FPM(2023) June Paper 01 Question 7/8 and solution

Question 7

Problem:

The equation of a curve is

$ y = \sqrt{\frac{\mathrm{e}^{4x}}{2x - 3}} $

When $x$ is increased to $x + \delta x$, $y$ increases to $y + \delta y$, where $\delta x$ and $\delta y$ are small.

1. (a) Show that $ \delta y \approx \frac{\mathrm{e}^{2x}(4x - 7)}{(2x - 3)^{\frac{3}{2}}}\delta x $
2. (b) Given that $x = 2.5$, find an estimate (2 s.f.) of $\delta y$ when $x$ increases by $0.2\%$.

Solution:

(a)

$ y = \left(\frac{\mathrm{e}^{4x}}{2x - 3}\right)^{\frac{1}{2}} $

$ \frac{dy}{dx} = \frac{1}{2} \left(\frac{\mathrm{e}^{4x}}{2x - 3}\right)^{-\frac{1}{2}} \cdot \frac{d}{dx}\left(\frac{\mathrm{e}^{4x}}{2x - 3}\right) $

$ \frac{d}{dx}\left(\frac{\mathrm{e}^{4x}}{2x - 3}\right) = \frac{(2x - 3)(4\mathrm{e}^{4x}) - \mathrm{e}^{4x}(2)}{(2x - 3)^2} = \frac{\mathrm{e}^{4x}(8x - 14)}{(2x - 3)^2} $

$ \left(\frac{\mathrm{e}^{4x}}{2x - 3}\right)^{-\frac{1}{2}} = \frac{(2x - 3)^{\frac{1}{2}}}{\mathrm{e}^{2x}} $

$ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{(2x - 3)^{\frac{1}{2}}}{\mathrm{e}^{2x}} \cdot \frac{\mathrm{e}^{4x}(8x - 14)}{(2x - 3)^2} $

$ \frac{dy}{dx} = \frac{\mathrm{e}^{2x}(4x - 7)}{(2x - 3)^{\frac{3}{2}}} $

Hence,

$ \delta y \approx \frac{\mathrm{e}^{2x}(4x - 7)}{(2x - 3)^{\frac{3}{2}}}\delta x $

(b)

$ \delta x = 0.2\% \text{ of } 2.5 = 0.002 \times 2.5 = 0.005 $

$ \delta y \approx \frac{\mathrm{e}^{5}(3)}{2^{\frac{3}{2}}} \cdot 0.005 $

$ \mathrm{e}^5 \approx 148.413 $

$ \delta y \approx \frac{148.413 \times 3}{2.828} \times 0.005 \approx 0.79 $

Final Answer: $0.79$

Question 8

Problem:

The derivative of a function is given by:

$ f'(x) = 18x^2 - 2x + 13. $

Given that $(2x - 1)$ is a factor of $f(x)$, show that the curve with equation $y = f(x)$ has only one intersection with the $x$-axis.

Solution:

Step 1: Expression for $f(x)$ by integration:

$ f(x) = \int (18x^2 - 2x + 13)\,dx $

$ f(x) = 6x^3 - x^2 + 13x + C $

where $C$ is a constant.

Step 2: Use $f\left(\frac{1}{2}\right) = 0$ to find $C$:

$ f\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 + 13\left(\frac{1}{2}\right) + C $

$ = \frac{3}{4} - \frac{1}{4} + \frac{13}{2} + C $

$ = \frac{2}{4} + \frac{26}{4} + C = 7 + C $

$ C = -7 $

$ f(x) = 6x^3 - x^2 + 13x - 7 $

Step 3: Perform synthetic division to confirm $(2x - 1)$ is a factor:

Rewrite $(2x - 1)$ as $x - \frac{1}{2}$. The coefficients are $6, -1, 13, -7$:

$ \begin{array}{r|rrrr} \frac{1}{2} & 6 & -1 & 13 & -7 \\ & & 3 & 1 & 7 \\ \hline & 6 & 2 & 14 & 0 \end{array} $

The remainder is $0$, so $(2x - 1)$ is a factor.

Step 4: Factorize $f(x)$:

$ f(x) = \left(x - \frac{1}{2}\right)(6x^2 + 2x + 14) = (2x - 1)(3x^2 + x + 7) $

Discriminant:

$ \Delta = 1^2 - 4(3)(7) = -83 $

Since $\Delta \lt 0$, there are no real roots from the quadratic.

Step 5: Conclusion:

$ x = \frac{1}{2} $

Therefore, the curve intersects the $x$-axis exactly once.