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Question 1

Topic 1.6 – Series

An arithmetic progression has fourth term 15 and eighth term 25.

Find the 30th term of the progression.

▶️Answer/Explanation

Solution :-

Let the first term be $a$ and the common difference be $d$.

Fourth term: $a + 3d = 15$
Eighth term: $a + 7d = 25$

Subtract the first equation from the second:
$(a + 7d) – (a + 3d) = 25 – 15$
$4d = 10$
$d = 2.5$

Substitute $d = 2.5$ into $a + 3d = 15$:
$a + 3(2.5) = 15$
$a + 7.5 = 15$
$a = 7.5$

30th term: $a + 29d = 7.5 + 29(2.5) = 7.5 + 72.5 = 80$

30th term: 80

Question 2

Topic 1.5 – Trigonometry

Find the exact solution of the equation

$ \cos\frac{1}{6}\pi + \tan 2x + \frac{\sqrt{3}}{2} = 0 $

for $-\frac{1}{4}\pi < x < \frac{1}{4}\pi$.

▶️Answer/Explanation

Solution :-

First, compute $\cos(\pi/6)$:
\[
\cos(\pi/6) = \sqrt{3}/2
\]
So the equation is:
\[
\frac{\sqrt{3}}{2} + \tan(2x) + \frac{\sqrt{3}}{2} = 0
\]
\[
\tan(2x) + \sqrt{3} = 0
\]
\[
\tan(2x) = -\sqrt{3}
\]
\[
2x = -\frac{\pi}{3} + k\pi, \quad k \in \mathbb{Z}
\]
\[
x = -\frac{\pi}{6} + \frac{k\pi}{2}
\]
For $-\frac{\pi}{4} < x < \frac{\pi}{4}$:
– Try $k = 0$: $x = -\pi/6 \approx -0.523$, which is in the interval.
– Try $k = 1$: $x = -\pi/6 + \pi/2 = \pi/3 \approx 1.047$, which is outside.
– Try $k = -1$: $x = -\pi/6 – \pi/2 = -2\pi/3 \approx -2.094$, which is outside.

So, the exact solution is $x = -\pi/6$.

Question 3

(a) Topic 1.6 – Series

(b) Topic 1.6 – Series

(a) Find the coefficients of $x^3$ and $x^4$ in the expansion of $(3 – ax)^5$, where $a$ is a constant. Give your answers in terms of $a$.

(b) Given that the coefficient of $x^4$ in the expansion of $(ax+7)(3-ax)^{5}$ is 240, find the positive value of $a$.

▶️Answer/Explanation

Solution :-

(a) Coefficients of $x^3$ and $x^4$ in $(3 – ax)^5$

Use the binomial theorem: $(3 – ax)^5 = \sum_{r=0}^5 \binom{5}{r} 3^{5-r} (-ax)^r = \sum_{r=0}^5 \binom{5}{r} 3^{5-r} (-a)^r x^r$.

Coefficient of $x^3$ (when $r = 3$):
\[
\binom{5}{3} 3^{5-3} (-a)^3 = 10 \cdot 9 \cdot (-a^3) = -90a^3
\]

Coefficient of $x^4$ (when $r = 4$):
\[
\binom{5}{4} 3^{5-4} (-a)^4 = 5 \cdot 3 \cdot a^4 = 15a^4
\]

Coefficients: $x^3$ is $-90a^3$, $x^4$ is $15a^4$.

(b) Coefficient of $x^4$ in $(ax + 7)(3 – ax)^5$ is 240, find positive $a$

Expand $(ax + 7)(3 – ax)^5$. The coefficient of $x^4$ comes from:
$ax \cdot (\text{coefficient of } x^3 \text{ in } (3 – ax)^5)$
$7 \cdot (\text{coefficient of } x^4 \text{ in } (3 – ax)^5)$

From (a):
Coefficient of $x^3$ in $(3 – ax)^5$ is $-90a^3$.
Coefficient of $x^4$ in $(3 – ax)^5$ is $15a^4$.

Coefficient of $x^4$:
\[
ax \cdot (-90a^3) + 7 \cdot 15a^4 = -90a^4 x + 105a^4 = a^4(-90x + 105)
\]
Set equal to 240:
\[
a^4(-90x + 105) = 240
\]
Focus on the constant part (coefficient of $x^4$):
\[
-90a^4 + 105a^4 = 15a^4 = 240
\]
\[
a^4 = \frac{240}{15} = 16
\]
\[
a = \sqrt[4]{16} = 2 \quad (\text{positive value})
\]

Positive $a$: 2

Question 4

Topic 1.5 – Trigonometry

Solve the equation $4\sin^{4}\theta + 12\sin^{2}\theta – 7 = 0$ for $0^{\circ} \leq \theta \leq 360^{\circ}$.

▶️Answer/Explanation

Solution :-

Let $u = \sin^2\theta$, so the equation becomes:
\[
4u^2 + 12u – 7 = 0
\]
Solve using the quadratic formula:
\[
u = \frac{-12 \pm \sqrt{12^2 – 4 \cdot 4 \cdot (-7)}}{8} = \frac{-12 \pm \sqrt{144 + 112}}{8} = \frac{-12 \pm \sqrt{256}}{8} = \frac{-12 \pm 16}{8}
\]
\[
u = \frac{4}{8} = 0.5 \quad \text{or} \quad u = \frac{-28}{8} = -3.5
\]
Since $u = \sin^2\theta \geq 0$, discard $u = -3.5$. So:
\[
\sin^2\theta = 0.5
\]
\[
\sin\theta = \pm \sqrt{0.5} = \pm \frac{\sqrt{2}}{2}
\]
For $0^\circ \leq \theta \leq 360^\circ$:
$\sin\theta = \frac{\sqrt{2}}{2}$: $\theta = 45^\circ, 135^\circ$
$\sin\theta = -\frac{\sqrt{2}}{2}$: $\theta = 225^\circ, 315^\circ$

Solutions: $45^\circ, 135^\circ, 225^\circ, 315^\circ$

Question 5

(a) Topic 1.2 – Functions

(b) Topic 1.2 – Functions

In the diagram, the graph with equation $y=f(x)$ is shown with solid lines and the graph with equation $y=g(x)$ is shown with broken lines.

(a) Describe fully a sequence of three transformations which transforms the graph of $y=f(x)$ to the graph of $y=g(x).$

(b) Find an expression for $g(x)$ in the form $af(bx+c)$ where a, b and c are integers.

▶️Answer/Explanation

Solution :-

$5(a) \text{ Reflection [in] y-axis}$

$\text{ Translation or shift } \begin{pmatrix} -1 \\ 0 \end{pmatrix}$

$\text{ Stretch, factor 2, parallel to y-axis}$

$\text{ Correct order and three correctly named transformations only}$

$\text{ Alternative Solution for first 3 marks}$

$\text{ Translation or shift } \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

$\text{ Reflection [in] y-axis}$

$\text{ Alternative solutions}$

$\text{ There are alternative solutions which can be marked in the same way}$
$\text{ e.g. the given stretch, translation } \begin{pmatrix} -4 \\ 0 \end{pmatrix} \text{, reflect in } x=-2.5$

(b) $g(x)=2f(-x-1)$ or $a=2, b=-1, c=-1$

Question 6

Topic 1.6 – Series

The first term of a convergent geometric progression is 10. The sum of the first 4 terms of the progression is $p$ and the sum of the first 8 terms of the progression is $q$. It is given that $\frac{q}{p} = \frac{17}{16}$.

Find the two possible values of the sum to infinity.

▶️Answer/Explanation

Solution :-
\[ S_n = a \frac{1 – r^n}{1 – r} \]
So:
– $ p = S_4 = 10 \frac{1 – r^4}{1 – r} $
– $ q = S_8 = 10 \frac{1 – r^8}{1 – r} $

Given $ \frac{q}{p} = \frac{17}{16} $, we set up the equation:
\[ \frac{q}{p} = \frac{10 \frac{1 – r^8}{1 – r}}{10 \frac{1 – r^4}{1 – r}} = \frac{1 – r^8}{1 – r^4} = \frac{17}{16} \]

Since $ 1 – r^8 = (1 – r^4)(1 + r^4) $, substitute this in:
\[ \frac{1 – r^8}{1 – r^4} = \frac{(1 – r^4)(1 + r^4)}{1 – r^4} = 1 + r^4 \]
Thus:
\[ 1 + r^4 = \frac{17}{16} \]

Solve for $ r^4 $:
\[ r^4 = \frac{17}{16} – 1 = \frac{17}{16} – \frac{16}{16} = \frac{1}{16} \]
\[ r^4 = \frac{1}{16} \]
\[ r = \pm \sqrt[4]{\frac{1}{16}} = \pm \left( \frac{1}{2} \right) = \pm \frac{1}{2} \]

Since the progression converges, $ |r| < 1 $, and both $ r = \frac{1}{2} $ and $ r = -\frac{1}{2} $ satisfy this.

Now, calculate the sum to infinity, $ S_\infty = \frac{a}{1 – r} $, for each $ r $:
1. If $ r = \frac{1}{2} $:
\[ S_\infty = \frac{10}{1 – \frac{1}{2}} = \frac{10}{\frac{1}{2}} = 10 \cdot 2 = 20 \]

2. If $ r = -\frac{1}{2} $:
\[ S_\infty = \frac{10}{1 – (-\frac{1}{2})} = \frac{10}{1 + \frac{1}{2}} = \frac{10}{\frac{3}{2}} = 10 \cdot \frac{2}{3} = \frac{20}{3} \]

Thus, the two possible values of the sum to infinity are $ 20 $ and $ \frac{20}{3} $.

Final Answer: The two possible values are 20 and $ \frac{20}{3} $ (approximately 6.67).

Question 7

(a) Topic 1.4 – Circular measure

(b) Topic 1.4 – Circular measure

The diagram shows a metal plate ABCDEF consisting of five parts. The parts BCD and DEF are semicircles. The part BAFO is a sector of a circle with centre O and radius 20 cm, and D lies on this circle. The parts OBD and ODF are triangles. Angles BOD and DOF are both $\theta$ radians.

(a) Given that $\theta = 1.2$, find the area of the metal plate. Give your answer correct to 3 significant figures.

(b) Given instead that the area of each semicircle is $50\pi$ cm$^{2}$, find the exact perimeter of the metal plate.

▶️Answer/Explanation

Solution :-

(a) $Area~of~sector~BOF = \frac{1}{2}\times20^{2}\times(2\pi-2.4) [=776.63…]$

$Length~BD=DF=2\times20~sin~0.6~or~\sqrt{20^{2}+20^{2}-2\times20\times20~cos~1.2} [=22.58…]$

$Area~of~two~semicircles = \pi\times(20~sin~0.6)^{2}[=400.64…]$

$Area~of~triangles = 2\times\frac{1}{2}\times20\times20~sin~1.2 [=372.81…]$

$Total~area = 1550 [cm²]$

7(b) $\frac{1}{2}\pi r^{2}=50\pi \Rightarrow r=10$

$\Rightarrow\theta=\frac{\pi}{3}$

$Arc~length~of~sector~BOF = 20\times2\pi-\text{their}~\frac{2\pi}{3}$

$Total~perimeter = 20\times(2\pi-\text{their}~\frac{2\pi}{3})+2\times\text{their}~10$

$\frac{140\pi}{3}~or~46\frac{2}{3}\pi$

Question 8

(a) Topic 1.1 – Quadratics

(b) Topic 1.2 – Functions

(d) Topic 1.2 – Functions

(a) Express $3x^{2}-12x+14$ in the form $3(x+a)^{2}+b$ where $a$ and $b$ are constants to be found.

The function $f(x)=3x^{2}-12x+14$ is defined for $x\ge k$ where $k$ is a constant.

(b) Find the least value of $k$ for which the function $f^{-1}$ exists.

For the rest of this question, you should assume that $k$ has the value found in part (b).

(d) Hence or otherwise solve the equation $ff(x)=29$.

▶️Answer/Explanation

(a) Express $3x^2 – 12x + 14$ as $3(x + a)^2 + b$

Rewrite by completing the square:

\[ 3x^2 – 12x + 14 = 3(x^2 – 4x) + 14 \]
\[ x^2 – 4x = (x – 2)^2 – 4 \]
\[ 3[(x – 2)^2 – 4] + 14 = 3(x – 2)^2 – 12 + 14 = 3(x – 2)^2 + 2 \]

So, $a = -2$, $b = 2$.

Answer: $3(x – 2)^2 + 2$

(b) Least $k$ for $f^{-1}$ to exist

$f(x) = 3(x – 2)^2 + 2$ has a vertex at $x = 2$. For $f^{-1}$ to exist, $f$ must be one-to-one. Since it’s increasing for $x \geq 2$, the smallest $k$ is 2.

Answer: $k = 2$

With domain $x \geq 2$, set $y = 3(x – 2)^2 + 2$:

\[ y – 2 = 3(x – 2)^2 \]
\[ (x – 2)^2 = \frac{y – 2}{3} \]
\[ x – 2 = \sqrt{\frac{y – 2}{3}} \quad (\text{positive since } x \geq 2\text{)} \]
\[ x = 2 + \sqrt{\frac{y – 2}{3}} \]

So, $f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}$, for $x \geq 2$.

Answer: $f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}}$

(d) Solve $f(f(x)) = 29$

\[ f(x) = 3(x – 2)^2 + 2 \]
\[ f(f(x)) = 3[f(x) – 2]^2 + 2 = 3[3(x – 2)^2]^2 + 2 = 27(x – 2)^4 + 2 \]

\[ 27(x – 2)^4 + 2 = 29 \]
\[ 27(x – 2)^4 = 27 \]
\[ (x – 2)^4 = 1 \]
\[ x – 2 = \pm 1 \]
\[ x = 3 \text{ or } 1 \]

Since $x \geq 2$, $x = 3$.

Answer: $x = 3$

Final Answers:
(a) $3(x – 2)^2 + 2$
(b) 2
(c) $2 + \sqrt{\frac{x – 2}{3}}$
(d) 3

Question 9

(a) Topic 1.1 – Quadratics

b) Topic 1.8 – Integration

The diagram shows the curves with equations $y=x^{3}-3x+3 and y=2x^{3}-4x^{2}+3$

(a) Find the x-coordinates of the points of intersection of the curves.

(b) Find the area of the shaded region.

▶️Answer/Explanation

Solution :-

(a) $y=x^{3}-3x+3 \text{ and } y=2x^{3}-4x^{2}+3x^{3}-4x^{2}+3x[=0]$

$[x](x-1)(x-3)[=0]$

x=0, 1 and 3 {x=0 may be seen in the working}

(b) Attempt at integration of both functions. Can be before or after subtraction of the functions or integrals

$=\pm\left(\frac{x^{4}}{4}+\frac{4x^{3}}{3}-\frac{3x^{2}}{2}\right) \text{ or } \left[\pm\left(\frac{x^{4}}{4}-\frac{3}{2}x^{2}+3x\right)-\left(\frac{2}{4}x^{4}-\frac{4}{3}x^{3}+3x\right)\right]$

$=\left[\left(-\frac{81}{4}+\frac{108}{3}-\frac{27}{2}\right)-\left(-\frac{1}{4}+\frac{4}{3}-\frac{3}{2}\right)\right],$

$\left(\frac{81}{4}-\frac{27}{2}+9\right)-\left(\frac{1}{4}-\frac{3}{2}+3\right)-\left\{\left(\frac{81}{2}-\frac{108}{3}+9\right)-\left(\frac{1}{2}-\frac{4}{3}+3\right)\right\}$

$=\frac{8}{3}$

Question 10

(a) Topic 1.3 – Coordinate geometry

(b) Topic 1.3 – Coordinate geometry

Points A and B have coordinates (4, 3) and (8, -5) respectively. A circle with radius 10 passes through the points A and B.

(a) Show that the centre of the circle lies on the line $y=\frac{1}{2}x-4$.

(b) Find the two possible equations of the circle.

▶️Answer/Explanation:

(a) Show the centre lies on $y = \frac{1}{2}x – 4$

Points A(4, 3) and B(8, -5) lie on a circle with radius 10. The centre C(h, k) must be equidistant from A and B, and the distance from C to each point is 10 (radius).

Distance CA = 10:
\[ \sqrt{(h – 4)^2 + (k – 3)^2} = 10 \]
\[ (h – 4)^2 + (k – 3)^2 = 100 \quad (1) \]

Distance CB = 10:
\[ \sqrt{(h – 8)^2 + (k + 5)^2} = 10 \]
\[ (h – 8)^2 + (k + 5)^2 = 100 \quad (2) \]

Expand both:
1. $h^2 – 8h + 16 + k^2 – 6k + 9 = 100$
\[ h^2 + k^2 – 8h – 6k + 25 = 100 \]
\[ h^2 + k^2 – 8h – 6k – 75 = 0 \quad (3) \]

2. $h^2 – 16h + 64 + k^2 + 10k + 25 = 100$
\[ h^2 + k^2 – 16h + 10k + 89 = 100 \]
\[ h^2 + k^2 – 16h + 10k – 11 = 0 \quad (4) \]

Subtract (3) from (4):
\[ (h^2 + k^2 – 16h + 10k – 11) – (h^2 + k^2 – 8h – 6k – 75) = 0 \]
\[ -16h + 10k – 11 + 8h + 6k + 75 = 0 \]
\[ -8h + 16k + 64 = 0 \]
\[ -8h + 16k = -64 \]
\[ h – 2k = 8 \]
\[ k = \frac{h – 8}{2} = \frac{1}{2}h – 4 \]

The centre (h, k) satisfies $k = \frac{1}{2}h – 4$, or $y = \frac{1}{2}x – 4$. This is the perpendicular bisector of AB, which makes sense geometrically.

Answer for (a): The centre lies on $y = \frac{1}{2}x – 4$.

(b) Find the two possible equations of the circle

Substitute $k = \frac{1}{2}h – 4$ into equation (3):
\[ h^2 + \left(\frac{1}{2}h – 4\right)^2 – 8h – 6\left(\frac{1}{2}h – 4\right) – 75 = 0 \]

Expand:
\[ \left(\frac{1}{2}h – 4\right)^2 = \frac{1}{4}h^2 – 4h + 16 \]
\[ -6\left(\frac{1}{2}h – 4\right) = -3h + 24 \]

\[ h^2 + \frac{1}{4}h^2 – 4h + 16 – 8h – 3h + 24 – 75 = 0 \]
\[ h^2 + \frac{1}{4}h^2 – 15h + 16 + 24 – 75 = 0 \]
\[ \frac{5}{4}h^2 – 15h – 35 = 0 \]

Multiply by 4:
\[ 5h^2 – 60h – 140 = 0 \]
\[ h^2 – 12h – 28 = 0 \]

Solve:
\[ h = \frac{12 \pm \sqrt{144 + 112}}{2} = \frac{12 \pm \sqrt{256}}{2} = \frac{12 \pm 16}{2} \]
\[ h = 14 \quad \text{or} \quad h = -2 \]

If $h = 14$:
\[ k = \frac{1}{2}(14) – 4 = 7 – 4 = 3 \]
Centre: (14, 3). Equation:
\[ (x – 14)^2 + (y – 3)^2 = 100 \]

If $h = -2$:
\[ k = \frac{1}{2}(-2) – 4 = -1 – 4 = -5 \]
Centre: (-2, -5). Equation:
\[ (x + 2)^2 + (y + 5)^2 = 100 \]

Check:
Centre (14, 3) to A(4, 3): $\sqrt{(14-4)^2 + (3-3)^2} = 10$
Centre (14, 3) to B(8, -5): $\sqrt{(14-8)^2 + (-5-3)^2} = \sqrt{36 + 64} = 10$
Centre (-2, -5) to A(4, 3): $\sqrt{(-2-4)^2 + (-5-3)^2} = \sqrt{36 + 64} = 10$
Centre (-2, -5) to B(8, -5): $\sqrt{(8+2)^2 + (-5+5)^2} = 10$

Both work.

Answer for (b):

\[ (x – 14)^2 + (y – 3)^2 = 100 \]
\[ (x + 2)^2 + (y + 5)^2 = 100 \]

Final Answers:
(a) Shown that centre lies on $y = \frac{1}{2}x – 4$
(b) $(x – 14)^2 + (y – 3)^2 = 100$ and $(x + 2)^2 + (y + 5)^2 = 100$

Question 11

(a) Topic 1.7 – Differentiation

(b) Topic 1.7 – Differentiation

The equation of a curve is $y = kx^{\frac{1}{2}} – 4x^2 + 2$, where $k$ is a constant.

(a) Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in terms of $k$.

(b) It is given that $k = 2$.

Find the coordinates of the stationary point and determine its nature.

(c) Points A and B on the curve have x-coordinates 0.25 and 1 respectively. For a different value of k, the tangents to the curve at the points A and B meet at a point with x-coordinate 0.6.

Find this value of k.

▶️Answer/Explanation

(a) Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in terms of $k$

The curve is $y = kx^{\frac{1}{2}} – 4x^2 + 2$.

First derivative:
\[ \frac{dy}{dx} = k \cdot \frac{1}{2}x^{-\frac{1}{2}} – 4 \cdot 2x = \frac{k}{2}x^{-\frac{1}{2}} – 8x \]

Second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{k}{2}x^{-\frac{1}{2}} – 8x \right) \]
\[ = \frac{k}{2} \cdot \left(-\frac{1}{2}\right)x^{-\frac{3}{2}} – 8 = -\frac{k}{4}x^{-\frac{3}{2}} – 8 \]

Answer for (a):
\[ \frac{dy}{dx} = \frac{k}{2}x^{-\frac{1}{2}} – 8x \]
\[ \frac{d^2y}{dx^2} = -\frac{k}{4}x^{-\frac{3}{2}} – 8 \]

(b) $k = 2$, find coordinates of stationary point and its nature

With $k = 2$, the curve is $y = 2x^{\frac{1}{2}} – 4x^2 + 2$.

Stationary point: $\frac{dy}{dx} = 0$
\[ \frac{dy}{dx} = \frac{2}{2}x^{-\frac{1}{2}} – 8x = x^{-\frac{1}{2}} – 8x \]
\[ x^{-\frac{1}{2}} – 8x = 0 \]
\[ x^{-\frac{1}{2}} = 8x \]
\[ \frac{1}{x^{\frac{1}{2}}} = 8x \]
\[ 1 = 8x \cdot x^{\frac{1}{2}} = 8x^{\frac{3}{2}} \]
\[ x^{\frac{3}{2}} = \frac{1}{8} \]
\[ x = \left(\frac{1}{8}\right)^{\frac{2}{3}} = (8^{-1})^{\frac{2}{3}} = 8^{-\frac{2}{3}} = (2^3)^{-\frac{2}{3}} = 2^{-2} = \frac{1}{4} \]

Find $y$:
\[ y = 2\left(\frac{1}{4}\right)^{\frac{1}{2}} – 4\left(\frac{1}{4}\right)^2 + 2 = 2 \cdot \frac{1}{2} – 4 \cdot \frac{1}{16} + 2 = 1 – \frac{1}{4} + 2 = 1 + 1.75 = 2.75 \]

Coordinates: $\left(\frac{1}{4}, 2.75\right)$.

Nature: Use $\frac{d^2y}{dx^2}$
\[ \frac{d^2y}{dx^2} = -\frac{2}{4}x^{-\frac{3}{2}} – 8 = -\frac{1}{2}x^{-\frac{3}{2}} – 8 \]
At $x = \frac{1}{4}$:
\[ x^{-\frac{3}{2}} = \left(\frac{1}{4}\right)^{-\frac{3}{2}} = (4^{-1})^{-\frac{3}{2}} = 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8 \]
\[ \frac{d^2y}{dx^2} = -\frac{1}{2} \cdot 8 – 8 = -4 – 8 = -12 \]

Since $\frac{d^2y}{dx^2} < 0$, it’s a maximum.

Answer for (b):$\left(\frac{1}{4}, 2.75\right)$, maximum

Point A ($x = 0.25$):
\[ y = k(0.25)^{\frac{1}{2}} – 4(0.25)^2 + 2 = k \cdot \frac{1}{2} – 4 \cdot \frac{1}{16} + 2 = \frac{k}{2} – \frac{1}{4} + 2 = \frac{k}{2} + 1.75 \]
Slope:
\[ \frac{dy}{dx} = \frac{k}{2}(0.25)^{-\frac{1}{2}} – 8 \cdot 0.25 = \frac{k}{2} \cdot 2 – 2 = k – 2 \]
Tangent: $y – \left(\frac{k}{2} + 1.75\right) = (k – 2)(x – 0.25)$

Point B ($x = 1$):
\[ y = k(1)^{\frac{1}{2}} – 4(1)^2 + 2 = k – 4 + 2 = k – 2 \]
Slope:
\[ \frac{dy}{dx} = \frac{k}{2}(1)^{-\frac{1}{2}} – 8 \cdot 1 = \frac{k}{2} – 8 \]
Tangent: $y – (k – 2) = \left(\frac{k}{2} – 8\right)(x – 1)$

Intersection at $x = 0.6$:
Tangent at A:
\[ y = (k – 2)(0.6 – 0.25) + \frac{k}{2} + 1.75 = (k – 2) \cdot 0.35 + \frac{k}{2} + 1.75 \]
\[ = 0.35k – 0.7 + \frac{k}{2} + 1.75 = 0.85k + 1.05 \]

Tangent at B:
\[ y = \left(\frac{k}{2} – 8\right)(0.6 – 1) + (k – 2) = \left(\frac{k}{2} – 8\right)(-0.4) + k – 2 \]
\[ = -0.2k + 3.2 + k – 2 = 0.8k + 1.2 \]

Set equal:
\[ 0.85k + 1.05 = 0.8k + 1.2 \]
\[ 0.05k = 0.15 \]
\[ k = 3 \]

Answer for (c):$k = 3$

Final Answers:
(a) $\frac{dy}{dx} = \frac{k}{2}x^{-\frac{1}{2}} – 8x$, $\frac{d^2y}{dx^2} = -\frac{k}{4}x^{-\frac{3}{2}} – 8$
(b) $\left(\frac{1}{4}, 2.75\right)$, maximum
(c) 3

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