Mathematics: Model Question Sets and Predicted Questions

This chapter does three things. It lays out the pattern of the Grade 11 Mathematics paper and shows how the teaching hours of the CDC syllabus translate into a sensible spread of marks. It then gives two complete model papers (11\times1+8\times5+3\times8=75 marks, three hours), each followed by a full solution. Finally it gives a ranked list of predicted questions, each with the reason for its rank.

13.1 Pattern and weighting of the paper

13.1.1 The grid

Group	Questions	Marks	What the examiner does
A	11 multiple-choice questions	11\times1=11	one mark each
B	8 short-answer questions	8\times5=40	five marks, often split 2+3 or 1+4
C	3 long-answer questions	3\times8=24	usually three parts, for example 3+3+2

The paper is out of 75; internal assessment adds 25. The last chapter is a choice: Computational Methods or Mechanics (Unit 20 or Unit 21); the question set on it carries an "Or".

13.1.2 Hours as a guide to marks

The CDC syllabus allots 160 working hours. The table converts each area's hours into a proportional share of 75 marks. It is our guide, not the official specification grid.

Area (chapters)	Hours	Share of 75 marks
Algebra (1–8)	44	about 20.6
Trigonometry (9–10)	12	about 5.6
Analytic geometry (11–13)	20	about 9.4
Vectors (14)	12	about 5.6
Statistics and probability (15–16)	12	about 5.6
Calculus (17–19)	48	about 22.5
Computational methods or mechanics (20 or 21)	12	about 5.6

Calculus and algebra together are nearly three-fifths of the paper. A student who is secure in derivatives and integration and in quadratic equations, sequences, matrices and complex numbers has already met the bulk of the marks. Derivatives alone is 22 hours, the largest chapter of the book.

13.1.3 What the two model papers do

Both papers give Group A three algebra, three trigonometry/geometry/vectors, one statistics-or-probability and four calculus-or-methods questions, Group B one question from each major strand, and Group C two calculus questions plus one mixed question. The two papers do not repeat a question form, so working both covers more ground than working one twice.

13.2 Model Question Set 1

Time: 3 hours. Full marks: 75. Attempt all questions.

Group A — Multiple choice (11 × 1 = 11)

1. If n(A)=30, n(B)=25 and n(A\cup B)=45, then n(A\cap B) is [1]a65b55c10d35
2. The 8th term of the G.P. 3,6,12,\dots is [1]a192b384c768d48
3. If the roots of x^2-6x+k=0 are equal, then k is [1]a3b6c9d36
4. The value of \sin^{-1}\left(-\frac{1}{2}\right) is [1]a\frac{\pi}{6}b-\frac{\pi}{6}c\frac{5\pi}{6}d\frac{7\pi}{6}
5. The distance of the origin from the line 3x+4y-10=0 is [1]a10b5c2d1
6. The vectors 3\mathbf{i}+2\mathbf{j} and 6\mathbf{i}+m\mathbf{j} are collinear when m equals [1]a2b3c4d6
7. If the mean is 30 and the standard deviation is 6, the coefficient of variation is [1]a5%b20%c25%d180%
8. \lim_{x\to3}\frac{x^2-9}{x-3} equals [1]a0b3c6d9
9. \frac{d}{dx}(\sin3x) equals [1]a\cos3xb3\cos3xc-3\cos3xd3\sin3x
10. \int\sec^2x\,dx equals [1]a\tan x+Cb\sec x\tan x+Cc\cot x+Cd\sec x+C

11. In the bisection method for a root in [2,3], the first midpoint is [1](a) 2.25(b) 2.5(c) 2.75(d) 3

    Or (Mechanics): The resultant of forces 6 N and 8 N acting at right angles is (a) 14 N (b) 2 N (c) 10 N (d) 48 N

Group B — Short answer (8 × 5 = 40)

12. Using a truth table, prove that \sim(p\wedge q)\equiv\sim p\vee\sim q. [5]
13. If f(x)=\frac{x+1}{x-1}, show that f(f(x))=x and hence write f^{-1}(x). [3+2]
14. The 3rd term of an A.P. is 11 and the 7th term is 23. Find the A.P. and the sum of its first 20 terms. [2+3]
15. Solve the system 3x+5y=13, x+2y=5 by the matrix method. [5]
16. If \alpha,\beta are the roots of 2x^2-5x+1=0, find \alpha^2+\beta^2 and \frac{\alpha}{\beta}+\frac{\beta}{\alpha}. [3+2]
17. Prove that \tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{8}=\frac{\pi}{4}. [5]
18. Show that x^2-3xy+2y^2=0 represents two straight lines and find the angle between them. [2+3]
19. The mean of a distribution is 45, its median is 42 and its standard deviation is 9. Find the mode, Karl Pearson's coefficient of skewness and the coefficient of variation. [1+2+2]

Group C — Long answer (3 × 8 = 24)

20. For f(x)=x^3-3x^2-9x+5: (i) find the intervals in which f is increasing and decreasing; (ii) find its local maximum and minimum values; (iii) find its point of inflection. [3+3+2]
21. (i) Evaluate \int x\cos x\,dx. (ii) Find the area enclosed by y=x^2 and the line y=4. (iii) Evaluate \int_0^2(3x^2+2x)\,dx. [3+3+2]
22. (i) Evaluate \lim_{x\to0}\frac{\sqrt{1+x}-1}{x}. (ii) Express 7\mathbf{i}+5\mathbf{j} as a linear combination of 2\mathbf{i}+\mathbf{j} and \mathbf{i}+2\mathbf{j}. (iii) Find the distance between the points (1,2,2) and (-1,0,3). [3+3+2]

13.2.1 Solutions to Model Set 1

Group A — answer key with reasons

1. (c) — n(A\cap B)=30+25-45=10.

2. (b) — r=2, t_8=3\times2^7=384.

3. (c) — equal roots need 36-4k=0, so k=9.

4. (b) — the principal value of \sin^{-1} lies in [-\frac{\pi}{2},\frac{\pi}{2}], and \sin(-\frac{\pi}{6})=-\frac{1}{2}.

5. (c) — \frac{|-10|}{\sqrt{9+16}}=2.

6. (c) — \frac{3}{6}=\frac{2}{m}\Rightarrow m=4.

7. (b) — \frac{6}{30}\times100=20\%.

8. (c) — \frac{(x-3)(x+3)}{x-3}=x+3\to6.

9. (b) — chain rule: \cos3x\cdot3.

10. (a) — standard integral.

11. (b) — \frac{2+3}{2}=2.5. Or: (c) — \sqrt{36+64}=10 N.

Group B — full solutions

12. Truth table for all four cases:

p	q	p\wedge q	\sim(p\wedge q)	\sim p	\sim q	\sim p\vee\sim q
T	T	T	F	F	F	F
T	F	F	T	F	T	T
F	T	F	T	T	F	T
F	F	F	T	T	T	T

The fourth and seventh columns are identical, so \sim(p\wedge q)\equiv\sim p\vee\sim q.

13. f(f(x))=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}=\frac{\frac{(x+1)+(x-1)}{x-1}}{\frac{(x+1)-(x-1)}{x-1}}=\frac{2x}{2}=x. Since f\circ f is the identity, f is its own inverse: f^{-1}(x)=\frac{x+1}{x-1}.

14. a+2d=11 and a+6d=23. Subtract: 4d=12, d=3, a=5. The A.P. is 5,8,11,14,\dots S_{20}=\frac{20}{2}[2(5)+19(3)]=10(67)=670.

15. In matrix form AX=B with A having rows (3,5),(1,2) and B=(13,5). |A|=6-5=1, so A^{-1} has rows (2,-5) and (-1,3). x=2(13)-5(5)=1 and y=-1(13)+3(5)=2. Check: 3(1)+5(2)=13 ✓, 1+2(2)=5 ✓. So x=1, y=2.

16. \alpha+\beta=\frac{5}{2}, \alpha\beta=\frac{1}{2}. \alpha^2+\beta^2=\left(\frac{5}{2}\right)^2-2\cdot\frac{1}{2}=\frac{25}{4}-1=\frac{21}{4}. \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{21/4}{1/2}=\frac{21}{2}.

17. \tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{5}=\tan^{-1}\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{10}}=\tan^{-1}\frac{7/10}{9/10}=\tan^{-1}\frac{7}{9}. Adding \tan^{-1}\frac{1}{8}: \tan^{-1}\frac{\frac{7}{9}+\frac{1}{8}}{1-\frac{7}{72}}=\tan^{-1}\frac{65/72}{65/72}=\tan^{-1}1=\frac{\pi}{4}.

18. The equation factorises as (x-y)(x-2y)=0, giving the lines x-y=0 and x-2y=0 (homogeneous, so both pass through the origin). Here a=1, h=-\frac{3}{2}, b=2: \tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|}=\frac{2\sqrt{\frac{9}{4}-2}}{3}=\frac{2\times\frac{1}{2}}{3}=\frac{1}{3}. So \theta=\tan^{-1}\frac{1}{3}\approx18.4^\circ.

19. Mode =3\times42-2\times45=126-90=36. S_k=\frac{45-36}{9}=1 (positively skewed). C.V. =\frac{9}{45}\times100=20\%.

Group C — full solutions

20. f'(x)=3x^2-6x-9=3(x-3)(x+1), zero at x=-1,3. (i) f'>0 for x<-1 and x>3; f'<0 for -1<x<3. So f is increasing on (-\infty,-1) and (3,\infty) and decreasing on (-1,3). (ii) f''(x)=6x-6. At x=-1: f''=-12<0, a local maximum, f(-1)=-1-3+9+5=10. At x=3: f''=12>0, a local minimum, f(3)=27-27-27+5=-22. (iii) f''(x)=0\Rightarrow x=1, and f'' changes sign there. f(1)=1-3-9+5=-6. The point of inflection is (1,-6).

21. (i) By parts, u=x, dv=\cos x\,dx: \int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x+\cos x+C. (ii) The curves meet where x^2=4, i.e. x=\pm2. The line is above the parabola, so area =\int_{-2}^{2}(4-x^2)dx=2\int_0^2(4-x^2)dx=2\left[4x-\frac{x^3}{3}\right]_0^2=2\left(8-\frac{8}{3}\right)=\frac{32}{3} square units. (iii) \int_0^2(3x^2+2x)dx=[x^3+x^2]_0^2=8+4=12.

22. (i) Rationalise: \frac{\sqrt{1+x}-1}{x}\cdot\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}=\frac{x}{x(\sqrt{1+x}+1)}=\frac{1}{\sqrt{1+x}+1}\to\frac{1}{2}. (ii) Let 7\mathbf{i}+5\mathbf{j}=p(2\mathbf{i}+\mathbf{j})+q(\mathbf{i}+2\mathbf{j}). Then 2p+q=7 and p+2q=5. Solving: p=3, q=1. So 7\mathbf{i}+5\mathbf{j}=3(2\mathbf{i}+\mathbf{j})+(\mathbf{i}+2\mathbf{j}). (iii) d=\sqrt{(1+1)^2+(2-0)^2+(2-3)^2}=\sqrt{4+4+1}=3.

13.3 Model Question Set 2

Time: 3 hours. Full marks: 75. Attempt all questions.

Group A — Multiple choice (11 × 1 = 11)

1. The negation of p\vee q is [1]a\sim p\vee\sim qb\sim p\wedge\sim qcp\wedge qd\sim p\vee q
2. If f(x)=2x-5, then f^{-1}(3) is [1]a1b4c11d6
3. The sum of the first 10 terms of an A.P. with a=2 and d=3 is [1]a155b145c165d310
4. The determinant of the matrix with rows (2,-1) and (3,4) is [1]a5b11c-11d8
5. The sum of the roots of 3x^2-12x+5=0 is [1]a-4b4c\frac{5}{3}d12
6. The modulus of (1+i)^2 is [1]a\sqrt{2}b2c4d1
7. The general solution of \tan x=1 is [1]a2n\pi+\frac{\pi}{4}bn\pi+\frac{\pi}{4}cn\pi\pm\frac{\pi}{4}dn\pi+\frac{\pi}{2}
8. The distance of the point (1,2,2) from the origin is [1]a3b5c9d2
9. If A and B are independent with P(A)=0.6 and P(B)=0.3, then P(A\cup B) is [1]a0.9b0.18c0.72d0.78
10. \lim_{x\to0}\frac{e^{3x}-1}{x} equals [1]a1b3ce^3d0

11. For f(x)=x^2-9, the Newton–Raphson value x_1 starting from x_0=4 is [1](a) 3.0(b) 3.125(c) 3.25(d) 2.5

    Or (Mechanics): A particle starting from rest with acceleration 4\ \text{m s}^{-2} covers in 3 s a distance of (a) 12 m (b) 18 m (c) 36 m (d) 24 m

Group B — Short answer (8 × 5 = 40)

12. For A=\{1,2,3,4,5\}, B=\{2,4,6\} and C=\{3,4,7\}, verify that A-(B\cup C)=(A-B)\cap(A-C). [5]
13. Find the inverse of f(x)=\frac{3x-1}{2x+5}. [5]
14. The sum to infinity of a G.P. is 15 and the sum of the squares of its terms is 45. Find the first term and common ratio. [5]
15. Evaluate the determinant with rows (2,0,1), (1,3,2), (1,1,1) and state whether the matrix is invertible. [3+2]
16. If one root of 2x^2-5x+k=0 is the reciprocal of the other, find k and the roots. [2+3]
17. Solve \sin2x=\cos x for 0\le x\le 2\pi. [5]
18. Find the distance of the point (-2,1) from the line 5x-12y+13=0, and the distance between the parallel lines 5x-12y+13=0 and 5x-12y-26=0. [2+3]
19. If P(A)=\frac{1}{3}, P(B)=\frac{1}{4} and P(A\cap B)=\frac{1}{12}, show that A and B are independent and find P(A\cup B) and P(\text{neither}). [2+2+1]

Group C — Long answer (3 × 8 = 24)

20. (i) If x=a\cos\theta and y=b\sin\theta, find \frac{dy}{dx}. (ii) Find the point on y=x^2-4x+5 at which the tangent is parallel to the x-axis. (iii) Find the slope of the normal to y=x^2 at x=3. [3+3+2]
21. (i) Evaluate \int x\ln x\,dx. (ii) Evaluate \int_0^{\pi/2}\sin2x\,dx. (iii) Find the area under y=3x^2 from x=0 to x=2. [3+3+2]
22. (i) Prove that the point dividing the join of A(\mathbf{a}) and B(\mathbf{b}) internally in the ratio m:n has position vector \frac{n\mathbf{a}+m\mathbf{b}}{m+n}. (ii) A bag holds 4 red and 6 black balls; two are drawn one after the other without replacement. Find the probability that one is red and one is black. (iii) A body falls freely from rest (g=10\ \text{m s}^{-2}); find the distance it falls in 2 s. [3+3+2]

13.3.1 Solutions to Model Set 2

Group A — answer key with reasons

1. (b) — De Morgan: \sim(p\vee q)\equiv\sim p\wedge\sim q.

2. (b) — solve 2x-5=3: x=4.

3. (a) — S_{10}=\frac{10}{2}[2(2)+9(3)]=5(31)=155.

4. (b) — 2\times4-(-1)(3)=11; (d) forgets the sign.

5. (b) — -\frac{b}{a}=\frac{12}{3}=4.

6. (b) — (1+i)^2=2i, whose modulus is 2. Alternatively |1+i|^2=(\sqrt{2})^2=2.

7. (b) — tangent has period \pi.

8. (a) — \sqrt{1+4+4}=3.

9. (c) — 0.6+0.3-0.18=0.72.

10. (b) — \frac{e^{3x}-1}{3x}\cdot3\to3.

11. (b) — x_1=4-\frac{16-9}{8}=4-0.875=3.125. Or: (b) — s=\frac{1}{2}at^2=\frac{1}{2}(4)(9)=18 m.

Group B — full solutions

12. B\cup C=\{2,3,4,6,7\}, so A-(B\cup C)=\{1,5\}. A-B=\{1,3,5\} and A-C=\{1,2,5\}, whose intersection is \{1,5\}. Both sides equal \{1,5\}.

13. Let y=\frac{3x-1}{2x+5}. Then 2xy+5y=3x-1\Rightarrow x(2y-3)=-1-5y\Rightarrow x=\frac{5y+1}{3-2y}. So f^{-1}(x)=\frac{5x+1}{3-2x}, defined for x\ne\frac{3}{2}.

14. \frac{a}{1-r}=15 and \frac{a^2}{1-r^2}=45. Dividing the second by the first: \frac{a}{1+r}=3, so a=3(1+r). From the first, a=15(1-r). Equate: 3+3r=15-15r\Rightarrow18r=12\Rightarrow r=\frac{2}{3}, and a=15\times\frac{1}{3}=5. Check: \frac{5}{1/3}=15 ✓; \frac{25}{1-\frac{4}{9}}=45 ✓.

15. Expand along row 1: 2(3\cdot1-2\cdot1)-0+1(1\cdot1-3\cdot1)=2(1)+1(-2)=0. Since the determinant is 0, the matrix is singular and has no inverse.

16. Product of roots =\frac{k}{2}=1\Rightarrow k=2. The equation is 2x^2-5x+2=0, so (2x-1)(x-2)=0 and the roots are 2 and \frac{1}{2} (reciprocals ✓).

17. 2\sin x\cos x-\cos x=0\Rightarrow\cos x(2\sin x-1)=0. \cos x=0\Rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}. \sin x=\frac{1}{2}\Rightarrow x=\frac{\pi}{6},\frac{5\pi}{6}. Solutions: \frac{\pi}{6},\ \frac{\pi}{2},\ \frac{5\pi}{6},\ \frac{3\pi}{2}.

18. Distance of (-2,1): \frac{|5(-2)-12(1)+13|}{\sqrt{25+144}}=\frac{|-9|}{13}=\frac{9}{13}. Distance between the parallel lines: \frac{|13-(-26)|}{13}=\frac{39}{13}=3.

19. P(A)P(B)=\frac{1}{3}\cdot\frac{1}{4}=\frac{1}{12}=P(A\cap B), so A and B are independent. P(A\cup B)=\frac{1}{3}+\frac{1}{4}-\frac{1}{12}=\frac{4+3-1}{12}=\frac{1}{2}. P(\text{neither})=1-\frac{1}{2}=\frac{1}{2}.

Group C — full solutions

20. (i) \frac{dx}{d\theta}=-a\sin\theta, \frac{dy}{d\theta}=b\cos\theta. \frac{dy}{dx}=-\frac{b\cos\theta}{a\sin\theta}=-\frac{b}{a}\cot\theta. (ii) \frac{dy}{dx}=2x-4. A horizontal tangent has slope 0: x=2, y=4-8+5=1. The point is (2,1). (iii) Slope of the tangent at x=3 is 2x=6. The normal is perpendicular, so its slope is -\frac{1}{6}.

21. (i) By parts with u=\ln x, dv=x\,dx: \int x\ln x\,dx=\frac{x^2}{2}\ln x-\int\frac{x^2}{2}\cdot\frac{1}{x}\,dx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C. (ii) \int_0^{\pi/2}\sin2x\,dx=\left[-\frac{\cos2x}{2}\right]_0^{\pi/2}=-\frac{\cos\pi}{2}+\frac{\cos0}{2}=\frac{1}{2}+\frac{1}{2}=1. (iii) Area =\int_0^2 3x^2\,dx=[x^3]_0^2=8 square units.

22. (i) Let P divide AB internally in the ratio m:n, so AP:PB=m:n and \overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}. With O as origin, \mathbf{p}-\mathbf{a}=\frac{m}{m+n}(\mathbf{b}-\mathbf{a}). Hence \mathbf{p}=\mathbf{a}+\frac{m(\mathbf{b}-\mathbf{a})}{m+n}=\frac{(m+n)\mathbf{a}+m\mathbf{b}-m\mathbf{a}}{m+n}=\frac{n\mathbf{a}+m\mathbf{b}}{m+n}. (ii) Red then black: \frac{4}{10}\cdot\frac{6}{9}=\frac{24}{90}. Black then red: \frac{6}{10}\cdot\frac{4}{9}=\frac{24}{90}. P=\frac{48}{90}=\frac{8}{15}. (iii) s=\frac{1}{2}gt^2=\frac{1}{2}\times10\times4=20 m.

13.4 Predicted Questions

The list below ranks the question forms most worth practising. Rank comes from the chapter's hours (13.1.2) and from how standard the question form is. Tier A forms are in chapters that are both heavy and rich in routine five-mark and eight-mark questions. Tier C forms are plausible but less certain. None is a promise.

13.4.1 Tier A — practise first

Rank	Topic and likely form	Marks	Why it is ranked here
1	Maxima and minima, increasing or decreasing intervals, point of inflection of a cubic	8	Derivatives is the largest chapter (22 hours) and this is its standard long question.
2	Integration: by parts, by substitution, area between curves	8	Anti-derivatives has 16 hours, and "evaluate, evaluate, find the area" fits the 3+3+2 split.
3	Limits: factorise, rationalise, standard limits; continuity with a constant k	5	Limits and continuity has 10 hours; both the form and the method are routine.
4	Differentiation: first principle, quotient rule, implicit, parametric, tangent	5	Many separate five-mark forms come from the same 22-hour chapter.
5	Matrices: inverse of a 2\times2 matrix, Cramer's rule, system by the matrix method	5 or 8	Matrices (6 hours) lends itself to a computed answer that is easy to check.
6	Quadratic equation: relation of roots and coefficients, forming a new equation, common root	5	A very standard form in a short chapter, and algebra is a heavy area.

13.4.2 Tier B — likely, but rotate

Rank	Topic and likely form	Marks	Why it is ranked here
7	Sequences: A.P. and G.P. problems, three numbers in G.P., A.M./G.M./H.M.	5	Six hours, and word problems give a clean unique answer.
8	Complex numbers: square root, modulus and conjugate properties	5	Four hours, but the forms are fixed and easy to set.
9	Inverse circular functions: proofs by the addition formula	5	Six hours; proofs such as \tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}=\frac{\pi}{4} are standard.
10	Trigonometric equations: factorise, general solution	5	Six hours, and a general solution is a clear-cut answer.
11	Pair of straight lines: angle, bisectors, condition for a pair	5	Seven hours, and each result is a formula that can be tested directly.
12	Vectors: collinearity, linear combination, midpoint and medians	5 or 8	Vectors has 12 hours, and proofs fit a long question.
13	Probability: addition law, independence, drawing without replacement	5	Six hours, and the answer is a fraction that can be checked.

13.4.3 Tier C — worth a quick pass

Rank	Topic and likely form	Marks	Why it is ranked here
14	Dispersion: standard deviation, coefficient of variation, Karl Pearson's skewness	5	Six hours, and grouped-data calculation is mechanical.
15	Computational methods (bisection, Newton–Raphson, trapezoidal and Simpson's rules) or mechanics	5	Twelve hours but a choice chapter; the "Or" keeps its weight modest.
16	Logic: truth tables, equivalences; sets: De Morgan and Venn problems	5	Eight hours, but only a few question forms.
17	Straight line: perpendicular distance, bisectors, area of a triangle	5	Seven hours, with a simple standard formula.
18	Coordinates in space: distance, direction cosines	5	Six hours, and it is easy to set but less often a full question.
19	Functions: inverse and composite, domain and range	5	Seven hours; usually appears as a short part of a question.
20	Real numbers and curve sketching	5	Four and five hours: the smallest weights in the syllabus.

Chapter summary

• The paper is 11\times1+8\times5+3\times8=75; calculus and algebra together carry nearly three-fifths of the hours.
• Plan the three hours: Group A 15 minutes, Group B 85, Group C 65, check 15.
• Calculus long questions follow a fixed pattern: find f', set it to zero, test with f''; evaluate integrals by parts or substitution; area =\int(\text{upper}-\text{lower})\,dx.
• Memorise the quick tools: \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta; \tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}; d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}; mode =3\,\text{median}-2\,\text{mean}.
• Section formula: \frac{n\mathbf{a}+m\mathbf{b}}{m+n}; Newton–Raphson x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.
• A prediction is a guide to what to practise first, never a promise of what will be set.