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

$\cos2x+x(-2\sin2x)$ oe, isw, nfww
M1 for $\frac{d}{dx}(\cos2x)=-2\sin2x$ soi, or M1 FT for correct product-rule structure.
A1

$\frac{dy}{dx}=1-\frac{2x+2}{x^2+2x}$
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.

$\frac{dv}{dt}=(2t-3)^2$ oe, soi, or $\frac{d}{dt}(\text{their }v)=k(2t-3)^2$, where $k\ne0$
B1 FT their expression for v if necessary.
Correct small changes relationship: $\frac{\delta v}{0.02}=\text{their }(2\times2.5-3)^2$ oe
M1 FT their $\left.\frac{dv}{dt}\right|_{t=2.5}$.
$\delta v=0.08$
A1

Correct use of $x^{2} h=5$ to find an expression that
can be used to eliminate h
M1 for $x^{2} h=5$ soi

Surface area: $x^{2}+4 x\left(\frac{5}{x^{2}}\right)$ oe
Derivative of the surface area: $2 x-20 x^{-2}$ oe
FT their surface area of form $a x^{2}+\frac{b}{x}$

Equates their $2 x-20 x^{-2}$ to 0 and solves to find
a value of x
FT their derivative of form $a x+\frac{b}{x^{2}}$ oe

x=2.15 or $2.154[\ldots] h=1.08$ or $1.077[\ldots]$ nfww
x=2.15 or $2.154[\ldots]$
h=1.08 or $1.077[\ldots]$ nfww
dep on all previous marks awarded

Alternative

Correct use of $x^{2} h=5$ to find an expression that
can be used to eliminate x
M1 for $x^{2} h=5$ soi

Surface area: $\left(\sqrt{\frac{5}{h}}\right)^{2}+4 h \times \sqrt{\frac{5}{h}}$ oe
Derivative of the surface area: $4 \sqrt{5} \times \frac{1}{2} h^{-\frac{1}{2}}-\frac{5}{h^{2}}$
oe
FT their surface area of form $\frac{a}{h}+b \sqrt{h}$

Equates their $4 \sqrt{5} \times \frac{1}{2} h^{-\frac{1}{2}}-\frac{5}{h^{2}}$ to 0 and solves
to find a value of h
FT their derivative of form $\frac{a}{\sqrt{h}}+\frac{b}{h^{2}}$ oe

h=1.08 or $1.077[\ldots] x=2.15$ or $2.154[\ldots]$ nfww
h=1.08 or $1.077[\ldots]$
x=2.15 or $2.154[\ldots]$ nfww
dep on all previous marks awarded

Correct derivative: $4\pi r-660r^{-2}$, isw
B1 for one correct term.
$4\pi r-\frac{660}{r^2}=0$ and solves for r
M1 FT their derivative, providing at least one term is correct.
$r=\sqrt[3]{\frac{660}{4\pi}}$ oe, isw, or 3.74 or 3.74494... rot to 3 or more dp
Second derivative $\frac{d^2S}{dr^2}=4\pi+1320r^{-3}$, and when r=3.74, $\frac{d^2S}{dr^2}>0$ oe, hence minimum.
A1 dep on previous mark.

$-\frac12x\cos2x+\frac14\sin2x+c$ oe, isw, nfww
M3 for $-\frac12x\cos2x+\frac14\sin2x$ or equivalent integration-by-parts working;
M2/M1 for partial correct working as in markscheme.
A1

$\dfrac{\ln3x}{x^{2}}=\displaystyle\int\dfrac{1-2\ln3x}{x^{3}}\,\mathrm{d}x$, so $2\displaystyle\int\dfrac{\ln3x}{x^{3}}\,\mathrm{d}x=\displaystyle\int\dfrac{1}{x^{3}}\,\mathrm{d}x-\dfrac{\ln3x}{x^{2}}$ oe. M1 for using integration as reverse of differentiation from part (a); allow their A,B, no extra terms.
$\displaystyle\int\dfrac{1}{x^{3}}\,\mathrm{d}x=-\dfrac{1}{2x^{2}}$ oe, nfww. B1, FT on their A.
$-\dfrac{1}{4x^{2}}-\dfrac{\ln3x}{2x^{2}}+c$ oe. B2; M1 dep for rearranging simplification and attempting to integrate $ax^{-3}$. A1 must include +c.

$y: \quad-\frac{1}{3} \cos (3 x+\pi)+c$
B1 for $-\frac{1}{3} \cos (3 x+\pi)$ or for
$k \cos (3 x+\pi)+c$ where k is $\frac{1}{3}$ or k<0

$\frac{4}{3}=-\frac{1}{3} \cos \left(\frac{3 \pi}{9}+\pi\right)+c$
FT their k providing B1 has been awarded

$$\left[y=-\frac{1}{3} \cos \left(\frac{5 \pi}{4}+\pi\right)+\frac{7}{6}=\right] \frac{7-\sqrt{2}}{6}$$

or exact equivalent

Writes as a 3-term quadratic in $e^x$: $3e^{2x}+5e^x-2=0$ soi
Solves/factorises their quadratic in $e^x$$e^x=\frac13$ [$e^x=-2$]
$x=\ln\frac13=-\ln3$ only solution, nfww
Correct plan: $\int_{\ln(1/3)}^0(3e^{2x}+5e^x-2)\,dx$
Integral $\frac32e^{2x}+5e^x-2x$
Correct substitution of limits
$\frac{14}{3}-2\ln3$
M2 dep on all previous marks;
M1 for one correct final term.