3. Pure Mathematics 3

For |f(x)|=g(x), first require g(x)≥0, then solve f(x)=g(x) or f(x)=−g(x). For inequalities, split according to whether the comparison side is non-negative.

Nested or variable-right-side moduli may need several intervals. Keep endpoints and test each candidate in the unsquared original equation.

|x−2|=x gives x≥0 and branches x−2=x (impossible) or 2−x=x, giving x=1.

Squaring both sides without checking the sign of the right side can add invalid solutions.

Use a known root a to divide P(x) by x−a, obtaining a lower-degree quotient. Then factor or solve the quotient and retain only roots satisfying the original polynomial.

Use the remainder theorem to test candidates, include missing powers as zero coefficients, and verify by multiplication. Repeated division reveals multiplicity.

If P(−1)=0 for a quartic, divide by x+1; if the quotient also vanishes at −1, the factor is repeated.

A factor found by trial does not guarantee the remaining quotient is quadratic; degree bookkeeping must be preserved.

For polynomial P, P(a)=0 exactly when x−a is a factor. Dividing by x−a leaves remainder P(a), so a quick substitution tests a candidate root.

After finding one factor, divide it out and solve the quotient. Repeated roots can be detected by testing the quotient or by checking derivatives when appropriate.

If P(−2)=0, x+2 divides P; a cubic then becomes a quadratic quotient that can be solved exactly.

The sign is tied to the factor: root a gives x−a, while root −a gives x+a.

Decompose a proper rational function into simpler fractions whose denominators are the linear or irreducible factors of the original denominator. Repeated factors require repeated terms.

Factor the denominator first, write the correct unknown numerators, clear denominators and equate coefficients. If the fraction is improper, divide before decomposing.

1/[x(x+1)] = 1/x − 1/(x+1), making its integral ln|x|−ln|x+1|+C.

A repeated factor such as (x−1)² needs A/(x−1)+B/(x−1)²; one term cannot represent both powers.

In (a+b)^n the term containing a^{n−r}b^r is C(n,r)a^{n−r}b^r. The general term is useful for coefficients, approximations and selected powers.

Match the requested power before simplifying, include signs in b, and check whether n is a non-negative integer for a finite expansion.

The coefficient of x³ in (2−x)^5 is C(5,3)·2²·(−1)³=−40.

The row of Pascal’s triangle supplies only C(n,r); powers and signs still belong to the term.

For positive arguments, log(ab)=log a+log b, log(a/b)=log a−log b and log(a^p)=p log a. Change of base converts any legal base to a calculator base.

Record argument restrictions before combining or expanding. A logarithm base must be positive and not equal to 1.

ln[(x−1)^2] can be written 2ln|x−1| for x≠1, not 2ln(x−1) without the domain restriction.

Logarithms do not distribute over addition, and dropping absolute values can exclude valid negative arguments.

e^{ln x}=x for x>0 and ln(e^x)=x for real x. Exponential and logarithmic graphs reflect in y=x, with domains and ranges exchanged.

When solving, isolate the exponential or logarithm, then apply the inverse and preserve positivity. For models, identify whether the exponent is linear in the independent variable.

e^{2x−1}=7 gives x=(1+ln7)/2; a logarithm of a non-positive quantity is not defined over the reals.

ln(e^{x}) and e^{ln x} have different domain conditions even though both simplify algebraically.

Solve logarithmic equations by combining legal logs, converting to an exponential equation and checking every solution in the original domain.

If a quadratic appears after exponentiating, both algebraic roots are only candidates; reject any that make a log argument zero or negative.

log₃(x)+log₃(x−2)=1 requires x>2. It becomes x(x−2)=3, whose roots are 3 and −1; only x=3 survives.

Exponentiating an equation preserves equivalence only when both sides were defined; it cannot legalise an invalid log argument.

Rewrite a nonlinear relation as Y=mX+c, then plot the transformed variables. The gradient and intercept must be converted back to the original parameters.

Choose transformations from the model, not from the data alone; label axes and preserve units. A straight plot tests compatibility with the model over the measured range.

For y=a/x+b, plotting y against 1/x gives gradient a and intercept b.

A high correlation after transformation does not prove the mechanism, and the fitted intercept may be a transformed constant.

cosec x=1/sin x, sec x=1/cos x and cot x=cos x/sin x. They are undefined where the denominator is zero and never equal zero themselves.

Use the original function’s sign and period to sketch reciprocal branches. Mark vertical asymptotes before plotting turning points.

sec x has asymptotes at π/2+nπ and reaches ±1 at x=nπ.

Reciprocal notation is not inverse-function notation: sec x is not cos⁻¹x.

Use fundamental, compound-angle and double-angle identities to replace a difficult expression with an equivalent one. The replacement is valid wherever the original and new expressions are defined.

Choose a target form first—for example, a single sine or a factorised square—and simplify one side only when proving an identity.

sin(π/2−x)=cos x converts a complementary-angle expression immediately.

A numerical agreement at a few angles is not proof, and cancellation can hide excluded denominator values.

For a composite y=f(g(x)), y′=f′(g(x))g′(x). Exponential, logarithmic and trigonometric outer functions all use this chain-rule structure.

Label the inner function and its derivative explicitly. For products or quotients, combine the product/quotient rule with the chain rule rather than expanding unnecessarily.

d[sin((2x+1)²)]/dx=4(2x+1)cos((2x+1)²).

The inner derivative cannot be omitted just because the outer derivative looks familiar.

For y=uv use u′v+uv′; for y=u/v use (u′v−uv′)/v². The formulas account for both numerator and denominator variation.

Choose factors that keep algebra manageable, and simplify only after applying the rule. A logarithmic derivative can be shorter for a product of powers.

d[x sin x]/dx=sin x+x cos x; d[(x+1)/(x−1)]/dx=−2/(x−1)².

Differentiating top and bottom independently is invalid for a quotient.

Power, exponential, logarithmic, trigonometric, product, quotient and chain rules are building blocks; a single expression may require several in sequence.

Rewrite constants and nested powers, mark the outermost operation, and check a derivative numerically at one simple point if the algebra is long.

For y=(x²+1)^2e^x, use product plus chain: y′=e^x[(x²+1)^2+4x(x²+1)].

A familiar-looking power does not override a product or composite structure; missing one rule changes the result.

Use substitution to expose an inner derivative, integration by parts to reduce a product, and partial fractions to split a rational function into standard terms.

Inspect the integrand before starting: ask what becomes simpler after differentiating one factor or replacing a repeated expression. Change limits and differentials consistently.

For ∫x ln x dx, take u=ln x and dv=x dx; for ∫2x/(1+x²)dx, use u=1+x².

A method is not justified because it produces more terms; the check is whether the new integral is genuinely simpler.

Rewrite products or powers of sin and cos using identities, then integrate each term. For an inner multiple angle, include the reciprocal derivative factor.

For even powers use half-angle identities; for odd powers save one factor and substitute. Keep absolute values in logarithmic results.

∫cos²x dx=x/2+sin2x/4+C; ∫sin(4x)dx=−cos(4x)/4+C.

A power on sin or cos is not an instruction to raise its antiderivative to a power.

After factoring the denominator, decompose a proper rational function into linear terms and irreducible quadratic terms. Integrate each using logarithm or arctangent forms.

Complete the square in a quadratic denominator before integrating. Repeated factors need one numerator for each power.

∫1/(x²+1)dx=tan⁻¹x+C, while ∫(2x)/(x²+1)dx=ln(x²+1)+C.

A quadratic denominator does not always produce a logarithm; if its numerator is constant after completing the square, an inverse tangent may appear.

For ∫x/(ax²+bx+c)dx, write the numerator as a multiple of the denominator derivative plus a constant remainder. The first part gives a logarithm; the remainder may require completing the square.

Differentiate the denominator first, solve a simple linear identity for the numerator split, and keep the denominator positive conditions for logs.

For ∫x/(x²+1)dx, x is half of (x²+1)′, so the result is ½ln(x²+1)+C.

The numerator need not equal the denominator derivative exactly; it may be a scaled derivative plus a leftover constant.

Use ∫du/(u²+a²)=(1/a)tan⁻¹(u/a)+C after rewriting the denominator as a square plus a positive constant.

Complete the square, factor the constant scale correctly, and use a substitution if the numerator contains the derivative of the shifted variable.

∫dx/[(x−2)²+9]=⅓tan⁻¹((x−2)/3)+C.

The coefficient is 1/a, not a; omitting it gives a derivative three times too large in the example.

Set u=g(x) when g′(x) appears, replace dx and every occurrence of x, integrate in u, then substitute back. A definite integral can instead use transformed limits.

Do not mix x and u in the same unfinished integral. Choose a substitution that removes the inner function rather than merely renaming it.

∫x√(x²+1)dx with u=x²+1 gives ½∫u^{1/2}du=(1/3)(x²+1)^{3/2}+C.

Substitution is incomplete if the differential factor is missing; the presence of g(x) alone is not enough.

For a continuous f, opposite signs at a and b guarantee at least one root between them. A root-location interval is evidence of existence, not automatically uniqueness.

Evaluate f at the endpoints accurately, keep the interval ordered, and report its width or decimal bounds as the requested accuracy.

If f(1)=−0.4 and f(2)=0.3, continuity gives a root in (1,2); bisection can reduce this to a bracket of width below 0.01.

A sign change can contain several roots, and a root exactly at an endpoint should be stated separately.

With x_{n+1}=g(x_n), a fixed point α satisfies g(α)=α. Starting from x₀ creates a sequence intended to approach α.

Choose a valid starting value, retain guard digits, and stop when successive values or the residual meet the tolerance. Verify the final value in f(x)=0.

For x=cosx, x₀=1 gives iterates that approach approximately 0.739; reporting the last iterate without a tolerance is incomplete.

Iteration can oscillate or diverge even when a root exists; apparent agreement of early digits is not a proof.

For x_{n+1}=g(x_n), a fixed point is locally attractive when |g′(α)|<1; a smaller magnitude generally means faster local error reduction.

Derive g from the original equation, check the derivative on the intended interval and compare stopping errors with the function residual.

Two algebraically equivalent rearrangements of x³+x−1=0 can have different g′ near the root; choose the one with smaller local magnitude.

Algebraic equivalence of equations does not imply numerical equivalence of iterations.

A vector has size and direction; in 2D or 3D write components as a column or i,j,k combination. The magnitude of (a,b,c) is √(a²+b²+c²).

Keep vectors distinct from points, which describe position relative to an origin. A unit vector has magnitude 1 and preserves direction.

The vector (3,4) has magnitude 5; its unit vector is (3/5,4/5).

The coordinates of a point and the components of a displacement are not interchangeable unless the origin and direction are specified.

Add and subtract vectors componentwise, multiply by a scalar to change magnitude or reverse direction, and use the dot product a·b=|a||b|cosθ to test perpendicularity.

A zero vector has no direction; scalar multiples are parallel. Keep dimensions and coordinate order consistent in every operation.

(2,−1)+(−3,4)=(−1,3); vectors are perpendicular when their dot product is zero.

The dot product is a scalar, not a vector, and a negative scalar reverses direction rather than merely reducing length.

The position vector of P is OP, the directed vector from the origin O to P. The displacement from A to B is OB−OA, and the midpoint is the average of the two position vectors.

Choose one origin, preserve direction in subtraction, and translate geometric statements into vector equations such as a point dividing AB in a given ratio.

If OA=(2,1) and OB=(8,4), then AB=(6,3) and the midpoint has position (5,2.5).

AB is not the same as BA; reversing the order changes the sign of the displacement.

A line through point a in direction d is r=a+λd. Different parameter values move along the same line, and a non-zero scalar multiple of d gives the same direction.

Use the position vector and direction vector separately; compare lines by solving component equations for a common point and compatible parameters.

r=(1,2)+λ(3,−1) passes through (1,2) and has direction (3,−1).

A point on a line is not itself a direction vector, and changing the parameter origin changes a but not the geometric line.

Two vector lines are parallel when their direction vectors are scalar multiples. They are identical only if they also share a point; otherwise they are distinct parallel lines.

Compare direction components first, then test a point from one line in the other. In 3D, skew lines are neither parallel nor intersecting.

r=(1,0)+λ(2,3) and r=(4,5)+μ(−4,−6) are parallel because directions are proportional; checking the point decides whether they coincide.

Equal gradients in a 2D graph are not enough to prove coincident lines; intercepts matter.

For vectors a and b, a·b=|a||b|cosθ and in components a₁b₁+a₂b₂(+a₃b₃). A zero scalar product means perpendicular non-zero vectors.

Use the dot product to find an angle, a projection or a perpendicular condition. Check that the angle is the requested principal angle and that neither vector is zero.

(1,2)·(2,−1)=0, so the vectors are perpendicular; their magnitudes do not need to be equal.

The scalar product is not a vector and a negative value means an obtuse angle, not an impossible result.

For dy/dx=f(x)g(y), separate into dy/g(y)=f(x)dx and integrate. A linear equation dy/dx+P(x)y=Q(x) uses an integrating factor e^{∫Pdx}.

State where division is legal and retain any constant solutions lost by dividing by g(y). Differentiate the result to verify the form.

dy/dx=xy separates as dy/y=x dx, giving ln|y|=x²/2+C and y=Ae^{x²/2}; y=0 is also included by A=0.

Dividing by y can hide the zero solution, and an integrating factor is not needed for every separable equation.

Integrating an ODE introduces arbitrary constants. The order of the equation indicates how many independent conditions are normally needed to select one solution.

Keep the general solution visible, use conditions at the specified point, and check the resulting function in both the differential equation and the conditions.

Solving y′=2x gives y=x²+C; y(0)=3 selects y=x²+3.

A general solution is not one trajectory, and a condition on y′ cannot replace a missing condition on y when only one constant exists.

An initial condition such as y(x₀)=y₀ is substituted into the integrated family to determine its constant. For a second-order equation, conditions on y and y′ may be needed.

Apply the condition after integrating, not before; if the problem gives a physical initial velocity, translate it into y′(x₀) explicitly.

If y′=3x²−2 and y(1)=4, then y=x³−2x+C and C=5.

The initial point is not automatically x=0; use the coordinate stated in the question.

A solution of an ODE is a function whose derivatives satisfy the equation on the stated interval. Solve using the appropriate method, then apply initial or boundary information.

Check differentiability and any division restrictions. Substitution into the original equation is the strongest final check, especially after squaring or taking logarithms.

For y′=y, y=Ce^x satisfies the equation; y(0)=2 selects C=2.

A formula that matches one initial point is not enough; it must also satisfy the differential relation throughout the domain.

Write z=a+bi, where a,b are real and i²=−1. The conjugate is a−bi, modulus is |z|=√(a²+b²), and z z̄=|z|².

Collect real and imaginary parts separately, rationalise denominators with the conjugate, and use modulus as a distance from the origin.

(2+3i)(2−3i)=13, so the modulus of 2+3i is √13.

i is not a variable with arbitrary square root; i² is fixed as −1 and real/imaginary parts must match independently.

Add complex numbers componentwise, multiply by expanding and replace i² with −1, and divide by multiplying numerator and denominator by the denominator’s conjugate.

After simplifying, write the answer in a+bi form. Equality of complex numbers requires equality of both components.

(1+2i)/(3−i) =[(1+2i)(3+i)]/10=(1+7i)/10.

A complex fraction is not simplified by dividing real parts and imaginary parts separately.

A polynomial of degree n with complex coefficients has exactly n complex roots counted with multiplicity. Real roots are included among these; non-real roots occur in conjugate pairs for real-coefficient polynomials.

Use a known root or factor to reduce degree, then use coefficient relationships or conjugacy to find remaining roots. Check multiplicities rather than counting distinct values only.

If a real cubic has root 2 and root 1+i, it must also have 1−i; the three roots account for the degree.

“Three roots” can include a repeated root; distinct-root counting can undercount the degree.

z=a+bi is rectangular form. In polar form z=r(cosθ+i sinθ)=r cisθ, where r=|z| and θ is an argument measured from the positive real axis.

Use rectangular form for addition and polar form for multiplication, division and powers. Choose an argument branch consistently and identify the quadrant.

1+i=√2 cis(π/4); its square is 2 cis(π/2)=2i.

The argument is not determined by tan⁻¹(b/a) alone; the quadrant may require adding π.

Multiplying by r cisθ scales a modulus by r and adds θ to the argument. Dividing subtracts arguments and divides moduli; powers use De Moivre’s theorem.

Reduce angles modulo 2π, keep all roots when solving zⁿ=w, and convert back to rectangular coordinates only at the end if needed.

The cube roots of 8 are 2 cis(2kπ/3) for k=0,1,2, giving three equally spaced points on a circle.

Taking only the principal root misses the other n−1 roots; complex roots are distributed by equal angular steps.

Represent z=a+bi by the point (a,b). Addition is vector addition, modulus is distance from the origin and conjugation reflects the point in the real axis.

Use geometry to check algebraic results: conjugates have the same modulus, and z+w is the diagonal endpoint of the parallelogram formed by z and w.

The points 1+i and 1−i are reflections with modulus √2; their sum is the real point 2.

The Argand-plane coordinates are not a graph of y=f(x); the imaginary axis is a second coordinate, not an output scale.

|z−a|=r is a circle centred at a with radius r; arg(z−a)=θ is a ray from a. Inequalities give interiors, exteriors or angular regions.

Sketch the reference point a first, then apply the distance or angle condition. Intersections of loci must satisfy every condition simultaneously.

|z−(2+i)|=3 is a circle centred at (2,1); |z|<2 is its interior.

An argument condition describes a direction, not a distance, and arg is defined modulo 2π unless a principal range is stated.

[r(cosθ+i sinθ)]^n=r^n(cos nθ+i sin nθ). To solve z^n=w, take the nth root of the modulus and add arguments (arg w+2kπ)/n for k=0,…,n−1.

List all distinct roots, then convert to rectangular form only if needed. Their Argand points lie equally spaced on a circle.

z³=8 has roots 2 cis(2kπ/3), k=0,1,2.

Taking only the principal argument yields one root and misses the remaining equally spaced solutions.