CAIE IGCSE Additional Math 4 Equations Inequalities and Graphs Topic Practice

$2 x^{2}+x-10=5$

leading to $2 x^{2}+x-15=0$ with a valid attempt to solve to obtain 2 values for x M1
$x=\frac{5}{2}, \quad-3$ A1 For both
Mark final answer for this equation,
A0 if - 3 is rejected
$$x^{2}+x-10=-5$$

leading to $2 x^{2}+x-5=0$ M1 Must be a correct 3-term quadratic
equated to zero
$x=\frac{-1 \pm \sqrt{41}}{4}$ oe 2 Dep M1 for a valid method of
solution to obtain 2 values for x.
Mark final answer for this equation,
A0 if negative root is rejected
Denominator of final answer must be positive

Correct graph of y=|x-2| and $x \leqslant -2.5,\ x \geqslant -0.25$
B1 for correct graph of y=|x-2|.
B1 for $x \leqslant -2.5$ or $x \geqslant -0.25$.
B1 STRICT FT for their critical values from the two intersections of their straight-line section of graph, providing it has negative gradient.

$$\begin{aligned} e^{3y-7+3y-1}+4e^{-3+3y-1}&=5 \quad\text{oe}\\ \text{or } e^{3y-4+3y-1}+4e^{3y-1}&=5e^3\\ \text{or } e^{3y-7}+\frac{4}{e^3}&=5e^{1-3y}\\ \text{or } e^{6y}+4e^{4+3y}&=5e^8 \end{aligned}$$$$\begin{aligned} e^{6y-8}+4e^{3y-4}-5&=0\\ \text{or } e^{3y-4}+4-5e^{4-3y}&=0\\ \text{or } e^{-7}(e^{3y})^2+4e^{-3}e^{3y}-5e&=0\\ \text{or } (e^{3y-1})^2+4e^3e^{3y-1}-5e^6&=0 \end{aligned}$$
B1 dep for arranging as 3-term quadratic.
$$\begin{aligned} (e^{3y-4}-1)(e^{3y-4}+5)&=0\\ \text{or }(e^{-3}e^{3y}-e)(e^{-4}e^{3y}+5)&=0\\ \text{or }(e^{3y-1}+5e^3)(e^{3y-1}-e^3)&=0 \end{aligned}$$
M1 dep for an attempt at solution.
$e^{3y-4}=1$ or $e^{3y-1}=e^3$$3y-4=\ln 1$ oe soi or 3y-1=3$y=\frac{4}{3}$