2. Pure Mathematics 2

|f(x)|=a with a≥0 means f(x)=a or f(x)=−a. If a<0 there is no real solution. For inequalities, the sign of the comparison determines whether the solution lies inside or outside an interval.

Solve each branch, intersect with any domain, and test boundary points. Nested moduli require a second case split rather than one global square.

|2x−3|=5 gives 2x−3=5 or −5, so x=4 or x=−1.

Squaring can introduce solutions when the right side is not known non-negative; always substitute candidates into the original modulus equation.

Divide a polynomial by x−a using long division or synthetic division. The remainder theorem gives remainder P(a), and P(a)=0 means x−a is a factor.

Write descending powers, include zero coefficients, and verify by divisor×quotient+remainder. Use a known factor to reduce the degree before solving the remaining polynomial.

For P(x)=x³−2x²−x+2, P(1)=0, so x−1 is a factor; division gives x²−x−2=(x−2)(x+1).

A zero remainder proves divisibility by the chosen factor only; it does not make every nearby value a root.

The factor theorem says x−a is a factor of P(x) exactly when P(a)=0. The remainder theorem says division by x−a leaves remainder P(a).

Test simple candidates first, divide out a confirmed factor, then solve the lower-degree quotient. A repeated root also requires P′(a)=0.

For P(x)=x³−4x²+x+6, P(2)=0, so x−2 is a factor; division reduces the cubic before finding the remaining roots.

P(a)=0 identifies a factor x−a, not x+a; substitute the sign carefully.

For a,b>0 and base k>0,k≠1, log_k(ab)=log_k a+log_k b, log_k(a/b)=log_k a−log_k b and log_k(a^p)=p log_k a.

Check every argument is positive before combining logs. Change of base gives log_k a=ln a/ln k and is useful when the calculator uses natural logs.

log₂(8x)−log₂x=3 for x>0, because the ratio is 8.

log(a+b) is not log a+log b; the product law does not apply to sums.

For a>0,a≠1, y=a^x has inverse x=log_a y. The natural logarithm ln x is log_e x, and ln(e^x)=x for real x while e^{ln x}=x only for x>0.

Use the inverse relation to solve for a variable, and preserve positivity when taking logarithms. Exponential growth has constant proportional rate; logarithmic growth has decreasing gradient.

3e^{2x}=12 gives e^{2x}=4 and x=½ln4.

ln(x²)=2lnx only when x>0; for x≠0 the safe statement is ln(x²)=2ln|x|.

To solve a logarithmic equation, combine logs only when arguments are positive, then exponentiate or change base. Every candidate must satisfy the original log domains.

If log terms have different bases, convert consistently. Squaring or exponentiating can create algebraic candidates that the original equation rejects.

ln(x−1)+ln(x+1)=ln3 becomes ln(x²−1)=ln3 with x>1, so x=2; x=−2 is rejected by the domain.

Equality of log expressions does not allow non-positive arguments, even if a later algebraic step produces a real number.

Transform variables so the model becomes Y=mX+c. For y=ab^x, plotting ln y against x gives gradient ln b and intercept ln a; for y=ax^n, plotting ln y against ln x gives gradient n.

Transform measured uncertainties and units consistently, then interpret the gradient and intercept in the original parameters. A straight plot supports the model but does not prove causation.

If ln y=0.7x+1.2, then a=e^{1.2} and b=e^{0.7} for y=ab^x.

The intercept is not always the original constant; it may be ln a or another transformed quantity.

cosec x=1/sin x, sec x=1/cos x and cot x=1/tan x=cos x/sin x wherever the denominator is non-zero. Their graphs inherit zeros and asymptotes from the original functions.

Use reciprocal values rather than inventing new triangle rules. Mark undefined angles and use the original sine or cosine sign to determine the reciprocal sign.

sec(π/3)=2 because cos(π/3)=1/2; sec x is undefined where cos x=0.

cosec x is not sin⁻¹x, and reciprocal graphs do not cross the x-axis because a reciprocal cannot equal zero.

Use compound-angle, double-angle and half-angle identities to rewrite expressions into a form suited to the question. For example, sin2x=2sinx cosx and cos2x=1−2sin²x.

Choose one direction of simplification, keep domain restrictions, and avoid replacing an identity with a numerical check at a few angles.

1−cos2x=2sin²x converts a cosine expression into a square that is easier to integrate or solve.

An identity is true for every allowed x; a relation that works only at selected angles is not an identity.

Differentiate e^{g(x)} as g′(x)e^{g(x)}, ln(g(x)) as g′(x)/g(x), and sin or cos of an inner function using the chain rule.

Identify the outer function first, differentiate it, then multiply by the derivative of the inner expression. Simplify only after all factors are present.

d[ln(1+x²)]/dx=2x/(1+x²), while d[e^{3x}sin x]/dx=e^{3x}(3sin x+cos x).

The derivative of ln g is not 1/ln g, and forgetting g′ changes the scale of the answer.

For y=uv, y′=u′v+uv′. For y=u/v, y′=(u′v−uv′)/v². Both rules follow from the product rule and reciprocal structure.

Label u and v before differentiating, keep the denominator squared in the quotient rule, and use a logarithmic derivative when products of powers are cumbersome.

d[x²e^x]/dx=e^x(x²+2x); d[(sin x)/x]/dx=(x cosx−sinx)/x².

Differentiating numerator and denominator separately is not the quotient rule.

A composite expression may require chain, product, quotient or implicit differentiation. Read its structure before applying a familiar rule.

Rewrite constants and powers clearly, differentiate one layer at a time, and check the result by estimating the sign or scale of the gradient at a simple point.

For y=(x²+1)^4/(x−1), treat the numerator and denominator as factors and apply product/quotient plus chain rules rather than expanding blindly.

There is no single “power rule” shortcut for a quotient or a composite power without the required extra factors.

Use substitution when an inner derivative is present, and integration by parts ∫u dv=uv−∫v du when a product becomes simpler after differentiating one factor.

For substitution, change the differential and limits consistently. For parts, choose u to become simpler and dv to be easy to integrate; differentiate the final answer to verify.

∫2x cos(x²)dx=sin(x²)+C by u=x². For ∫x e^x dx, take u=x and dv=e^x dx.

Substitution is not complete until every x and dx is replaced, and integration by parts does not remove the need for C in an indefinite integral.

Rewrite powers or products using identities such as sin²x=(1−cos2x)/2, then integrate term by term. If an inner angle is present, substitute it with its differential.

Use the identity that reduces the power or creates a derivative factor; keep absolute values in logarithmic antiderivatives such as ∫tan x dx=−ln|cos x|+C.

∫sin²x dx=x/2−sin2x/4+C, while ∫sec²(3x)dx=tan(3x)/3+C.

∫sin²x is not −cos³x/3; the power is on the function, not on its differential pattern.

With equal spacing h and ordinates y₀,…,yₙ, the trapezium estimate is h/2[y₀+yₙ+2(y₁+…+yₙ₋₁)]. It approximates the definite integral by straight chords.

Use n subintervals and h=(b−a)/n. For a curve that is concave up, the chord estimate lies above or below according to the curvature, so compare with an exact value when available.

Using h=0.5 with ordinates at x=0,0.5,1 gives area ≈0.25[y₀+2y₁+y₂].

The number of ordinates is n+1, not n, and the endpoint weights are 1 rather than 2.

If f is continuous and f(a) and f(b) have opposite signs, at least one root lies in (a,b). A bracket locates a root; it does not necessarily prove uniqueness.

Check continuity and sign values, then narrow the interval by bisection or another allowed method. State the interval and its width when reporting accuracy.

f(1)<0 and f(2)>0 implies a root in (1,2) for continuous f; repeated bisection gives a guaranteed bracket.

A sign change can hide an odd number of roots, and a zero endpoint must be treated separately rather than called an interior root.

Iteration replaces x by x_{n+1}=g(x_n). A fixed point α satisfies g(α)=α, corresponding to a root of the rearranged equation.

Choose a starting value in the stated interval, compute enough figures during iteration, and stop using a tolerance on successive values or the residual. Different rearrangements can converge differently.

For x=cos x, starting x₀=0 gives a sequence approaching about 0.739; starting values should remain in a region where g behaves stably.

A few stable-looking digits do not prove convergence, and iteration can diverge or enter a cycle even when the equation has a root.

For x_{n+1}=g(x_n), convergence near a fixed point is encouraged when |g′(x)|<1 there; the derivative also indicates how quickly errors shrink locally.

Derive g from the original equation without changing the root set, select an admissible initial value, and verify the final approximation in the original function.

Rearranging x³+x−1=0 as x=(1−x)^{1/3} may behave differently from another rearrangement; compare the local slope before iterating.

An iteration formula is not automatically valid just because it is algebraically equivalent; convergence is a separate question.