CAIE IGCSE Additional Math 14 Calculus Topic Practice

Attempts to find $y=4 x-5 x^{-2}$
at least one term correct

$\frac{\mathrm{d} y}{\mathrm{~d} x}=4+10 x^{-3}$
Correct completion to given answer $\frac{2\left(2 x^{3}+5\right)}{x^{3}}$
Alternative version

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{12 x^{4}-8 x^{4}+10 x}{x^{4}}$ or equivalent
unsimplified form

M1 for an attempt at the quotient rule with correct structure
$$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x^{2}\left(\text { their } 12 x^{2}\right)-\left(4 x^{3}-5\right)(\text { their } 2 x)}{\left(x^{2}\right)^{2}}$$

Correct completion to given answer $\frac{2\left(2 x^{3}+5\right)}{x^{3}}$

[When x=5 ] y=9
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2}{3}(5 x+2)^{-\frac{1}{3}} \times 5$ oe
B1 for $\frac{\mathrm{d} y}{\mathrm{~d} x}=k(5 x+2)^{-\frac{1}{3}}, k \neq \frac{10}{3}$$\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=5}=\frac{10}{9}$
FT their $\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=5}$ providing previous B1
awarded

Equation of normal e.g.
$$y-\text { their } 9=-\frac{1}{\text { their } \frac{10}{9}}(x-5) \text { oe, soi }$$
FT $\frac{-1}{\text { their }\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x=5}}$ and their y-coordinate

Eliminates one variable e.g.
$$11-x-\text { their } 9=-\frac{1}{\text { their } \frac{10}{9}}(x-5)$$
dep on previous M mark

For Q: x=-25, y=36
A1 for each

Coordinates of R:(35,-18)
or ((10-their(-25)), (18-their36))
FT their coordinates of Q

$\frac{dy}{dx}=1-\frac{2x+2}{x^2+2x}$
B1 for $\frac{d(-\ln(x^2+2x))}{dx}=\frac{1}{x^2+2x}\times f(x)$ soi.
Equates their first derivative to 0 and simplifies as far as 2(x+1)=x(x+2) oe
M1 FT their $\frac{dy}{dx}$.
$x^2-2=0$ or $x^2=2$$x=\sqrt{2}$ [$x=-\sqrt{2}$]
Justification that the negative solution should be rejected, e.g. $x=-\sqrt{2}$ gives $y=2-\ln(2-2\sqrt{2})$, which is impossible.