Common mistakes

Check every element of the underlying set A, including those that appear in no pair at all.

They are independent. R = \{(1,2),(2,1)\} on \{1,2\} is symmetric but not transitive, since (1,2) and (2,1) are present but (1,1) is not.

It need not be. The empty relation on a non-empty set is the standard counter-example.

Examples can only disprove. To prove a property, argue with arbitrary a, b, c.

Compare the range with the codomain given in the question. If they differ, the function is not onto.

On \mathbb{R} it is not: f(2) = f(-2) = 4. It becomes one-one only after the domain is restricted to [0,\infty) or (-\infty,0].

The equivalence holds only for functions from a finite set to itself. For infinite domains, check each property separately.

\cos^{-1}(-x) = \pi - \cos^{-1}x.

Check the branch first. If x \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], reduce to the angle in the branch with the same sine.

\sin^{-1} denotes the inverse function. Write \csc x or (\sin x)^{-1} if the reciprocal is meant.

A principal value is a single number inside the stated range. General solutions belong to trigonometric equations, not to this chapter.

Their domain is |x| \geq 1. An argument like \sec^{-1}(0.5) is undefined.

The range is the open interval \left(-\frac{\pi}{2}, \frac{\pi}{2}\right); \frac{\pi}{2} is an asymptote, never a value.

Expand (A+B)^2 as A^2 + AB + BA + B^2. It equals A^2 + 2AB + B^2 only if A and B commute.

Not valid for matrices. Non-zero matrices can multiply to the zero matrix.

(AB)^{T} = B^{T}A^{T} — the order reverses.

For AB, the number of columns of A must equal the number of rows of B. Write the orders down before starting.

Recompute — the diagonal of \frac{1}{2}(A - A^T) is always zero, so a non-zero entry there signals an arithmetic error.

Without it the two parts add to 2A. Verify by adding them back together.

Transpose it. \operatorname{adj}(A) = [C_{ij}]^{T}.

Write out the sign grid \begin{bmatrix} + & - & + \\ - & + & - \\ + & - & +\end{bmatrix} before starting.

|kA| = k^{n}|A| where n is the order of A.

The determinant may come out negative; area is \frac{1}{2} of its absolute value.

It has no unique solution. Test (\operatorname{adj} A)B to decide between no solution and infinitely many.

Evaluate |A| first. If it is zero, state that A^{-1} does not exist and stop.

Compute the left-hand limit, the right-hand limit and the function value, then compare all three.

The implication runs the other way only. Corners and cusps are continuous but not differentiable.

y is a function of x, so the chain rule applies: \dfrac{d}{dx}(\ln y) = \dfrac{1}{y}\cdot\dfrac{dy}{dx}.

Neither rule applies — the base and the exponent vary. Use logarithms: \dfrac{dy}{dx} = x^{x}(1 + \ln x).

\dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)\cdot\dfrac{dt}{dx}.

Substitute the original expression for y to give the answer in x.

Substitute the critical point back into the original function to obtain the value.

Compare f at both endpoints and at every critical point inside the interval.

Classify it. f(x) = x^3 at x = 0 has zero derivative but no extremum.

Use the constraint to eliminate one variable first, then differentiate.

Differentiate the general relation with respect to t first, then substitute the instantaneous values.

Every indefinite integral carries +\,C.

If u = g(x), the limits become g(a) and g(b). Alternatively substitute back to x before applying the original limits.

\ln|x| + C.

Apply ILATE. Choosing badly usually produces an integral harder than the original.

Divide first to get a polynomial plus a proper fraction, then decompose the proper part.

The first gives a logarithm; the second gives an inverse tangent. The sign changes the answer completely.

Take the absolute value. Area is a magnitude and cannot be negative.

Split at the crossing and add \left|\int\right| for each piece.

Test a point inside the interval to see which curve is genuinely on top there.

Set the curves equal and solve; the roots are the x-coordinates of the limits.

Switch to \int x\,dy with horizontal strips, which crosses such a region cleanly.

\dfrac{dy}{dx} = v + x\dfrac{dv}{dx} — y = vx is a product.

Put the equation into the standard form \frac{dy}{dx} + Py = Q first.

Use \text{I.F.} = e^{\int P\,dx} with no constant; the constant enters once, at the final integration.

Substitute v = \frac{y}{x} back to express the solution in x and y.

State that the degree is not defined. Order is still defined.

Include +C; apply an initial condition only when the question supplies one.

\vec{a}\cdot\vec{b} is a number; \vec{a}\times\vec{b} is a vector. Check the type before writing the final line.

\vec{a}\times\vec{b} = -\left(\vec{b}\times\vec{a}\right).

Triangle: \frac{1}{2}\left|\vec{a}\times\vec{b}\right|. Parallelogram: \left|\vec{a}\times\vec{b}\right|.

The \hat{j} component carries a minus sign: \vec{a}\times\vec{b} = (a_2b_3-a_3b_2)\hat{i} - (a_1b_3-a_3b_1)\hat{j} + (a_1b_2-a_2b_1)\hat{k}.

Direction cosines are the components of the unit vector; divide the direction ratios by the magnitude.

For parallel lines \vec{b_1}\times\vec{b_2} = \vec{0}, so use d = \frac{\left|\vec{b}\times(\vec{a_2}-\vec{a_1})\right|}{|\vec{b}|} instead.

In \frac{x+2}{3} = \frac{y-1}{4} = \frac{z}{5}, the point is (-2, 1, 0) — set each numerator to zero.

The angle between two lines is the acute one; take the absolute value of the cosine.

\vec{a_1}, \vec{a_2} are points on the lines; \vec{b_1}, \vec{b_2} are the directions. Label them before substituting.

Divide the ratios by \sqrt{a^2+b^2+c^2} to obtain direction cosines, then check l^2+m^2+n^2=1.

Substitute the corner coordinates into Z and state the value, along with where it occurs.

Apply the open half-plane test. An unbounded region may have no maximum at all.

Include them; they usually form two of the boundaries of the feasible region.

Substitute (0,0) into the inequality. If it holds, shade the side containing the origin.

Solve the two boundary equations simultaneously for an exact answer.

The condition — the event known to have occurred — is written after the bar. Identify it before writing the fraction.

If A and B are mutually exclusive with non-zero probabilities, they are dependent: knowing A occurred tells you B did not.

The denominator is the total probability of the observed event — sum over every cause.

Check that the prior probabilities sum to 1 before proceeding.

Use P(A\cap B) = P(A)P(B\mid A) unless independence has been established or stated.