Key Points

Prove a property with arbitrary elements; disprove it with a single concrete counter-example.

On a non-empty set, the empty relation is symmetric and transitive but not reflexive; the universal relation \(A\times A\) is an equivalence relation.

Equivalence classes are either identical or disjoint, and their union is the whole set.

\(x\mapsto x^2\): neither on \(\mathbb{R}\to\mathbb{R}\); one-one but not onto on \(\mathbb{N}\to\mathbb{N}\); bijective on \([0,\infty)\to[0,\infty)\).

One-one: every horizontal line cuts the graph at most once. Onto: every horizontal line at a codomain height cuts it at least once.

\(\sin^{-1}(\sin\frac{5\pi}{6})=\frac\pi6\), because \(\frac{5\pi}{6}\notin[-\frac\pi2,\frac\pi2]\) and \(\sin\frac{5\pi}6=\sin\frac\pi6\).

Solve \(-1\le g(x)\le 1\). For \(\sec^{-1}(g(x))\) solve \(|g(x)|\ge1\).

\(\sin^{-1},\tan^{-1}\) increase; \(\cos^{-1},\cot^{-1}\) decrease. Extreme values of expressions follow from the endpoints of the ranges.

Always state order as (rows) × (columns). A \(2\times3\) and a \(3\times2\) matrix are different.

Never assume matrices commute. Keep the order of factors in every step.

\(AB = AC\) does not imply \(B = C\), and \(AB = O\) does not imply \(A = O\) or \(B = O\).

From \(a_{ii} = -a_{ii}\), every diagonal entry of a skew-symmetric matrix is 0.

An \(n\times n\) symmetric matrix is fixed by \(\frac{n(n+1)}{2}\) entries; a skew-symmetric one by \(\frac{n(n-1)}{2}\) entries (its diagonal is zero).

Inverse needs AB = BA = I, which requires square matrices of the same order.

If \(A^2-3A+2I=O\) then \(A(3I-A)=2I\), so \(A^{-1}=\tfrac12(3I-A)\). Multiply the equation by \(A^{-1}\) whenever the constant term is non-zero.

A square matrix is invertible exactly when \(|A| \ne 0\).

Choose the row or column with the most zeros – fewer calculations, fewer slips.

When the area is given, solve \(\det = +2\Delta\) and \(\det = -2\Delta\). Expect two values.

Before finding an inverse, evaluate |A|. If it is zero, the inverse does not exist – say so and stop.

After solving AX = B, substitute the values back into one or two original equations. It takes 30 seconds and catches most errors.

• \(|A|\neq0\): unique solution \(X=A^{-1}B\).
• \(|A|=0\) and \((\operatorname{adj}A)B\neq O\): no solution (inconsistent).
• \(|A|=0\) and \((\operatorname{adj}A)B=O\): infinitely many solutions or none – check further.

LHL = RHL = \(f(c)\). For piecewise functions, test only the joining points.

\(|x-c|\) at \(c\); \(x^{1/3}\) at 0 (vertical tangent). Differentiability ⇒ continuity, never the reverse.

Recognise multiple-angle formulas: with \(x=\tan\theta\), \(\frac{2x}{1-x^2}=\tan2\theta\) and \(\frac{1-x^2}{1+x^2}=\cos2\theta\); with \(x=\sin\theta\), \(3x-4x^3=\sin3\theta\). Always check that the resulting angle lies in the principal branch.

Mark the zeros of \(f\,'\) on a number line and test one point in each interval; the sign of \(f\,'\) gives increasing (+) or decreasing (−).

• Fixed perimeter ⇒ maximum area rectangle is a square.
• Closed cylinder of given volume with least surface area: h = 2r (height = diameter).
• Open box from an a × a sheet: cut squares of side a/6.
• Sum of two positive numbers fixed ⇒ product is greatest when they are equal.

Compare the values at critical points inside the interval and at the two end points.

\(\int\frac{f\,'(x)}{f(x)}dx=\log|f(x)|+C\), e.g. \(\int\frac{2x+3}{x^2+3x+1}dx=\log|x^2+3x+1|+C\).

In \(\int_0^{\pi/2}\sin^2x\cos x\,dx\) put \(u=\sin x\): limits become 0 and 1, giving \(\int_0^1u^2du=\frac13\).

Replace \(x\) by \(a+b-x\) and add the two forms of the integral – many trigonometric integrals collapse to \(\int_a^b1\,dx\).

A circle or ellipse centred at the origin is symmetric in both axes: compute the first-quadrant area and multiply by 4. A parabola y² = 4ax is symmetric about the x-axis: double the upper half.

If a derivative appears inside sin, cos, e, log (e.g. \(y''+e^{y'}=0\)), the equation is not a polynomial in derivatives and its degree is not defined.

The general solution of an nth-order equation has exactly n arbitrary constants.

For \(\vec r=x\hat i+y\hat j+z\hat k\): \(l=\frac xr,\ m=\frac yr,\ n=\frac zr\) and \(l^2+m^2+n^2=1\).

\(|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b\); \((\vec a+\vec b)\cdot(\vec a-\vec b)=|\vec a|^2-|\vec b|^2\).

On \(\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z}{4}=\lambda\), every point is \((1+2\lambda,\ -3-\lambda,\ 4\lambda)\).

Two non-parallel lines intersect exactly when their shortest distance is 0 (they are coplanar).

To decide which side of \(ax+by=c\) to shade, substitute a point not on the line (usually the origin). If it satisfies the inequality, shade its side.

If \(Z\) has the same optimal value at two adjacent corners, all points on the edge joining them are optimal.

For independent events, \(P(\text{at least one})=1-\prod P(A_i')\). This is usually quicker than adding cases.