0.3136)^2 \approx 1 – 0.0982 = 0.9018 \neq 0$, and $1 – 0.3136 \approx 0.6864 \neq 0$. It’s valid.
For $\theta \approx 117.7^\circ$, $\tan 117.7^\circ = \tan(180^\circ – 62.3^\circ) = -\tan 62.3^\circ \approx -1.9136$, $1 – (-1.9136)^2 \approx 1 – 3.662 = -2.662 \neq 0$, and $1 – (-1.9136) \approx 2.9136 \neq 0$. It’s valid.

Thus, the solutions are:
\[
\theta \approx 17.4^\circ \quad \text{and} \quad \theta \approx 117.7^\circ.
\]

Question 5

Topic – ALV: 3.4

A curve has equation $y = \frac{1+e^{2x}}{1+3x}$. The curve has exactly one stationary point $P$.

(a) Find $\frac{dy}{dx}$ and hence show that the x-coordinate of $P$ satisfies the equation $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$.

(b) Show by calculation that the x-coordinate of P lies between 0.35 and 0.45 .

(c) Use an iterative formula based on the equation in part (a) to find the x-coordinate of P correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

▶️Answer/Explanation

Solution: –

(a) Find $\frac{dy}{dx}$ and show that the x-coordinate of $P$ satisfies $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$

The curve is $y = \frac{1 + e^{2x}}{1 + 3x}$. To find the stationary point $P$, we need $\frac{dy}{dx} = 0$.

Use the quotient rule:
\[
\frac{dy}{dx} = \frac{\frac{d}{dx}(1 + e^{2x}) \cdot (1 + 3x) – (1 + e^{2x}) \cdot \frac{d}{dx}(1 + 3x)}{(1 + 3x)^2}.
\]
Derivative of the numerator $1 + e^{2x}$: $\frac{d}{dx}(1 + e^{2x}) = 0 + 2e^{2x} = 2e^{2x}$.
Derivative of the denominator $1 + 3x$: $\frac{d}{dx}(1 + 3x) = 3$.

So,
\[
\frac{dy}{dx} = \frac{(2e^{2x})(1 + 3x) – (1 + e^{2x}) \cdot 3}{(1 + 3x)^2}.
\]
Simplify the numerator:
\[
2e^{2x}(1 + 3x) – 3(1 + e^{2x}) = 2e^{2x} + 6x e^{2x} – 3 – 3e^{2x} = (2e^{2x} – 3e^{2x}) + 6x e^{2x} – 3 = -e^{2x} + 6x e^{2x} – 3.
\]
\[
\frac{dy}{dx} = \frac{-e^{2x} + 6x e^{2x} – 3}{(1 + 3x)^2}.
\]
Set $\frac{dy}{dx} = 0$ for the stationary point:
\[
e^{2x} + 6x e^{2x} – 3 = 0.
\]
Factor out $e^{2x}$ (since $e^{2x} \neq 0$):
\[
e^{2x}(6x – 1) – 3 = 0.
\]
\[
e^{2x}(6x – 1) = 3.
\]
\[
6x – 1 = 3e^{-2x}.
\]
\[
6x = 1 + 3e^{-2x}.
\]
\[
x = \frac{1}{6} + \frac{3}{6}e^{-2x} = \frac{1}{6} + \frac{1}{2}e^{-2x}.
\]
This matches the given equation, so the x-coordinate of $P$ satisfies $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$.

(b) Show that the x-coordinate of $P$ lies between 0.35 and 0.45

We need to check if there’s a solution to $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$ in the interval $0.35 < x < 0.45$.

Define the function:
\[
f(x) = x – \frac{1}{6} – \frac{1}{2}e^{-2x}.
\]
We need to show $f(x) = 0$ for some $x$ in $(0.35, 0.45)$.

Evaluate $f(x)$ at the endpoints and check the behavior:
At $x = 0.35$:
\[
e^{-2 \cdot 0.35} = e^{-0.7} \approx 0.4966.
\]
\[
\frac{1}{2} \cdot 0.4966 \approx 0.2483.
\]
\[
\frac{1}{6} \approx 0.1667.
\]
\[
f(0.35) = 0.35 – 0.1667 – 0.2483 = 0.35 – 0.415 = -0.065.
\]
At $x = 0.45$:
\[
e^{-2 \cdot 0.45} = e^{-0.9} \approx 0.4066.
\]
\[
\frac{1}{2} \cdot 0.4066 \approx 0.2033.
\]
\[
f(0.45) = 0.45 – 0.1667 – 0.2033 = 0.45 – 0.37 = 0.08.
\]
$f(0.35) < 0$ and $f(0.45) > 0$, so by the Intermediate Value Theorem, there’s a root in $(0.35, 0.45)$.

To confirm it’s the only root (since the curve has exactly one stationary point), check the derivative of $f(x)$:
\[
f'(x) = 1 – \frac{1}{2} \cdot (-2) e^{-2x} = 1 + e^{-2x}.
\]
Since $e^{-2x} > 0$, $f'(x) > 1 > 0$, so $f(x)$ is strictly increasing. Thus, there’s exactly one root, and it lies in $(0.35, 0.45)$.

(c) Use an iterative formula to find the x-coordinate of $P$ correct to 3 significant figures, giving each iteration to 5 significant figures

From (a), $x = \frac{1}{6} + \frac{1}{2}e^{-2x}$. Rearrange for iteration:
\[
x_{n+1} = \frac{1}{6} + \frac{1}{2}e^{-2x_n}.
\]
Start with an initial guess within $(0.35, 0.45)$, say $x_0 = 0.4$ (midpoint for convergence).

$x_0 = 0.40000$:
\[
e^{-2 \cdot 0.40000} = e^{-0.8} \approx 0.44933.
\]
\[
\frac{1}{2} \cdot 0.44933 = 0.22467.
\]
\[
x_1 = 0.16667 + 0.22467 = 0.39134.
\]
$x_1 = 0.39134$:
\[
e^{-2 \cdot 0.39134} = e^{-0.78268} \approx 0.45726.
\]
\[
\frac{1}{2} \cdot 0.45726 = 0.22863.
\]
\[
x_2 = 0.16667 + 0.22863 = 0.39530.
\]
$x_2 = 0.39530$:
\[
e^{-2 \cdot 0.39530} = e^{-0.79060} \approx 0.45370.
\]
\[
\frac{1}{2} \cdot 0.45370 = 0.22685.
\]
\[
x_3 = 0.16667 + 0.22685 = 0.39352.
\]
$x_3 = 0.39352$:
\[
e^{-2 \cdot 0.39352} = e^{-0.78704} \approx 0.45515.
\]
\[
\frac{1}{2} \cdot 0.45515 = 0.22758.
\]
\[
x_4 = 0.16667 + 0.22758 = 0.39425.
\]
$x_4 = 0.39425$:
\[
e^{-2 \cdot 0.39425} = e^{-0.78850} \approx 0.45437.
\]
\[
\frac{1}{2} \cdot 0.45437 = 0.22719.
\]
\[
x_5 = 0.16667 + 0.22719 = 0.39386.
\]
$x_5 = 0.39386$:
\[
e^{-2 \cdot 0.39386} = e^{-0.78772} \approx 0.45482.
\]
\[
\frac{1}{2} \cdot 0.45482 = 0.22741.
\]
\[
x_6 = 0.16667 + 0.22741 = 0.39408.
\]
$x_6 = 0.39408$:
\[
e^{-2 \cdot 0.39408} = e^{-0.78816} \approx 0.45452.
\]
\[
\frac{1}{2} \cdot 0.45452 = 0.22726.
\]
\[
x_7 = 0.16667 + 0.22726 = 0.39393.
\]

The values are converging. To 3 significant figures:
$x_6 = 0.39408$, $x_7 = 0.39393$, both round to 0.394.
So, the x-coordinate of $P$ is 0.394 (to 3 significant figures).

Iterations to 5 significant figures:
$x_0 = 0.40000$
$x_1 = 0.39134$
$x_2 = 0.39530$
$x_3 = 0.39352$
$x_4 = 0.39425$
$x_5 = 0.39386$
$x_6 = 0.39408$
$x_7 = 0.39393$

Question 6

Topic – ALV: 1.8

The diagram shows the curve with equation $y = \sqrt{\sin 2x + \sin^2 2x}$ for $0 \le x \le \frac{1}{6}\pi$. The shaded region is bounded by the curve and the straight lines $x = \frac{1}{6}\pi$ and $y = 0$.

(a) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region.

Give your answer correct to 2 significant figures.

(b) The shaded region is rotated completely about the x-axis.
Find the exact volume of the solid produced.

▶️Answer/Explanation

Solution: –

6(a) Use y-values (0), $\sqrt{\sin\frac{\pi}{6}+\sin^{2}\frac{\pi}{6}}$, $\sqrt{\sin\frac{\pi}{3}+\sin^{2}\frac{\pi}{3}}$, or decimal equivalents

Use correct formula, or equivalent, with $h=\frac{\pi}{6}$

Obtain 0.39

6(b) Use $(\pi)\int(\sin 2x + \sin^2 2x) dx$

Express integrand in the form $k_1 \sin 2x + k_2 + k_3 \cos 4x$

Obtain correct $\sin 2x + \frac{1}{2} – \frac{1}{2} \cos 4x$

Integrate to obtain $-\frac{1}{2} \cos 2x + \frac{1}{2}x – \frac{1}{8} \sin 4x$

Apply limits correctly to obtain exact value

Obtain volume = $\frac{1}{4}\pi + \frac{1}{12}\pi^2 – \frac{1}{16}\pi\sqrt{3}$

Question 7

Topic – ALV: 1.8

(a) The polynomial p(x) is defined by

$p(x) = 9x^3 + 6x^2 + 12x + k$,

where k is a constant.

(a) Find the quotient when p(x) is divided by (3x + 2) and show that the remainder is (k – 8).

(b) It is given that $\int_1^6 \frac{p(x)}{3x+2} dx = a + \ln 64$, where $a$ is an integer.

Find the values of $a$ and $k$.

▶️Answer/Explanation

Solution: –

7(a) Carry out division at least as far as $3x^{2}+n_{1}$

Obtain quotient $3x^{2}+4$

Confirm remainder is $k-8$

Alternative Method for Question 7(a)

Synthetic division

Obtain quotient $3x^{2}+4$

Confirm remainder is $k-8$

7(b) Integrate to obtain at least a term in $x^3$ and term of form $n_{1} \ln(3x+2)$

Obtain $x^3 + 4x + \frac{1}{3}(k-8) \ln(3x+2)$

Apply limits correctly to expression with three terms

Obtain $a=235$

Equate logarithm term to $\ln 64$ and apply appropriate logarithm properties

Obtain $k=17$

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