Unit P2: Pure Mathematics AS 2

A mathematical proof starts from stated assumptions and uses valid logical steps to show that the conclusion must follow for every object in the stated domain.

To prove $2x^2+8\ge8x$ for every real $x$, move everything to one side: $$2x^2-8x+8=2(x-2)^2.$$ Because a real square is non-negative, $2(x-2)^2\ge0$, so the stated inequality follows for every real $x$.

When multiplying, dividing, taking roots or cancelling factors, state the condition that makes the step valid. For example, dividing an inequality by a positive quantity preserves its direction; dividing by a quantity of unknown sign does not.

Proof by exhaustion works when the allowed possibilities can be divided into a finite, complete set of cases. Each case must be checked, and the cases must leave no allowed value uncovered.

To prove $n^2+n$ is even for every integer $n$, exhaust the two parity cases. If $n=2k$, then $n^2+n=2k(2k+1)$, which has a factor 2. If $n=2k+1$, then $n^2+n=(2k+1)(2k+2)=2(2k+1)(k+1)$, also even. Every integer is even or odd, so the proof is complete.

For a literal finite set, the single-digit primes are exactly $2,3,5,7$. The values of $p^2+p$ are respectively $6,12,30,56$, all even. Naming the complete allowed set is what turns these checks into a proof.

A universal claim says a conclusion holds for every allowed input. To disprove it, one counterexample is enough: choose an input that satisfies the claim's conditions but makes its conclusion false.

Consider the claim ‘$n^2-n+1$ is prime for every positive integer $n$’. Take $n=5$, which is in the stated domain. Then $$5^2-5+1=21=3\times7,$$ so the result is composite. Therefore $n=5$ is a counterexample and the universal claim is false.

A useful search targets values likely to expose the weakness, but the final disproof must show the calculation and the failed property. A valid counterexample need not be the smallest one.

Dividing a polynomial $f(x)$ by a linear expression gives a quotient $Q(x)$ and constant remainder $R$. One identity connects the calculation to both the Remainder and Factor Theorems.

For long division, order descending powers and insert zero coefficients for missing terms. Divide the leading terms, multiply the whole divisor by the new quotient term, subtract, and repeat until the remainder is constant. Verify with dividend = divisor × quotient + remainder.

For $f(x)=2x^3+x^2-7x+2$, division by $x-2$ gives $$f(x)=(x-2)(2x^2+5x+3)+8.$$ The theorem check agrees: $f(2)=16+4-14+2=8$, so $x-2$ is not a factor.

For $g(x)=2x^3-3x^2-8x+12$, $g(2)=0$, so $x-2$ is a factor. Division gives $2x^2+x-6=(2x-3)(x+2)$; hence $$g(x)=(x-2)(2x-3)(x+2).$$

A circle is the set of points a fixed distance $r$ from its centre $(a,b)$. Its coordinate equation records that distance using Pythagoras:

Read the centre as $(a,b)$ and the radius as the positive square root of the right-hand side. The signs inside the brackets are reversed. To build an equation, substitute the known centre and radius; if a diameter's endpoints are known, its midpoint is the centre and half its length is the radius.

For $x^2+y^2-6x+4y-12=0$, move the constant and add $9$ and $4$ to both sides: $$x^2-6x+9+y^2+4y+4=12+9+4,$$ so $$(x-3)^2+(y+2)^2=25.$$ Thus the centre is $(3,-2)$ and the radius is $5$.

In the example, $P=(7,1)$ lies on the circle because $4^2+3^2=25$. The radius $CP$ has gradient $3/4$, so the tangent gradient is $-4/3$ and its equation is $y-1=-\frac43(x-7)$.

A sequence is an ordered list whose position matters. A rule can give a term directly from its position, or generate each term from the previous one.

For $u_n=3n-1$, the sequence begins $2,5,8,11,\ldots$ and $u_7=3(7)-1=20$ without finding terms 5 or 6.

For $x_1=2$ and $x_{n+1}=x_n^2-1$, $$x_2=3,\qquad x_3=8,\qquad x_4=63.$$ Each input is the preceding term, so the starting value is part of the definition.

An arithmetic sequence has a constant difference $d$. Starting from first term $a$, moving to term $n$ adds $d$ exactly $n-1$ times.

The sum formula follows by writing the same finite sum forwards and backwards: $$S_n=a+(a+d)+\cdots+[a+(n-1)d],$$ $$S_n=[a+(n-1)d]+\cdots+(a+d)+a.$$ Each of the $n$ paired columns totals $2a+(n-1)d$, so doubling and then halving gives the formula.

Sigma notation compresses an addition: $\sum_{r=1}^{n}u_r$ means $u_1+u_2+\cdots+u_n$. Setting $a=1,d=1$ gives $$1+2+\cdots+n=\frac{n(n+1)}2.$$

For $7,11,15,\ldots$, $a=7$ and $d=4$. Thus $u_{20}=83$ and $$S_{20}=\frac{20}{2}(7+83)=900.$$

Classify a sequence by comparing consecutive terms or by identifying an exact repeating cycle.

For an explicit sequence, inspect $u_{n+1}-u_n$: a value always positive proves increasing, while one always negative proves decreasing. For $u_n=n^2+1$, $$u_{n+1}-u_n=2n+1>0$$ for positive integer $n$, so the sequence is increasing.

The sequence $4,\frac34,-\frac13,4,\frac34,-\frac13,\ldots$ repeats every three terms, so it is periodic of order $3$. With a one-step deterministic recurrence, returning to an earlier value makes the same following cycle repeat.

A geometric sequence multiplies by the same common ratio $r$ each time. With first term $a$, its terms and finite sum are

Write $S_n=a+ar+\cdots+ar^{n-1}$ and multiply by $r$. Subtracting $rS_n$ from $S_n$ cancels the middle terms, leaving $$(1-r)S_n=a-ar^n,$$ which gives the finite-sum formula.

Only when $|r|<1$ does $r^n\to0$, so only then does the geometric sum converge to $$S_\infty=\frac{a}{1-r}.$$ A negative ratio may converge while the terms alternate signs.

For $a=5,r=0.8$, requiring $S_n>24$ gives $25(1-0.8^n)>24$, hence $0.8^n<0.04$. Therefore $$n>\frac{\log(0.04)}{\log(0.8)}\approx14.43,$$ so the smallest integer is $15$.

For a positive integer $n$, the binomial theorem gives every term of $(a+bx)^n$ without repeated multiplication.

Here $n!=n(n-1)\cdots2\cdot1$ and $0!=1$. The index $r$ is the power of $x$, so choosing $r=0,1,2,\ldots$ produces terms in ascending powers.

For $(2-3x)^4$, $$\begin{aligned}(2-3x)^4={}&2^4+\binom41 2^3(-3x)+\binom42 2^2(-3x)^2\\&+\binom43 2(-3x)^3+(-3x)^4\\={}&16-96x+216x^2-216x^3+81x^4.\end{aligned}$$

In $y=a^x$, the variable is in the exponent. The restrictions $a>0$ and $a\ne1$ give a real exponential curve rather than a sign-changing or constant rule.

Every permitted base gives $a^0=1$, so the graph crosses the $y$-axis at $(0,1)$. Its domain is all real $x$, its range is $y>0$, and $y=0$ is a horizontal asymptote: the curve approaches but never meets the $x$-axis.

Known graph transformations preserve this structure. For $y=3\cdot2^x+4$, the $y$-intercept is $(0,7)$, the horizontal asymptote is $y=4$, and the curve is increasing above that line.

A logarithm is an exponent: $\log_a x=y$ means $a^y=x$. Therefore $a>0$, $a\ne1$, and every logarithm argument must be positive.

These laws are the index laws read through the inverse operation: multiplying powers adds exponents, dividing subtracts them, and raising a power multiplies its exponent.

Solve $\log_2(x-1)+\log_2(x-3)=3$. The domain requires $x>3$. Combining gives $$\log_2[(x-1)(x-3)]=\log_2 8,$$ so $(x-1)(x-3)=8$ and $(x-5)(x+1)=0$. Only $x=5$ satisfies the domain.

To solve $a^x=b$, apply a logarithm to both sides. The power law moves the unknown exponent in front, where ordinary algebra can isolate it.

This requires $a>0$, $a\ne1$, and $b>0$. The change-of-base formula allows any convenient valid base $c$, commonly $10$ or $e$.

For $5^{2x-1}=17$, take natural logarithms: $$(2x-1)\ln5=\ln17.$$ Hence $$x=\frac12\left(1+\frac{\ln17}{\ln5}\right)\approx1.380.$$ Substitution restores $5^{2x-1}=17$, providing a check.

A trigonometric identity is true for every angle where both sides are defined. These two identities let you replace one function by another without changing the relationship.

Useful rearrangements are $\sin^2\theta=1-\cos^2\theta$ and $\cos^2\theta=1-\sin^2\theta$. Choose the version that leaves only one trigonometric function.

If $\cos\theta=-\frac35$ and $\frac\pi2<\theta<\pi$, then $$\sin^2\theta=1-\frac9{25}=\frac{16}{25}.$$ Sine is positive in this interval, so $\sin\theta=\frac45$ and $\tan\theta=-\frac43$. The identity gives the magnitude; the interval determines the sign.

For example, $8\tan\theta=3\cos\theta$ becomes $8\sin\theta=3\cos^2\theta=3(1-\sin^2\theta)$, hence $$3\sin^2\theta+8\sin\theta-3=0.$$

A trigonometric equation is complete only when every valid branch in the stated interval has been found. Keep the angle unit and interval visible throughout.

Use this sequence: rewrite in one trigonometric function; factor or solve the resulting algebraic equation; reject values outside the function's range; generate every periodic angle; undo any shift or multiple; then apply the original endpoints.

Here $n\in\mathbb Z$ and $\alpha$ is the appropriate inverse-trigonometric value. If $u=2x$ or $u=x+\frac\pi2$, solve over the corresponding $u$-interval before converting back to $x$.

Solve $6\cos^2x+\sin x-5=0$ for $0\le x<360^\circ$. Let $s=\sin x$: $$6(1-s^2)+s-5=0\quad\Longrightarrow\quad(3s+1)(2s-1)=0.$$ Thus $\sin x=\frac12$ or $\sin x=-\frac13$, giving $$x=30^\circ,\ 150^\circ,\ 199.5^\circ,\ 340.5^\circ$$ to one decimal place where needed.

The derivative $f'(x)$ is the gradient of a curve. Its sign tells whether the function rises or falls, while a stationary point is a point where $f'(x)=0$.

When $f'(c)=0$, the second derivative gives a quick local test: $f''(c)<0$ indicates a maximum and $f''(c)>0$ a minimum. If $f''(c)=0$, this test is inconclusive, so use the sign of $f'$.

For $f(x)=x^3-3x^2-9x+5$, $$f'(x)=3(x+1)(x-3).$$ Thus the stationary points are $(-1,10)$ and $(3,-22)$. The derivative is positive for $x<-1$, negative for $-1<x<3$, and positive for $x>3$; therefore the first point is a local maximum and the second a local minimum.

For a curve sketch, combine stationary coordinates and increasing/decreasing intervals with intercepts and end behaviour. In a practical optimisation on a restricted domain, compare the objective value at every feasible stationary point and included endpoint, then state the maximum or minimum with its units and context.

A definite integral gives the accumulated signed value of a function between two bounds. First find an antiderivative $F$ with $F'(x)=f(x)$, then evaluate upper bound minus lower bound.

For example, $$\int_1^3(2x^2-4x+5)\,dx=\left[\frac23x^3-2x^2+5x\right]_1^3=15-\frac{11}{3}=\frac{34}{3}.$$ Write the substitution line before simplifying; this makes the order of subtraction visible.

Reversing the bounds changes the sign: $\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$. Adjacent intervals add, so $\int_a^c f=\int_a^b f+\int_b^c f$; this can reveal a missing integral without integrating again.

A definite integral is signed, but geometric area is non-negative. Sketch or compare the boundaries first, find their intersections, and integrate the vertical height of the region with respect to x.

The curves $y=6x-x^2$ and $y=2x$ meet where $6x-x^2=2x$, so $x=0$ or $x=4$. Between these values the quadratic is above the line. Hence $$A=\int_0^4\big[(6x-x^2)-2x\big]dx=\left[2x^2-\frac{x^3}{3}\right]_0^4=\frac{32}{3}.$$

If the upper boundary changes, or a curve crosses the x-axis inside the region, split the integral at that x-value and make each piece positive. Straight-line boundaries are handled by the same upper-minus-lower rule.

The trapezium rule replaces a curve by straight chords over equal-width strips. With n strips from a to b, calculate the width and list all n+1 ordinates before applying the endpoint weights.

Here $h=0.25$, so $$T_4=\frac{0.25}{2}\{1.0000+1.7321+2(1.2247+1.4142+1.5811)\}\approx1.3965.$$ Keep extra calculator digits until the final rounding.

More strips make h smaller and usually improve the approximation. If a trusted exact or more accurate value is available, the estimated absolute error is the absolute difference from the trapezium estimate. The chord picture also explains direction: chords below the curve give an underestimate; chords above it give an overestimate.