Calculus

Calculus is, with algebra, the heaviest block of the paper (our estimate: about 8 questions).

The official booklet's model questions include a hyperbolic-inverse derivative and a first-order linear differential equation, so the syllabus is read broadly: standard limits, derivatives of inverse and hyperbolic functions, tangent / rate / extremum problems, a definite integral or an area, and a first-order differential equation.

Speed comes from knowing the standard results cold.

5.1 Limit and continuity of functions, indeterminate forms, L'Hospital's rule

\lim_{x\to a}f(x)=L exists iff the left and right limits both equal L. f is continuous at a if \lim_{x\to a}f(x)=f(a). Polynomials, e^x, \sin x are continuous everywhere; rational functions except at zeros of the denominator.

Standard limit (x\to0)	Value
\dfrac{\sin x}{x},\ \dfrac{\tan x}{x},\ \dfrac{\sin^{-1}x}{x},\ \dfrac{\tan^{-1}x}{x}	1
\dfrac{1-\cos x}{x^2}	\frac{1}{2}
\dfrac{e^x-1}{x},\ \dfrac{\ln(1+x)}{x}	1
\dfrac{a^x-1}{x}	\ln a
\dfrac{(1+x)^n-1}{x}	n
(1+x)^{1/x}, and \big(1+\frac{1}{x}\big)^x as x\to\infty	e
\dfrac{x^n-a^n}{x-a} (x\to a)	na^{n-1}

Indeterminate forms: \frac{0}{0},\frac{\infty}{\infty},0\cdot\infty,\infty-\infty,1^\infty,0^0,\infty^0. L'Hôpital: for \frac{0}{0} or \frac{\infty}{\infty}, \lim\frac{f}{g}=\lim\frac{f'}{g'} (repeat if still indeterminate).

Convert 0\cdot\infty to a quotient, \infty-\infty to a single fraction, and take logarithms for 1^\infty,0^0,\infty^0.

Example \lim_{x\to0}\frac{\tan x-\sin x}{x^3}=\lim\frac{\tan x(1-\cos x)}{x^3}=\frac{1}{2}. As x\to\infty the highest power of a rational function decides the limit.

5.2 Derivatives, rules, applications

Rules: sum, product (uv)'=u'v+uv', quotient \big(\frac{u}{v}\big)'=\frac{u'v-uv'}{v^2}, chain \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}; parametric \frac{dy}{dx}=\frac{dy/dt}{dx/dt}; implicit: differentiate both sides; logarithmic differentiation for x^x, u^v: \frac{d}{dx}x^x=x^x(1+\ln x).

f(x)	f'(x)	f(x)	f'(x)
x^n	nx^{n-1}	\sin^{-1}x	\frac{1}{\sqrt{1-x^2}}
e^x, a^x	e^x, a^x\ln a	\tan^{-1}x	\frac{1}{1+x^2}
\ln x, \log_ax	\frac{1}{x}, \frac{1}{x\ln a}	\sec^{-1}x	\frac{1}{|x|\sqrt{x^2-1}}
\sin x, \cos x	\cos x, -\sin x	\sinh x, \cosh x	\cosh x, \sinh x
\tan x, \cot x	\sec^2x, -\csc^2x	\tanh x	\mathrm{sech}^2x
\sec x, \csc x	\sec x\tan x, -\csc x\cot x	\sinh^{-1}x	\frac{1}{\sqrt{x^2+1}}
\ln(\sec x+\tan x)	\sec x	\cosh^{-1}x (x>1)	\frac{1}{\sqrt{x^2-1}}
\sqrt{x}	\frac{1}{2\sqrt{x}}	\tanh^{-1}x (|x|<1)	\frac{1}{1-x^2}
		\mathrm{sech}^{-1}x (0<x<1)	\frac{-1}{x\sqrt{1-x^2}}

Hyperbolic: \sinh x=\frac{e^x-e^{-x}}{2}, \cosh x=\frac{e^x+e^{-x}}{2}, \cosh^2x-\sinh^2x=1; \sinh^{-1}x=\ln(x+\sqrt{x^2+1}), \cosh^{-1}x=\ln(x+\sqrt{x^2-1}), \tanh^{-1}x=\frac{1}{2}\ln\frac{1+x}{1-x}.

The derivative of \mathrm{sech}^{-1}x=\cosh^{-1}(1/x) exists only inside its domain 0<x<1 — the reason the booklet's model answer is "x<1".

Higher derivatives and Leibniz: \frac{d^n}{dx^n}e^{ax}=a^ne^{ax}, \frac{d^n}{dx^n}\sin x=\sin(x+\frac{n\pi}{2}). Rolle (continuous on [a,b], differentiable inside, f(a)=f(b) gives f'(c)=0) and Mean value theorem f'(c)=\frac{f(b)-f(a)}{b-a}.

Applications

• Tangent at (x_1,y_1): slope f'(x_1); normal slope -\frac{1}{f'(x_1)}. A horizontal tangent has f'=0.
• Rate of change: differentiate the relation with respect to time. Sphere: V=\frac{4}{3}\pi r^3\Rightarrow\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}. Ladder: x^2+y^2=L^2\Rightarrow x\dot x+y\dot y=0.
• Maxima and minima. Critical points: f'(x)=0. Second-derivative test: f''<0 local max, f''>0 local min. For a closed interval compare the critical values with the endpoint values. x^2-4x+7 has minimum 3 at x=2. Given perimeter or sum, a rectangle / product is maximal when the parts are equal.
• Increasing where f'>0, decreasing where f'<0; concave up where f''>0; a point of inflection where f'' changes sign.

5.3 Integration, definite integral, area

Standard integrals (+C omitted): \int x^n=\frac{x^{n+1}}{n+1} (n\ne-1), \int\frac{1}{x}=\ln|x|, \int e^{ax}=\frac{e^{ax}}{a}, \int a^x=\frac{a^x}{\ln a}, \int\sin=-\cos, \int\cos=\sin, \int\sec^2=\tan, \int\csc^2=-\cot, \int\sec\tan=\sec, \int\tan x=\ln|\sec x|, \int\sec x=\ln|\sec x+\tan x|.

\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}. \int\frac{f'(x)}{f(x)}dx=\ln|f|.

Methods. Substitution (\int\frac{x}{x^2+1}dx=\frac{1}{2}\ln(x^2+1)). By parts \int uv'=uv-\int u'v, with u chosen by ILATE (Inverse, Log, Algebraic, Trig, Exponential): \int xe^xdx=xe^x-e^x; \int\ln x\,dx=x\ln x-x.

Special \int e^x[f(x)+f'(x)]dx=e^xf(x). Partial fractions for rational functions.

Definite integral. \int_a^bf=F(b)-F(a).

Properties: \int_a^bf(x)dx=\int_a^bf(a+b-x)dx; \int_{-a}^af=2\int_0^af if f even, 0 if odd; \int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx=\frac{\pi}{4} (also with \sin^n,\cos^n); Wallis: \int_0^{\pi/2}\sin^nx\,dx=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots (times \frac{\pi}{2} if n even). \int_0^{\pi/2}\sin^2x\,dx=\frac{\pi}{4}.

Area. Under y=f(x) from a to b: \int_a^b|f|\,dx; between curves: \int_a^b(f-g)\,dx where f\ge g; with respect to y: \int x\,dy. Circle area \pi r^2, area of an ellipse \pi ab. Area between y^2=4ax and x^2=4ay: \frac{16a^2}{3}.

Intersect the curves first to find the limits.

5.4 Differential equations

Order = highest derivative present; degree = its power after removing radicals and fractions of derivatives. \big(\frac{d^2y}{dx^2}\big)^3+\big(\frac{dy}{dx}\big)^4+y=0 has order 2, degree 3.

Type	Form	Method
separable	\frac{dy}{dx}=f(x)g(y)	\int\frac{dy}{g}=\int f\,dx
homogeneous	\frac{dy}{dx}=F\big(\frac{y}{x}\big)	put y=vx
linear	\frac{dy}{dx}+Py=Q	IF =e^{\int P\,dx}; y\cdot\text{IF}=\int Q\cdot\text{IF}\,dx
exact	M\,dx+N\,dy=0, \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}	\int M\,dx (y const) + terms of N free of x =C

If M\,dx+N\,dy=0 is not exact, try an integrating factor: e^{\int\frac{M_y-N_x}{N}dx} when that depends only on x. Examples: \frac{dy}{dx}+\frac{2}{x}y=x^2 has IF x^2; \frac{dy}{dx}+y=e^{-x} gives y=(x+C)e^{-x}.

The number of arbitrary constants in the general solution equals the order.

(a) Archimedes: polygons inside and outside a circle

(b) Perimeter bounds for pi from n = 6 to 96 sides

(a) A speedometer shows the derivative of distance at this instant

(b) Difference quotients of s = 0.8 t^2 + 2 t at t = 5

(a) A ladder leaning on a wall: the top slides down as the foot moves out

(b) Ladder of 5 m: top speed at five foot positions

(a) A planimeter: a mechanical integrator that measures area

(b) Left sums and trapezoids for the area under y = x^2

(a) A hot drink cools towards room temperature

(b) Cooling curve: exact solution and Euler steps

Chapter summary

• Standard limits plus L'Hôpital settle every limit; check the form before using L'Hôpital.
• Chain rule, parametric and logarithmic differentiation cover all derivative types.
• Critical points from f'=0; the sign of f'' decides max or min; compare endpoints on closed intervals.
• Integrate by substitution, by parts (ILATE) or by the e^x(f+f') pattern; use symmetry for definite integrals.
• Areas: find the intersections, then integrate top minus bottom.
• Classify a first-order equation (separable, homogeneous, linear, exact) before solving.

Practice questions

Set A — Recall (14 questions)

1. The value of \lim_{x\to0}\dfrac{\sin3x}{x} isa1b3c1/3d0
2. The value of \lim_{x\to0}(1+2x)^{1/x} isaeb2ece^2d1
3. \dfrac{d}{dx}\tan^{-1}x equalsa\dfrac{1}{1+x^2}b\dfrac{1}{\sqrt{1-x^2}}c-\dfrac{1}{1+x^2}d\sec^2x
4. \displaystyle\int e^{2x}dx equalsa2e^{2x}+Cbe^{2x}+Cc\dfrac{e^x}{2}+Cd\dfrac{e^{2x}}{2}+C
5. The value of \displaystyle\int_0^{\pi}\sin x\,dx isa2b0c1d\pi
6. The order and degree of \big(\frac{d^2y}{dx^2}\big)^3+\big(\frac{dy}{dx}\big)^4+y=0 area3, 2b2, 3c2, 1d3, 3
7. The general solution of \dfrac{dy}{dx}=y isay=Cxby=x+Ccy=Ce^xdy=C\ln x
8. The integrating factor of \dfrac{dy}{dx}+\dfrac{2}{x}y=x^2 isa2xbe^{2x}c\ln x^2dx^2
9. \dfrac{d}{dx}\big(e^x\sin x\big) equalsae^x\cos xbe^x(\sin x+\cos x)ce^x(\sin x-\cos x)d-e^x\sin x
10. The slope of the tangent to y=x^3-3x at x=2 isa3b6c9d12
11. The local maximum value of f(x)=3x-x^3 isa2b-2c0d4
12. The value of \lim_{x\to0}\dfrac{1-\cos x}{x^2} isa0b1c2d1/2
13. The area under y=x^2 from x=0 to x=3 isa3b9c27d6
14. The value of \displaystyle\int_0^1xe^x\,dx isa0bec1de-1

Answers and explanations — Set A

1. (b) — \frac{\sin3x}{x}=3\cdot\frac{\sin3x}{3x}\to3.

2. (c) — (1+2x)^{1/x}=[(1+2x)^{1/2x}]^2\to e^2.

3. (a) — standard derivative.

4. (d) — \frac{e^{2x}}{2}; (a) differentiates instead.

5. (a) — [-\cos x]_0^\pi=1+1=2.

6. (b) — highest derivative is the second (order 2), raised to the power 3 (degree 3).

7. (c) — separate: \frac{dy}{y}=dx, \ln y=x+c.

8. (d) — e^{\int\frac{2}{x}dx}=e^{2\ln x}=x^2.

9. (b) — product rule.

10. (c) — 3x^2-3=12-3=9.

11. (a) — f'=3-3x^2=0 at x=\pm1; f''=-6x<0 at x=1, f(1)=2.

12. (d) — \frac{1-\cos x}{x^2}\to\frac{1}{2}.

13. (b) — \frac{x^3}{3}\Big|_0^3=9.

14. (c) — by parts: [xe^x-e^x]_0^1=(e-e)-(0-1)=1.

Set B — Understanding (10 questions)

1. The value of \lim_{x\to0}\dfrac{\tan x-\sin x}{x^3} isa1/2b1c0d1/3
2. The function f(x)=\dfrac{x^2-4}{x-2} (x\ne2) with f(2)=k is continuous at 2 when k=a0b2c4dno value
3. If y=\ln(\sec x+\tan x), then \dfrac{dy}{dx} equalsa\tan xb\sec^2xc\sec x\tan xd\sec x
4. If x=t^2 and y=t^3, then \dfrac{dy}{dx} at t=2 isa6b3c12d3/2
5. The radius of a sphere grows at 2 cm/s. When r=5 cm the volume grows ata100\pib50\pic200\pi cm³/sd400\pi
6. The minimum value of x^2-4x+7 isa7b4c2d3
7. \displaystyle\int\frac{x}{x^2+1}dx equalsa\ln(x^2+1)+Cb\tfrac{1}{2}\ln(x^2+1)+Cc\tan^{-1}x+Cd2\ln(x^2+1)+C
8. The value of \displaystyle\int_0^{\pi/2}\sin^2x\,dx isa\pi/4b\pi/2c1d\pi/8
9. The area enclosed between y=x and y=x^2 isa1/2b1/3c1/6d1/12
10. The solution of \dfrac{dy}{dx}=\dfrac{x+y}{x} isay=x(\ln x+C)by=x\ln x+Ccy=x+C\ln xdy=Cx^2

Answers and explanations — Set B

1. (a) — \tan x-\sin x=\tan x(1-\cos x)\approx x\cdot\frac{x^2}{2}.

2. (c) — \frac{x^2-4}{x-2}=x+2\to4; continuity needs k=4.

3. (d) — \frac{\sec x\tan x+\sec^2x}{\sec x+\tan x}=\sec x.

4. (b) — \frac{3t^2}{2t}=\frac{3t}{2}=3 at t=2.

5. (c) — \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi\cdot25\cdot2=200\pi.

6. (d) — (x-2)^2+3.

7. (b) — substitute u=x^2+1.

8. (a) — \int\frac{1-\cos2x}{2}=\frac{1}{2}\big[\frac{\pi}{2}\big]=\frac{\pi}{4}.

9. (c) — \int_0^1(x-x^2)dx=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}.

10. (a) — homogeneous: y=vx gives xv'=1, v=\ln x+C.

Set C — Application (8 questions)

1. The value of \lim_{x\to0}\dfrac{e^x-1-x}{x^2} isa0b1c2d1/2
2. The maximum value of f(x)=xe^{-x} for x>0 isa1/ebec1d1/(2e)
3. The normal to y=x^2 at (1,1) isa2x+y-3=0bx+2y-3=0cx-2y+1=0d2x-y-1=0
4. The value of \displaystyle\int_0^{\pi/2}\dfrac{\sin x}{\sin x+\cos x}dx isa\pi/2b1c\pi/8d\pi/4
5. The area enclosed by y^2=4x and y=x isa16/3b4c8/3d32/3
6. A 5 m ladder slides with its foot moving away from the wall at 3 m/s. When the foot is 3 m from the wall, the top slides down at (m/s)a4/3b9/4c3/4d4
7. The general solution of \dfrac{dy}{dx}+y=e^{-x} isay=(x+C)e^{-x}by=(x+C)e^{x}cy=e^{-x}+Cdy=xe^x+C
8. The integrating factor of \dfrac{dy}{dx}+\dfrac{x}{1+x}y=1+x isae^xb(1+x)e^xc\dfrac{e^x}{1+x}d\dfrac{e^x}{x}

Answers and explanations — Set C

1. (d) — e^x-1-x=\frac{x^2}{2}+\frac{x^3}{6}+\cdots.

2. (a) — f'=e^{-x}(1-x)=0 at x=1; f(1)=1/e and f''(1)<0.

3. (b) — tangent slope 2, normal slope -\frac{1}{2}: y-1=-\frac{1}{2}(x-1).

4. (d) — I=\int_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}dx by x\to\frac{\pi}{2}-x; 2I=\frac{\pi}{2}.

5. (c) — meet at x=0,4: \int_0^4(2\sqrt{x}-x)dx=\frac{32}{3}-8=\frac{8}{3}.

6. (b) — x^2+y^2=25: x\dot x+y\dot y=0 with x=3, y=4: 9+4\dot y=0, |\dot y|=\frac{9}{4}.

7. (a) — IF =e^x: (ye^x)'=1, so ye^x=x+C.

8. (c) — \int\frac{x}{1+x}dx=x-\ln(1+x), so IF =\frac{e^x}{1+x}.

Group D — Model answers for long questions (10 marks each, for deeper revision)

1. Evaluate limits (standard forms, L'Hospital's rule), test continuity, and solve an optimisation and a tangent problem. [10]
2. Evaluate integrals (by parts), find an area, and solve first-order differential equations including Newton's law of cooling. [10]

Model answer D1: Limits, derivatives and applications (10 marks)

Introduction. A function is continuous at a if \lim_{x\to a}f(x)=f(a). Standard limits: \lim_{x\to0}\frac{\sin x}{x}=1, \lim_{x\to0}\frac{e^x-1}{x}=1, \lim_{x\to0}(1+x)^{1/x}=e.

Limits.

1. \lim_{x\to0}\frac{1-\cos x}{x^2}=\lim\frac{2\sin^2(x/2)}{x^2}=\frac12.
2. L'Hospital's rule for \frac00 or \frac\infty\infty: \lim_{x\to0}\frac{x-\sin x}{x^3}=\lim\frac{1-\cos x}{3x^2}=\lim\frac{\sin x}{6x}=\frac16.

Optimisation (open box). A square sheet of side 12\ \text{cm} has corner squares of side x cut off and the sides are folded up.

1. Volume V=x(12-2x)^2.
2. \frac{dV}{dx}=(12-2x)^2-4x(12-2x)=(12-2x)(12-6x).
3. Setting it to zero: x=6 (no box) or x=2.
4. \frac{d^2V}{dx^2}<0 at x=2, so it is a maximum: V_{\max}=2(8)^2=128\ \text{cm}^3.

Tangent. For y=x^3-3x+2 at x=2: y=4, \frac{dy}{dx}=3x^2-3=9, so the tangent is y-4=9(x-2), i.e. y=9x-14. The normal has slope -\frac19.

Topic	Rule	Result
Limit	L'Hospital	\frac16
Maximum volume	V'=0, V''<0	128\ \text{cm}^3 at x=2
Tangent	slope =y'	y=9x-14

Conclusion. Derivatives give slopes and extreme values, and L'Hospital's rule resolves indeterminate forms.

Model answer D2: Integration and differential equations (10 marks)

Introduction. Integration reverses differentiation, and a definite integral gives an area: \int_a^bf(x)\,dx=F(b)-F(a).

1. Integration by parts. \int u\,dv=uv-\int v\,du. For \int xe^x\,dx let u=x, dv=e^xdx: =xe^x-e^x+C=e^x(x-1)+C.

2. Area. The area between y=x and y=x^2 (they meet at x=0 and x=1) is \int_0^1(x-x^2)\,dx=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac12-\frac13=\frac16 square units. Also \int_0^1\frac{dx}{1+x^2}=\tan^{-1}1=\frac\pi4.

3. Separable equation. \frac{dy}{dx}=xy\Rightarrow\int\frac{dy}{y}=\int x\,dx\Rightarrow y=Ce^{x^2/2}.

4. Linear equation. \frac{dy}{dx}+2y=e^{-x} with y(0)=2.

1. Integrating factor =e^{\int2\,dx}=e^{2x}.
2. \frac{d}{dx}(ye^{2x})=e^{x}, so ye^{2x}=e^x+C.
3. y=e^{-x}+Ce^{-2x}, and y(0)=2\Rightarrow1+C=2\Rightarrow C=1.
4. Solution: y=e^{-x}+e^{-2x}.

5. Newton's law of cooling. \frac{dT}{dt}=-k(T-T_a) gives T=T_a+(T_0-T_a)e^{-kt}. A drink at 100^\circ\text{C} in a 20^\circ\text{C} room: T=20+80e^{-kt}.

If T=50 at t=10\ \text{min}: 30=80e^{-10k}, so k=\frac{\ln(8/3)}{10}=0.0981\ \text{min}^{-1}. At t=20: T=20+80e^{-1.962}=20+80(0.1406)=31.25^\circ\text{C}.

Problem	Method	Answer
\int xe^x	By parts	e^x(x-1)+C
Area between curves	\int(\text{top}-\text{bottom})	\frac16
Linear ODE	Integrating factor	e^{-x}+e^{-2x}
Cooling	Separate variables	31.25^\circ\text{C}

Conclusion. The choice between substitution, parts, separation and an integrating factor depends on the form of the integrand or equation.