1. **State the problem:** We have a curve $C$ with equation $y = f(x)$ where $$f(x) = 2x^3 - kx^2 + 14x + 24,$$ and $k$ is a constant. We need to: (a)(i) Find the first derivative $f'(x)$. (a)(ii) Find the second derivative $f''(x)$. (b) Given that the curve $y = f'(x)$ intersects $y = f''(x)$ at points $A$ and $B$, and the $x$-coordinate of $A$ is 5, find $k$. (c) Find the coordinates of point $B$. 2. **Find the first derivative $f'(x)$:** Using the power rule, $\frac{d}{dx} x^n = nx^{n-1}$, $$f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(kx^2) + \frac{d}{dx}(14x) + \frac{d}{dx}(24) = 6x^2 - 2kx + 14.$$ 3. **Find the second derivative $f''(x)$:** Differentiate $f'(x)$ again, $$f''(x) = \frac{d}{dx}(6x^2) - \frac{d}{dx}(2kx) + \frac{d}{dx}(14) = 12x - 2k.$$ 4. **Find $k$ using the intersection at $x=5$:** At intersection points, $f'(x) = f''(x)$. Substitute $x=5$: $$f'(5) = 6(5)^2 - 2k(5) + 14 = 150 - 10k + 14 = 164 - 10k,$$ $$f''(5) = 12(5) - 2k = 60 - 2k.$$ Set equal: $$164 - 10k = 60 - 2k,$$ Rearranged: $$164 - 60 = 10k - 2k,$$ $$104 = 8k,$$ $$k = \frac{104}{8} = 13.$$ 5. **Find coordinates of point $B$:** At $B$, $f'(x) = f''(x)$. Using $k=13$, set: $$6x^2 - 2(13)x + 14 = 12x - 2(13),$$ Simplify: $$6x^2 - 26x + 14 = 12x - 26,$$ Bring all terms to one side: $$6x^2 - 26x + 14 - 12x + 26 = 0,$$ $$6x^2 - 38x + 40 = 0,$$ Divide entire equation by 2: $$3x^2 - 19x + 20 = 0.$$ Solve quadratic: $$x = \frac{19 \pm \sqrt{(-19)^2 - 4(3)(20)}}{2 \times 3} = \frac{19 \pm \sqrt{361 - 240}}{6} = \frac{19 \pm \sqrt{121}}{6} = \frac{19 \pm 11}{6}.$$ Two solutions: $$x = \frac{19 + 11}{6} = 5,$$ $$x = \frac{19 - 11}{6} = \frac{8}{6} = \frac{4}{3}.$$ Since $x=5$ corresponds to point $A$, point $B$ has $x=\frac{4}{3}$. Find $y$ coordinate of $B$ by substituting into $f'(x)$: $$f'\left(\frac{4}{3}\right) = 6\left(\frac{4}{3}\right)^2 - 2(13)\left(\frac{4}{3}\right) + 14 = 6 \times \frac{16}{9} - \frac{104}{3} + 14 = \frac{96}{9} - \frac{104}{3} + 14 = \frac{32}{3} - \frac{104}{3} + 14 = -\frac{72}{3} + 14 = -24 + 14 = -10.$$ So, coordinates of $B$ are $$\left(\frac{4}{3}, -10\right).$$