Unit P3: Pure Mathematics A2 3

A rational expression is simplified by factorising numerator and denominator, cancelling common factors, and using algebraic division when the numerator's degree is at least the denominator's. Record excluded denominator values before cancelling.

For the specification example, $$\frac{x^3+1}{x^2-1}=\frac{(x+1)(x^2-x+1)}{(x+1)(x-1)}=\frac{x^2-x+1}{x-1},\qquad x\ne-1,1.$$ Cancelling removes a common factor, not the original restriction $x\ne-1$.

Now divide: $x^2-x+1=x(x-1)+1$, so $$\frac{x^3+1}{x^2-1}=x+\frac1{x-1},\qquad x\ne-1,1.$$ The quotient has lower-degree remainder, which is the useful stopping form.

For an unknown quotient and remainder, write an identity such as $$\frac{4x^3+2x^2+3x+8}{x^2+4}=Ax+B+\frac{Cx+D}{x^2+4},$$ multiply through by the denominator, then equate coefficients or substitute convenient values.

A function assigns exactly one output to each input in its domain. Different inputs may share an output, so a many-one mapping is still a function; an inverse function exists only after the mapping is one-one on its chosen domain.

Let $f(x)=2x+3$ on $\mathbb R$ and $g(x)=x^2$ for $x\ge0$. Then $$fg(x)=f(g(x))=2x^2+3,\qquad x\ge0.$$ The restriction makes $g$ one-one, so $g^{-1}(x)=\sqrt{x}$ for $x\ge0$. Domain and range exchange when a function is inverted.

The graphs of $y=f(x)$ and $y=f^{-1}(x)$ are reflections in $y=x$. If a horizontal line meets the original graph more than once, restrict its domain before claiming an inverse function.

Modulus makes a quantity non-negative: $|u|=u$ for $u\ge0$ and $|u|=-u$ for $u<0$. This piecewise rule controls both sketches and algebraic solutions.

For $|2x-1|=x+5$, use the two linear branches: $$2x-1=x+5\Rightarrow x=6,$$ $$-(2x-1)=x+5\Rightarrow x=-\frac43.$$ Both satisfy their corresponding sign condition, so both are solutions.

The same expressions meet at $x=-\frac43$ and $x=6$. Testing the intervals, or comparing the two graphs, gives $$|2x-1|>x+5\quad\Longrightarrow\quad x<-\frac43\text{ or }x>6.$$ Strict inequality excludes the intersection points.

Transformations outside f change output coordinates; transformations inside f change input coordinates in the opposite way. Map a known point rather than relying on a verbal left/right guess.

A negative outside factor also reflects in the x-axis; a negative inside factor reflects in the y-axis. Magnitudes greater than 1 stretch vertically outside but compress horizontally inside.

If $(6,-2)$ lies on $y=f(x)$, then on $$y=2f(3x)+5$$ the input coordinate becomes $6/3=2$ and the output becomes $2(-2)+5=1$. Thus the mapped point is $(2,1)$.

For a permitted combination, apply every horizontal coordinate change to u and every vertical change to v, then use intercepts and mapped features to complete the sketch. For example, $y=f(-x)+1$ maps $(u,v)$ to $(-u,v+1)$.

Secant, cosecant and cotangent are reciprocal or quotient functions, so their graphs inherit zeros, signs and forbidden inputs from cosine and sine. Inverse trigonometric functions instead return a principal angle from a restricted one-one branch.

A forbidden input is a vertical asymptote. Secant and cosecant form branches outside the horizontal band between -1 and 1; cotangent decreases between consecutive asymptotes. The same structure can be read in degrees by replacing a full turn with 360 degrees.

For $0\le x<2\pi$, $\sec x=-2$ becomes $\cos x=-\frac12$, giving $x=\frac{2\pi}{3},\frac{4\pi}{3}$. The calculator value $\arccos(-\frac12)=\frac{2\pi}{3}$ is the principal angle; symmetry supplies the second interval solution.

The two further identities are versions of the Pythagorean identity written entirely in tangent/secant or cotangent/cosecant. Deriving them reveals both the algebra and their domain restrictions.

Divide $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ to obtain $\tan^2\theta+1=\sec^2\theta$. Dividing instead by $\sin^2\theta$ gives $1+\cot^2\theta=\operatorname{cosec}^2\theta$.

For example, $$\tan^2\theta=3\sec\theta-3$$ becomes $$\sec^2\theta-1=3\sec\theta-3,$$ so $(\sec\theta-1)(\sec\theta-2)=0$. Thus $\sec\theta=1$ or $2$ before any stated interval is applied.

Compound-angle formulae convert sums and differences of angles into products, or combine a sine-cosine pair into one shifted function. Choose the direction that reduces the number of trigonometric terms.

To combine $$a\cos\theta+b\sin\theta=R\cos(\theta-\alpha),$$ expand the right side and match coefficients: $R\cos\alpha=a$, $R\sin\alpha=b$, so $R=\sqrt{a^2+b^2}$. Choose the sign and quadrant of $\alpha$ from the coefficients; an equivalent shifted sine form is also valid.

For $0\le\theta<2\pi$, $$3\cos\theta+4\sin\theta=2$$ becomes $5\cos(\theta-\alpha)=2$, where $\alpha=\arctan(4/3)$. Hence $\theta-\alpha=\pm\arccos(2/5)+2\pi n$, giving $\theta\approx2.09$ or $6.05$ radians in the interval.

For an identity, work from one side and look for a compound angle. For example, $$\cos x\cos2x+\sin x\sin2x=\cos(2x-x)=\cos x.$$ This uses the cosine-difference formula rather than checking selected values.

The graph of $y=e^x$ is always positive, passes through $(0,1)$ and approaches the horizontal asymptote $y=0$ as $x\to-\infty$. Its domain is $\mathbb R$ and its range is $(0,\infty)$.

Writing $ax+b=a(x+b/a)$ shows that the inside change moves and horizontally scales the base graph; adding $c$ moves every output vertically and therefore moves the asymptote from $y=0$ to $y=c$.

For $$y=e^{-2x+1}+3,$$ the asymptote is $y=3$, the range is $y>3$, and the curve decreases because the coefficient of $x$ is negative. At $x=0$, $y=e+3$, so $(0,e+3)$ fixes its vertical position.

The natural logarithm is the inverse of the exponential function: $\ln(e^x)=x$ and $e^{\ln x}=x$ for $x>0$. Therefore the graphs of $y=e^x$ and $y=\ln x$ are reflections in $y=x$.

For $e^{3x-1}=7$, take natural logs to get $3x-1=\ln7$, so $x=(1+\ln7)/3$. For $\ln(5-2x)=1$, exponentiate to obtain $5-2x=e$, hence $x=(5-e)/2$; this satisfies the required domain $5-2x>0$.

A logarithmic plot turns a nonlinear relationship into a straight line. The chosen horizontal axis determines which model is being tested, while the gradient and intercept recover the original parameters.

Using base-10 logarithms, an intercept $c$ means $a=10^c$ or $k=10^c$. For the exponential plot, a gradient $m$ means $b=10^m$. With natural logarithms, use $e^c$ and $e^m$ instead; keep one base throughout.

Suppose a graph of $\log_{10}y$ against $\log_{10}x$ has gradient $-\frac12$ and intercept $2$. Then $$\log_{10}y=-\frac12\log_{10}x+2,$$ so $$y=10^2x^{-1/2}=100x^{-1/2}.$$

Estimate the gradient from two well-separated points on the straight line of best fit, not automatically from two raw data points. Read the vertical intercept at horizontal coordinate zero, then convert the logged parameters back to the original scale.

Differentiate a sum or difference term by term. For each exponential, logarithmic or trigonometric term, multiply by the derivative of its linear inner expression.

For $$y=3e^{-2x}+4\ln(5x)-2\cos(3x),\qquad x>0,$$ differentiate each term: $$\frac{dy}{dx}=-6e^{-2x}+\frac4x+6\sin(3x).$$ The domain condition comes from the logarithm.

These trigonometric derivatives assume angles are measured in radians. A derivative may then be evaluated at a point to obtain a tangent gradient or set equal to zero to locate stationary candidates.

The structure of the expression selects the rule: multiplication needs the product rule, division needs the quotient rule, and a function inside another function needs the chain rule. More than one rule may be nested in the same derivative.

Examples expose the chosen structure: $$\frac{d}{dx}(2x^4\sin x)=8x^3\sin x+2x^4\cos x,$$ $$\frac{d}{dx}\left(\frac{e^{3x}}x\right)=\frac{e^{3x}(3x-1)}{x^2},$$ $$\frac{d}{dx}\big(\tan^2(2x)\big)=4\tan(2x)\sec^2(2x).$$

For a nested expression, identify the outermost operation first, write its rule without simplifying, and then differentiate each inner part. Factor common terms afterwards when solving a stationary-point equation.

If a curve is given as $x=f(y)$, differentiate with respect to $y$ first and then take the reciprocal. This works on a local inverse branch wherever $dx/dy\ne0$.

Use this order: find dx/dy; invert it; simplify in terms of y; then, only if required, use the original relation and the stated interval to rewrite the result in terms of x. The interval controls any square-root sign.

Suppose $x=\sin(3y)$ on $-\frac\pi6<y<\frac\pi6$. Then $$\frac{dx}{dy}=3\cos(3y),\qquad \frac{dy}{dx}=\frac1{3\cos(3y)}.$$ On this branch cosine is positive and $\cos(3y)=\sqrt{1-x^2}$, so $$\frac{dy}{dx}=\frac1{3\sqrt{1-x^2}}.$$

After evaluating $dy/dx$ at a point, the normal gradient is its negative reciprocal. Keep the point coordinates with the gradient when forming a tangent or normal equation.

An exponential model describes change at a rate proportional to its current displacement from a limiting level. Read the initial value at t=0, the sign of the rate, and the long-term behaviour before trusting a numerical prediction.

For a temperature model $$T=20+80e^{-0.3t},\qquad t\ge0,$$ the initial temperature is $100$, the limiting temperature is $20$, and $$\frac{dT}{dt}=-24e^{-0.3t}=-0.3(T-20).$$ Thus the temperature falls quickly at first and the magnitude of its rate tends to zero.

A model's algebra may be correct but its prediction may be inappropriate outside the observed time range: unlimited growth may exceed a physical capacity, while a decay limit may be unrealistic. Compare an improved model by its fit in the relevant range and by whether its initial value, limit and allowed outputs make sense.

Integrate a sum or difference term by term. For an exponential or trigonometric function with a linear inner expression, divide by the inner coefficient because differentiation would multiply by it.

For example, $$\int\left(4e^{-2x}+\frac3x-5\sin3x+2\cos4x\right)dx$$ $$=-2e^{-2x}+3\ln|x|+\frac53\cos3x+\frac12\sin4x+C.$$ Differentiating the result term by term restores the integrand.

For a definite integral, find one antiderivative $F$ and calculate $F(b)-F(a)$. The arbitrary constant cancels, so it is not included in the final definite value.

Integration by recognition reverses the chain rule. Identify a repeated inner function f(x), compare the remaining factor with its derivative, and correct only by a constant multiplier.

Since $\frac{d}{dx}(5x^2+4)=10x$, $$\int10x(5x^2+4)^{1/2}dx=\frac23(5x^2+4)^{3/2}+C.$$ Likewise, $$\int\frac{6x}{3x^2+5}dx=\ln(3x^2+5)+C.$$

Known inverse derivatives also give $$\int\frac{dx}{a^2+x^2}=\frac1a\arctan\left(\frac xa\right)+C,qquad \int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\left(\frac xa\right)+C,$$ for $a>0$ on their real domains.

If a function is continuous on an interval and its values at the two endpoints have opposite signs, its graph must cross the x-axis somewhere between them. Therefore the equation f(x)=0 has at least one root in that interval.

State a suitable function f, calculate f(a) and f(b), show that one is negative and the other positive, then state both the sign change and continuity before concluding that a root lies between a and b.

For $f(x)=x^3-x-1$, $f(1.3)=-0.103$ and $f(1.4)=0.344$. Their product is negative and this polynomial is continuous, so $f(x)=0$ has a root in $(1.3,1.4)$. To confirm a value rounds to 3 decimal places, apply the same test at the two rounding boundaries, such as $1.1335$ and $1.1345$ for $1.134$.

An iteration repeatedly substitutes the latest approximation into a given recurrence. For $x_{n+1}=g(x_n)$, start from the stated $x_1$ and use each calculated value to produce the next one; a settled value is a fixed point satisfying $x=g(x)$.

First rearrange the original equation into the required fixed-point form if asked. Enter the given starting value, evaluate the recurrence with the correct brackets, and carry the unrounded calculator value into the next step. Continue for the requested number of iterations or until successive values agree to the requested accuracy.

To approximate the positive solution of $x^2+x-5=0$, use $x_{n+1}=\sqrt{5-x_n}$ with $x_1=2$. Then $$x_2=\sqrt3=1.732050\ldots,$$ $$x_3=\sqrt{5-1.732050\ldots}=1.807747\ldots.$$ Repeating gives values that settle near $1.7913$; substituting this value into the original equation provides a check.