FPM(2023) June Paper 01 Question 3/4 and solution

Question 3

Figure 1 shows part of the curve with equation $$ y=\frac{x}{2}+\frac{4}{x^2} $$ in the interval $0.8\lt x\lt 7$ By drawing a suitable straight line on the grid, obtain an estimate, to one decimal place, of the roots of the equation $$ 3x^3-12x^2+8=0 $$ in the interval $0.8\lt x\lt 7$

(5)

Solution:

The graph of the function $ y = \frac{x}{2} + \frac{4}{x^2} $ for $ 0.8 \lt x \lt 7 $ is given.

$ \begin{aligned} 3x^3 - 12x^2 + 8 &= 0 \\ \text{divided by } x^2:\quad 3x - 12 + \frac{8}{x^2} &= 0 \\ \frac{8}{x^2} &= -3x + 12 \\ \frac{4}{x^2} &= -\frac{3}{2}x + 6 \\ \frac{x}{2} + \frac{4}{x^2} &= -x + 6 \\ y &= -x + 6 \end{aligned} $

Thus the equation $ 3x^3 - 12x^2 + 8 = 0 $ is equivalent to finding the intersection between the curve $ y = \frac{x}{2} + \frac{4}{x^2} $ and the straight line $ y = -x + 6 $.

Step 1: Add the straight line.

We plot the straight line $ y = -x + 6 $ together with the curve.

Step 2: Graphical estimation.

Step 3: Estimate the intersection points.

From the graph, the intersections occur approximately at:

• $ x \approx 0.9 $
• $ x \approx 3.8 $

Final Answer:

$ x \approx 0.9 \quad \text{and} \quad x \approx 3.8 $

Question 4

A particle $P$ is moving along the $x$-axis. At time $t$ seconds, $t \geqslant 0$, the velocity, $v \mathrm{~m}/\mathrm{s}$, of $P$ is given by $v=2t^2-16t+30$

(a) Find the acceleration, in $\mathrm{m}/\mathrm{s}^2$, of $P$ when $t=5$ [2 Marks]

$P$ comes to instantaneous rest at the points $M$ and $N$ at times $t_1$ seconds and $t_2$ seconds where $t_2>t_1$

(b) Find the exact distance $MN$ [8 Marks]

Problem:

A particle $P$ is moving along the $x$-axis. At time $t$ seconds, $t \ge 0$, the velocity, $v \, \mathrm{m/s}$, of $P$ is given by:

$ v = 2t^2 - 16t + 30 $

1. (a) Find the acceleration, in $ \mathrm{m/s}^2 $, of $P$ when $t = 5$.
2. (b) $P$ comes to instantaneous rest at the points $M$ and $N$ at times $t_1$ and $t_2$, where $t_2 > t_1$. Find the exact distance $MN$.

Solution:

(a) Acceleration at $t = 5$:

The acceleration $a$ is:

$ a = \frac{dv}{dt} $

$ a = \frac{d}{dt}(2t^2 - 16t + 30) = 4t - 16 $

At $t = 5$:

$ a = 4(5) - 16 = 4 \, \mathrm{m/s}^2 $

Answer: $4 \, \mathrm{m/s}^2$

(b) Distance $MN$:

The particle is at rest when $v = 0$:

$ 2t^2 - 16t + 30 = 0 $

$ t^2 - 8t + 15 = 0 $

$ (t - 5)(t - 3) = 0 $

$ t_1 = 3, \quad t_2 = 5 $

Displacement:

$ x = \int_3^5 (2t^2 - 16t + 30)\,dt $

$ x = \left[\frac{2t^3}{3} - 8t^2 + 30t\right]_3^5 $

$ x(3) = 36 $

$ x(5) = \frac{100}{3} $

$ MN = \left| x(5) - x(3) \right| = \left| \frac{100}{3} - 36 \right| = \frac{8}{3} $

Final Answer: $ \frac{8}{3} \, \mathrm{m} $