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(a) Factorise the numerator:
$$2x^2 - 5x - 3 = (2x + 1)(x - 3)$$

Factorise the denominator as a difference of two squares:
$$4x^2 - 1 = (2x - 1)(2x + 1)$$

Divide both numerator and denominator by the common factor $(2x + 1)$:
$$\frac{(2x+1)(x-3)}{(2x-1)(2x+1)} = \frac{x-3}{2x-1}$$

(b) Multiply both sides by $(x+1)(x+2)$:
$$2(x+2) + 3(x+1) = 2(x+1)(x+2)$$

Expand the brackets:
$$2x + 4 + 3x + 3 = 2(x^2 + 3x + 2)$$
$$5x + 7 = 2x^2 + 6x + 4$$

Rearrange into a quadratic equation:
$$2x^2 + x - 3 = 0$$

Factorise the quadratic:
$$(2x + 3)(x - 1) = 0$$
So, $x = 1$ or $x = -1.5$.

(c) Write over a common denominator:
$$\frac{x(x+4) + 2(x-3)}{(x-3)(x+4)}$$

Expand and simplify the numerator:
$$\frac{x^2 + 4x + 2x - 6}{(x-3)(x+4)} = \frac{x^2 + 6x - 6}{(x-3)(x+4)}$$

(a) M1 for $(2x+1)(x-3)$,
M1 for $(2x-1)(2x+1)$,
A1 for final answer $\frac{x-3}{2x-1}$. (4 marks total for a)

(b) M1 for clearing fractions: $2(x+2) + 3(x+1) = 2(x+1)(x+2)$,
M1 for expanding to get $5x + 7 = 2x^2 + 6x + 4$,
M1 for solving their 3-term quadratic $2x^2+x-3=0$,
A1 for $x = 1$,
A1 for $x = -1.5$. (5 marks total for b)

(c) M1 for expressing with a common denominator: $x(x+4) + 2(x-3)$,
A1.81 for final simplified answer $\frac{x^2+6x-6}{(x-3)(x+4)}$ or $\frac{x^2+6x-6}{x^2+x-12}$. (2.81 marks total for c)

(a) Differentiate $y = x^3 - 3x^2 - 9x + 5$:
$$\frac{dy}{dx} = 3x^2 - 6x - 9$$

Set $\frac{dy}{dx} = 0$ for turning points:
$$3x^2 - 6x - 9 = 0 \implies x^2 - 2x - 3 = 0 \implies (x-3)(x+1) = 0$$
So $x = 3$ or $x = -1$.

Find the corresponding $y$-coordinates:
When $x = 3$:
$$y = 3^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22$$
When $x = -1$:
$$y = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$$
So the turning points are $(3, -22)$ and $(-1, 10)$.

(b) When $x = 2$, the $y$-coordinate is:
$$y = 2^3 - 3(2)^2 - 9(2) + 5 = 8 - 12 - 18 + 5 = -17$$

The gradient of the tangent is the value of $\frac{dy}{dx}$ at $x = 2$:
$$m = 3(2)^2 - 6(2) - 9 = 12 - 12 - 9 = -9$$

Using $y - y_1 = m(x - x_1)$:
$$y - (-17) = -9(x - 2)$$
$$y + 17 = -9x + 18 \implies y = -9x + 1$$

(c) Solve the equation:
$$x^3 - 3x^2 - 9x + 5 = -9x + 1$$
$$x^3 - 3x^2 + 4 = 0$$

Since $y = -9x + 1$ is tangent at $x = 2$, $x = 2$ must be a repeated root. Thus, $(x-2)^2 = x^2 - 4x + 4$ is a factor.
Divide $x^3 - 3x^2 + 4$ by $x^2 - 4x + 4$:
$$(x^2 - 4x + 4)(x + 1) = 0$$
So the roots are $x = 2$ and $x = -1$.

Find the corresponding points of intersection:
For $x = 2$, $y = -9(2) + 1 = -17$, giving the point $(2, -17)$.
For $x = -1$, $y = -9(-1) + 1 = 10$, giving the point $(-1, 10)$.

(a) M1 for correct derivative: $3x^2 - 6x - 9$,
M1 for setting derivative to 0 and finding $x = 3, x = -1$,
A1 for $(3, -22)$,
A1 for $(-1, 10)$. (5 marks total for a)

(b) M1 for substituting $x = 2$ into $y$ to get $y = -17$,
M1 for substituting $x = 2$ into $\frac{dy}{dx}$ to get $m = -9$,
A1.81 for $y = -9x + 1$. (4.81 marks total for b)

(c) M1 for simplifying to $x^3 - 3x^2 + 4 = 0$,
A1 for finding $x = -1$,
A1 for points $(2, -17)$ and $(-1, 10)$. (2 marks total for c)

(a) The total height of the solid is the height of the cylinder plus the radius of the hemisphere:
$$h + r = 14 \implies h = 14 - r$$

The volume of the cylinder is $V_{\text{cylinder}} = \pi r^2 h = \pi r^2 (14 - r)$.
The volume of the hemisphere is $V_{\text{hemisphere}} = \frac{2}{3}\pi r^3$.

The total volume $V$ is:
$$V = \pi r^2 (14 - r) + \frac{2}{3}\pi r^3$$
$$V = \pi r^2 \left(14 - r + \frac{2}{3}r\right) = \pi r^2 \left(14 - \frac{1}{3}r\right)$$

(b) Set the volume equal to $432\pi$:
$$\pi r^2 \left(14 - \frac{1}{3}r\right) = 432\pi$$

Divide both sides by $\pi$:
$$14r^2 - \frac{1}{3}r^3 = 432$$

Multiply the entire equation by $-3$:
$$-42r^2 + r^3 = -1296$$
$$r^3 - 42r^2 + 1296 = 0$$

(c) When $r = 6$:
The height of the cylinder is $h = 14 - 6 = 8\text{ cm}$.

The total surface area consists of:
- The circular base of the cylinder: $A_{\text{base}} = \pi r^2 = \pi (6)^2 = 36\pi$
- The curved surface area of the cylinder: $A_{\text{cylinder}} = 2\pi r h = 2\pi (6)(8) = 96\pi$
- The curved surface area of the hemisphere: $A_{\text{hemisphere}} = 2\pi r^2 = 2\pi (6)^2 = 72\pi$

Total surface area is:
$$36\pi + 96\pi + 72\pi = 204\pi\text{ cm}^2$$

(a) B1 for $h = 14 - r$,
M1 for $V = \pi r^2 h + \frac{2}{3}\pi r^3$,
A1 for completing the algebraic steps to show $V = \pi r^2 \left(14 - \frac{1}{3}r\right)$. (3 marks total for a)

(b) M1 for setting their expression equal to $432\pi$ and cancelling $\pi$,
M1 for multiplying by $-3$ to eliminate fractions,
A1 for obtaining $r^3 - 42r^2 + 1296 = 0$ clearly. (3 marks total for b)

(c) B1 for $h = 8$,
M1 for area of base = $36\pi$ or curved cylinder area = $96\pi$,
M1 for area of hemisphere = $72\pi$,
M1 for summing all three components: $36\pi + 96\pi + 72\pi$,
A1.81 for $204\pi$. (5.81 marks total for c)

(a) Time taken by Alice is $\frac{\text{Distance}}{\text{Speed}} = \frac{12}{x}$ hours.

(b) Time taken by Bob is $\frac{\text{Distance}}{\text{Speed}} = \frac{12}{x - 1.5}$ hours.

(c) Alice's time is 24 minutes less than Bob's time.
$24\text{ minutes} = \frac{24}{60}\text{ hours} = 0.4\text{ hours}$.
$$\frac{12}{x - 1.5} - \frac{12}{x} = 0.4$$

Multiply by $x(x - 1.5)$:
$$12x - 12(x - 1.5) = 0.4x(x - 1.5)$$
$$12x - 12x + 18 = 0.4x^2 - 0.6x$$
$$18 = 0.4x^2 - 0.6x$$

Divide by $0.4$:
$$45 = x^2 - 1.5x \implies x^2 - 1.5x - 45 = 0$$

(d) Solve the equation:
$$2x^2 - 3x - 90 = 0$$
$$(2x + 15)(x - 6) = 0$$

So $x = 6$ or $x = -7.5$.
Since speed must be positive, Alice's speed is $6\text{ km/h}$.

(e) Bob's speed is $6 - 1.5 = 4.5\text{ km/h}$.
Time taken by Bob = $\frac{12}{4.5} = \frac{8}{3}\text{ hours} = 160\text{ minutes}$.

(a) B1 for $\frac{12}{x}$. (1 mark total for a)

(b) B1 for $\frac{12}{x - 1.5}$. (1 mark total for b)

(c) M1 for $\frac{12}{x-1.5} - \frac{12}{x} = \frac{24}{60}$,
M1 for clearing fractions: $12x - 12(x-1.5) = 0.4x(x-1.5)$,
M1 for expanding brackets: $18 = 0.4x^2 - 0.6x$,
A1.81 for completing algebraic steps to show $x^2 - 1.5x - 45 = 0$. (4.81 marks total for c)

(d) M1 for factorising or using the formula: $(2x+15)(x-6)=0$,
A1 for $x = 6$,
A1 for rejecting $x = -7.5$. (3 marks total for d)

(e) M1 for speed of Bob = $4.5$ or time = $12 / 4.5$,
A1 for 160 (or 2 hours 40 minutes). (2 marks total for e)

(a) Intersections with $x$-axis occur when $y = 0$:
$$(x-2)(x+1)(x-3) = 0 \implies x = 2, -1, 3$$
So points are $(2, 0)$, $(-1, 0)$, and $(3, 0)$.

Intersection with $y$-axis occurs when $x = 0$:
$$y = (0-2)(0+1)(0-3) = (-2)(1)(-3) = 6$$
So point is $(0, 6)$.

(b) Expand the first two brackets:
$$(x-2)(x+1) = x^2 - x - 2$$

Now multiply by the third bracket:
$$(x^2 - x - 2)(x - 3) = x(x^2 - x - 2) - 3(x^2 - x - 2)$$
$$= x^3 - x^2 - 2x - 3x^2 + 3x + 6$$
$$= x^3 - 4x^2 + x + 6$$

(c) Differentiate $y = x^3 - 4x^2 + x + 6$:
$$\frac{dy}{dx} = 3x^2 - 8x + 1$$

Set $\frac{dy}{dx} = 0$:
$$3x^2 - 8x + 1 = 0$$

Using the quadratic formula:
$$x =
\frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(1)}}{2(3)} = \frac{8 \pm \sqrt{64 - 12}}{6} = \frac{8 \pm \sqrt{52}}{6}$$
$$x_1 \approx 2.54, \quad x_2 \approx 0.13$$

Find the corresponding $y$-coordinates:
For $x \approx 0.13$:
$$y \approx (0.13)^3 - 4(0.13)^2 + 0.13 + 6 \approx 6.07$$

For $x \approx 2.54$:
$$y \approx (2.54)^3 - 4(2.54)^2 + 2.54 + 6 \approx -0.88$$

So the turning points are $(0.13, 6.07)$ and $(2.54, -0.88)$.

(a) B1 for $x$-intercepts: $(-1,0), (2,0), (3,0)$,
B1 for $y$-intercept: $(0,6)$,
B1.81 for writing clearly as coordinates. (3.81 marks total for a)

(b) M1 for expanding two brackets to get $x^2 - x - 2$,
M1 for multiplying by the third bracket with at least 4 terms correct,
A1 for completing expansion to get $x^3 - 4x^2 + x + 6$. (3 marks total for b)

(c) M1 for derivative $\frac{dy}{dx} = 3x^2 - 8x + 1$,
M1 for setting derivative to 0 and using quadratic formula,
A1 for $x \approx 2.54$ and $x \approx 0.13$,
A1 for $y \approx -0.88$ and $y \approx 6.07$,
A1 for writing answers as coordinates: $(0.13, 6.07)$ and $(2.54, -0.88)$. (5 marks total for c)

(a) (i) Line $AC$ passes through $(2, 1)$ and $(4, 5)$:
$$\text{Gradient } m = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2$$
Using the point-slope form with $(2, 1)$:
$$y - 1 = 2(x - 2) \implies y = 2x - 3$$

(ii) Line $BC$ passes through $(8, 1)$ and $(4, 5)$:
$$\text{Gradient } m = \frac{5 - 1}{4 - 8} = \frac{4}{-4} = -1$$
Using the point-slope form with $(8, 1)$:
$$y - 1 = -1(x - 8) \implies y = -x + 9$$

(b) The horizontal line forming the base of the triangle is the line passing through $A(2, 1)$ and $B(8, 1)$, which is $y = 1$.

The region $R$ is:
- Above or on the line $y = 1$:
$$y \ge 1$$
- Below or on the line $AC$, $y = 2x - 3$. Testing a point inside the triangle, say $(4, 2)$:
$$2 \le 2(4) - 3 = 5 \quad (\text{True})$$
So:
$$y \le 2x - 3$$
- Below or on the line $BC$, $y = -x + 9$. Testing $(4, 2)$:
$$2 \le -4 + 9 = 5 \quad (\text{True})$$
So:
$$y \le -x + 9$$

Therefore, the three inequalities defining region $R$ are $y \ge 1$, $y \le 2x - 3$, and $y \le -x + 9$.

(a)(i) M1 for finding gradient = 2,
A1 for $y = 2x - 3$. (3 marks total for a(i))

(a)(ii) M1 for finding gradient = -1,
A1 for $y = -x + 9$. (3 marks total for a(ii))

(b) B1 for $y \ge 1$,
M1 for identifying the correct directions of the inequalities using a test point,
A1 for $y \le 2x - 3$,
A1.81 for $y \le -x + 9$. (5.81 marks total for b)

(a) (i) Apply the exponent $\frac{2}{3}$ to each factor:
$$\left(64x^6 y^{-3}\right)^{\frac{2}{3}} = (64)^{\frac{2}{3}} \times (x^6)^{\frac{2}{3}} \times (y^{-3})^{\frac{2}{3}}$$
$$(64)^{\frac{2}{3}} = (4^3)^{\frac{2}{3}} = 16$$
$$(x^6)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4$$
$$(y^{-3})^{\frac{2}{3}} = y^{-3 \times \frac{2}{3}} = y^{-2}$$
So, the simplified form is $16x^4 y^{-2}$ or $\frac{16x^4}{y^2}$.

(ii) Simplify the numerator first:
$$3x^2 y \times 4x^{-3} y^5 = 12 x^{2-3} y^{1+5} = 12 x^{-1} y^6$$

Now divide by the denominator:
$$\frac{12 x^{-1} y^6}{6x y^2} = 2 x^{-1-1} y^{6-2} = 2 x^{-2} y^4 = \frac{2y^4}{x^2}$$

(b) Express all bases as powers of 5:
$$25 = 5^2, \quad 125 = 5^3$$

Substitute these into the equation:
$$5^{2x-1} \times (5^2)^{x+2} = (5^3)^{x-1}$$
$$5^{2x-1} \times 5^{2(x+2)} = 5^{3(x-1)}$$
$$5^{2x-1} \times 5^{2x+4} = 5^{3x-3}$$

Add the exponents on the left-hand side:
$$5^{2x-1 + 2x+4} = 5^{3x-3}$$
$$5^{4x+3} = 5^{3x-3}$$

Equate the exponents:
$$4x + 3 = 3x - 3$$
$$x = -6$$

(a)(i) M1 for evaluating $64^{2/3} = 16$,
M1 for correct application of index laws to variables: $x^4$ or $y^{-2}$,
A1 for $16x^4 y^{-2}$ or $\frac{16x^4}{y^2}$. (3 marks total for a(i))

(a)(ii) M1 for simplifying the numerator to $12x^{-1}y^6$,
M1 for correct division of coefficients and indices,
A1 for $2x^{-2}y^4$ or $\frac{2y^4}{x^2}$. (3 marks total for a(ii))

(b) M1 for writing 25 and 125 as powers of 5: $5^2$ and $5^3$,
M1 for LHS as single power of 5: $5^{2x-1 + 2x+4}$,
M1 for RHS power of 5: $5^{3x-3}$,
M1 for setting up linear equation $4x+3 = 3x-3$,
A1.81 for $x = -6$. (5.81 marks total for b)

(a) Equate the curve and the line to find their points of intersection:
$$\frac{12}{x} = 2x + 2$$

Multiply by $x$ (since $x
eq 0$):
$$12 = 2x^2 + 2x$$
$$2x^2 + 2x - 12 = 0$$

Divide by 2:
$$x^2 + x - 6 = 0$$

Factorise the quadratic:
$$(x + 3)(x - 2) = 0$$
So, $x = 2$ or $x = -3$.

Find the corresponding $y$-coordinates:
When $x = 2$, $y = \frac{12}{2} = 6$. The point is $(2, 6)$.
When $x = -3$, $y = \frac{12}{-3} = -4$. The point is $(-3, -4)$.
So the intersection points are $(2, 6)$ and $(-3, -4)$.

(b) The intersection point with the positive $x$-coordinate is $(2, 6)$.
The gradient of the line $y = 2x + 2$ is $m_1 = 2$.

The gradient of the perpendicular line is:
$$m_2 = -\frac{1}{m_1} = -\frac{1}{2} = -0.5$$

Using the point-slope form with $(2, 6)$:
$$y - 6 = -0.5(x - 2)$$
$$y - 6 = -0.5x + 1 \implies y = -0.5x + 7$$

(c) A translation by the vector $\begin{pmatrix} 1 \\ -3 \end{pmatrix}$ shifts the graph 1 unit to the right and 3 units downwards.
So, replace $x$ with $x - 1$ and subtract 3 from the function:
$$y = \frac{12}{x-1} - 3$$

(a) M1 for setting $\frac{12}{x} = 2x+2$,
M1 for forming quadratic: $2x^2 + 2x - 12 = 0$,
M1 for factoring to get $x = 2, x = -3$,
A1 for $(2, 6)$,
A1.81 for $(-3, -4)$. (5.81 marks total for a)

(b) M1 for perpendicular gradient = $-0.5$,
M1 for using point $(2, 6)$ in equation of line,
A1 for $y = -0.5x + 7$. (3 marks total for b)

(c) M1 for replacing $x$ with $x-1$ or shifting down by subtracting 3,
A2 for correct final equation $y = \frac{12}{x-1} - 3$. (3 marks total for c)

(a) Substitute \(x = 0\) into the equation of the curve: \(y = 0^3 - 4(0)^2 - 3(0) + 12 = 12\). The coordinates are \((0, 12)\). (b) Substituting \(x = 4\) into the equation: \(4^3 - 4(4)^2 - 3(4) + 12 = 64 - 64 - 12 + 12 = 0\). Since \(y = 0\), the curve intersects the x-axis at \(x = 4\). We can factorize the cubic by grouping: \(x^3 - 4x^2 - 3x + 12 = x^2(x - 4) - 3(x - 4) = (x^2 - 3)(x - 4)\). Setting \(x^2 - 3 = 0\) gives \(x = \pm \sqrt{3}\). Thus, the other two intersection points are \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\). (c) Find the derivative: \(dy/dx = 3x^2 - 8x - 3\). Set \(dy/dx = 0\) to find stationary points: \(3x^2 - 8x - 3 = 0 \implies (3x + 1)(x - 3) = 0\). This gives \(x = -1/3\) and \(x = 3\). Substituting \(x = -1/3\) into the original equation: \(y = (-1/3)^3 - 4(-1/3)^2 - 3(-1/3) + 12 = -1/27 - 4/9 + 1 + 12 = 338/27 \approx 12.52\). Substituting \(x = 3\): \(y = 3^3 - 4(3)^2 - 3(3) + 12 = 27 - 36 - 9 + 12 = -6\). Correct to 2 decimal places, the stationary points are \((-0.33, 12.52)\) and \((3, -6)\). (d) The gradient of the tangent at \(x = 0\) is the value of \(dy/dx\) at \(x = 0\): \(dy/dx = 3(0)^2 - 8(0) - 3 = -3\). The tangent passes through the point \((0, 12)\). So the equation is \(y = -3x + 12\).

(a) B1 for \((0, 12)\). (b) M1 for showing \(y = 0\) when \(x = 4\), M1 for factorizing to \((x^2-3)(x-4)\), A1 for \(x = \pm \sqrt{3}\), A1 for writing coordinates \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\). (c) M1 for differentiating correctly to get \(3x^2 - 8x - 3\), M1 for setting their derivative to 0 and solving to get \(x = -1/3\) and \(x = 3\), A1 for \(x\)-coordinates, A1 for \(y = 12.52\), A1 for \(y = -6\). (d) M1 for substituting \(x = 0\) into their derivative, A1 for \(y = -3x + 12\).

(a) Factorize the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). Factorize the denominator: \(4x^2 - 1 = (2x + 1)(2x - 1)\). Simplify by cancelling the common factor \((2x + 1)\): \(\frac{(2x + 1)(x - 3)}{(2x + 1)(2x - 1)} = \frac{x - 3}{2x - 1}\). (b) Multiply the entire equation by the common denominator \((y + 2)(2y - 1)\): \(3(2y - 1) + 5(y + 2) = 2(y + 2)(2y - 1)\). Expand both sides: \(6y - 3 + 5y + 10 = 2(2y^2 + 3y - 2)\) which simplifies to \(11y + 7 = 4y^2 + 6y - 4\). Rearrange into standard quadratic form: \(4y^2 - 5y - 11 = 0\). Use the quadratic formula: \(y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(4)(-11)}}{2(4)} = \frac{5 \pm \sqrt{25 + 176}}{8} = \frac{5 \pm \sqrt{201}}{8}\). This gives \(y = \frac{5 + 14.177}{8} \approx 2.40\) and \(y = \frac{5 - 14.177}{8} \approx -1.15\).

(a) M2 for factorizing numerator to \((2x + 1)(x - 3)\) (M1 for \((2x+a)(x+b)\) where \(ab=-3\) or \(2b+a=-5\)), M1 for factorizing denominator to \((2x + 1)(2x - 1)\), A1 for final fraction \(\frac{x - 3}{2x - 1}\). (b) M1 for clear algebraic step of multiplying by common denominator, M2 for correct expansion to \(11y + 7\) and \(4y^2 + 6y - 4\) (M1 for one side correct), A1 for writing standard quadratic form \(4y^2 - 5y - 11 = 0\), M1 for correct substitution into quadratic formula, A1 for \(\sqrt{201}\) or equivalent, A2 for both answers correct to 2 decimal places (A1 for one correct).

(a) The volume of a hemisphere is \(V_{hemi} = \frac{2}{3}\pi r^3\). The volume of a cone is \(V_{cone} = \frac{1}{3}\pi r^2 h\). Here \(h = 3r\), so \(V_{cone} = \frac{1}{3}\pi r^2 (3r) = \pi r^3\). The total volume \(V\) is the sum of these volumes: \(V = \frac{2}{3}\pi r^3 + \pi r^3 = \frac{5}{3}\pi r^3\). (b)(i) We are given \(V = 360\), so \(\frac{5}{3}\pi r^3 = 360\). Rearranging for \(r^3\): \(r^3 = \frac{360 \times 3}{5\pi} = \frac{216}{\pi} \approx 68.755\). Taking the cube root gives \(r = \sqrt[3]{68.755} \approx 4.0967\) cm, which is \(4.10\) cm correct to 3 significant figures. (b)(ii) The total surface area of the toy is the sum of the curved surface area of the hemisphere and the curved surface area of the cone. Curved surface area of the hemisphere: \(A_{hemi} = 2\pi r^2\). Curved surface area of the cone: \(A_{cone} = \pi r l\), where \(l\) is the slant height of the cone. \(l = \sqrt{r^2 + h^2} = \sqrt{r^2 + (3r)^2} = \sqrt{10r^2} = r\sqrt{10}\). Total surface area \(A = 2\pi r^2 + \pi r (r\sqrt{10}) = \pi r^2 (2 + \sqrt{10})\). Using \(r \approx 4.0967\): \(A = \pi (4.0967)^2 (2 + \sqrt{10}) \approx \pi (16.783) (5.1623) \approx 272.18\) \(\text{cm}^2\), which is \(272\) \(\text{cm}^2\) correct to 3 significant figures.

(a) M1 for \(\frac{2}{3}\pi r^3\), M1 for \(\frac{1}{3}\pi r^2(3r)\) or \(\pi r^3\), A1 for adding and obtaining \(\frac{5}{3}\pi r^3\) with clear working shown. (b)(i) M1 for setting \(\frac{5}{3}\pi r^3 = 360\), M1 for \(r^3 = \frac{216}{\pi}\), A1 for \(r = 4.10\) or \(4.097\). (b)(ii) M1 for slant height of cone formula \(l = \sqrt{r^2 + (3r)^2}\), M1 for \(l = r\sqrt{10}\) or numerical value \(12.95\), M1 for curved surface area of cone \(\pi r l\), M1 for curved surface area of hemisphere \(2\pi r^2\), A1 for adding both surface areas, A1 for final answer \(272\) (accept \(272 - 273\)).