Title: InkMaths / Further Maths: definitions and slips
On every note of this deck, unless a note says otherwise:
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Front: CHECK YOURSELF
Evaluate (1 + i)⁴ without expanding four brackets.
Back: (1 + i)² = 2i, and (2i)² = −4. Two squarings beat a binomial expansion; De Moivre gives the same via modulus (√2)⁴ = 4 and argument 4 × π/4 = π.

Front: SPOT THE ERROR
Say what is wrong with this claim:
Differentiating cosh introduces a minus sign, just as cos does.
Back: Both derivatives are positive. cosh' = sinh and sinh' = cosh, with no minus anywhere in the pair. The minus already sits inside sinh's definition, doing the job the sign flip does in trig.

Front: RECALL
State the three ways to write a plane, and how to move between them.
Back: Vector form is r = a + λb + μc. Scalar product form is r · n = a · n, where n is normal to both direction vectors and comes from their cross product. Cartesian form is ax + by + cz = d, and its coefficients are that same normal, which is what makes angle and distance questions quick.

Front: SPOT THE ERROR
Say what is wrong with this claim:
Once the inductive step is proved, the statement holds for all n.
Back: The step only passes truth along. Without a verified base case there is nothing to pass. A false formula can satisfy the step perfectly and fail at n = 1, so the base case carries a mark of its own and the examiner looks for it first.

Front: DEFINE
Define: improper integral.
Back: An integral with an infinite limit or an integrand unbounded at a limit, defined as the limit of an ordinary integral, and said to converge when that limit is finite.

Front: DEFINE
Define: invariant line.
Back: A line whose every point maps to a point of the same line under the transformation. Points may move along it; none leaves it.

Front: CHECK YOURSELF
Find the mean value of f(x) = x³ on [0, 2].
Back: ∫x³ dx from 0 to 2 is 4; dividing by the width 2 gives mean 2. Integral over width, always.

Front: CHECK YOURSELF
Find the distance from the origin to the plane 2x + 2y + z = 9.
Back: |0 + 0 + 0 − 9|/√(4 + 4 + 1) = 9/3 = 3. Substitute the point, subtract the constant, divide by the length of the normal vector.

Front: DEFINE
Define: principal argument.
Back: For z ≠ 0, the angle the position vector of z makes with the positive real axis, measured anticlockwise, in the interval (−π, π]. The origin has modulus 0 and no argument at all.

Front: RECALL
State the trick for an induction divisibility proof.
Back: Write f(k + 1) − f(k), or f(k + 1) − m f(k) for a suitable multiple m, so the assumed term drops out. What is left must show the divisor as an explicit factor. Finish with the conclusion sentence naming true for n = 1, assumed for n = k, shown for n = k + 1.

Front: SPOT THE ERROR
Say what is wrong with this claim:
Polar area is the integral of r dθ, by analogy with area under a graph.
Back: Polar regions are swept by sectors, not strips, and a sector's area is ½r² dθ. The r comes in squared, with a half in front.