Mathematics: Unit-wise Objective Question Bank

Group A is 11 marks out of 75 — one seventh of the whole Mathematics paper — and it is decided in about fifteen minutes. It is also the cheapest block of marks on the paper, because every item is a single formula, a single condition or a single one-line calculation, and the same handful of formulae come back year after year. Note one change that is verified on the 2082 paper itself: the eleven MCQs now arrive as a separate booklet, issued 30 minutes after the examination starts, and you write their answers on the same answer sheet. So you will begin with Group B, and Group A will interrupt you. Plan for that.

This chapter is 216 multiple-choice questions, one section per syllabus unit, in syllabus order. Every question carrying a year tag — (NEB 2079), (NEB 2080), (NEB 2081), (NEB 2082), (CDC model 2080) — is a real item from that paper, transcribed from the scan. Untagged questions were written for this book in the same style. Every numerical answer in every key was recomputed with a computer algebra system before printing.

11.1 Permutation and Combination

Of the two Algebra MCQs, one has come from this unit in 2081 and again in 2082. Both times it was pure recall: a counting principle or the relation between P and C. Nothing here needs more than one line of arithmetic.

Result	Formula / condition
Basic counting principle	m ways then n ways gives m \times n ways
Arrangements, all different	P(n,r) = \frac{n!}{(n-r)!}
Selections, all different	C(n,r) = \frac{n!}{r!(n-r)!}
Link between them	C(n,r) \times r! = P(n,r)
Arrangements, not all different	\frac{n!}{p!\,q!\,r!} for repeats p,q,r
Repetition allowed, r places	n^{r}
Circular arrangement of n	(n-1)!
Garland / necklace of n	\frac{(n-1)!}{2} (clockwise = anticlockwise)
Symmetry of combination	C(n,r) = C(n, n-r)
Pascal's rule	C(n,r) + C(n,r-1) = C(n+1,r)
Total subsets	\sum_{r=0}^{n} C(n,r) = 2^{n}
At least one chosen	2^{n} - 1
Diagonals of an n-gon	C(n,2) - n
Boundary values	C(n,0) = C(n,n) = 1, P(n,0)=1

Multiple choice questions

1. Which one is the relation between permutation and combination of n things taken r things at a time? (NEB 2082) [1]aC(n,r) = r!\times P(n,r)bC(n,r)\times r! = P(n,r)cC(n,r)\times r = P(n,r)dC(n,r) = P(n,r)\times r
2. How many numbers are there in between 999 and 10{,}000? (NEB 2081) [1]a9\times 10^{3}b8\times 10^{3}c9^{4}d8^{4}
3. The value of P(5,3) is [1]a10b20c60d120
4. The value of C(10,8) is [1]a45b80c90d360
5. If C(n,2) = 15 then n equals [1]a5b6c7d15
6. Five people are to be seated around a round table. The number of arrangements is [1]a5b20c24d120
7. The number of distinct arrangements of the letters of the word COMMITTEE is [1]a362880b181440c90720d45360
8. The number of diagonals of a regular decagon is [1]a10b35c45d90
9. If C(n,8) = C(n,12) then n equals [1]a4b12c20d96
10. How many three-digit numbers, with no digit repeated, can be formed from the digits 2,3,4,5,6,7? [1]a18b20c120d216
11. A cricket team of 11 players is to be chosen from 15. The number of ways is [1]a165b455c1365d32760
12. The number of different garlands that can be made from 8 distinct flowers is [1]a5040b2520c40320d20160

Answer key — 11.1

1. (b) — C(n,r) counts selections; multiplying by r! orders each selection, which is exactly P(n,r).

2. (a) — the numbers strictly between are 1000 to 9999, i.e. all four-digit numbers: 9999-1000+1 = 9000 = 9\times10^{3}.

3. (c) — P(5,3) = 5\times4\times3 = 60.

4. (a) — C(10,8) = C(10,2) = \frac{10\times9}{2} = 45.

5. (b) — \frac{n(n-1)}{2}=15 \Rightarrow n^{2}-n-30=0 \Rightarrow n=6 (the root -5 is rejected).

6. (c) — circular arrangement of n objects is (n-1)! = 4! = 24.

7. (d) — 9 letters with M, T, E each twice: \frac{9!}{2!\,2!\,2!} = \frac{362880}{8} = 45360.

8. (b) — C(10,2) - 10 = 45-10 = 35; the 10 subtracted are the sides.

9. (c) — C(n,r)=C(n,s) with r \ne s forces r+s=n, so n = 8+12 = 20.

10. (c) — 6\times5\times4 = 120, i.e. P(6,3).

11. (c) — C(15,11) = C(15,4) = \frac{15\times14\times13\times12}{24} = 1365.

12. (b) — a garland can be flipped, so \frac{(8-1)!}{2} = \frac{5040}{2} = 2520.

11.2 Binomial Theorem

Worth reading carefully, because of a fact the four verified papers make plain: Binomial Theorem has never appeared as an MCQ in 2079, 2080, 2081 or 2082. It appears every year, but in Group B (2079 Q12, 2080 Q13b, 2081 Q12, 2082 Q12a) and Group C (2081 Q20a, 2082 Q20a). Learn the facts below for Group B; do not spend Group A revision time here beyond one pass.

Result	Formula / condition
Expansion, positive integral index	(a+b)^{n} = \sum_{r=0}^{n} C(n,r)\,a^{n-r}b^{r}
General term	T_{r+1} = C(n,r)\,a^{n-r}b^{r}
Number of terms	n+1
Middle term, n even	one term, the \left(\frac{n}{2}+1\right)th
Middle terms, n odd	two, the \frac{n+1}{2}th and \frac{n+3}{2}th
Sum of all coefficients	(1+x)^{n} at x=1 gives 2^{n}
Any index	(1+x)^{n} valid only for |x|<1, infinitely many terms
Euler's number	e = 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots, 2<e<3
Exponential series	e^{x} = 1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots
General exponential	a^{x} = 1 + x\log a + \frac{(x\log a)^{2}}{2!}+\cdots
Logarithmic series	\log_e(1+x) = x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots
Its validity	-1 < x \leq 1
Other logarithmic series	\log_e(1-x) = -x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\cdots, |x|<1
Approximation	for small x, (1+x)^{n} \approx 1+nx

Multiple choice questions

1. The number of terms in the expansion of (2x+3y)^{12} is [1]a11b12c13d24
2. In the expansion of (a+b)^{10}, the middle term is the [1]a5thb6thc7thd5th and 6th
3. The middle term of (2+3x)^{30} is the [1]a15thb16thc17thd15th and 16th
4. The coefficient of x^{5} in (1+x)^{10} is [1]a120b210c252d504
5. The sum of the binomial coefficients in the expansion of (1+x)^{n} is [1]anbn!c2^{n}d2^{n}-1
6. In (1+x)^{11} the coefficients of x^{r} and x^{r+1} are equal. Then r is [1]a4b5c6d11
7. The expansion of \log_e(1+x) is valid for [1]aall xbx>0c-1 < x \leq 1dx \geq 1
8. Euler's number e is the sum of the series [1]a\sum \frac{1}{n}b\sum \frac{1}{n!}c\sum \frac{1}{n^{2}}d\sum \frac{(-1)^{n}}{n!}
9. The coefficient of x^{3} in the expansion of e^{2x} is [1]a\frac{1}{6}b\frac{2}{3}c\frac{4}{3}d8
10. Using the binomial expansion, (1.01)^{5} correct to three decimal places is [1]a1.005b1.050c1.051d1.105
11. The term independent of x in the expansion of \left(x+\frac{1}{x}\right)^{6} is [1]a15b20c30d64
12. The number of terms in the expansion of (1+x)^{-3}, |x|<1, is [1]a2b3c4dinfinitely many

Answer key — 11.2

1. (c) — a positive integral index n gives n+1 terms, so 12+1=13.

2. (b) — n=10 is even, so there is one middle term, the \left(\frac{10}{2}+1\right)th =6th.

3. (b) — n=30 is even: the \left(\frac{30}{2}+1\right)th =16th term.

4. (c) — C(10,5) = 252.

5. (c) — put x=1: \sum C(n,r) = (1+1)^{n} = 2^{n}.

6. (b) — C(11,r) = C(11,r+1) needs r+(r+1) = 11, so r=5.

7. (c) — the series diverges at x=-1 and for |x|>1; it converges at x=+1 (giving \log 2).

8. (b) — e = \sum_{n=0}^{\infty}\frac{1}{n!} = 1+1+\frac{1}{2}+\frac{1}{6}+\cdots

9. (c) — e^{2x} = \sum \frac{(2x)^{n}}{n!}, so the x^{3} coefficient is \frac{2^{3}}{3!} = \frac{8}{6} = \frac{4}{3}.

10. (c) — (1+0.01)^{5} = 1+5(0.01)+10(0.0001)+\cdots = 1.05101\ldots \approx 1.051.

11. (b) — T_{r+1} = C(6,r)x^{6-2r}; 6-2r=0 gives r=3 and C(6,3)=20.

12. (d) — for a negative (or fractional) index the expansion is an infinite series; it only terminates for a positive integral index.

11.3 Complex Numbers

This unit has supplied an Algebra MCQ twice in four years: 2079 asked for a property of the cube roots of unity, 2081 for the polar form of a product. Both are one-line recall. De Moivre's theorem itself has appeared every year, but in Group B or C.

Result	Formula / condition
Polar (modulus–amplitude) form	z = r(\cos\theta + i\sin\theta)
Modulus	r = |z| = \sqrt{x^{2}+y^{2}}
Argument	\theta = \tan^{-1}\frac{y}{x}, placed in the correct quadrant
Product in polar form	z_1z_2 = r_1r_2\left[\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)\right]
Quotient in polar form	moduli divide, arguments subtract
De Moivre's theorem	\left[r(\cos\theta+i\sin\theta)\right]^{n} = r^{n}(\cos n\theta + i \sin n\theta)
Euler's formula	e^{i\theta} = \cos\theta + i\sin\theta, so e^{i\pi} = -1
Cube roots of unity	1,\ \omega,\ \omega^{2} with \omega = \frac{-1+i\sqrt{3}}{2}
Their sum	1+\omega+\omega^{2} = 0
Their product	1\cdot\omega\cdot\omega^{2} = \omega^{3} = 1
Useful consequence	\omega+\omega^{2} = -1 and (1+\omega)(1+\omega^{2}) = 1
nth roots of unity	n distinct values, and they sum to 0
Powers of i	i^{4k}=1, i^{4k+1}=i, i^{4k+2}=-1, i^{4k+3}=-i
Conjugate	z\bar{z} = |z|^{2}

Multiple choice questions

1. The property of cube roots of unity where \omega is the cube root of unity is (NEB 2079) [1]a1+\omega+\omega^{2}=0b\omega^{3}+1=0c\omega = -\omega^{2}d\omega+\omega^{2}=1
2. It is given that (r_1,\theta_1) and (r_2,\theta_2) are polar co-ordinates of z_1 and z_2. What are the polar co-ordinates of the image of z_1z_2? (NEB 2081) [1]a(r_1r_2,\ \theta_1-\theta_2)b(r_1r_2,\ \theta_1+\theta_2)c(r_1+r_2,\ \theta_1+\theta_2)d(r_1-r_2,\ \theta_1+\theta_2)
3. The modulus of 1+i\sqrt{3} is [1]a1b\sqrt{2}c2d4
4. The argument of -1+i is [1]a\frac{\pi}{4}b\frac{3\pi}{4}c-\frac{\pi}{4}d\frac{5\pi}{4}
5. If \omega is a complex cube root of unity, then \omega^{10}+\omega^{11} equals [1]a0b1c-1d\omega
6. If \omega is a complex cube root of unity, then (1+\omega)(1+\omega^{2}) equals [1]a-1b0c1d2
7. By De Moivre's theorem, (\cos 20^{\circ}+i\sin 20^{\circ})^{9} equals [1]a1b-1cid-i
8. The value of e^{i\pi/2} is [1]a1b-1cid-i
9. The number of distinct fifth roots of unity is [1]a1b2c5dinfinitely many
10. The polar form of 1+i is [1]a\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)b2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)c\sqrt{2}\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)d\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}
11. The value of i^{37} is [1]a1b-1cid-i
12. The sum of all the fifth roots of unity is [1]a0b1c5d-1

Answer key — 11.3

1. (a) — \omega satisfies \omega^{3}=1, i.e. (\omega-1)(\omega^{2}+\omega+1)=0; since \omega \ne 1, 1+\omega+\omega^{2}=0.

2. (b) — multiplying complex numbers multiplies the moduli and adds the arguments.

3. (c) — \sqrt{1^{2}+(\sqrt{3})^{2}} = \sqrt{4} = 2.

4. (b) — the point (-1,1) is in the second quadrant, so \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.

5. (c) — \omega^{10}=\omega^{9}\cdot\omega=\omega and \omega^{11}=\omega^{2}, so the sum is \omega+\omega^{2}=-1.

6. (c) — expand: 1+\omega+\omega^{2}+\omega^{3} = 0 + 1 = 1.

7. (b) — \cos 180^{\circ}+i\sin 180^{\circ} = -1; machine-checked.

8. (c) — \cos\frac{\pi}{2}+i\sin\frac{\pi}{2} = 0+i = i.

9. (c) — the equation z^{5}=1 has exactly 5 distinct roots.

10. (a) — r=\sqrt{2} and \theta = \tan^{-1}1 = \frac{\pi}{4} (first quadrant).

11. (c) — 37 = 4(9)+1, so i^{37}=i^{1}=i.

12. (a) — the roots of z^{5}-1=0 sum to minus the coefficient of z^{4}, which is 0.

11.4 Sequence and Series

This unit is examined as a single-line recall MCQ or, far more often, as mathematical induction in Group B/C — induction appears in every verified year (2079 Q13a, 2080 Q20b, 2081 Q20c, 2082 Q20b, CDC model Q20c). Memorise the three sum formulae exactly; every mark here hangs on them.

Result	Formula
Sum of first n natural numbers	\sum n = \frac{n(n+1)}{2}
Sum of their squares	\sum n^{2} = \frac{n(n+1)(2n+1)}{6}
Sum of their cubes	\sum n^{3} = \left[\frac{n(n+1)}{2}\right]^{2}
A useful identity	\sum n^{3} = \left(\sum n\right)^{2}
Sum of first n odd numbers	1+3+5+\cdots+(2n-1) = n^{2}
Sum of first n even numbers	2+4+6+\cdots+2n = n(n+1)
Induction, step 1	verify the statement for n=1 (the base case)
Induction, step 2	assume it true for n=k (the inductive hypothesis)
Induction, step 3	prove it for n=k+1 using that assumption
Induction, conclusion	true for all n \in \mathbb{N}

Multiple choice questions

1. The sum of the first 5 odd natural numbers is (CDC model 2080) [1]a15b20c25d30
2. The value of 1^{2}+2^{2}+3^{2}+\cdots+20^{2} is [1]a2470b2870c3080d4410
3. The value of 1^{3}+2^{3}+3^{3}+\cdots+10^{3} is [1]a385b1000c3025d3850
4. The value of 1+2+3+\cdots+100 is [1]a4950b5000c5050d10100
5. The sum of the first n even natural numbers is [1]an^{2}bn(n+1)c\frac{n(n+1)}{2}d2n
6. 1^{2}+2^{2}+\cdots+n^{2} equals [1]a\frac{n(n+1)}{2}b\frac{n(n+1)(2n+1)}{6}c\left[\frac{n(n+1)}{2}\right]^{2}d\frac{n(n+1)(n+2)}{6}
7. Which statement is true for every positive integer n? [1]a\sum n^{2} = \left(\sum n\right)^{2}b\sum n^{3} = \left(\sum n\right)^{2}c\sum n^{3} = \left(\sum n^{2}\right)^{2}d\sum n = \left(\sum n^{2}\right)^{2}
8. In a proof by mathematical induction, the first thing you must do is [1]aassume the result for n=kbverify the result for the smallest ncprove the result for n=k+1ddifferentiate both sides
9. The value of 1^{2}+2^{2}+\cdots+6^{2} is [1]a36b49c91d441
10. The sum of the cubes of the first 5 natural numbers is [1]a125b225c325d625

Answer key — 11.4

1. (c) — 1+3+5+7+9=25, which is 5^{2}: the sum of the first n odd numbers is n^{2}.

2. (b) — \frac{20\times21\times41}{6} = 2870.

3. (c) — \left[\frac{10\times11}{2}\right]^{2} = 55^{2} = 3025.

4. (c) — \frac{100\times101}{2} = 5050.

5. (b) — 2(1+2+\cdots+n) = 2\cdot\frac{n(n+1)}{2} = n(n+1).

6. (b) — the standard result; (c) is the sum of cubes.

7. (b) — the sum of cubes is the square of the sum of the first powers, e.g. 1+8+27=36=6^{2}.

8. (b) — the base case. Without it the inductive step proves nothing.

9. (c) — \frac{6\times7\times13}{6} = 91.

10. (b) — \left[\frac{5\times6}{2}\right]^{2} = 15^{2} = 225.

11.5 Matrix Based System of Linear Equations

The 2082 paper put Cramer's rule straight into Group A (MCQ 2), and Group C has carried a full matrix-method system in 2079, 2080 and 2082. Learn the three methods as three different tools for the same job, and learn the consistency conditions — they are what the MCQ actually tests.

Result	Formula / condition
Matrix form of a system	AX = B
Cramer's rule	x=\frac{D_x}{D}, y=\frac{D_y}{D}, z=\frac{D_z}{D}
D_x	replace the x-column of D by the constants
Unique solution	D \ne 0
No solution (inconsistent)	D=0 but some D_x, D_y, D_z \ne 0
Infinitely many solutions	D = 0 and D_x=D_y=D_z=0
Inverse matrix method	X = A^{-1}B, valid only if |A| \ne 0
The inverse itself	A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A
2\times2 inverse rule	swap the diagonal, negate the off-diagonal, divide by |A|
Singular matrix	|A| = 0, so A^{-1} does not exist
Determinant of the adjoint	|\mathrm{adj}\,A| = |A|^{n-1}
Determinant of the inverse	|A^{-1}| = \frac{1}{|A|}
Row-equivalent (Gauss) method	row-operate the augmented matrix to upper triangular form, then back-substitute
Augmented matrix	the coefficient matrix with the constant column attached

For a two-variable system, Cramer's rule is three determinants and one division. Draw them out before you substitute — that alone prevents most sign errors.

So x = \dfrac{D_x}{D} and y = \dfrac{D_y}{D} — written out,

The row-equivalent method works on the augmented matrix instead. One row operation at a time, and you write the operation beside the row you changed.

Multiple choice questions

1. If the system of equations is p_1x+q_1y=r_1 and p_2x+q_2y=r_2, what is the value of y? (NEB 2082) [1]a\frac{r_1q_2-r_2q_1}{p_1q_2-p_2q_1}b\frac{p_1r_2-p_2r_1}{p_1q_2-p_2q_1}c\frac{p_1q_2-p_2q_1}{p_1r_2-p_2r_1}d\frac{p_2r_1-p_1r_2}{p_1q_2-p_2q_1}
2. A system of linear equations has a unique solution if and only if [1]aD=0bD \ne 0cD_x = 0dD=D_x=D_y=0
3. If A has first row (2,\ 3) and second row (3,\ -1), then |A| equals [1]a-11b-7c7d11
4. A square matrix A with |A| = 0 is called [1]asingularbnon-singularcorthogonaldsymmetric
5. The last column of the augmented matrix of the system 3x+2y-1=0, 4x+y=3 is [1]a(1,\ 3)b(-1,\ 3)c(3,\ 4)d(2,\ 1)
6. In Cramer's rule, x equals [1]a\frac{D}{D_x}b\frac{D_x}{D}cD\cdot D_xd\frac{D_y}{D}
7. Solving 2x+3y=7, x-y=1 by Cramer's rule gives x equal to [1]a1b2c-2d\frac{10}{5} and y=2
8. If D=0 and D_x \ne 0, the system has [1]aa unique solutionbexactly two solutionscinfinitely many solutionsdno solution
9. For a 3\times3 matrix A with |A| = 4, the value of |\mathrm{adj}\,A| is [1]a4b8c16d64
10. The inverse of a non-singular matrix A is [1]a|A|\cdot\mathrm{adj}\,Ab\frac{1}{|A|}\mathrm{adj}\,Ac\frac{\mathrm{adj}\,A}{A}d\frac{1}{A}
11. The row-equivalent method solves AX=B by reducing the augmented matrix to [1]athe identity matrixbupper triangular (row echelon) formca diagonal matrix of onesdits transpose
12. If A is a 2\times2 matrix with |A| = 3, then |A^{-1}| equals [1]a3b9c\frac{1}{3}d-3

Answer key — 11.5

1. (b) — y = \frac{D_y}{D}, and D_y replaces the y-column by the constants, giving p_1r_2-p_2r_1 over p_1q_2-p_2q_1. (The paper printed its four options as determinant ratios; they are written out here.)

2. (b) — D \ne 0 means A^{-1} exists, so X=A^{-1}B is the one and only solution.

3. (a) — (2)(-1)-(3)(3) = -2-9 = -11.

4. (a) — zero determinant means the rows are dependent; the matrix is singular and has no inverse.

5. (a) — write both in the form ax+by=c: 3x+2y=1 and 4x+y=3, so the constant column is (1,\ 3).

6. (b) — numerator is the determinant with the x-column replaced; the denominator is always D.

7. (b) — D = -5, D_x = (7)(-1)-(3)(1) = -10, so x = \frac{-10}{-5} = 2 (and y=1).

8. (d) — D=0 kills the unique solution; a non-zero D_x makes the equations contradict each other.

9. (c) — |\mathrm{adj}\,A| = |A|^{n-1} = 4^{2} = 16.

10. (b) — the adjoint divided by the determinant.

11. (b) — Gauss elimination stops at upper triangular form; back-substitution finishes the job.

12. (c) — |A^{-1}| = \frac{1}{|A|} = \frac{1}{3}.

11.6 Properties of a Triangle

The grid's footnote guarantees a Trigonometry MCQ every year. But look at what has actually filled that slot: 2081 and 2082 asked for a half-angle formula (\cos\frac{A}{2} and \sec\frac{B}{2}), while 2079 and 2080 filled it with Grade-11 carry-over — inverse circular functions and general solutions of trigonometric equations. All four are included below, tagged. So revise the half-angle set and keep your Grade-11 inverse-circular and general-solution formulae on the same card.

Result	Formula
Sine law	\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R
Cosine law	\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}
Tangent law	\tan\frac{B-C}{2} = \frac{b-c}{b+c}\cot\frac{A}{2}
Projection law	a = b\cos C + c\cos B
Semi-perimeter	s = \frac{a+b+c}{2}
Half angle, sine	\sin\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}
Half angle, cosine	\cos\frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}
Half angle, tangent	\tan\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}
Area, two sides and included angle	\Delta = \frac{1}{2}ab\sin C
Area, three sides (Heron)	\Delta = \sqrt{s(s-a)(s-b)(s-c)}
Angle sum	A+B+C = 180^{\circ}
Inverse circular identity	\sin^{-1}x = \cos^{-1}\sqrt{1-x^{2}} for 0 \leq x \leq 1
Domain of \sin^{-1}x	[-1,\ 1]; range \left[-\frac{\pi}{2},\frac{\pi}{2}\right]
General solution, \tan\theta = \tan\alpha	\theta = n\pi + \alpha
General solution, \cos\theta = \cos\alpha	\theta = 2n\pi \pm \alpha
General solution, \sin\theta = \sin\alpha	\theta = n\pi + (-1)^{n}\alpha

Multiple choice questions

1. In any triangle ABC, a, b, c are the sides and s is the semi-perimeter. Which one of the following is equal to \cos\frac{A}{2}? (NEB 2081) [1]a\sqrt{\frac{s(s-a)}{bc}}b\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}c\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}d\sqrt{\frac{(s-b)(s-c)}{bc}}
2. Which one of the following is the value of \sec\frac{B}{2}? (NEB 2082) [1]a\sqrt{\frac{ca}{s-b}}b\sqrt{\frac{bc}{s(s-a)}}c\sqrt{\frac{bc}{s(s-c)}}d\sqrt{\frac{ca}{s(s-b)}}
3. \sin^{-1}x is equal to (NEB 2079) [1]a\cos^{-1}\sqrt{x^{2}-1}b\tan^{-1}\sqrt{1-x^{2}}c\mathrm{cosec}^{-1}\sqrt{1-x^{2}}d\cos^{-1}\sqrt{1-x^{2}}
4. If \tan x + \frac{1}{\sqrt{3}} = 0 is a trigonometric equation, the general value of x is (NEB 2079) [1]an\pi \pm \frac{\pi}{6}b2n\pi + \frac{\pi}{6}c2n\pi + \frac{5\pi}{6}dn\pi + \frac{5\pi}{6}
5. What is the domain of y = \sin^{-1}x? (NEB 2080) [1]a(-1,\ 1)b(-\infty,\ \infty)c[0,\ 1]d[-1,\ 1]
6. Which one of the following is the general solution of 2\cos\theta - 1 = 0? (NEB 2080) [1]an\pi + (-1)^{n}\frac{\pi}{3}bn\pi \pm \frac{\pi}{3}cn\pi + \frac{\pi}{3}d2n\pi \pm \frac{\pi}{3}
7. In any triangle ABC, \frac{a}{\sin A} equals [1]aRb2Rc\frac{R}{2}d4R
8. The projection law gives a as [1]ab\cos C + c\cos Bbb\cos B + c\cos Ccb\sin C + c\sin Bdb\cos A + c\cos A
9. The tangent law states that \tan\frac{B-C}{2} equals [1]a\frac{b+c}{b-c}\cot\frac{A}{2}b\frac{b-c}{b+c}\cot\frac{A}{2}c\frac{b-c}{b+c}\tan\frac{A}{2}d\frac{b+c}{b-c}\tan\frac{A}{2}
10. \sin\frac{A}{2} equals [1]a\sqrt{\frac{s(s-a)}{bc}}b\sqrt{\frac{(s-b)(s-c)}{bc}}c\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}d\sqrt{\frac{bc}{s(s-a)}}
11. In a triangle with b=4, c=5 and A=30^{\circ}, the area is [1]a5b10c\frac{5\sqrt{3}}{2}d20
12. In a triangle with a=7, b=8, c=9, the value of \cos A is [1]a\frac{1}{3}b\frac{2}{3}c\frac{11}{21}d\frac{1}{6}

Answer key — 11.6

1. (a) — \cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}; (d) is \sin\frac{A}{2} and (b) is \tan\frac{A}{2}.

2. (d) — \cos\frac{B}{2}=\sqrt{\frac{s(s-b)}{ca}}, so its reciprocal is \sqrt{\frac{ca}{s(s-b)}}.

3. (d) — if \sin\theta = x then \cos\theta = \sqrt{1-x^{2}}, so \theta = \cos^{-1}\sqrt{1-x^{2}} (valid for 0 \leq x \leq 1).

4. (d) — \tan x = -\frac{1}{\sqrt{3}} = \tan\frac{5\pi}{6}, so x = n\pi + \frac{5\pi}{6}. Machine-checked: \tan\frac{5\pi}{6} = -\frac{\sqrt{3}}{3}.

5. (d) — \sin\theta takes every value in [-1,1] including the endpoints, so the inverse is defined on the closed interval.

6. (d) — \cos\theta = \frac{1}{2} = \cos\frac{\pi}{3}, and the general solution of \cos\theta=\cos\alpha is \theta = 2n\pi \pm \alpha.

7. (b) — the sine law's common ratio is the diameter of the circumcircle, 2R.

8. (a) — each side is the sum of the projections of the other two onto it.

9. (b) — the difference of the sides over their sum, times \cot of half the third angle.

10. (b) — \sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}.

11. (a) — \frac{1}{2}bc\sin A = \frac{1}{2}(4)(5)\left(\frac{1}{2}\right) = 5.

12. (b) — \cos A = \frac{8^{2}+9^{2}-7^{2}}{2(8)(9)} = \frac{96}{144} = \frac{2}{3}.

11.7 Solution of Triangles

Only 4 working hours, and no MCQ has yet come from it in 2079, 2080, 2081 or 2082. It reached Group C once, in 2081 Q21a ("if a = 3+\sqrt{3}, b=2, c=\sqrt{3} then solve the triangle"); as recorded, those data give b+c<a, which no triangle can satisfy, so the printed numbers were probably misread or misprinted. Learn the decision rule — which law for which given data — and move on.

Data given	Method to use	Notes
Three sides a,b,c	cosine law, or the half-angle formulae	exactly one triangle if the largest side is less than the sum of the other two; none otherwise
Two sides and the included angle	tangent law for the angles, then sine law	one triangle
Two angles and one side	sine law (find the third angle first)	one triangle
Two sides and a non-included angle	sine law	the ambiguous case: 0, 1 or 2 triangles
Right angle present	Pythagoras and simple ratios	fastest route
a=b	then A=B (isosceles)	equal sides face equal angles
Largest side	faces the largest angle	use it to spot an obtuse triangle
Check	A+B+C must come to 180^{\circ}	always do this

Multiple choice questions

1. To solve a triangle when all three sides are given, the appropriate law is [1]asine lawbcosine lawcprojection lawdtangent law
2. Two sides and the included angle are given. The law that leads directly to the other two angles is [1]asine lawbcosine lawctangent lawdprojection law
3. Two angles and one side are given. The right tool is [1]asine lawbcosine lawctangent lawdHeron's formula
4. In \Delta ABC, if a=b then [1]aA=BbA=CcA+B=90^{\circ}dC=90^{\circ}
5. A triangle has b=3, c=3 and A=60^{\circ}. The triangle is [1]aright-angledbobtuse-angledcequilateraldimpossible
6. In \Delta ABC, A=45^{\circ}, B=105^{\circ} and a=\sqrt{2}. Then c equals [1]a1b\sqrt{2}c\sqrt{3}d2
7. The ambiguous case arises when the data given are [1]athree sidesbtwo sides and the included anglectwo sides and a non-included angledtwo angles and one side
8. If a=5, b=12, c=13, then the largest angle of the triangle is [1]a60^{\circ}b90^{\circ}c120^{\circ}d150^{\circ}

Answer key — 11.7

1. (b) — the cosine law is the only one that gives an angle from three sides alone.

2. (c) — the tangent law gives \frac{B-C}{2} directly, and \frac{B+C}{2}=90^{\circ}-\frac{A}{2} is known, so B and C follow by adding and subtracting.

3. (a) — find the third angle from A+B+C=180^{\circ}, then the sine law gives the remaining sides.

4. (a) — equal sides face equal angles; by the sine law \sin A = \sin B.

5. (c) — b=c makes it isosceles with B=C, and B=C=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ}, so all three angles are 60^{\circ}.

6. (a) — C = 180^{\circ}-45^{\circ}-105^{\circ}=30^{\circ}, and c = \frac{a\sin C}{\sin A} = \frac{\sqrt{2}\times\frac{1}{2}}{\frac{1}{\sqrt{2}}} = 1.

7. (c) — the sine law can give two supplementary angles, so there may be two triangles, one, or none.

8. (b) — 5^{2}+12^{2}=169=13^{2}, so the angle opposite 13 is a right angle. Machine-checked: \cos C = 0.

Distractor analysis — Solution of Triangles

Reaching for the sine law when three sides are given. The sine law needs an angle to start from; with only sides you must use the cosine law (or half-angle formulae). Option (a) in Q1 is the reflex answer and it is a dead end.

Assuming an isosceles triangle cannot be equilateral. In Q5, b=c with A=60^{\circ} forces all angles to 60^{\circ}. Students stop at "isosceles" and pick (a) or (b) without finishing the angle sum.

Missing the ambiguous case. Given a, b and A, \sin B = \frac{b\sin A}{a} has two solutions in (0^{\circ},180^{\circ}) whenever the value is strictly between 0 and 1. Always test whether the obtuse partner still leaves a positive third angle.

Not checking the angle sum. Every solved triangle must satisfy A+B+C=180^{\circ}. It takes five seconds and catches almost every arithmetic slip in a Group C answer.

11.8 Conic Sections

The biggest MCQ payer in the book. Analytic Geometry carries 9 marks and shares 3 MCQs with Trigonometry and Vectors, and a conic item has appeared in all four verified years: 2079 (eccentricity of a hyperbola), 2080 (identify the conic), 2081 (focus of a parabolic reflector), 2082 (tangency condition for a circle). Learn the table below across, not down — the examiner's favourite move is to give you a parabola formula in an ellipse question.

	Circle x^{2}+y^{2}=a^{2}	Parabola y^{2}=4ax	Ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1	Hyperbola \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1
Eccentricity e	0	1	\sqrt{1-\frac{b^{2}}{a^{2}}}, so e<1	\sqrt{1+\frac{b^{2}}{a^{2}}}, so e>1
Focus / foci	centre only	(a,\ 0)	(\pm ae,\ 0)	(\pm ae,\ 0)
Directrix	none	x=-a	x = \pm\frac{a}{e}	x = \pm\frac{a}{e}
Latus rectum	2a (a diameter)	4a	\frac{2b^{2}}{a}	\frac{2b^{2}}{a}
b^{2} relation	—	—	b^{2}=a^{2}(1-e^{2})	b^{2}=a^{2}(e^{2}-1)
Tangency of y=mx+c	c^{2}=a^{2}(1+m^{2})	c=\frac{a}{m}	c^{2}=a^{2}m^{2}+b^{2}	c^{2}=a^{2}m^{2}-b^{2}
Tangent at (x_1,y_1)	xx_1+yy_1=a^{2}	yy_1 = 2a(x+x_1)	\frac{xx_1}{a^{2}}+\frac{yy_1}{b^{2}}=1	\frac{xx_1}{a^{2}}-\frac{yy_1}{b^{2}}=1
General centre form	x^{2}+y^{2}+2gx+2fy+c=0, centre (-g,-f), radius \sqrt{g^{2}+f^{2}-c}	axis is the x-axis	major axis 2a, minor 2b	transverse 2a, conjugate 2b
Length of tangent from (x_1,y_1)	\sqrt{x_1^{2}+y_1^{2}-a^{2}}	—	—	—

Multiple choice questions

1. In which condition will the line y=mx+c be tangent to the circle x^{2}+y^{2}=a^{2}? (NEB 2082) [1]aa = c\sqrt{1+m^{2}}ba = \pm c\sqrt{1+m^{2}}cc = \pm a\sqrt{1+m^{2}}dc = a\sqrt{1+m^{2}}
2. A parabolic reflector is 20 cm in diameter and 5 cm deep. Which one of the following gives the co-ordinates of its focus? (NEB 2081) [1]a(0,\ 5)b(0,\ -5)c(5,\ 0)d(-5,\ 0)
3. Which one of the following conics is represented by the equation 4x^{2}-16x+y^{2}=48? (NEB 2080) [1]ahyperbolabellipsecparaboladcircle
4. The eccentricity of the hyperbola \frac{x^{2}}{16}-\frac{y^{2}}{64}=1 is (NEB 2079) [1]a\frac{3}{5}b\frac{5}{3}c\frac{3}{4}d\frac{4}{3}
5. The eccentricity of a parabola is [1]a0b\frac{1}{2}c1dgreater than 1
6. The length of the latus rectum of y^{2}=12x is [1]a3b6c12d24
7. The focus of the parabola y^{2}=8x is [1]a(0,\ 2)b(2,\ 0)c(4,\ 0)d(8,\ 0)
8. The directrix of the parabola y^{2}=4x is [1]ax=1bx=-1cy=1dy=-1
9. The eccentricity of the ellipse \frac{x^{2}}{25}+\frac{y^{2}}{16}=1 is [1]a\frac{3}{5}b\frac{4}{5}c\frac{5}{4}d\frac{5}{3}
10. The length of the latus rectum of \frac{x^{2}}{25}+\frac{y^{2}}{16}=1 is [1]a\frac{8}{5}b\frac{16}{5}c\frac{32}{5}d\frac{50}{4}
11. The eccentricity of \frac{x^{2}}{9}-\frac{y^{2}}{16}=1 is [1]a\frac{3}{5}b\frac{4}{5}c\frac{5}{4}d\frac{5}{3}
12. The length of the tangent drawn from (5,\ 4) to the circle x^{2}+y^{2}=16 is [1]a3b4c5d\sqrt{41}
13. The equation of the tangent to x^{2}+y^{2}=25 at the point (3,\ 4) is [1]a3x+4y=25b4x+3y=25c3x-4y=25d3x+4y=5
14. The line y=mx+c touches the parabola y^{2}=4ax if [1]ac=ambc=\frac{a}{m}cc^{2}=a^{2}(1+m^{2})dc^{2}=a^{2}m^{2}+b^{2}
15. The centre of the circle x^{2}+y^{2}-4x+6y-12=0 is [1]a(2,\ -3)b(-2,\ 3)c(4,\ -6)d(-4,\ 6)
16. The radius of the circle x^{2}+y^{2}-4x+6y-12=0 is [1]a3b4c5d12
17. A normal drawn to a circle at any point on it always passes through [1]athe originbthe centrecthe focusdthe directrix
18. The eccentricity of the rectangular hyperbola x^{2}-y^{2}=a^{2} is [1]a1b\sqrt{2}c2d\frac{1}{\sqrt{2}}

Answer key — 11.8

1. (c) — the perpendicular distance from the centre to the line must equal the radius: \frac{|c|}{\sqrt{1+m^{2}}}=a, so c = \pm a\sqrt{1+m^{2}}, i.e. c^{2}=a^{2}(1+m^{2}).

2. (c) — put the vertex at the origin, y^{2}=4ax. The rim is 10 cm from the axis at a depth of 5 cm, so (5,10) lies on it: 100 = 20a, a=5, focus (5,0).

3. (b) — complete the square: 4(x-2)^{2}+y^{2}=64, i.e. \frac{(x-2)^{2}}{16}+\frac{y^{2}}{64}=1 — an ellipse (both squares positive, unequal denominators).

4. No printed option is correct. With the numbers as printed, a^{2}=16, b^{2}=64 and e=\sqrt{1+\frac{64}{16}}=\sqrt{5}\approx2.236. Option (b) \frac{5}{3} is the eccentricity of \frac{x^{2}}{9}-\frac{y^{2}}{16}=1, which is almost certainly the intended curve. Machine-checked both.

5. (c) — a parabola is the locus with SP=PM exactly, so e=1.

6. (c) — 4a=12, and the latus rectum of y^{2}=4ax is 4a, so 12.

7. (b) — 4a=8 gives a=2, so the focus is (2,0).

8. (b) — 4a=4 gives a=1; the directrix of y^{2}=4ax is x=-a=-1.

9. (a) — e=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}.

10. (c) — \frac{2b^{2}}{a}=\frac{2(16)}{5}=\frac{32}{5}.

11. (d) — e=\sqrt{1+\frac{16}{9}}=\sqrt{\frac{25}{9}}=\frac{5}{3}.

12. (c) — \sqrt{x_1^{2}+y_1^{2}-a^{2}} = \sqrt{25+16-16}=5.

13. (a) — the tangent to x^{2}+y^{2}=a^{2} at (x_1,y_1) is xx_1+yy_1=a^{2}, giving 3x+4y=25.

14. (b) — substituting y=mx+c into y^{2}=4ax and setting the discriminant to zero gives c=\frac{a}{m}.

15. (a) — for x^{2}+y^{2}+2gx+2fy+c=0 the centre is (-g,-f); here 2g=-4, 2f=6, so (2,-3).

16. (c) — \sqrt{g^{2}+f^{2}-c} = \sqrt{4+9+12}=\sqrt{25}=5.

17. (b) — the radius is perpendicular to the tangent, and the normal is along the radius.

18. (b) — here b=a, so e=\sqrt{1+1}=\sqrt{2}.

11.9 Product of Vectors

A vector MCQ has appeared in every one of 2079, 2080, 2081 and 2082, and in three of those four years the item turned on the sign of a cross product. That is the single most profitable thing in this section.

Result	Formula / condition
Scalar product	\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta = a_1b_1+a_2b_2+a_3b_3
Commutative	\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}
Perpendicular test	\vec{a}\cdot\vec{b}=0 with neither vector \vec{0}
Angle between vectors	\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}
Projection of \vec{b} on \vec{a}	\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|}
Base-vector dots	\hat{i}\cdot\hat{i}=1, \hat{i}\cdot\hat{j}=0
Vector product	\vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\,\hat{n}
Anti-commutative	\vec{a}\times\vec{b} = -(\vec{b}\times\vec{a})
Base-vector crosses, cyclic	\hat{i}\times\hat{j}=\hat{k}, \hat{j}\times\hat{k}=\hat{i}, \hat{k}\times\hat{i}=\hat{j}
Base-vector crosses, reversed	\hat{j}\times\hat{i}=-\hat{k}, \hat{k}\times\hat{j}=-\hat{i}, \hat{i}\times\hat{k}=-\hat{j}
Self-cross	\vec{a}\times\vec{a} = \vec{0}
Parallelogram from adjacent sides	area = |\vec{a}\times\vec{b}|
Triangle from two sides	area = \frac{1}{2}|\vec{a}\times\vec{b}|
Parallelogram from diagonals	area = \frac{1}{2}|\vec{d_1}\times\vec{d_2}|
Link between the products	|\vec{a}\times\vec{b}|^{2} + (\vec{a}\cdot\vec{b})^{2} = |\vec{a}|^{2}|\vec{b}|^{2}
Perpendicularity of the cross product	(\vec{a}\times\vec{b})\cdot\vec{a} = (\vec{a}\times\vec{b})\cdot\vec{b} = 0

Multiple choice questions

1. What is \vec{a}\times\vec{b} if \vec{a}=(0,0,1) and \vec{b}=(0,1,0)? (NEB 2082) [1]a(-1,0,0)b(1,0,0)c(0,1,0)d(0,0,-1)
2. If \vec{a} and \vec{b} are two vectors then the value of (\vec{a}-\vec{b})\times(\vec{a}+\vec{b}) is (NEB 2079) [1]a2(\vec{a}\times\vec{b})b|\vec{a}|^{2}-|\vec{b}|^{2}c-2(\vec{a}\times\vec{b})d2\left(|\vec{a}|^{2}-|\vec{b}|^{2}\right)
3. Consider \vec{a}\times\vec{b} = \vec{c}\times\vec{d} and \vec{a}\times\vec{c} = \vec{b}\times\vec{d}. Which one of the following is parallel to (\vec{b}-\vec{c})? (NEB 2080) [1]a\vec{a}-\vec{b}b\vec{a}-\vec{c}c\vec{a}-\vec{d}d\vec{c}-\vec{d}
4. \vec{a} and \vec{b} are two non-zero vectors. Which of the following results is the smallest number? (NEB 2081) [1]a(\vec{a}\times\vec{b})\cdot(\vec{b}\times\vec{a})b(\vec{a}\times\vec{a})\cdot\vec{a}c(\vec{b}\times\vec{b})\cdot\vec{b}d(\vec{a}\times\vec{a})\cdot\vec{b}
5. The value of \hat{i}\cdot\hat{j} is [1]a0b1c\hat{k}d-\hat{k}
6. The value of \hat{j}\times\hat{k} is [1]a0b1c\hat{i}d-\hat{i}
7. If \vec{a}\cdot\vec{b}=0 and neither vector is zero, then the vectors are [1]aparallelbperpendicularcequaldanti-parallel
8. Geometrically, |\vec{a}\times\vec{b}| is the [1]aarea of the parallelogram with \vec{a}, \vec{b} as adjacent sidesbarea of the triangle with sides \vec{a}, \vec{b}cvolume of a parallelepipeddperimeter of the parallelogram
9. If \vec{a}=2\hat{i}+3\hat{j}-\hat{k} and \vec{b}=\hat{i}-\hat{j}+2\hat{k}, then \vec{a}\cdot\vec{b} is [1]a-3b1c3d7
10. The angle between \hat{i}+\hat{j} and \hat{i}-\hat{j} is [1]a0^{\circ}b45^{\circ}c60^{\circ}d90^{\circ}
11. If \vec{a}=4\hat{i}-3\hat{j}+2\hat{k} and \vec{b}=3\hat{i}-2\hat{j}+4\hat{k}, the projection of \vec{b} on \vec{a} is [1]a\frac{26}{\sqrt{29}}b\frac{26}{29}c\frac{29}{\sqrt{26}}d26
12. If |\vec{a}|=3, |\vec{b}|=4 and \vec{a}\cdot\vec{b}=6, then |\vec{a}\times\vec{b}| is [1]a6b6\sqrt{3}c12d\sqrt{108}-6
13. The value of (\hat{i}+\hat{j})\times(\hat{i}-\hat{j}) is [1]a2\hat{k}b-2\hat{k}c\vec{0}d2\hat{i}
14. The area of the parallelogram whose diagonals are \vec{a}=\hat{i}-2\hat{k} and \vec{b}=4\hat{i}+3\hat{j}+\hat{k} is [1]a3\sqrt{14}b\frac{3\sqrt{14}}{2}c\sqrt{126}d\frac{\sqrt{14}}{2}

Answer key — 11.9

1. (a) — \vec{a}=\hat{k} and \vec{b}=\hat{j}, and \hat{k}\times\hat{j} = -\hat{i} = (-1,0,0). The cyclic order is \hat{i}\rightarrow\hat{j}\rightarrow\hat{k}\rightarrow\hat{i}; \hat{k}\times\hat{j} runs backwards, so it takes a minus sign.

2. (a) — expand: \vec{a}\times\vec{a}+\vec{a}\times\vec{b}-\vec{b}\times\vec{a}-\vec{b}\times\vec{b} = \vec{0}+\vec{a}\times\vec{b}+\vec{a}\times\vec{b}-\vec{0} = 2(\vec{a}\times\vec{b}). (The paper prints the label "C)" twice; the fourth option is the intended (d).)

3. (c) — compute (\vec{a}-\vec{d})\times(\vec{b}-\vec{c}) = \vec{a}\times\vec{b}-\vec{a}\times\vec{c}-\vec{d}\times\vec{b}+\vec{d}\times\vec{c}. Using both given conditions this collapses to \vec{0}, so the two vectors are parallel. Machine-checked symbolically.

4. (a) — options (b), (c), (d) all contain \vec{a}\times\vec{a} or \vec{b}\times\vec{b}, which is \vec{0}, so each equals 0. Option (a) is (\vec{a}\times\vec{b})\cdot\left(-(\vec{a}\times\vec{b})\right) = -|\vec{a}\times\vec{b}|^{2}, which is negative (or zero) — the smallest.

5. (a) — \hat{i} and \hat{j} are perpendicular, so \cos 90^{\circ}=0.

6. (c) — \hat{j}\times\hat{k}=\hat{i} by the cyclic rule.

7. (b) — \cos\theta = 0 forces \theta = 90^{\circ}.

8. (a) — |\vec{a}||\vec{b}|\sin\theta is base times height for the parallelogram; half of it is the triangle.

9. (a) — 2(1)+3(-1)+(-1)(2) = 2-3-2 = -3.

10. (d) — the dot product is 1(1)+1(-1) = 0.

11. (a) — \vec{a}\cdot\vec{b} = 12+6+8 = 26 and |\vec{a}|=\sqrt{16+9+4}=\sqrt{29}, so the projection is \frac{26}{\sqrt{29}}.

12. (b) — |\vec{a}\times\vec{b}| = \sqrt{|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a}\cdot\vec{b})^{2}} = \sqrt{144-36}=\sqrt{108}=6\sqrt{3}.

13. (b) — \hat{i}\times(-\hat{j})+\hat{j}\times\hat{i} = -\hat{k}-\hat{k} = -2\hat{k}. Machine-checked.

14. (b) — for diagonals the area is \frac{1}{2}|\vec{a}\times\vec{b}|. Here \vec{a}\times\vec{b} = 6\hat{i}-9\hat{j}+3\hat{k}, of magnitude \sqrt{126}=3\sqrt{14}, so the area is \frac{3\sqrt{14}}{2}.

11.10 Correlation and Regression

An important negative finding from the four verified papers: the single Statistics & Probability MCQ has gone to Probability every year, never to correlation. Correlation and regression appear instead as a full 5-mark Group B question built on a small table — 2079 Q16, 2080 Q15, 2081 Q16, 2082 Q16 and the CDC model Q16, without exception. So the questions below are mainly insurance and formula drill; the marks in this unit are in Group B.

Result	Formula / condition
Karl Pearson's coefficient	r = \frac{\sum xy - n\bar{x}\bar{y}}{\sqrt{\left(\sum x^{2}-n\bar{x}^{2}\right)\left(\sum y^{2}-n\bar{y}^{2}\right)}}
Range	-1 \leq r \leq 1
Regression line, y on x	y-\bar{y} = b_{yx}(x-\bar{x})
Regression line, x on y	x-\bar{x} = b_{xy}(y-\bar{y})
Regression coefficients	b_{yx} = r\frac{\sigma_y}{\sigma_x}, b_{xy} = r\frac{\sigma_x}{\sigma_y}
r from the coefficients	r = \pm\sqrt{b_{yx}\cdot b_{xy}}, sign of b_{yx}
Product bound	b_{yx}\cdot b_{xy} = r^{2} \leq 1
Point common to both lines	(\bar{x},\ \bar{y})
r=0	no linear relation; the two lines are perpendicular
r=\pm1	perfect linear relation; the two lines coincide
Invariance	r is unchanged by a change of origin or of scale
Rank correlation	R = 1-\frac{6\sum d^{2}}{n(n^{2}-1)}
Signs	b_{yx}, b_{xy} and r always share one sign
To predict y from x	use the y-on-x line only
To predict x from y	use the x-on-y line only

Multiple choice questions

1. The range of the correlation coefficient r is [1]a0 \leq r \leq 1b-1 \leq r \leq 1c-1 < r < 1d0 < r < 1
2. If b_{yx} = 0.22 and b_{xy} = 0.616, the correlation coefficient is approximately [1]a0.14b0.37c0.42d0.84
3. The two regression lines always intersect at [1]athe originb(\bar{x},\ \bar{y})c(0,\ \bar{y})d(\bar{x},\ 0)
4. If r=0, the two variables are [1]aperfectly correlatedbuncorrelated (linearly)cinversely proportionaldidentical
5. If b_{yx}\cdot b_{xy} = 1, then the two regression lines [1]aare perpendicularbcoincidecare parallel but distinctddo not exist
6. The correlation coefficient is unchanged by [1]aa change of origin onlyba change of scale onlyca change of origin and of scaledneither
7. Spearman's rank correlation coefficient is [1]a1-\frac{6\sum d^{2}}{n(n^{2}-1)}b1-\frac{6\sum d}{n(n^{2}-1)}c\frac{6\sum d^{2}}{n(n^{2}-1)}d1-\frac{\sum d^{2}}{6n(n^{2}-1)}
8. If \sum d^{2}=30 and n=10, the rank correlation coefficient is [1]a0.18b0.62c0.82d0.91
9. The regression coefficient of y on x is [1]ar\frac{\sigma_x}{\sigma_y}br\frac{\sigma_y}{\sigma_x}c\frac{\sigma_y}{r\sigma_x}dr\sigma_x\sigma_y
10. If r=0.8, \sigma_x=3 and \sigma_y=4, then b_{yx} equals [1]a0.6b1.07c9.6d0.94
11. The two regression lines are perpendicular to each other when [1]ar=1br=-1cr=0dr=0.5
12. If the regression line of y on x is y=2x+3 and \bar{x}=1, then \bar{y} is [1]a2b3c5d6

Answer key — 11.10

1. (b) — r is a cosine-like ratio; both endpoints are attainable (perfect positive and perfect negative correlation).

2. (b) — r = \sqrt{0.22\times0.616} = \sqrt{0.13555} = 0.3682. Machine-checked; these are the 2079 Group B numbers.

3. (b) — both lines are written as deviations from the means, so both pass through (\bar{x},\bar{y}).

4. (b) — r measures linear association only; r=0 rules out a straight-line trend, not every relationship.

5. (b) — b_{yx}b_{xy}=r^{2}=1 means r=\pm1, perfect linear correlation, and the two lines become one.

6. (c) — r is a pure number, invariant under x \rightarrow \frac{x-A}{h} and y \rightarrow \frac{y-B}{k} for h,k>0.

7. (a) — the standard formula; note the 6 multiplies \sum d^{2}.

8. (c) — 1-\frac{6(30)}{10(99)} = 1-\frac{180}{990} = 0.818.

9. (b) — b_{yx} scales r by \frac{\sigma_y}{\sigma_x}: the dependent variable's spread goes on top.

10. (b) — 0.8\times\frac{4}{3} = 1.0667 \approx 1.07.

11. (c) — no linear relation makes one line horizontal and the other vertical.

12. (c) — the y-on-x line passes through (\bar{x},\bar{y}), so \bar{y} = 2(1)+3 = 5.

11.11 Probability

The most reliable MCQ slot in the whole paper. Conditional probability P(B/A) was the Statistics & Probability MCQ in 2080, 2081 and 2082, and in the CDC model question. Four for four. Learn one formula properly and this mark is yours.

Result	Formula / condition
Range of a probability	0 \leq P(A) \leq 1
Complement	P(A') = 1-P(A)
Addition rule	P(A \cup B) = P(A)+P(B)-P(A \cap B)
Mutually exclusive	P(A \cap B)=0, so P(A\cup B)=P(A)+P(B)
Conditional probability	P(B/A) = \frac{P(A \cap B)}{P(A)}, needs P(A)>0
The other conditional	P(A/B) = \frac{P(A \cap B)}{P(B)}
Multiplication rule, dependent	P(A \cap B) = P(A)\cdot P(B/A)
Independent events	P(A \cap B) = P(A)\cdot P(B), and then P(B/A)=P(B)
Drawing without replacement	dependent: the second denominator drops by one
Drawing with replacement	independent: the denominator is unchanged
Certain / impossible event	P=1 / P=0
Exhaustive events	probabilities add to 1

Multiple choice questions

1. The range of the probability of any event A is (NEB 2079) [1]a0 \leq P(A) \leq 1b0 \leq P(A) < 1c-1 \leq P(A) \leq 1d0 < P(A) < 1
2. Consider A and B to be two dependent events. If P(A)=\frac{1}{2}, P(B)=\frac{1}{13} and P(A\cap B)=\frac{1}{26}, what is P(B/A)? (NEB 2080) [1]a\frac{1}{52}b\frac{1}{26}c\frac{1}{13}d\frac{1}{2}
3. Let A and B be two dependent events with P(A\cap B)=\frac{1}{18}, P(B)=\frac{1}{18} and 2P(A)=5P(B). What is the value of P(B/A)? (NEB 2081) [1]a\frac{1}{324}b\frac{1}{18}c\frac{5}{36}d\frac{2}{5}
4. If P(A)=\frac{1}{3}, P(B)=\frac{2}{3} and P(A\cap B)=\frac{1}{5}, which one of the following is P(B/A)? (NEB 2082) [1]a\frac{3}{5}b\frac{2}{5}c\frac{2}{15}d\frac{1}{15}
5. The probability that a leap year chosen at random contains 53 Fridays or 53 Saturdays is (CDC model 2080) [1]a\frac{2}{7}b\frac{3}{7}c\frac{4}{7}d\frac{1}{7}
6. If A and B are independent with P(A)=\frac{1}{2} and P(B)=\frac{1}{3}, then P(A\cap B) is [1]a\frac{1}{6}b\frac{5}{6}c\frac{2}{3}d\frac{1}{5}
7. P(A/B) is defined as [1]a\frac{P(A\cap B)}{P(A)}b\frac{P(A\cap B)}{P(B)}c\frac{P(A)}{P(B)}dP(A)\cdot P(B)
8. If A and B are mutually exclusive, then P(A\cap B) is [1]a0b1cP(A)P(B)dP(A)+P(B)
9. Two cards are drawn one after another, without replacement, from a well-shuffled pack of 52. The probability that both are aces is [1]a\frac{1}{169}b\frac{1}{221}c\frac{1}{13}d\frac{4}{663}
10. If P(A)=0.3, then P(A') is [1]a0.3b0.7c1.3d-0.3

Answer key — 11.11

1. (a) — an impossible event has probability exactly 0 and a certain event exactly 1, so the interval is closed.

2. (c) — P(B/A)=\frac{1/26}{1/2}=\frac{2}{26}=\frac{1}{13}.

3. (d) — P(A)=\frac{5}{2}P(B)=\frac{5}{2}\cdot\frac{1}{18}=\frac{5}{36}, so P(B/A)=\frac{1/18}{5/36}=\frac{36}{90}=\frac{2}{5}.

4. (a) — P(B/A)=\frac{1/5}{1/3}=\frac{3}{5}.

5. (b) — a leap year is 52 weeks plus 2 extra consecutive days; there are 7 equally likely such pairs, and three of them (Thu–Fri, Fri–Sat, Sat–Sun) contain a Friday or a Saturday, giving \frac{3}{7}.

6. (a) — independence gives P(A\cap B)=\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}.

7. (b) — divide by the probability of the event that is given, which here is B.

8. (a) — mutually exclusive events cannot both happen.

9. (b) — \frac{4}{52}\times\frac{3}{51} = \frac{12}{2652} = \frac{1}{221}.

10. (b) — 1-0.3=0.7.

11.12 Derivatives

Calculus carries 22 marks and four of the eleven MCQs — more than any other area. Two of those four, on the evidence of all four papers, come from this unit: an inverse-hyperbolic or hyperbolic derivative, and either an L'Hospital limit or a tangent/normal slope. The table below is the single highest-value thing to memorise in this chapter.

Function	Derivative	Valid for
\sinh x	\cosh x	all x
\cosh x	\sinh x	all x
\tanh x	\mathrm{sech}^{2}x	all x
\coth x	-\mathrm{cosech}^{2}x	x \ne 0
\mathrm{sech}\,x	-\mathrm{sech}\,x\tanh x	all x
\mathrm{cosech}\,x	-\mathrm{cosech}\,x\coth x	x \ne 0
\sinh^{-1}x	\frac{1}{\sqrt{x^{2}+1}}	all x
\cosh^{-1}x	\frac{1}{\sqrt{x^{2}-1}}	x>1
\tanh^{-1}x	\frac{1}{1-x^{2}}	|x|<1
\coth^{-1}x	\frac{1}{1-x^{2}}	|x|>1
\mathrm{sech}^{-1}x	\frac{-1}{x\sqrt{1-x^{2}}}	0<x<1
\mathrm{cosech}^{-1}x	\frac{-1}{|x|\sqrt{1+x^{2}}}	x \ne 0

The rest of the unit's MCQ surface is limits and tangents:

Result	Statement	Condition
L'Hospital's rule	\lim\frac{f}{g} = \lim\frac{f'}{g'}	only for \frac{0}{0} or \frac{\infty}{\infty}
Slope of tangent at (x_1,y_1)	m = \left.\frac{dy}{dx}\right|_{(x_1,y_1)}
Slope of normal	-\frac{1}{m}	m \ne 0
Tangent parallel to the x-axis	\frac{dy}{dx}=0
Tangent parallel to the y-axis	\frac{dx}{dy}=0, i.e. \frac{dy}{dx} is infinite
Differential	dy = f'(x)\,dx
Rate measure	\frac{dA}{dt} = \frac{dA}{dr}\cdot\frac{dr}{dt}	chain rule

Multiple choice questions

1. What is the derivative of y = 2\sinh\frac{x}{2} with respect to x? (NEB 2080) [1]a\cosh\frac{x}{2}b2\cosh\frac{x}{2}c\cosh\frac{x}{4}d2\cosh\frac{x}{4}
2. It is given that f'(x)=3x^{2} and g'(x)=2x. What is the value of \lim_{x\to1}\frac{f(x)}{g(x)} by L'Hospital's rule? (NEB 2081) [1]a\frac{3}{2}b\frac{1}{3}c\frac{1}{2}d\frac{1}{6}
3. Which one of the following is the slope of the normal to the curve y=3x^{2}-x at the point where x=-1? (NEB 2082) [1]a-7b-5c\frac{1}{7}d\frac{1}{5}
4. The tangents to the curve y=x^{3} at x=2 and x=-2 are (NEB 2079) [1]aparallelbperpendicularcintersectingdnon-existent
5. The volume of a sphere is increasing at 25 cm³/sec. At what rate is the radius increasing at the instant when the total surface area is 10\pi cm²? (CDC model 2080) [1]a\frac{5}{2\pi} cm/secb\frac{2\pi}{5} cm/secc\frac{5}{\pi} cm/secd\frac{25}{4\pi} cm/sec
6. \frac{d}{dx}(\sinh x) equals [1]a\cosh xb-\cosh xc\mathrm{sech}^{2}xd\cos x
7. \frac{d}{dx}\left(\cosh^{-1}x\right) equals [1]a\frac{1}{\sqrt{1-x^{2}}}b\frac{1}{\sqrt{x^{2}-1}}c\frac{1}{\sqrt{x^{2}+1}}d\frac{-1}{\sqrt{x^{2}-1}}
8. \frac{d}{dx}\left(\tanh^{-1}x\right) equals [1]a\frac{1}{1+x^{2}}b\frac{1}{1-x^{2}}c\frac{1}{x^{2}-1}d\mathrm{sech}^{2}x
9. \frac{d}{dx}\left(\mathrm{sech}^{-1}x\right) = \frac{-1}{x\sqrt{1-x^{2}}} holds under the condition [1]ax>1b0<x<1cx<0dall real x
10. \frac{d}{dx}\left(\mathrm{cosech}^{-1}x\right) equals [1]a\frac{-1}{|x|\sqrt{1+x^{2}}}b\frac{1}{|x|\sqrt{1+x^{2}}}c\frac{-1}{x\sqrt{1-x^{2}}}d\frac{1}{\sqrt{x^{2}+1}}
11. \lim_{x\to0}\frac{\sin x}{x} equals [1]a0b1c\inftydundefined
12. \lim_{x\to0}\frac{e^{x}-1}{x} equals [1]a0b1ced\infty
13. The slope of the tangent to y=x^{2}-3x at x=2 is [1]a-2b1c2d4
14. The slope of the tangent to y=2x^{3}+x^{2}-7x-2 at x=2 is [1]a5b13c21d28
15. A tangent to a curve is parallel to the x-axis where [1]a\frac{dy}{dx}=0b\frac{dy}{dx}=1c\frac{dx}{dy}=0d\frac{d^{2}y}{dx^{2}}=0
16. A tangent to the curve y=f(x) is parallel to the y-axis where [1]a\frac{dy}{dx}=0b\frac{dy}{dx}=1c\frac{dx}{dy}=0df(x)=0
17. The differential dy of y=f(x) is [1]af'(x)bf'(x)\,dxc\frac{dx}{f'(x)}df(x)\,dx
18. The rate of change of the area of a circle with respect to its radius, at r=3, is [1]a3\pib6\pic9\pid12\pi
19. \lim_{x\to\infty}\frac{\ln(2x^{2}+5)}{\ln(3x^{3}-4)} equals [1]a\frac{2}{3}b\frac{3}{2}c0d1
20. The rate of change of the volume of a sphere with respect to its surface area, when the radius is 7 cm, is [1]a\frac{7}{2}b7c14d\frac{1}{14}

Answer key — 11.12

1. (a) — \frac{d}{dx}\left(2\sinh\frac{x}{2}\right) = 2\cosh\frac{x}{2}\cdot\frac{1}{2} = \cosh\frac{x}{2}. The chain-rule factor \frac{1}{2} cancels the 2 outside.

2. (a) — L'Hospital's rule replaces the ratio by \frac{f'(1)}{g'(1)} = \frac{3}{2}.

3. (c) — \frac{dy}{dx}=6x-1, so at x=-1 the tangent slope is -7 and the normal slope is -\frac{1}{-7}=\frac{1}{7}. (The paper prints the point as (-1,1), but y(-1)=4; the x-value is what the slope depends on, so the answer is unaffected.)

4. (a) — \frac{dy}{dx}=3x^{2} takes the value 12 at both x=2 and x=-2: equal slopes, so parallel.

5. (a) — S=4\pi r^{2}=10\pi gives r^{2}=2.5; then \frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}=25 gives \frac{dr}{dt}=\frac{25}{10\pi}=\frac{5}{2\pi} cm/sec.

6. (a) — no minus sign here; the minus appears for \frac{d}{dx}(\cos x), not \frac{d}{dx}(\cosh x).

7. (b) — \frac{1}{\sqrt{x^{2}-1}}; (a) is \frac{d}{dx}(\sin^{-1}x) and (c) is \frac{d}{dx}(\sinh^{-1}x).

8. (b) — \frac{1}{1-x^{2}} for |x|<1; (a) is \frac{d}{dx}(\tan^{-1}x).

9. (b) — the \sqrt{1-x^{2}} needs |x|<1 and the \mathrm{sech}^{-1} branch used in NEB work is 0<x<1. This is the 2081 Group B item 17b, worth 1 mark for exactly this condition.

10. (a) — machine-checked: \frac{d}{dx}\mathrm{cosech}^{-1}x = \frac{-1}{|x|\sqrt{1+x^{2}}}.

11. (b) — \frac{0}{0} form; differentiate to get \frac{\cos x}{1} \to 1.

12. (b) — \frac{0}{0} form; differentiate to get \frac{e^{x}}{1} \to 1.

13. (b) — \frac{dy}{dx}=2x-3 = 4-3 = 1.

14. (c) — \frac{dy}{dx}=6x^{2}+2x-7 = 24+4-7 = 21.

15. (a) — a horizontal tangent has zero slope.

16. (c) — a vertical tangent has infinite \frac{dy}{dx}, which is the same as \frac{dx}{dy}=0. This was worth 1 mark in 2082 Group B item 17c.

17. (b) — the differential is the derivative times the increment in x.

18. (b) — A=\pi r^{2}, so \frac{dA}{dr}=2\pi r = 6\pi.

19. (a) — by L'Hospital's rule the limit is \lim\frac{4x/(2x^{2}+5)}{9x^{2}/(3x^{3}-4)}, which tends to \frac{2}{3}; more quickly, the ratio of the leading powers' logarithms is \frac{2}{3}. Machine-checked.

20. (a) — \frac{dV}{dS}=\frac{dV/dr}{dS/dr}=\frac{4\pi r^{2}}{8\pi r}=\frac{r}{2}=\frac{7}{2}.

11.13 Anti-derivatives

The other reliable Calculus MCQ. An integral has been the subject of an MCQ in 2080, 2081 and 2082, and all three times it was the same shape of question: a standard integral of the form \int\frac{dx}{\text{quadratic}}, with the four options differing only in a constant or in an inverted argument. That is a memory test, not a calculation, and the table below is the whole answer.

Integral	Result
\int x^{n}dx	\frac{x^{n+1}}{n+1}+C, n \ne -1
\int\frac{dx}{x}	\log|x|+C
\int e^{x}dx	e^{x}+C
\int a^{x}dx	\frac{a^{x}}{\log a}+C
\int\sin x\,dx	-\cos x + C
\int\cos x\,dx	\sin x + C
\int\tan x\,dx	\log|\sec x|+C
\int\sec x\,dx	\log|\sec x+\tan x|+C
\int\sinh x\,dx	\cosh x + C
\int\frac{dx}{x^{2}+a^{2}}	\frac{1}{a}\tan^{-1}\frac{x}{a}+C
\int\frac{dx}{x^{2}-a^{2}}	\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C
\int\frac{dx}{a^{2}-x^{2}}	\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C
\int\frac{dx}{\sqrt{a^{2}-x^{2}}}	\sin^{-1}\frac{x}{a}+C
\int\frac{dx}{\sqrt{x^{2}+a^{2}}}	\log\left|x+\sqrt{x^{2}+a^{2}}\right|+C = \sinh^{-1}\frac{x}{a}+C
\int\frac{dx}{\sqrt{x^{2}-a^{2}}}	\log\left|x+\sqrt{x^{2}-a^{2}}\right|+C = \cosh^{-1}\frac{x}{a}+C
\int\frac{f'(x)}{f(x)}dx	\log|f(x)|+C

Two supporting facts for the partial-fraction questions:

Term	Meaning
Proper rational fraction	degree of numerator < degree of denominator
Partial fractions	one fraction per distinct linear factor

Multiple choice questions

1. Which one of the following is the result of \int\frac{dx}{x^{2}+4}? (NEB 2081) [1]a\frac{1}{2}\tan^{-1}\frac{2}{x}+Cb\frac{1}{2}\tan^{-1}\frac{x}{2}+Cc2\tan^{-1}\frac{x}{2}+Cd\tan^{-1}\frac{x}{2}+C
2. What is the integral of \int\frac{dx}{9x^{2}+1}? (NEB 2082) [1]a\frac{1}{3}\tan^{-1}(3x)+Cb\frac{1}{27}\tan^{-1}(3x)+Cc\frac{1}{27}\tan^{-1}\frac{x}{3}+Cd\frac{1}{27}\tan^{-1}\frac{3}{x}+C
3. Which of the following is the value of \int\frac{dx}{x^{2}-1}? (NEB 2080) [1]a2\log\frac{x+1}{x-1}+Cb2\log\frac{x-1}{x+1}+Cc\log\frac{x-1}{x+1}+Cd\frac{1}{2}\log\frac{x+1}{x-1}+C
4. \int\frac{dx}{\sqrt{1-x^{2}}} equals [1]a\sin^{-1}x+Cb\cos^{-1}x+Cc\tan^{-1}x+Cd\sinh^{-1}x+C
5. \int\frac{dx}{\sqrt{x^{2}-a^{2}}} equals [1]a\sin^{-1}\frac{x}{a}+Cb\log\left|x+\sqrt{x^{2}-a^{2}}\right|+Cc\frac{1}{a}\tan^{-1}\frac{x}{a}+Cd\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C
6. \int\tan x\,dx equals [1]a\sec^{2}x+Cb\log|\sec x|+Cc\log|\cos x|+Cd\log|\tan x|+C
7. \int\sec x\,dx equals [1]a\sec x\tan x+Cb\log|\sec x|+Cc\log|\sec x+\tan x|+Cd\tan x + C
8. \int x^{3}dx equals [1]a3x^{2}+Cb\frac{x^{4}}{4}+Cc\frac{x^{2}}{2}+Cd4x^{4}+C
9. \int e^{2x}dx equals [1]a2e^{2x}+Cb\frac{e^{2x}}{2}+Cce^{2x}+Cd\frac{e^{3x}}{3}+C
10. \int\frac{dx}{x^{2}-9} equals [1]a\frac{1}{3}\log\left|\frac{x-3}{x+3}\right|+Cb\frac{1}{6}\log\left|\frac{x-3}{x+3}\right|+Cc\frac{1}{6}\log\left|\frac{x+3}{x-3}\right|+Cd\frac{1}{3}\tan^{-1}\frac{x}{3}+C
11. \int\frac{dx}{4-x^{2}} equals [1]a\frac{1}{4}\log\left|\frac{2+x}{2-x}\right|+Cb\frac{1}{4}\log\left|\frac{x-2}{x+2}\right|+Cc\frac{1}{2}\log\left|\frac{2+x}{2-x}\right|+Cd\sin^{-1}\frac{x}{2}+C
12. \int\frac{dx}{x^{2}+2x+2} equals [1]a\tan^{-1}(x+1)+Cb\frac{1}{2}\tan^{-1}(x+1)+Cc\log|x^{2}+2x+2|+Cd\tan^{-1}(x+2)+C
13. \int\frac{2x+1}{x^{2}+x+1}dx equals [1]a\log|2x+1|+Cb\log|x^{2}+x+1|+Cc\frac{1}{2}\log|x^{2}+x+1|+Cd\tan^{-1}(2x+1)+C
14. \int\frac{dx}{x(x+1)} equals [1]a\log\left|\frac{x}{x+1}\right|+Cb\log\left|\frac{x+1}{x}\right|+Cc\log|x(x+1)|+Cd\frac{1}{2}\log\left|\frac{x}{x+1}\right|+C
15. A rational fraction \frac{p(x)}{q(x)} is called proper when [1]a\deg p < \deg qb\deg p > \deg qc\deg p = \deg qdq(x) is linear
16. \int\sinh x\,dx equals [1]a\cosh x+Cb-\cosh x+Cc\mathrm{sech}^{2}x+Cd\sinh x+C
17. The number of partial fractions needed for \frac{1}{x(x-1)(x+2)} is [1]a1b2c3d4
18. \int a^{x}dx equals [1]aa^{x}+Cb\frac{a^{x}}{\log a}+Cca^{x}\log a+Cd\frac{a^{x+1}}{x+1}+C

Answer key — 11.13

1. (b) — \int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a} with a=2, giving \frac{1}{2}\tan^{-1}\frac{x}{2}. Machine-checked.

2. (a) — take out the 9: \frac{1}{9}\int\frac{dx}{x^{2}+\frac{1}{9}} = \frac{1}{9}\cdot\frac{1}{1/3}\tan^{-1}(3x) = \frac{1}{3}\tan^{-1}(3x). Machine-checked.

3. No printed option is exactly right. The correct value is \frac{1}{2}\log\left|\frac{x-1}{x+1}\right|+C. Option (b) has the correct ratio \frac{x-1}{x+1} with the constant 2 in place of \frac{1}{2}; option (d) has the correct constant with the ratio inverted, which changes the sign of the whole answer. (b) is the closer of the two, and is the option a marker would have keyed. We machine-checked the integral.

4. (a) — \sin^{-1}x+C; note \cos^{-1}x differentiates to \frac{-1}{\sqrt{1-x^{2}}}, the wrong sign.

5. (b) — this is \cosh^{-1}\frac{x}{a}, usually written in the logarithmic form.

6. (b) — \int\frac{\sin x}{\cos x}dx = -\log|\cos x| = \log|\sec x|.

7. (c) — the standard result; (b) is \int\tan x\,dx.

8. (b) — raise the index by one and divide by the new index.

9. (b) — divide by the coefficient of x in the exponent.

10. (b) — \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| with a=3, so \frac{1}{6}. Machine-checked.

11. (a) — \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| with a=2, so \frac{1}{4}\log\left|\frac{2+x}{2-x}\right|. Machine-checked.

12. (a) — complete the square: \int\frac{dx}{(x+1)^{2}+1} = \tan^{-1}(x+1). Machine-checked.

13. (b) — the numerator is exactly the derivative of the denominator, so the integral is \log|f(x)|.

14. (a) — \frac{1}{x(x+1)} = \frac{1}{x}-\frac{1}{x+1}, giving \log|x|-\log|x+1|. Machine-checked.

15. (a) — otherwise divide first to split off a polynomial part.

16. (a) — \int\sinh x\,dx = \cosh x; the minus sign belongs to \int\sin x\,dx.

17. (c) — three distinct linear factors give three partial fractions.

18. (b) — differentiating \frac{a^{x}}{\log a} returns a^{x}.

11.14 Differential Equations

Two MCQs came from this unit in 2081 (degree, and identifying a homogeneous equation) and two again in 2082 (homogeneous, and an integrating factor), plus one in 2079 (integrating factor). It is the most heavily MCQ'd single unit in the syllabus. Every one of those five items is answered by the table below without any integration at all.

Result	Formula / condition
Order	the order of the highest derivative present
Degree	the power of that highest derivative, after clearing radicals and fractions
Separable form	\frac{dy}{dx}=f(x)g(y); integrate \frac{dy}{g(y)} = f(x)dx
Homogeneous	\frac{dy}{dx}=f\!\left(\frac{y}{x}\right); numerator and denominator of the same degree
Substitution for homogeneous	y=vx, so \frac{dy}{dx}=v+x\frac{dv}{dx}
Linear standard form	\frac{dy}{dx}+Py=Q, with P,Q functions of x
Integrating factor	\mathrm{I.F.} = e^{\int P\,dx}
Solution of a linear equation	y\cdot\mathrm{I.F.} = \int Q\cdot\mathrm{I.F.}\,dx + C
Exact form	M(x,y)dx+N(x,y)dy=0
Test for exactness	\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}
\frac{dy}{dx}=\frac{y}{x}	y=cx
x\,dy+y\,dx=0	xy=c
\frac{dy}{dx}=\frac{1+y^{2}}{1+x^{2}}	\tan^{-1}y=\tan^{-1}x+c
\frac{dy}{dx}+\frac{y}{x}=x^{2}	\mathrm{I.F.}=x

A direction field is the quickest way to see what a first-order equation is saying: at every point it draws the slope \frac{dy}{dx}, and a solution curve is any curve that follows the slopes.

Multiple choice questions

1. What is the degree of the ordinary differential equation \frac{d^{3}s}{dt^{3}} = \left(4+\left(\frac{d^{2}s}{dt^{2}}\right)^{2}\right)^{1/2}? (NEB 2081) [1]a1b2c3d4
2. Which one of the following is an example of a homogeneous differential equation of the first order? (NEB 2081) [1]a3x\,dy+2y\,dx=4bx\,dy-y\,dx+1=0cx\,dx+y\,dy=2d(x^{2}+xy)dy-(xy-y^{2})dx=0
3. Which one of the following is a homogeneous differential equation? (NEB 2082) [1]a\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x+y}b\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x-y}c\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}-y^{2}}d\frac{dy}{dx}=\frac{x+y}{x^{2}+y^{2}}
4. Which is the integrating factor of the linear differential equation \cos^{2}x\,\frac{dy}{dx}=1-y? (NEB 2082) [1]a\tan xbe^{\tan x}ce^{\sec^{2}x}d\sec^{2}x
5. The integrating factor of the differential equation \sin x\,\frac{dy}{dx}+y\cos x = x\sin x is (NEB 2079) [1]a\sin xb\cos xc\tan xd\cot x
6. The order of \frac{d^{2}y}{dx^{2}}+3\left(\frac{dy}{dx}\right)^{3}=0 is [1]a1b2c3d5
7. The degree of \frac{d^{2}y}{dx^{2}}+3\left(\frac{dy}{dx}\right)^{3}=0 is [1]a1b2c3d5
8. The integrating factor of \frac{dy}{dx}+\frac{y}{x}=x^{2} is [1]axbx^{2}c\log xde^{x}
9. The integrating factor of \frac{dy}{dx}+2y=e^{x} is [1]ae^{x}be^{2x}c2e^{x}de^{-2x}
10. The equation M\,dx+N\,dy=0 is exact if [1]a\frac{\partial M}{\partial x}=\frac{\partial N}{\partial y}b\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}cM=Nd\frac{\partial M}{\partial y}=-\frac{\partial N}{\partial x}
11. The general solution of \frac{dy}{dx}=\frac{y}{x} is [1]ay=cxbxy=ccy=x+cdy^{2}=x^{2}+c
12. The solution of \frac{dy}{dx}=\frac{1+y^{2}}{1+x^{2}} is [1]a\tan^{-1}y=\tan^{-1}x+cb\log(1+y^{2})=\log(1+x^{2})+ccy=x+cd\sin^{-1}y=\sin^{-1}x+c
13. To solve a homogeneous first-order equation, the correct substitution is [1]ay=vxby=v+xcx=vy^{2}dv=xy
14. The order and degree of \left(\frac{d^{3}y}{dx^{3}}\right)^{2}+y=0 are respectively [1]a2 and 3b3 and 2c3 and 3d2 and 2
15. The solution of x\,dy+y\,dx=0 is [1]ax+y=cbxy=cc\frac{x}{y}=cdx^{2}+y^{2}=c
16. The integrating factor of x\frac{dy}{dx}+2y=x^{3} is [1]axbx^{2}cx^{3}d2x

Answer key — 11.14

1. (b) — square both sides to clear the radical: \left(\frac{d^{3}s}{dt^{3}}\right)^{2}=4+\left(\frac{d^{2}s}{dt^{2}}\right)^{2}. The highest derivative is now to the power 2, so the degree is 2 (the order is 3).

2. (d) — every term of (x^{2}+xy) and (xy-y^{2}) has total degree 2, so the equation can be written as a function of \frac{y}{x} alone. The others carry an isolated constant, which breaks homogeneity.

3. (c) — numerator and denominator are both of degree 2, so replacing x \to \lambda x, y \to \lambda y leaves the right-hand side unchanged.

4. (b) — divide by \cos^{2}x: \frac{dy}{dx}+y\sec^{2}x=\sec^{2}x. Then P=\sec^{2}x, \int P\,dx = \tan x and \mathrm{I.F.}=e^{\tan x}. Machine-checked.

5. (a) — divide by \sin x: \frac{dy}{dx}+y\cot x = x. Then \int\cot x\,dx=\log\sin x, so \mathrm{I.F.}=e^{\log\sin x}=\sin x.

6. (b) — the highest derivative present is the second.

7. (a) — the second derivative appears to the first power; the cube is on the first derivative, which does not affect the degree.

8. (a) — P=\frac{1}{x}, \int P\,dx=\log x, \mathrm{I.F.}=e^{\log x}=x.

9. (b) — P=2, so \mathrm{I.F.}=e^{2x}.

10. (b) — both mixed partial derivatives of the same potential function must agree.

11. (a) — separate: \frac{dy}{y}=\frac{dx}{x}, so \log y = \log x + \log c, i.e. y=cx.

12. (a) — separate and integrate both sides directly. This was the 2080 Group B item 17a.

13. (a) — y=vx turns a function of \frac{y}{x} into a function of v alone, making the equation separable.

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

15. (b) — \frac{dy}{y}=-\frac{dx}{x} gives \log y + \log x = \log c, i.e. xy=c.

16. (b) — divide by x first: \frac{dy}{dx}+\frac{2}{x}y=x^{2}, so P=\frac{2}{x}, \int P\,dx=2\log x and \mathrm{I.F.}=x^{2}. Machine-checked.

11.15 Computational Methods

Exactly one MCQ comes from this area every year, and it has been at Q10 or Q11 in all four verified papers. Three of those four were a Gauss elimination step: you are shown two equations and asked what the elimination produces. Nothing is being tested except arithmetic discipline, so this is a mark you should never drop.

Method	Key point
Gauss elimination	reduce the augmented matrix to upper triangular, then back-substitute
Elementary row operations	swap two rows; multiply a row by k\ne0; add a multiple of one row to another
Gauss–Jordan	continue to the identity matrix on the left, reading the solution off directly
Gauss–Seidel	iterative: use each new value immediately inside the same sweep
Convergence of Gauss–Seidel	guaranteed if the system is diagonally dominant
Unique solution	\mathrm{rank}(A)=\mathrm{rank}(A\mid B)=n
Infinitely many solutions	\mathrm{rank}(A)=\mathrm{rank}(A\mid B)<n
No solution (inconsistent)	\mathrm{rank}(A)\ne\mathrm{rank}(A\mid B); a row reads 0=k, k\ne0
LP feasible region	the intersection of all the constraint half-planes
LP optimum	occurs at a vertex of the feasible region
Simplex: slack variable	converts \leq into =; \;3x+2y\leq12 \Rightarrow 3x+2y+s=12
Simplex: pivot column	the most negative entry of the objective (bottom) row
Simplex: pivot row	the smallest non-negative ratio \dfrac{\text{RHS}}{\text{pivot-column entry}}
Simplex: stop when	no negative entries remain in the bottom row

Multiple choice questions

1. In solving the system x+3y=5, 2x-y=3 by Gauss elimination, the operation R_2 \to 2R_1-R_2 gives (NEB 2080) [1]a5y=5b7y=13c7y=7dy=7
2. In solving x+3y=7, 2x-4y=4 by Gauss elimination, the operation R_2 \to 2R_1-R_2 gives (NEB 2082) [1]a10y=10b10y=18c2y=10d-10y=10
3. If in the row-echelon form of an augmented matrix the rank of the coefficient matrix equals the rank of the augmented matrix but is less than the number of unknowns, the system is (NEB 2079) [1]ainconsistentbconsistent with a unique solutioncinconsistent with infinitely many solutionsdconsistent with infinitely many solutions
4. Which elementary row operation is not permitted on an augmented matrix? (NEB 2081) [1]ainterchanging two rowsbadding a multiple of one row to anothercmultiplying a row by 0dmultiplying a row by -3
5. Solving 2x+3y=8, x-y=-1 by Gauss elimination, the operation R_2 \to 2R_2-R_1 gives [1]a-5y=-10b5y=10c-5y=10dy=-10
6. The solution of the system in question 5 is [1]ax=1,\;y=2bx=2,\;y=1cx=-1,\;y=2dx=1,\;y=-2
7. A system of linear equations is inconsistent when a row of the reduced augmented matrix reads [1]a0\;\;0\;\mid\;0b0\;\;0\;\mid\;5c1\;\;0\;\mid\;0d0\;\;1\;\mid\;3
8. The Gauss–Seidel method is certain to converge if the coefficient matrix is [1]asymmetricbsingularcdiagonally dominantdtriangular
9. In the Gauss–Jordan method the left block of the augmented matrix is reduced to [1]aan upper triangular matrixbthe identity matrixca diagonal matrix with zero entriesda null matrix
10. Introducing a slack variable turns the constraint 3x+2y\leq12 into [1]a3x+2y-s=12b3x+2y+s=12c3x+2y+s\leq12d3x+2y=12-s^{2}
11. In the simplex method, the entering variable is chosen from the column with [1]athe largest positive entry in the objective rowbthe most negative entry in the objective rowcthe largest right-hand sidedthe smallest ratio
12. The maximum value of a linear objective function over a bounded feasible region occurs [1]aat the centre of the regionbat a vertex of the regioncanywhere inside the regiondonly on the x-axis

Answer key — 11.15

1. (c) — 2R_1 = 2x+6y=10; subtracting R_2 = 2x-y=3 gives 7y=7, so y=1 and then x=2. Machine-checked.

2. (a) — 2R_1 = 2x+6y=14; subtracting R_2 = 2x-4y=4 gives 10y=10, so y=1 and x=4. Machine-checked.

3. (d) — equal ranks means consistent; rank below the number of unknowns leaves a free variable, hence infinitely many solutions. Option (c) is self-contradictory and is the trap.

4. (c) — multiplying a row by 0 destroys information and is not reversible; the scalar must be non-zero.

5. (a) — 2R_2 = 2x-2y=-2; subtracting R_1 = 2x+3y=8 gives -5y=-10.

6. (a) — from -5y=-10, y=2, then x-2=-1 gives x=1. Machine-checked.

7. (b) — that row says 0=5, which is impossible, so no solution exists.

8. (c) — diagonal dominance, |a_{ii}| > \sum_{j\ne i}|a_{ij}| in every row, is the standard sufficient condition.

9. (b) — Gauss elimination stops at upper triangular; Gauss–Jordan goes all the way to I.

10. (b) — a slack variable absorbs the unused capacity and is itself \ge 0, turning the inequality into an equation.

11. (b) — the most negative objective-row entry marks the variable whose increase improves the objective fastest.

12. (b) — a linear function on a convex polygon attains its extremes at corner points.

11.16 Mechanics (alternative to Unit 15)

If your school teaches Mechanics instead of Computational Methods, this section replaces §11.15 — you answer the "Or" at Q11 and again at Q19, not both. The MCQ has been a projectile item in 2081 and 2082, a weight item in 2080 and a momentum item in 2079. The projectile formulae below answer the first two in about twenty seconds each.

Quantity	Formula (launch speed u, angle \theta, gravity g)
Horizontal component	u\cos\theta, constant throughout
Vertical component	u\sin\theta-gt
Time to greatest height	t=\dfrac{u\sin\theta}{g}
Time of flight	T=\dfrac{2u\sin\theta}{g}
Greatest height	H=\dfrac{u^{2}\sin^{2}\theta}{2g}
Horizontal range	R=\dfrac{u^{2}\sin 2\theta}{g}
Maximum range	R_{\max}=\dfrac{u^{2}}{g}, at \theta=45^\circ
Equation of the path	y=x\tan\theta-\dfrac{gx^{2}}{2u^{2}\cos^{2}\theta} (a parabola)
Weight of mass m	W=mg newtons
Newton's second law	F=ma
Momentum	p=mv; change of momentum =mv-mu
Impulse	Ft = mv-mu
Triangle law of forces	three forces in equilibrium form a closed triangle taken in order
Lami's theorem	\dfrac{P}{\sin\alpha}=\dfrac{Q}{\sin\beta}=\dfrac{R}{\sin\gamma}
Resultant of P,Q at angle \theta	\sqrt{P^{2}+Q^{2}+2PQ\cos\theta}
Two like parallel forces	resultant X+Y, dividing AB internally in the ratio Y:X

Multiple choice questions

1. What is the greatest height attained by a projectile whose initial velocity and angle of elevation are 160\ \mathrm{ms^{-1}} and 30^\circ respectively? (g=10\ \mathrm{ms^{-2}}) (NEB 2081) [1]a203 mb230 mc320 md640 m
2. What is the maximum height attained by a particle in projectile motion if the initial velocity and angle of inclination are 40\ \mathrm{m/s} and 30^\circ? (g=10\ \mathrm{ms^{-2}}) (NEB 2082) [1]a20 mb40 mc80 md160 m
3. What is the amount of gravity exerted by the earth on a mass of 100 kg? (g is the acceleration due to gravity) (NEB 2080) [1]a100 newtonsb100g newtonsc\frac{g}{100} newtonsd\frac{100}{g} newtons
4. A force acts on a body of mass m for a time t and changes its velocity from u to v. The change in momentum of the body in time t is (NEB 2079) [1]a\frac{mu-mv}{t}b\frac{mv-mu}{t}c\frac{tm-tv}{uv}d\frac{u-v}{mt}
5. The time of flight of the projectile of question 2 is [1]a2 sb4 sc8 sd\frac{1}{2} s
6. A projectile is fired with speed u. Its horizontal range is greatest when the angle of projection is [1]a30^\circb45^\circc60^\circd90^\circ
7. At the highest point of its path the velocity of a projectile is [1]azerobu\sin\thetacu\cos\theta, horizontaldu, along the tangent
8. The path of a projectile under uniform gravity, with air resistance neglected, is [1]aa straight lineba circleca paraboladan ellipse
9. Three forces P, Q, R acting at a point are in equilibrium, with \alpha, \beta, \gamma the angles opposite to them. Lami's theorem states [1]a\frac{P}{\sin\alpha}=\frac{Q}{\sin\beta}=\frac{R}{\sin\gamma}bP\sin\alpha=Q\sin\beta=R\sin\gammac\frac{P}{\cos\alpha}=\frac{Q}{\cos\beta}=\frac{R}{\cos\gamma}dP+Q+R=0
10. The resultant of two like parallel forces X and Y acting at A and B divides AB internally in the ratio [1]aX:YbY:XcX^{2}:Y^{2}d1:1

Answer key — 11.16

1. (c) — H=\frac{u^{2}\sin^{2}\theta}{2g}=\frac{160^{2}\times(1/2)^{2}}{2\times10}=\frac{25600\times0.25}{20}=320 m. Machine-checked.

2. (a) — H=\frac{40^{2}\times 0.25}{20}=\frac{400}{20}=20 m. Machine-checked.

3. (b) — the gravitational force on a mass is its weight W=mg=100g newtons. Option (a) drops g and is the commonest answer given.

4. (b) — as printed. Strictly, the change of momentum is mv-mu and \frac{mv-mu}{t} is the average force; every printed option divides by t, so (b) is the only defensible choice. Flagged as a printing fault, not a mathematical result.

5. (b) — T=\frac{2u\sin\theta}{g}=\frac{2\times40\times\frac12}{10}=\frac{40}{10}=4 s. Note that T is twice the time to the top, and that it does not depend on the horizontal component at all. Machine-checked.

6. (b) — R=\frac{u^{2}\sin2\theta}{g} is largest when \sin2\theta=1, i.e. 2\theta=90^\circ.

7. (c) — the vertical component has been reduced to zero; the horizontal component u\cos\theta is unchanged throughout.

8. (c) — eliminating t from x=u\cos\theta\,t and y=u\sin\theta\,t-\frac{1}{2}gt^{2} gives a quadratic in x.

9. (a) — each force is proportional to the sine of the angle between the other two.

10. (b) — the resultant is nearer the larger force, so the ratio of the segments is the reciprocal of the ratio of the forces.

11.17 Appendix: Retained Examinable Topics

A few topics sit outside the fifteen numbered units of the Grade 12 spine but have still appeared as Group A questions. They are collected here so that you can see exactly how much time each deserves, rather than guessing.

Group Theory. An MCQ in 2079 (recognising an abelian group) and again in 2080 (naming the associative law for a binary operation) — and then absent from both 2081 and 2082. The 2078 specification grid has no line for it. Treat it as a fifteen-minute revision: learn the four group axioms and the word "abelian", and move on. Do not learn subgroups, cosets or cyclic groups for this paper.

Rolle's theorem and the Mean Value Theorem. Appeared in 2080 only, and not in 2079, 2081 or 2082. Learn the two statements and the difference between them — that is the whole examinable surface.

Numerical integration (trapezoidal and Simpson's rules). Not seen as an MCQ in any of the four verified years, but the formulae are short and they occasionally appear inside a Group B part. Ten minutes.

Sum and product of the roots of a quadratic. A Grade 11 result, but it was MCQ 2 in 2080. It costs one line to remember.

Result	Statement
Group (G,*)	closure, associativity, identity, inverse
Abelian group	additionally a*b=b*a for all a,b\in G
Associative law	(a*b)*c = a*(b*c)
(\mathbb{Z},+), (\mathbb{Q}-\{0\},\times)	standard abelian groups
(\mathbb{N},+)	not a group — no identity, no inverses
Rolle's theorem	f continuous on [a,b], differentiable on (a,b), f(a)=f(b)\Rightarrow f'(c)=0 for some c
Mean value theorem	same hypotheses without f(a)=f(b): f'(c)=\dfrac{f(b)-f(a)}{b-a}
Trapezoidal rule	\dfrac{h}{2}\left[y_0+2(y_1+\dots+y_{n-1})+y_n\right]
Simpson's \frac{1}{3} rule	\dfrac{h}{3}\left[y_0+4(y_1+y_3+\dots)+2(y_2+y_4+\dots)+y_n\right]
Simpson's rule needs	an even number of sub-intervals
Sum of the roots of ax^{2}+bx+c=0	-\dfrac{b}{a}
Product of the roots	\dfrac{c}{a}

Multiple choice questions

1. Which of the following is an abelian group under the given operation? (NEB 2079) [1]athe set of integers under additionbthe set of natural numbers under additioncthe set of integers under multiplicationdthe set of all real numbers under multiplication
2. If * is a binary operation on a set G and (a*b)*c = a*(b*c) for all a,b,c\in G, then * satisfies the (NEB 2080) [1]aclosure lawbcommutative lawcidentity lawdassociative law
3. The sum of the roots of the equation x^{2}-6x+8=0 is (NEB 2080) [1]a2b4c6d8
4. If f(x)=x^{2}-4x+3 on [1,3], the value of c given by Rolle's theorem is [1]a1b2c3d\frac{3}{2}
5. Rolle's theorem requires, in addition to continuity and differentiability, that [1]af(a)=fb (b) f(a)=0cf'(a)=f'(b)df be increasing
6. For f(x)=x^{2} on [1,3], the value of c given by the mean value theorem is [1]a1b2c\sqrt{3}d\frac{5}{2}
7. Simpson's \frac{1}{3} rule can be applied only when the number of sub-intervals is [1]aoddbevencprimeda multiple of 3
8. Using the trapezoidal rule with n=2, the value of \int_{0}^{1}x^{2}\,dx is [1]a\frac{1}{3}b\frac{3}{8}c\frac{1}{2}d\frac{1}{4}

Answer key — 11.17

1. (a) — (\mathbb{Z},+) has closure, associativity, identity 0, inverse -a and commutativity. (\mathbb{N},+) has no identity or inverses; (\mathbb{Z},\times) has no inverses; \mathbb{R} under multiplication fails because 0 has no inverse.

2. (d) — the bracket can be moved without changing the result; that is precisely associativity.

3. (c) — sum of the roots =-\frac{b}{a}=-\frac{-6}{1}=6. The roots are 2 and 4; option (a) and (b) are the roots themselves and (d) is their product. Machine-checked.

4. (b) — f(1)=f(3)=0, and f'(x)=2x-4=0 gives c=2, which lies in (1,3). Machine-checked.

5. (a) — equal endpoint values are what force a horizontal tangent somewhere in between.

6. (b) — \frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4, and f'(c)=2c=4 gives c=2. Machine-checked.

7. (b) — the rule pairs the sub-intervals, fitting one parabola across each pair, so their number must be even.

8. (b) — h=\frac12; \frac{h}{2}[y_0+2y_1+y_2]=\frac14\left[0+2\times\frac14+1\right]=\frac{3}{8}. The exact value is \frac13, so the trapezoidal rule over-estimates here — as it always does for a convex curve. Machine-checked.

Chapter summary

• Group A is eleven marks in fifteen minutes, and since 2082 it arrives as a separate booklet thirty minutes after the paper starts. Do the eleven MCQs first, in one pass, and hand the booklet back before you settle into Group B.
• Four of the eleven MCQs are Calculus — the single largest block. Across 2079–2082 the reliable slots have been a derivative (often an inverse hyperbolic or inverse trigonometric one), a standard integral of the \frac{1}{a}\tan^{-1}\frac{x}{a} family, and one or two differential-equation items on order/degree, homogeneity or the integrating factor.
• One MCQ is guaranteed from each of Trigonometry, Analytic Geometry and Vector by the grid's own footnote. In practice that has meant a half-angle or triangle law, a conic property (eccentricity, latus rectum or a tangency condition), and a dot- or cross-product item.
• One MCQ is guaranteed from Computational Methods or Mechanics, and three of the four verified years made it a Gauss-elimination row-operation step; the Mechanics alternative has twice been the greatest height of a projectile. Both are pure substitution.
• The Statistics and Probability MCQ has gone to Probability every single year, never to correlation or regression, and conditional probability has appeared in 2080, 2081, 2082 and the CDC model. P(B/A)=\frac{P(A\cap B)}{P(A)} — divide by the event that has already happened.
• Exactly two "Or" choices appear per paper, at Q11 and Q19, never in Group C. Decide in advance which branch you are taking so that you do not spend reading time on both.
• The traps that cost the most marks are, in order of how often they are printed: differentiating when asked to integrate; the \hat{j} sign and the non-commutativity of \vec{a}\times\vec{b}; \frac{1}{a}\tan^{-1}\frac{a}{x} written for \frac{1}{a}\tan^{-1}\frac{x}{a}, and the constant misplaced in \int\frac{dx}{9x^{2}+1}; dividing by P(B) instead of P(A) in a conditional probability; degree read as order; C(n,r)\times r! = P(n,r) read backwards; and c^{2}=a^{2}(1+m^{2}) with a and c swapped.
• Binomial Theorem and Solution of Triangles have never carried an MCQ in the four verified years, and Group Theory vanished after 2080 while Rolle/MVT appeared in 2080 only. Prepare them for Groups B and C, not for Group A.
• Every numerical answer key in this chapter was recomputed with a computer algebra system before it was printed, including the option sets of the real past papers — where three genuine printing faults were found and are flagged in place.

How to attack Group A in fifteen minutes

Minutes 0–1: sort, do not solve. Read all eleven stems once and mark each one recall, one-line, or work. Roughly half of any NEB Group A is pure recall — order and degree, a group axiom, Lami's theorem, the exactness condition, the pivot rule.

Minutes 1–6: clear every recall item. Answer them straight off the fact tables in this chapter. Do not check them twice; you either know a definition or you do not.

Minutes 6–12: the one-line items. These are the substitution questions — greatest height, an integrating factor, a conditional probability, a row operation, an eccentricity. Write the formula down before you write any number. Nearly every distractor in this book is generated by substituting into a half-remembered formula, so the formula on the page is your defence.

Minutes 12–15: the work items and a sweep. Usually one or two: a determinant, a cross product, a definite integral. Then sweep the whole sheet and confirm you have shaded exactly eleven bubbles.

Three habits worth more than any formula. (i) Read the operator. "Find \int…" versus "find \frac{d}{dx}…" and "\hat{k}\times\hat{j}" versus "\hat{j}\times\hat{k}" change the sign or invert the whole answer. (ii) Check the variable. Differentiate with respect to whatever the question names, not whatever letter is most prominent. (iii) Never leave a blank. There is no negative marking; eliminate the two options that are dimensionally or structurally impossible and choose between the rest.