1. Pure Mathematics 1

Rewrite ax²+bx+c as a(x−h)²+k by factoring a and adding and subtracting the required square. The vertex is (h,k), and the sign of a tells whether it is a minimum or maximum.

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.

The constant changes when the square is completed; do not write (x−3)²+5, which expands to a different quadratic.

For ax²+bx+c=0, the discriminant Δ=b²−4ac. Δ>0 gives two distinct real roots, Δ=0 one repeated real root, and Δ<0 no real roots.

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.

Δ>0 does not mean both roots are positive; root signs require additional information such as sum and product.

A quadratic equation can be solved by factorising, completing the square or using x=(−b±√(b²−4ac))/(2a). All methods should return the same roots.

Factorise when integer or simple factors are visible; use the formula for a general or parameterised quadratic; complete the square when the vertex or inequality range is needed.

For 2x²+3x−2=0, factorising gives (2x−1)(x+2)=0, so x=1/2 or −2. The formula is a reliable check.

After multiplying or dividing an equation, preserve equivalence; taking a square root too early can lose the ± branch.

For two equations, eliminate one variable by substitution or a linear combination, solve the resulting equation, then back-substitute and verify in both original equations.

If one equation is quadratic, each valid root may produce a different pair. Keep every algebraic branch until the original equations or stated domain reject it.

From y=2x+1 and x²+y²=25, substitute to obtain x²+(2x+1)²=25; solve for both possible x values, then calculate y for each.

Solving only the positive square-root branch can omit a valid intersection, while numerical rounding can make a true pair appear not to satisfy the equations.

If an equation contains x² and x⁴, set u=x² and solve the resulting quadratic in u. Only values u≥0 correspond to real x, and each positive u gives x=±√u.

Make the substitution explicit, solve in the new variable, apply its domain restriction, then reverse the substitution and check the original equation.

x⁴−5x²+4=0 becomes u²−5u+4=0, so u=1 or 4 and x=±1,±2.

A negative solution for u cannot produce a real x, and replacing x² by u does not mean x itself equals u.

A function f maps each permitted input x to exactly one output f(x). The domain is the allowed set of inputs; the range is the set of outputs actually produced.

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.

A relation can assign one input several outputs and then is not a function; domain restrictions are part of the definition, not optional notes.

The range is the set of outputs produced by a function on its stated domain. For (f∘g)(x)=f(g(x)), x must be allowed by g and g(x) must lie in f’s domain.

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.

The range of f∘g is not automatically the range of f; only the part of f reached by g is relevant.

A function is one-one if f(a)=f(b) implies a=b. Equivalently, no horizontal line meets its graph more than once. This allows an inverse function to undo f on its 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.

A function can have an inverse relation but not an inverse function if it is not one-one on the stated domain.

If a graph folds back, restrict the original domain to a one-one branch before defining an inverse. The chosen branch determines which inverse values are allowed.

For x², choosing x≥0 gives inverse √x; choosing x≤0 gives inverse −√x. State the restriction as part of the function, not as an afterthought.

On [0,∞), f(x)=x² has f⁻¹(x)=√x with domain x≥0. On (−∞,0], the inverse is −√x.

Writing ±√x gives two outputs and is an inverse relation, not a single-valued inverse function.

For y=f(x), y=f(x)+a shifts up a, y=f(x−a) shifts right a, y=bf(x) scales vertical values, and y=f(bx) scales horizontal coordinates by 1/b.

Read transformations from the formula, track a known point (x,y), and apply inside changes before outside changes when several are combined.

From y=x², y=2(x−3)²+1 has vertex (3,1), vertical stretch 2 and the same upward-opening shape.

The sign inside the bracket reverses the horizontal direction: f(x−3) moves right, not left.

The gradient m measures change in y per unit change in x. A line can be written y=mx+c, y−y₁=m(x−x₁), or ax+by+c=0 depending on the information given.

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.

The intercept is not the y-coordinate of every point, and a vertical line has undefined gradient rather than gradient zero.

Use y=mx+c for gradient/intercept, y−y₁=m(x−x₁) for a known point and gradient, and ax+by+c=0 for intersections, distances or a normal-vector description.

Convert between forms carefully and keep coefficients proportional when comparing the same line. For a line through two points, calculate its gradient before choosing the form.

2x−3y+6=0 becomes y=(2/3)x+2, revealing gradient 2/3 and y-intercept 2.

Multiplying all coefficients by a non-zero constant gives the same line; changing only one coefficient changes its geometry.

The circle with centre (a,b) and radius r has (x−a)²+(y−b)²=r². Expanding gives a general quadratic form, but completing squares recovers the geometry.

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.

The signs in the centre reverse when completing squares: (x−3)² gives centre x=3, while (y+2)² gives y=−2.

Substitute the line equation into the circle equation. The resulting quadratic has two roots for a secant, one repeated root for a tangent, and no real roots when the line misses the circle.

The repeated-root condition is equivalent to the perpendicular distance from the centre to the line equalling the radius. Use whichever route makes the parameter condition clearer.

Substituting y=mx+c into a circle gives Δ=0 at tangency; the repeated x-value then gives the contact point.

A discriminant classifies real algebraic intersections, but a point must also satisfy any stated segment or domain restriction.

An intersection of y=f(x) and y=g(x) has f(x)=g(x); solve for x, then substitute to find y. The number of real roots gives the number of intersections.

Factorisation may reveal exact roots; otherwise use a graph or numerical method only after locating intervals. Reject roots outside the domain or interval shown.

For y=x² and y=2x+3, x²=2x+3 gives x=3 or −1, producing points (3,9) and (−1,1).

Solving f(x)=g(x) finds x-coordinates only; reporting roots without their corresponding y-values is not a complete intersection answer.

One radian is the angle subtended by an arc equal to the radius. A full turn is 2π radians, and θ radians correspond to arc length s=rθ.

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.

A radian is not a physical length, and writing s=rθ with θ in degrees silently introduces a factor of π/180.

For radius r and angle θ in radians, arc length s=rθ and sector area A=½r²θ. Segment area requires subtracting the corresponding triangle from the sector.

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.

Sector area is not ½r²sinθ; that is the triangle area used when a segment is formed.

For y=a sin(bx+c)+d, |a| is amplitude, period is 2π/|b|, −c/b is the horizontal shift, and d is the midline. Equivalent rules apply to cosine.

Mark the midline and one cycle before sketching. Check the sign of a and the requested domain; transformations change coordinates, not just the appearance.

y=2cos(3x−π)+1 has amplitude 2, period 2π/3, midline y=1 and shift π/3 to the right.

The coefficient inside the bracket changes period inversely; multiplying x by 3 does not stretch the graph by 3.

The standard exact values sin, cos and tan at 0, π/6, π/4, π/3 and π/2 follow from 30–60–90 and 45–45–90 triangles, with signs set by the quadrant.

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.

The reference angle gives a magnitude, not automatically the sign; tan is undefined where cos is zero.

sin⁻¹, cos⁻¹ and tan⁻¹ are functions with restricted principal ranges: typically [−π/2,π/2], [0,π] and (−π/2,π/2). They undo the trig function only on those ranges.

For an equation, use the principal value first, then generate other angles with symmetry and periodicity in the stated interval.

tan⁻¹(1)=π/4, but tan x=1 also has x=π/4+nπ; the inverse notation alone gives only the principal answer.

sin⁻¹x is not 1/sin x, and swapping sin with sin⁻¹ without checking the domain changes the problem.

Identities such as sin²x+cos²x=1, tanx=sinx/cosx and 1+tan²x=sec²x follow from definitions and hold wherever both sides are defined.

To prove an identity, simplify one side to the other using a common denominator or a fundamental identity; do not assume the result you are trying to prove.

(1−cos²x)/sinx simplifies to sinx when sinx≠0 because 1−cos²x=sin²x.

Algebraic cancellation can remove values where a denominator is zero; record the excluded angles even if the simplified expression looks defined.

Solve a trig equation by reducing it to a principal angle, applying quadrant symmetry, then adding periods. The interval determines which solutions survive.

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.

One inverse-trig answer is not the complete solution, and the period of tan is π rather than 2π.

For a positive integer n, (a+b)^n=Σᵣ C(n,r)a^{n−r}b^r. The general term identifies the power and coefficient without writing every term.

Track powers of the variable and signs carefully; for a coefficient question, set the target power before simplifying. The expansion is finite only for non-negative integer n.

The x² term in (1+2x)^5 is C(5,2)·(2x)²=40x².

The coefficient C(n,r) is not the whole term; the powers of both factors still contribute.

An arithmetic progression has constant difference d, with uₙ=a+(n−1)d and Sₙ=n/2[2a+(n−1)d]. A geometric progression has constant ratio r, with uₙ=arⁿ⁻¹ and Sₙ=a(rⁿ−1)/(r−1) when r≠1.

Check consecutive differences or ratios before choosing a formula. For an infinite geometric sum, convergence requires |r|<1 and S∞=a/(1−r).

For 3,6,12,…, a=3 and r=2, so it has no finite infinite sum because |r|>1.

A sequence can have a convergent partial-sum pattern without being geometric; never infer r from non-consecutive terms alone.

Use uₙ=a+(n−1)d for an arithmetic progression and uₙ=arⁿ⁻¹ for a geometric progression. Their finite sums follow from adding terms or multiplying by the common ratio.

Identify a,d or r from the information given, keep n as the term number, and check whether a sum is finite or infinite before choosing Sₙ or S∞.

For 5,8,11,…, uₙ=3n+2 and Sₙ=n(3n+7)/2; the first term is included once, not n times.

The nth term and nth partial sum are different quantities: Sₙ is not obtained by substituting n into uₙ.

For first term a and common ratio r, S∞=a/(1−r) exists only if |r|<1. The partial sums approach a finite limit because later terms shrink to zero.

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.

A ratio close to 1 may converge slowly, while r=1 or −1 does not produce a finite infinite sum.

The derivative f′(x) is lim_{h→0}[f(x+h)−f(x)]/h when the limit exists. It is the gradient of the tangent and measures instantaneous rate of change.

A secant gradient uses two distinct points; the tangent is its limiting value. At a sharp corner or cusp the two-sided derivative may not exist.

For f(x)=x², the difference quotient is 2x+h, so letting h→0 gives f′(x)=2x.

Substituting h=0 into the quotient before simplifying causes division by zero and is not differentiation.

Use linearity, the power rule d(xⁿ)/dx=nxⁿ⁻¹, product and quotient rules, and the chain rule for composites. Differentiate with respect to the stated variable.

For a product, u′v+uv′; for a quotient, (u′v−uv′)/v²; for f(g(x)), multiply by g′(x). Simplify after differentiating to avoid losing factors.

d[(x²+1)^3]/dx=6x(x²+1)² by the chain rule.

The derivative of uv is not u′v′, and the chain-rule multiplier is required even when the inner expression looks simple.

f′(x) gives the gradient or rate at x; solve f′(x)=m for a specified tangent gradient and f′(x)=0 for stationary candidates. Units carry through the rate interpretation.

Translate the derivative back into the problem: distance gives velocity, velocity gives acceleration, and a stationary candidate still needs classification or endpoint comparison for an optimum.

If s(t)=t³−6t²+9t, then v=s′=3t²−12t+9; stationary position occurs where v=0, but the physical time interval decides which roots matter.

A zero derivative is not automatically a maximum or minimum and may be a stationary inflection.

At a stationary point f′(x)=0. If f″(x)>0 it is a local minimum; if f″(x)<0 a local maximum. When f″=0, inspect the sign of f′ or higher derivatives.

Solve f′=0, calculate coordinates, classify each point and check endpoints if the question asks for an absolute maximum or minimum.

For f=x³−3x, f′=3(x²−1), giving x=±1; f″=6x classifies (−1,2) as a maximum and (1,−2) as a minimum.

Local classification does not compare distant points or endpoints, and f″=0 is inconclusive rather than proof of no turning point.

Integrate powers term by term using ∫xⁿdx=xⁿ⁺¹/(n+1)+C for n≠−1. Integration is the family of antiderivatives whose derivative is the original function.

Rewrite expressions into powers first, then integrate each term. Include C for an indefinite integral and differentiate the result to check.

∫(3x²−4x+5)dx=x³−2x²+5x+C.

The constant is not optional in an indefinite integral: different C values represent different antiderivatives.

After integrating a derivative, +C represents the unknown vertical position. A condition such as y=4 when x=1 determines C.

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.

Setting C=0 assumes a particular origin that the question may not give; it is not a harmless simplification.

∫ₐᵇf(x)dx=F(b)−F(a) for an antiderivative F. It accumulates signed area: portions below the x-axis contribute negatively.

Substitute the upper and lower limits into the antiderivative, not the original integrand. Reverse limits change the sign, and equal limits give zero.

∫₀²(2x−1)dx=[x²−x]₀²=2; the graph has a small negative part but the net signed area is 2.

A definite integral is not always geometric area; use absolute values or split intervals when total area is requested.

Area between curves is ∫(upper−lower)dx or ∫(right−left)dy. Rotating a region about an axis uses washers or shells, with radius and thickness tied to the chosen variable.

Find intersections first, split where the order changes, and keep units squared or cubed. A volume formula must match the axis and region.

Rotating y=x from x=0 to 2 about the x-axis gives V=π∫₀²x²dx=8π/3.

Integrating one curve without subtracting the other can count the wrong region, and a graph crossing may require separate intervals.