1. Pure Mathematics AS 1

Keep the factor outside the square: ax²+bx+c=a[x²+(b/a)x]+c. Completing the square is also the quickest route to a quadratic equation’s exact roots and range.

x²−6x+5=(x−3)²−4, so the graph has minimum −4 at x=3 and range y≥−4.

Use the discriminant when the question asks how many roots, a tangent condition or a parameter range. It is often cleaner than applying the quadratic formula twice.

For x²−4x+k=0, Δ=16−4k. Two real roots require k<4; tangency occurs at k=4.

Example: $2x^2+3x-2=0$ gives $(2x-1)(x+2)=0$, so $x=\tfrac12$ or $x=-2$. Keep the $\pm$ branch when square roots arise and give exact roots unless approximation is requested.

For a quadratic inequality, first find the boundary roots, place them on a number line, then determine the sign in each interval from the parabola or a test value. Include a root only for $\leq$ or $\geq$.

Thus $y=3$ gives $x=-\tfrac72$, and $y=-2$ gives $x=-1$. The intersection points are $(-\tfrac72,3)$ and $(-1,-2)$; substitution in both originals verifies them.

Algebraic solution pairs are exactly the graph intersections. Zero, one or two real pairs correspond to no intersection, tangency or two intersections in the usual line-quadratic case.

Use four checks: identify the same repeated expression, substitute without changing coefficients, apply the range/domain of $g$, then reverse the substitution completely and verify in the original equation.

The same structure may use $u=x^3$, $u=1/x$, $u=\tan x$ or a stated expression such as $u=2x-3$. Its restrictions differ: $x^2\geq0$, $1/x\ne0$, and trigonometric reversal must respect the required interval.

Read f(a) as the output when input a is used. For composites, (f∘g)(x)=f(g(x)) and x must lie in g’s domain with g(x) in f’s domain.

For f(x)=√(x−2), the real domain is x≥2 and the range is y≥0; f(6)=2.

Find the inner range first, then apply the outer function. Restricting a domain can change the range and can make a formula that is usually invertible behave differently.

If g(x)=x² on [0,2] and f(u)=√u, then (f∘g)(x)=x on [0,2]; using all real x would describe a different domain and range.

To find f⁻¹, write y=f(x), interchange x and y, then solve for y. The inverse domain is the original range, and its range is the original domain.

f(x)=3x−2 is one-one on ℝ and f⁻¹(x)=(x+2)/3. The graphs reflect in y=x.

Draw $y=x$, reflect several defining points and preserve their order. An $x$-intercept $(a,0)$ becomes the $y$-intercept $(0,a)$; horizontal and vertical asymptotes swap; the domain and range swap.

The two graphs can meet only on $y=x$, because a reflected point is unchanged there. Such intersections satisfy $f(x)=x$, provided $x$ lies in both relevant domains.

Use the original graph's endpoints, open/closed points, turning behaviour and asymptotes rather than inventing a new shape. The inverse must still pass the vertical-line test because the original is one-one.

Track a known point, intercept, asymptote or turning point through the coordinate mapping. This is safer than relying on a memorised verbal direction, especially for inside transformations.

When two transformations share one axis, order can change the translation size. State a valid sequence explicitly and verify it by applying the operations to the function or to a reference point.

Calculate m=(y₂−y₁)/(x₂−x₁), then substitute a known point to determine the intercept. Parallel lines have equal gradients; perpendicular non-vertical lines satisfy m₁m₂=−1.

Through (2,5) with gradient −3: y−5=−3(x−2), so y=−3x+11.

Use $y=mx+c$ to expose gradient/intercept, $y-y_1=m(x-x_1)$ with a known point, and $ax+by+c=0$ when an integer-coefficient final form or simultaneous intersection is convenient.

A perpendicular bisector passes through midpoint $M$ and has gradient $-1/m_{AB}$. Compute those two facts first, then use point-gradient form; this is the standard route to a circle centre from a chord.

Use the radius as a distance, check r²>0, and substitute a point to test whether it lies on the circle. A zero radius is a single point, not a proper circle.

x²+y²−6x+4y−3=0 becomes (x−3)²+(y+2)²=16, so centre (3,−2) and radius 4.

Use elementary properties when they shorten the route: a radius is perpendicular to the tangent at contact; the perpendicular bisector of a chord passes through the centre; an angle in a semicircle is $90^\circ$; symmetry about a diameter pairs points.

For a tangent through a known point, either impose discriminant $=0$ after substitution or find the radius gradient and use its negative reciprocal. Then substitute back to obtain and verify the contact point.

A line is tangent exactly when the perpendicular distance from the circle centre to the line equals the radius. This may classify a parameter without solving both coordinates.

If one graph contains a parameter $k$, rearrange the intersection equation into a polynomial in $x$ whose coefficients depend on $k$. For a quadratic, use its discriminant to classify the graph positions.

Convert degrees before using calculus, arc length or sector formulas. Keep θ dimensionless in the formula and attach units to s or area.

A radius 5 cm and angle 1.2 rad give arc length 6.0 cm; the same numerical angle in degrees would give a wrong result.

Use the same angle units in every term, identify whether the question asks for sector or segment, and check that 0≤θ≤2π for an ordinary sector.

With r=4 and θ=π/3, s=4π/3 and sector area=8π/3. A minor segment would be this sector minus ½r²sinθ.

Mark the requested domain, midline or tangent asymptotes, then place one cycle from known exact points before repeating by the period. A negative outside coefficient reflects in the midline.

$y=2\cos(3x-\pi)+1$ has amplitude $2$, period $2\pi/3$, midline $y=1$ and shift $\pi/3$ right. Its greatest and least values are $3$ and $-1$.

Reduce angles using periodicity and reference angles before applying the table. Keep radicals exact until a decimal is explicitly requested.

sin(5π/6)=sin(π−π/6)=1/2, while cos(5π/6)=−√3/2 because cosine is negative in quadrant II.

When inverse notation appears inside an equation, evaluate the principal value with consistent degree/radian mode, then continue the algebra. Finding every solution of a trig equation is a separate next objective based on graphs and the stated interval.

Use the ratio identity to replace tangent by sine/cosine, especially when forming a common denominator. Use the Pythagorean identity to replace $1-\sin^2\theta$ by $\cos^2\theta$ or $1-\cos^2\theta$ by $\sin^2\theta$.

To prove an identity, work from one side and make valid algebraic substitutions until it matches the other. Do not begin by assuming both sides equal; show every cancellation or common denominator.

Factorise or use a substitution when expressions such as 2sin²x−sinx−1 appear. Check every candidate in the original equation, especially after squaring.

2sinx−1=0 gives sinx=1/2, so on [0,2π] the solutions are π/6 and 5π/6.

Write the general term T_{r+1}=inom nr a^{n-r}b^r. If a variable occurs in $a$ or $b$, equate its resulting power to the requested power, solve for the integer $r$, then simplify the entire term.

If the second term is negative, include its sign inside the power. For example, the sign of $(-3x)^r$ depends on whether $r$ is odd or even.

Calculate at least two consecutive differences or ratios. Classify only if the same value continues across all given adjacent pairs; then use that value as $d$ or $r$.

$11,7,3,-1,\ldots$ is AP with $d=-4$. $3,-6,12,-24,\ldots$ is GP with $r=-2$; its signs alternate because the ratio is negative.

Define the first term and difference/ratio for each progression. Turn every stated term, sum or three-term relationship into an equation, solve the simultaneous system, reject values that violate the original order or denominator conditions, and substitute back.

Check convergence before using the formula. A negative r gives alternating partial sums, but still converges when |r|<1.

3−1.5+0.75−… has a=3,r=−0.5 and S∞=3/1.5=2; the alternating signs do not prevent convergence.

Rewrite roots and reciprocals as rational powers before differentiating, for example $\sqrt{x}=x^{1/2}$ and $1/x^2=x^{-2}$. Simplify only when doing so preserves the function's domain.

Name the dependent variables, write the relation between them, differentiate with respect to the required variable (often time), substitute the specified instant only after differentiating, and state the signed result with units.

Solve $f'(x)=0$, substitute each solution into $f$ to obtain full coordinates, then classify. If $f''(a)>0$ the gradient is increasing through zero and the point is a local minimum; if $f''(a)<0$ it is a local maximum.

Place the stationary coordinates, use their nature to set the local turning direction, then combine them with intercepts, domain and end behaviour. A local classification alone does not determine the whole graph.

Rewrite roots and reciprocals as rational powers, split constant multiples, sums and differences, then integrate term by term. For a linear inner expression, divide by its gradient $a$.

Differentiate the antiderivative to check every coefficient and power. Include $+C$ for an indefinite integral because all constants have derivative zero.

Integrate first, then substitute the given coordinate or initial value. In a motion problem, use the condition on displacement or velocity at the stated time.

If dy/dx=6x−2 and y(1)=5, then y=3x²−2x+C, giving C=4 and y=3x²−2x+4.

Find an antiderivative, substitute the upper limit and subtract the value at the lower limit. Reverse limits reverse the sign; equal limits give zero.

A simple improper integral converges only if the one-sided limiting value is finite. Never substitute an endpoint where the integrand or antiderivative expression is undefined.

Sketch or compare the boundaries, solve intersections to obtain limits, identify which curve is upper/outer on each interval, split wherever that order or sign changes, then integrate and state square or cubic units.

For rotation about the $y$-axis, express the horizontal radius in terms of $y$ and integrate the corresponding disc/washer cross-sectional area with respect to $y$ when the syllabus problem requires it.