2026 and 2025 paper questions

Complete the square: $z^2+4z+5=(z+2{)}^2+1=(z+2{)}^2-i^2$, because $1=-i^2$. A difference of squares gives $(z+2-i)(z+2+i)$.

Check by expanding: $(z+2{)}^2-i^2=z^2+4z+4+1$. Options B and D have the wrong sign of the $2$ (they factorise $z^2-4z+\dots$). Options C and D use $2i$ instead of $i$: $(z+2{)}^2-(2i{)}^2=z^2+4z+8$, which is not the given expression.

The period of $\tan x$ is $\pi$, and the period of $\tan ax$ is $\frac{\pi }{|a|}$. Here $a=\frac{5}{7}$.

Option A, $\frac{5\pi }{7}$, multiplies by $a$ instead of dividing, and D, $7\pi$, ignores the $5$.

Here $x=-1<0$ and $y=\sqrt{3}>0$, so $z$ is in the second quadrant. The acute angle is $\alpha ={\tan }^{-1}\left|\frac{\sqrt{3}}{-1}\right|=\frac{\pi }{3}$, and in the second quadrant $\theta =\pi -\alpha =\frac{2\pi }{3}$.

Option B is only the acute angle $\alpha$; option A is an angle in the fourth quadrant. Option D, $\frac{5\pi }{3}$, points in the fourth quadrant and is outside the principal range $-\pi <\theta \le \pi$. Check: $2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)=2\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=-1+i\sqrt{3}$.

A function is periodic if there is a number $T>0$ with $f(x+T)=f(x)$ for every $x$ in the domain; the smallest such $T$ is the period.

Option A, $f(x)=f(-x)$, is the condition for an even function. B and C are constant functions, which are periodic only in a trivial way and are not the definition.

An inconsistent system has no solution at all ($\mathrm{Rank}A\ne \mathrm{Rank}{A}_b$). A system with at least one solution (B), including infinitely many (C), is consistent.

Option D cannot occur: a linear system has no solution, one solution or infinitely many, never exactly two.

Since $-1\le \sin 2\theta \le 1$, the term $b\sin 2\theta$ runs from $-|b|$ to $|b|$, so $a+b\sin 2\theta$ has minimum $a-|b|$ and maximum $a+|b|$.

The factor $2$ in $2\theta$ changes the period only, not the range. Option D multiplies the whole range by $2$, and B and C give only one end.

Multiplying a matrix by $3$ multiplies every row by $3$. Taking the common factor $3$ out of each of the three rows, one at a time, gives $|3A|=3\cdot 3\cdot 3|A|=3^3|A|=27|A|$. In general $|kA|=k^n|A|$ for an $n\times n$ matrix.

Option A multiplies only one row by $3$. Option C is $3^2$, which is the factor for a $2\times 2$ matrix, and B has no basis.

SUNDAY has $6$ different letters, and an arrangement uses all of them with order mattering, so the number of ways is ${}^6{P}_6=6!=720$.

Option A, $120$, is $5!$; it would be the answer for $5$ letters. The other values $480$ and $600$ are not factorials, so they cannot count arrangements of $6$ different letters.

$a_{15}=a_1+14d$, so $75=5+14d$, $14d=70$ and $d=5$.

There are 14 steps from the 1st term to the 15th. Dividing by 15 instead of 14 gives $\frac{70}{15}$, which is not an option.

Selecting players is a combination (order does not matter): ${}^{12}{C}_8={}^{12}{C}_4=\frac{12\cdot 11\cdot 10\cdot 9}{4!}=\frac{11880}{24}=495$.

The large options come from counting orders as well: ${}^{12}{P}_8=19958400$, far bigger than any selection count. The selection count is ${}^{12}{P}_8÷8!=495$.

Since $i^2=-1$, we have $289=-(17i{)}^2$, so $z^2+289=z^2-(17i{)}^2$. This is a difference of squares, $a^2-b^2=(a-b)(a+b)$ with $a=z$ and $b=17i$: $z^2+289=(z-17i)(z+17i)$.

Option A is the factorisation of $z^2-289$, a different expression. Options B and D are perfect squares: $(z+17i{)}^2=z^2+34iz-289$ has a middle term, so neither can equal $z^2+289$. Check of C: $(z+17i)(z-17i)=z^2-(17i{)}^2=z^2+289$.

The reciprocals of an H.P. are in A.P., so $2,x,5$ are in A.P. The middle term is the arithmetic mean: $x=\frac{2+5}{2}=\frac{7}{2}$.

Option A, $\frac{2}{7}$, is $\frac{1}{x}$: it is the middle term of the H.P., not the value of $x$.

Here $x=\sqrt{3}>0$ and $y=1>0$, so $z$ lies in the first quadrant and $\mathrm{Arg}z=\alpha ={\tan }^{-1}\left|\frac{y}{x}\right|={\tan }^{-1}\frac{1}{\sqrt{3}}=\frac{\pi }{6}$.

Option A, $\frac{\pi }{3}$, comes from inverting the ratio, ${\tan }^{-1}\frac{\sqrt{3}}{1}$. Remember that $\tan \theta =\frac{y}{x}$, imaginary part over real part. As a check, $|z|=2$ and $2\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)=2\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)=\sqrt{3}+i$.

$S_{\infty }=\frac{a_1}{1-r}$, so $a_1=S_{\infty }(1-r)=10(1-0.6)=10(0.4)=4$.

Multiplying $10$ by $0.6$ gives 6 (option C); that is $S_{\infty }r$, not $S_{\infty }(1-r)$.

The matrix is already in echelon form: the leading entry of row 1 is $1$, the leading entry of row 2 is $1$ and lies to the right of it, and the zero row is at the bottom. The rank is the number of non-zero rows, which is $2$.

Option C counts all three rows, including the zero row. Option D is impossible, since a $3\times 3$ matrix cannot have rank above $3$. Option A would need two rows to vanish.

Put $n=4$ in the given result: $S_4=4^2=16$. Direct addition agrees: $1+3+5+7=16$.

Option B, 9, is the sum of 3 terms ($n=3$). Option C, 15, is $16-1$, what is left if the $+1$ is dropped; the sum of 4 terms is exactly $4^2$.

A system of linear equations is consistent if it has at least one solution. When $\mathrm{Rank}A=\mathrm{Rank}{A}_b$, the solution is either unique (rank equals the number of unknowns) or there are infinitely many (rank less than the number of unknowns).

Option B is wrong because a consistent system may have exactly one solution. Option C describes an inconsistent system. Option D cannot happen: a linear system never has exactly two solutions.

$n=6$ is even, so there is one middle term, the $\left(\frac{6}{2}+1\right)=4$th term, $T_4=T_{3+1}$:

Options A and D have the wrong coefficient $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$, and options C and D have the wrong exponents $x^4{y}^2$ (the exponents must add up to $6$ and be equal here).

The first term is $a_1=a$, the second is $a+d$, the third is $a+2d$. The common difference has been added $n-1$ times by the $n$th term, so $a_n=a+(n-1)d$.

Option A, $a+nd$, would give the first term as $a+d$, so it is really the $(n+1)$th term. Option D multiplies instead of adding.

If P(-1)=0, then x=-1 is a zero (root) of P(x), because a zero of a polynomial is a number a with P(a)=0. By the Factor Theorem, (x+1) is then a factor of P(x).

Option B, C and D describe the signal, not the polynomial: the value P(-1)=0 says nothing about stability, periodicity or error. The only statement it proves is that P has a zero at x=-1.

\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0 gives \cos\theta=0, so \theta=90^\circ.

0^\circ would make the dot product maximum (|\vec a||\vec b|), and 180^\circ makes it -|\vec a||\vec b|; it is the cross product that vanishes at 0^\circ and 180^\circ.

The terms of an H.P. are the reciprocals of the terms of an A.P. For positive terms, a larger number has a smaller reciprocal. If the A.P. increases, for example 1,2,3,\dots, the H.P. 1,\dfrac12,\dfrac13,\dots decreases.

Option A is the opposite: a decreasing A.P. such as 3,2,1 gives an increasing H.P. \dfrac13,\dfrac12,1. Option C gives a constant H.P., and option D is impossible because \dfrac10 is undefined.

The volume of the parallelepiped with co-terminal edges \vec a,\vec b,\vec c is |\vec a\cdot\vec b\times\vec c|=|12|=12 (book 3.7.5).

Options A, C and D come from dividing or multiplying 12 by some factor. The \tfrac16 is only for the tetrahedron (which would give 2), so it is not needed here; the parallelepiped volume is the triple product itself.

The book gives two ways to expand (a+b)^n: the binomial theorem and Pascal's triangle, whose nth row supplies the binomial coefficients.

Factorization, roots and division belong to polynomials (Unit 5) and are not used to write out a binomial expansion.

W=\vec F\cdot\vec d=(2)(3)+(1)(4)+(3)(-2)=6+4-6=4.

Option B, 10, comes from ignoring the \hat k term (6+4+0), option D, 16, from using +6 instead of -6 for it, and option A, -4, from flipping the overall sign. The \hat k product is (3)(-2)=-6.

In a proof by induction we first prove the basis step P(1), then assume that the statement is true for n=k. This assumption is called the inductive hypothesis.

The base case (A) is the check at the first value of n, and the proof (B) and the result (D) come at the end, after P(k+1) has been shown.

In the division algorithm \text{dividend}=\text{divisor}\times\text{quotient}+\text{remainder}, the division stops only when the remainder is 0 or has a degree smaller than the degree of the divisor. So the remainder is always of lower degree than the divisor.

Options A and B would mean the remainder could still be divided by the divisor, so the division would not be finished. Option D is meaningless for a polynomial.

\tan\theta=2=\dfrac21 in the first quadrant: opposite side 2, adjacent side 1, so the hypotenuse is \sqrt{1^2+2^2}=\sqrt5.

Option B, \dfrac2{\sqrt5}, is \sin\theta (opposite over hypotenuse), the ratio most often confused with cosine here.

By the Factor Theorem, (x-a) is a factor of f(x) exactly when f(a)=0. Here a=2, so f(2)=0.

Option A, -2, is the tempting wrong choice: it would come from f(-2), which belongs to the factor (x+2). The factor (x-2) vanishes at x=2, not at x=-2.

\sin x=\dfrac35 with x acute gives \cos x=\sqrt{1-\dfrac9{25}}=\dfrac45.

Option A, \dfrac{12}{25}, forgets the factor 2, and C, \dfrac7{25}, is \cos2x=1-2\sin^2x.

\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta is largest when \cos\theta=1, that is \theta=0^\circ, so the vectors are parallel (same direction).

Mutually perpendicular vectors give \cos90^\circ=0, the smallest magnitude, not the largest; opposite directions give \cos180^\circ=-1, the minimum value.

A graph symmetric about the origin satisfies f(-x)=-f(x) and is an odd function; \sin x is the standard example, \sin(-x)=-\sin x.

An even function (A) has a graph symmetric about the y-axis.

\vec u\cdot\vec v=(2)(-2)+(3)(0)+(-4)(-1)=-4+0+4=0. The \hat j component of \vec v is 0, so the middle product vanishes.

The vectors are perpendicular. Option D, -2, comes from adding -4 and the +2 that is not there; the \hat k term is (-4)(-1)=+4, not -4.

The range of a\sin bx is [-|a|,|a|]. Here a=4, so -4\le4\sin4x\le4.

The factor 4 inside, in \sin4x, changes the period (to \dfrac\pi2) but not the range. Option C wrongly multiplies the two 4's to get 16.

Three vectors are coplanar when their scalar triple product is zero, \vec P\cdot(\vec Q\times\vec R)=0 (book 3.7.6): \vec Q\times\vec R is perpendicular to the plane of \vec Q and \vec R, and if \vec P lies in that plane it is perpendicular to \vec Q\times\vec R.

Options B and C are meaningless: \vec P\cdot\vec Q is a scalar, and a scalar has no cross or dot product with a vector. Option A is a vector (zero or not) and is a different condition.

Use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta) with \alpha=75^\circ, \beta=15^\circ.

Options B and D are cosine sums, which come from \cos\alpha\cos\beta or \sin\alpha\sin\beta, and C has the wrong sign. Check: \dfrac12(1+0.866)=0.933=0.966\times0.966.

The word MATHS has 5 different letters, and a 5-letter word uses all of them with order mattering. The number of arrangements is {}^{5}P_{5}=5!=120.

Option B, 60, would be {}^{5}P_{3} (3-letter words); option A, 20, is {}^{5}P_{2}. Here all 5 letters are used, so r=n=5.

This is the fundamental law of trigonometry: \cos(A-B)=\cos A\cos B+\sin A\sin B. The sign between the products is plus for the cosine of a difference.

Option A is \cos(A+B), and C and D are sine formulas (C is \sin(A+B) and D is \sin(B-A) with the wrong sign).

Choosing 6 numbers out of 49 for a lottery ticket is a selection: the order in which the numbers are picked does not matter. So the number of ways is the combination {}^{49}C_{6}.

Options B and C, \dfrac{49!}{6!} and {}^{49}P_{6}, count ordered selections ({}^{49}P_{6}=\dfrac{49!}{43!}), which count each group of 6 numbers 6! times.

For two sines use -2\sin\alpha\sin\beta=\cos(\alpha+\beta)-\cos(\alpha-\beta), that is \sin\alpha\sin\beta=\dfrac12\big[\cos(\alpha-\beta)-\cos(\alpha+\beta)\big].

Option B, \dfrac{\sqrt3}{2}, is \sin60^\circ alone and forgets the factor \sin30^\circ=\dfrac12. Check: 0.866\times0.5=0.433=\dfrac{\sqrt3}{4}.

Verify that {}^{n}C_{r}+{}^{n}C_{r-1}={}^{n+1}C_{r}, written \dbinom{n}{r}+\dbinom{n}{r-1}=\dbinom{n+1}{r}. Use {}^{n}C_{r}=\dfrac{n!}{r!\,(n-r)!}.

Hence verified.

Use {}^{n}P_{r}=n(n-1)(n-2)\cdots(n-r+1) for both permutations.

Check: {}^{9}P_{4}=3024 and {}^{8}P_{3}=336, and 3024\div336=9.

The last number is the remainder, 0, so x-1 is a factor (Factor Theorem). The other numbers are the coefficients of the quotient, which has degree one less than P.

Together with the factor x-1 from part (a), all the zeros of P(x) are 0,1,2,3, and P(x)=x(x-1)(x-2)(x-3).

Let f(x)=x^{3}-3x^{2}-4x+12. Try the divisors of the last term 12 as values of x until f(x)=0.

The roots are 2,\ 3 and -2.

A group is a selection, so order does not matter. There are 5+4=9 students in all, and 3 are chosen with no condition.

EDUCATION has 9 different letters: the vowels are E, U, A, I, O (5 of them) and the consonants are D, C, T, N (4 of them).

By the Factor Theorem, (x+2)=(x-(-2)) is a factor of f(x) if and only if f(-2)=0.

Since f(-2)=0, (x+2) is a factor of f(x).

The expansion gives back P(x), so the factorisation is confirmed.

There are 8+6=14 persons. A committee is a selection, so use combinations. "At least 2 women" means 2,\ 3,\ 4 or 5 women, and the rest are men.

The committee can be formed in 1526 ways.

Since P(-3)=0, (x+3) is a factor of P(x). Hence verified.

Since (x+3) is a factor, divide by x+3 with synthetic divisor c=-3. Coefficients: 2,\ 1,\ -13,\ 6.

The roots of P(x)=0 are the zeros of P(x). Use the factorisation of part (b).

No method is named, so use Gauss elimination: reduce the augmented matrix A_b to echelon form, then back-substitute.

Statement. For real angles \alpha>\beta in standard position, \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.

Statement

For any three vectors, \vec a\cdot\vec b\times\vec c=\vec b\cdot\vec c\times\vec a=\vec c\cdot\vec a\times\vec b. Moving the vectors one place round the cycle \vec a\to\vec b\to\vec c\to\vec a leaves the scalar triple product unchanged.

Proof

Choose a team of 5 from 7 engineers and 5 managers (order does not matter, so use combinations). Total persons: 12.

Parts (c) and (d) have the same answer because both mean 4 or 5 engineers.

By the Remainder Theorem, the remainder on division by x-2 is P(2). The remainder is zero, so P(2)=0.

From part (a), k=15, so P(x)=2x^{3}-15x^{2}+16x+12. Since P(2)=0, (x-2) is a factor. Divide by x-2 with synthetic division, c=2, coefficients 2,\ -15,\ 16,\ 12.

Since 169=-(13i)^2, we get z^2+169=z^2-(13i)^2=(z-13i)(z+13i).

Option A is the factorisation of z^2-169. Options C and D are perfect squares whose expansion has a middle term \mp26iz, so they cannot equal z^2+169.

9=-(3i)^2, so z^2+9=z^2-(3i)^2=(z-3i)(z+3i).

Option B factorises z^2-9. Options C and D are perfect squares and have a middle term when expanded, so they cannot equal z^2+9.

z=1+i lies in the first quadrant (x=1>0,\ y=1>0), so \operatorname{Arg}z=\tan^{-1}\left|\dfrac11\right|=\tan^{-1}1=\dfrac{\pi}{4}.

Option D, \dfrac\pi2, is the argument of the pure imaginary number i (when x=0), not of 1+i. Here the real and imaginary parts are equal, so the point lies on the line y=x at 45^\circ.

The angle \dfrac{2\pi}{3}=120^\circ is in the second quadrant, where cosine is negative and sine is positive: \cos\dfrac{2\pi}{3}=-\dfrac12 and \sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}. So x=4\left(-\dfrac12\right)=-2 and y=4\cdot\dfrac{\sqrt3}{2}=2\sqrt3.

Option C would need \cos\theta>0 (first quadrant, \theta=\dfrac\pi3), and option B would need both negative (third quadrant). The check |z|=\sqrt{4+12}=4 holds for all four options, so the signs must be decided from the quadrant.

The identity matrix I_n is already in reduced echelon form: every row has a leading 1, and there is no zero row. So it has n non-zero rows and \operatorname{Rank}I_n=n, its order.

Rank can never exceed the number of rows, so option C is impossible. Option A would need a zero row, and the rank of any matrix is defined, so D is wrong too.

An inconsistent system has no solution at all (\operatorname{Rank}A\neq\operatorname{Rank}A_b). A system with one solution (C) or at least one solution (B) is consistent.

Option D cannot occur: a linear system has no solution, one solution or infinitely many, never exactly two.

By definition a system of linear equations with no solution at all is inconsistent; in terms of rank, \operatorname{Rank}A\neq\operatorname{Rank}A_b.

A unique solution (B) and infinitely many solutions (C) both make the system consistent, since at least one solution exists. "Repeated solutions" (D) is not a category for linear systems.

Take the common factor 2 out of row 1 and the common factor 3 out of row 2. Each factor multiplies the determinant: the new determinant is 2\cdot3\cdot\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=6\cdot5=30.

Option B is 2\cdot5 (row 1 only) and option C is 3\cdot5 (row 2 only). Option A forgets to multiply by the original value 5.

a_5=a_1+(5-1)d=2+4(3)=14.

Option B, 11, is the 4th term (2+3\cdot3); it comes from using n-1 with n=4. Using 5d instead of 4d gives 17, which is not an option.

S_{15}=\dfrac{15}{2}\left[2(2)+14(4)\right]=\dfrac{15}{2}(60)=450.

Using 15d in place of (n-1)d=14d gives \dfrac{15}{2}(64)=480, the tempting option C.

A sequence is a harmonic sequence when the reciprocals of its terms form an A.P. So the H.P. made from the A.P. a_1,a_2,a_3,\dots is \dfrac1{a_1},\dfrac1{a_2},\dfrac1{a_3},\dots.

Options B and C are sums (series), not sequences, and C takes the reciprocal of the whole sum rather than of each term.

T_1=(3)(3^0)=3 and T_2=(5)(3^1)=15, so T_1+T_2=18.

Option A, 5, is only the arithmetic factor 2n+1 of the second term; the geometric factor 3^{n-1} must also be multiplied in.

The common ratio is r=2x. The series converges when |r|<1, that is |2x|<1, so -\dfrac12<x<\dfrac12. Of the options only x=\dfrac14 lies in this interval (then r=\dfrac12).

At x=\dfrac12 the ratio is r=1, which is not less than 1, so the series does not converge (option A). Options C and D give r=2 and r=4.

The coefficient of the middle term of (a+b)^n is a binomial coefficient: \dbinom{n}{n/2} for even n (or the two equal coefficients \dbinom{n}{(n-1)/2} for odd n). Its value changes with n, for example \dbinom{2}{1}=2, \dbinom{4}{2}=6, \dbinom{6}{3}=20.

So it is not always 1 (A), not independent of n (B), and never zero (D).

The first step of a proof by induction is the basis step: verify the statement for the first value, usually n=1, the base case.

The hypothesis (A) is assumed in the second step, and the conclusion (C) and result (B) come last.

A statement such as \"the number of ways to climb n steps is the nth term of a sequence\" is a claim about every positive integer n. Counting the 10-step case does not show it for all n; the general claim is proved by mathematical induction (the count for n+1 steps follows from the counts for n and n-1).

Empirical counting only checks particular cases, and Pascal's triangle and the binomial theorem are for expanding (a+b)^n.

The remainder has a degree smaller than the divisor. The divisor is linear (degree 1), so the remainder has degree less than 1, that is degree 0: it is a constant (possibly 0). For example P(x) divided by x-c leaves the constant P(c).

Option B, degree 1, is the tempting wrong choice because it is the degree of the divisor; a remainder of degree 1 could still be divided by the linear divisor.

n=6 is even, so the middle term is the \left(\dfrac62+1\right)=4th term:

Options A and D use the wrong coefficient \dbinom62, and C and D use the exponents x^4y^2.

The Factor Theorem says: if f(a)=0, then (x-a) is a factor of f(x) (and conversely).

Option D, (x+a), is the usual trap: it is the factor when f(-a)=0. Options A and B are not factors at all: f(a) is a number and a is a number, while a factor of a polynomial is itself a polynomial.

By the Remainder Theorem the remainder on dividing by x-2 is P(2).

The remainder is 1, not 0, so x-2 is not a factor (option B is wrong). Substituting x=-2 instead of x=2 would be the mistake of dividing by x+2.

\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0 with both magnitudes non-zero forces \cos\theta=0, so \theta=90^\circ: the vectors are mutually perpendicular.

Parallel vectors have \cos\theta=\pm1, so their dot product is \pm|\vec a||\vec b|, not zero. It is the cross product that is zero for parallel vectors.

\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0 gives \cos\theta=0, so \theta=90^\circ.

0^\circ would make the dot product maximum (|\vec a||\vec b|), and 180^\circ makes it -|\vec a||\vec b|; it is the cross product that vanishes at 0^\circ and 180^\circ.

Multiply matching components and add (book 3.5.2): \vec a\cdot\vec b=a_1a_2+b_1b_2, because \hat i\cdot\hat i=\hat j\cdot\hat j=1 and \hat i\cdot\hat j=0.

Option D, a_2b_1-a_1b_2, has the shape of a 2\times2 determinant, which belongs to the cross product, not the dot product.

The projection of \vec a on \vec b is \dfrac{\vec a\cdot\vec b}{|\vec b|} (book 3.5.11). If \vec a\perp\vec b then \vec a\cdot\vec b=0, so the projection is 0 (the vector has no component along \vec b).

It is not undefined: the formula only fails when |\vec b|=0, which is not the case for a perpendicular pair of non-zero vectors.

Orthogonal means perpendicular, so \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0, giving \cos\theta=0 and \theta=\dfrac{\pi}{2}.

\pi is the angle for opposite directions and 2\pi (a full turn) brings the vector back to parallel; neither is a right angle.

Option A, 0, would mean the vectors are coplanar; they are not, since the determinant is non-zero. Option B, 1, is the volume of the unit cube, whereas these three edges span a solid of volume 2.

\pi+\beta is an allied angle, and \pi is an even multiple of a right angle, so cosine keeps its name. The angle \pi+\beta is in the third quadrant, where cosine is negative.

Options C and D change cosine into sine, which happens only for \dfrac{\pi}{2} and \dfrac{3\pi}{2}.

Use 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta) with \alpha=75^\circ and \beta=15^\circ.

The other options are cosine sums or differences, or have the wrong sign.

Add the formulas \cos(A+B)=\cos A\cos B-\sin A\sin B and \cos(A-B)=\cos A\cos B+\sin A\sin B. The sine terms cancel.

Option D is the formula for \sin A\cos B, not \cos A\cos B.

\theta is acute, so \dfrac\theta2 is acute and \cos\dfrac\theta2>0. Use the half angle identity.

Option B, \dfrac12, is \cos\theta itself, and C is \cos45^\circ. Check: \cos\theta=\dfrac12 gives \theta=60^\circ, and \cos30^\circ=\dfrac{\sqrt3}2.

The period of \cos x is 2\pi, so the period of \cos ax is \dfrac{2\pi}{|a|}. Here a=\dfrac25.

Option A, \dfrac52\pi, is \dfrac{\pi}{2/5}, which is the period of a tangent, not a cosine; B is the period of \cos x itself.

An odd function satisfies f(-x)=-f(x). \tan(-x)=-\tan x, so \tan x is odd. \cos x and \sec x are even, and \sin x+\cos x is neither: f(-x)=-\sin x+\cos x, which is neither f(x) nor -f(x) (Example 9(i) of the book).

A function is even if f(-x)=f(x). Only \cos(-x)=\cos x satisfies this. The others are odd: (-x)^3=-x^3, \sin(-x)=-\sin x, and f(x)=x gives f(-x)=-x.