TMJC JC1 Promo 9758/2019/Q6

1. Rewrite the given logarithmic equation in exponential form and implicitly differentiate with respect to $x$.

$y = \ln(2 - e^{2x})$
$e^y = 2 - e^{2x}$

Differentiating both sides with respect to $x$:
$e^y \frac{dy}{dx} = -2e^{2x}$

2. Differentiate a second time to show the required identity.

Differentiating again with respect to $x$ using the product rule on the left-hand side:
$e^y \frac{d^2 y}{dx^2} + \left(e^y \frac{dy}{dx}\right) \frac{dy}{dx} = -4e^{2x}$
$e^y \left[ \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \right] = -4e^{2x}$ (Shown)

3. Evaluate $y$, $\frac{dy}{dx}$, and $\frac{d^2 y}{dx^2}$ at $x=0$.

When $x = 0$:
$e^y = 2 - e^0 = 1 \implies y = 0$

Substitute $x=0$ and $y=0$ into the first derivative equation:
$e^0 \frac{dy}{dx} = -2e^0 \implies \frac{dy}{dx} = -2$

Substitute $x=0$, $y=0$, and $\frac{dy}{dx}=-2$ into the second derivative equation:
$e^0 \left[ \frac{d^2 y}{dx^2} + (-2)^2 \right] = -4e^0$
$1 \cdot \left( \frac{d^2 y}{dx^2} + 4 \right) = -4 \implies \frac{d^2 y}{dx^2} = -8$

4. Differentiate the shown identity to find the third derivative at $x=0$.

Differentiating $e^y \left[ \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \right] = -4e^{2x}$ with respect to $x$:
$e^y \left[ \frac{d^3 y}{dx^3} + 2\frac{dy}{dx}\frac{d^2 y}{dx^2} \right] + e^y \frac{dy}{dx} \left[ \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \right] = -8e^{2x}$

Substituting the known values at $x=0$:
$1 \cdot \left[ \frac{d^3 y}{dx^3} + 2(-2)(-8) \right] + 1 \cdot (-2) \left[ -8 + (-2)^2 \right] = -8e^0$
$\frac{d^3 y}{dx^3} + 32 - 2(-4) = -8$
$\frac{d^3 y}{dx^3} + 40 = -8 \implies \frac{d^3 y}{dx^3} = -48$

5. Construct the Maclaurin series up to the term in $x^3$.

The Maclaurin series is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$

$y = 0 + (-2)x + \frac{-8}{2}x^2 + \frac{-48}{6}x^3 + \dots$
$y = -2x - 4x^2 - 8x^3 + \dots$

1. Rewrite the expression into a product of binomial expressions in the form $(1+ax)^n$.

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

2. Find the binomial expansion of $(1-2x)^{1/2}$ up to the term in $x^2$.

$(1-2x)^{\frac{1}{2}} = 1 + \left(\frac{1}{2}\right)(-2x) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(-2x)^2 + \dots$
$= 1 - x - \frac{1}{8}(4x^2) + \dots$
$\approx 1 - x - \frac{1}{2}x^2$

3. Factorise the constant out of $(2+3x)^{-1}$ to get the form $k(1+bx)^{-1}$ and expand up to the term in $x^2$.

$(2+3x)^{-1} = \left[2\left(1 + \frac{3}{2}x\right)\right]^{-1} = \frac{1}{2}\left(1 + \frac{3}{2}x\right)^{-1}$
$= \frac{1}{2}\left[1 + (-1)\left(\frac{3}{2}x\right) + \frac{(-1)(-2)}{2!}\left(\frac{3}{2}x\right)^2 + \dots\right]$
$\approx \frac{1}{2}\left(1 - \frac{3}{2}x + \frac{9}{4}x^2\right)$

4. Multiply the two series expansions together and simplify up to the $x^2$ term.

$\frac{\sqrt{1-2x}}{2+3x} \approx \frac{1}{2} \left(1 - x - \frac{1}{2}x^2\right) \left(1 - \frac{3}{2}x + \frac{9}{4}x^2\right)$
$= \frac{1}{2} \left[ 1\left(1 - \frac{3}{2}x + \frac{9}{4}x^2\right) - x\left(1 - \frac{3}{2}x\right) - \frac{1}{2}x^2(1) \right]$
$= \frac{1}{2} \left[ 1 - \frac{3}{2}x + \frac{9}{4}x^2 - x + \frac{3}{2}x^2 - \frac{1}{2}x^2 \right]$
$= \frac{1}{2} \left[ 1 - \frac{5}{2}x + \left(\frac{9}{4} + \frac{6}{4} - \frac{2}{4}\right)x^2 \right]$
$= \frac{1}{2} \left( 1 - \frac{5}{2}x + \frac{13}{4}x^2 \right)$
$= \frac{1}{2} - \frac{5}{4}x + \frac{13}{8}x^2$

5. State the range of values of $x$ for which the overall expansion is valid by taking the intersection of the validity ranges of the individual series.

For $(1-2x)^{\frac{1}{2}}$, valid when $|-2x| < 1 \implies |x| < \frac{1}{2}$.
For $\left(1+\frac{3}{2}x\right)^{-1}$, valid when $\left|\frac{3}{2}x\right| < 1 \implies |x| < \frac{2}{3}$.

The expansion is valid for the intersection of these two sets of values, which is $|x| < \frac{1}{2}$ (or $-\frac{1}{2} < x < \frac{1}{2}$).