Unit P1: Pure Mathematics AS 1

Index laws describe repeated multiplication and remain consistent for zero, negative and rational exponents. Before combining powers, check that they have the same base.

To solve $3^{2p-5}=3^{-1/2}$, equal positive bases give equal exponents: $2p-5=-\tfrac12$, so $p=\tfrac94$.

Rewrite composite bases such as 9 or 27 as powers of one base, simplify each side fully, then equate exponents only after the bases match.

A surd is an exact irrational root. Simplify it by extracting perfect-square factors, combine only like surds, and rationalise a denominator without changing the value.

Since $(2+\sqrt3)(2-\sqrt3)=1$, $\dfrac1{2+\sqrt3}=2-\sqrt3$. The conjugate removes the surd from the denominator while multiplying by 1.

For example, $3\sqrt8-\sqrt{18}=3(2\sqrt2)-3\sqrt2=3\sqrt2$. Simplify each root before deciding whether terms are like surds.

A quadratic function $f(x)=ax^2+bx+c$, $a\ne0$, has a parabolic graph. Its algebraic form can expose intercepts, symmetry or the turning point before any points are plotted.

For $y=(x-1)(x-5)$, the roots are 1 and 5, so the axis is $x=3$. Substitution gives $y(3)=-4$, hence the minimum is $(3,-4)$; the positive leading coefficient means the graph opens upward.

Mark the turning point, intercepts and symmetry axis, then join them with one smooth parabola. If there are no real roots, the graph remains entirely above or below the horizontal axis according to its turning point and opening.

For $x^2-4x+k=0$ to have no real roots, $\Delta=16-4k<0$, so $k>4$. The inequality on the discriminant becomes the condition on the parameter.

First rearrange the equation into quadratic form and identify $a$, $b$ and $c$ with their signs. For tangency use equality; for ‘at least one real root’ use $\Delta\ge0$.

Factor out the leading coefficient from the $x$ terms, take half the coefficient of $x$ inside the bracket, square it, and compensate for the added term. The resulting form gives the turning point and supports exact solution.

$2x^2-8x+3=2(x^2-4x)+3=2[(x-2)^2-4]+3=2(x-2)^2-5$. Hence $2x^2-8x+3=0$ gives $(x-2)^2=5/2$ and $x=2\pm\sqrt{10}/2$.

Use factorisation when factors are visible, completing the square when vertex form or an exact rearrangement is valuable, and the quadratic formula as a general exact method. A calculator can confirm but not replace required algebraic stages.

A simultaneous solution must satisfy both equations. Isolate one variable, substitute its complete expression into the other equation, solve for every possible value, then back-substitute to form ordered pairs.

For $y=x+1$ and $x^2+y^2=13$, substitution gives $x^2+(x+1)^2=13$, so $x^2+x-6=0$. Thus $x=2$ or $-3$, producing $(2,3)$ and $(-3,-2)$.

A line and a quadratic curve may meet twice, once or not at all. The number of real substituted roots matches the number of real intersection points.

To interpret $f(x)<g(x)$ graphically, find where the graphs meet and read the $x$-intervals on which the graph of $f$ lies below the graph of $g$. Equality determines whether intersection endpoints are included.

$x^2<x+2$ compares the parabola $y=x^2$ with the line $y=x+2$. They intersect at $x=-1$ and $x=2$; the parabola lies below the line between them, so $-1<x<2$.

Retain domain restrictions when reducing a fractional inequality. For example, multiplying $a/x<b$ by the always-positive $x^2$ gives $ax<bx^2$ together with $x\ne0$.

A graphical inequality is a region separated by a boundary curve or line. Draw the boundary first, decide whether it belongs to the region, then use a test point to choose the correct side.

For $y\ge x^2$ and $y<x+2$, draw the parabola solid and the line dotted. Shade points on or above the parabola and below the line; the common region exists between their intersection values $x=-1$ and $x=2$.

For several inequalities, shade only the overlap satisfying every condition. A vertical boundary such as $x\le3$ is handled by the same solid/dotted and test-point rules.

Move every term to one side so the problem becomes a sign question. Find the critical roots, split the number line at those roots, and determine where the expression has the required sign.

$(x-4)(x+1)\le0$ has roots $-1$ and $4$. The upward-opening product is non-positive between the roots, including zeros, so $-1\le x\le4$.

For a quadratic compared with a line, bring the line to the same side first: $x^2<3x+4$ becomes $x^2-3x-4<0$, then factor and apply the sign method.

Expanding, collecting and factorising are equivalent rewritings of a polynomial. Preserve every term and sign so the new expression has the same value for all $x$.

$x^3+4x^2+3x=x(x^2+4x+3)=x(x+1)(x+3)$. The factorised form shows zeros $0,-1,-3$, while expansion confirms the identity.

The notation $f(x)$ names the output of a polynomial rule. Solving $f(x)=g(x)$ requires bringing all terms to one side, simplifying, then factorising the resulting polynomial of degree at most 3 in this unit.

A function sketch records intercepts, symmetry, turning behaviour, periodicity and asymptotes. Intersections of $y=f(x)$ and $y=g(x)$ are the geometric solutions of $f(x)=g(x)$.

An asymptote is a line the curve approaches according to its limiting behaviour. It controls the sketch even when it is not part of the graph; reciprocal functions are undefined at $x=0$.

To find exact intersection coordinates, solve $f(x)=g(x)$ algebraically, then substitute each valid $x$ into either function for $y$. A sketch predicts how many real solutions to expect and helps detect missing roots.

A graph transformation moves each point of $y=f(x)$ by changing either its output or its input. Output changes act vertically; input changes act horizontally in the inverse-looking direction.

If $(3,5)$ lies on $y=f(x)$, then $(1,5)$ lies on $y=f(x+2)$, while $(3,10)$ lies on $y=2f(x)$. Transform intercepts, turning points and asymptotes by the same mapping.

Apply the one stated transformation to the whole curve, preserving its connections and characteristic shape. For periodic graphs, transform wavelength/period and amplitude consistently rather than moving isolated peaks.

A straight line has constant gradient $m$. Two points determine that gradient when $x_2\ne x_1$; substituting either point into $y-y_1=m(x-x_1)$ then fixes the unique line. Expand and collect terms only if a requested form such as $ax+by+c=0$ is needed.

For the line perpendicular to $3x+4y=18$ through $(2,3)$, first write the given line as $y=-\tfrac34x+\tfrac92$. Its perpendicular gradient is $\tfrac43$, so the required equation is $y-3=\tfrac43(x-2)$. Equivalently, $4x-3y+1=0$. Substituting $(2,3)$ checks the point lies on it.

Gradient records a line's direction. Lines with the same direction are parallel; a quarter-turn changes a non-zero finite gradient to its negative reciprocal.

A line through $P(-3,7)$ and $Q(9,11)$ has gradient $\tfrac{11-7}{9-(-3)}=\tfrac13$. A line through $Q(9,11)$ and $R(12,2)$ has gradient $\tfrac{2-11}{12-9}=-3$. Their product is $-1$, so the lines are perpendicular and $\angle PQR=90^\circ$.

For parallel lines, equal gradients establish equal direction. To show they are distinct rather than the same line, compare their intercepts or test whether a point on one lies on the other.

Lower-case sides are opposite their matching upper-case angles, and $K$ denotes the triangle's area. The useful rule is determined by which sides and angles are known, not by which formula looks shortest.

The sine rule's SSA case can produce two angles because $\sin B=\sin(\pi-B)$. If $a=8$, $b=10$ and $A=30^\circ$, then $\sin B=10\sin30^\circ/8=5/8$, so $B\approx38.7^\circ$ or $141.3^\circ$. Both give $A+B<180^\circ$, so both triangles are possible. Reject any alternative that makes the angle sum at least $180^\circ$ or conflicts with the side ordering.

An angle of $\theta$ radians subtends an arc whose length is $\theta$ times the radius. Thus radians make circular measure a direct proportional relationship: $\theta=s/r$.

For radius $6\text{ cm}$ and angle $1.2$ radians, the arc length is $s=6(1.2)=7.2\text{ cm}$ and the sector area is $A=\tfrac12(6^2)(1.2)=21.6\text{ cm}^2$. A sector perimeter would be $7.2+2(6)=19.2\text{ cm}$ because its two radii must also be included.

A trigonometric graph repeats a characteristic cycle. Sketch it by preserving its key values, symmetry and period, then apply any transformation to every key point and asymptote. In the table, $k$ denotes any integer.

Start with one complete base cycle, mark zeros, maxima, minima and any tangent asymptotes, then transform their coordinates. Repeat the transformed cycle using its new period; join sine and cosine points smoothly, but keep tangent branches separated at each asymptote.

For $y=f(x)$, the derivative $f'(x)$ or $\dfrac{dy}{dx}$ gives the instantaneous rate of change of $y$ with respect to $x$. Geometrically, its value at $x=a$ is the gradient of the tangent to the curve there.

The quotient is the gradient of the secant through inputs $a$ and $a+h$. As $h$ tends to zero, the second point approaches the first and the secant gradients approach the tangent gradient. This limit is why a derivative describes change at one point rather than only an average over an interval.

For $f(x)=x^2$, $\dfrac{f(a+h)-f(a)}h=\dfrac{(a+h)^2-a^2}{h}=2a+h$. Taking $h\to0$ gives $f'(a)=2a$, so the tangent gradient at $x=3$ is $6$.

Differentiating again gives $f''(x)=\dfrac{d^2y}{dx^2}$, the rate at which the first derivative changes. If $y$ is displacement and $x$ is time, for example, $dy/dx$ is velocity and $d^2y/dx^2$ is acceleration; the units change accordingly.

The power rule applies term by term to sums, differences and constant multiples. Before using it, expand products and rewrite roots or variables in denominators as powers of $x$; this exposes every term in the required form $ax^n$.

For $y=\dfrac{x^2+5x-3}{3\sqrt{x}}$, first write $y=\tfrac13x^{3/2}+\tfrac53x^{1/2}-x^{-1/2}$. Therefore $\dfrac{dy}{dx}=\tfrac12x^{1/2}+\tfrac56x^{-1/2}+\tfrac12x^{-3/2}$. Each coefficient is multiplied by its old power, then the power is reduced by one.

At a point on a curve, differentiation supplies the tangent gradient. The normal is the straight line through the same point that is perpendicular to that tangent.

For $y=x^2+2x$ at $x=1$, the point is $(1,3)$ and $dy/dx=2x+2$, so $m_T=4$. The tangent is $y-3=4(x-1)$. The normal gradient is $-\tfrac14$, so the normal is $y-3=-\tfrac14(x-1)$. Substitution of $(1,3)$ checks both equations pass through the correct point.

If a tangent must be parallel to a line of gradient $m$, solve $f'(x)=m$ for every valid $x$, then recover the corresponding point on the curve. For a perpendicular tangent, equate $f'(x)$ to the negative reciprocal of the given line's gradient.

Indefinite integration reverses differentiation. It finds every function whose derivative is the given integrand, so its result is a family rather than one unique curve.

The constant of integration $C$ is required because every constant differentiates to zero. Curves such as $F(x)+2$ and $F(x)-7$ have the same derivative, so derivative information alone cannot distinguish their vertical positions.

Since $\dfrac{d}{dx}(2x^3-2x^2+5x)=6x^2-4x+5$, $$\int(6x^2-4x+5)\,dx=2x^3-2x^2+5x+C.$$ Differentiating the result removes $C$ and reproduces the integrand, which is the most direct check.

For each term $ax^n$, increase the power by one and divide the coefficient by that new power. First expand brackets and rewrite roots or variables in denominators as powers, then integrate sums, differences and constant multiples term by term.

Therefore $$\int\frac{(x+2)^2}{\sqrt{x}}\,dx=\frac25x^{5/2}+\frac83x^{3/2}+8x^{1/2}+C.$$ Differentiating each term returns the rewritten integrand.

If $f'(x)=3x^2-4$ and the curve passes through $(2,5)$, then $f(x)=x^3-4x+C$. Substitution gives $5=8-8+C$, so $C=5$ and $f(x)=x^3-4x+5$.