Fundamentals
The core facts every aspirant should own — each a titled nugget with a real-world story, the concept in plain words, and a memory trick. Works even when the internet doesn't.
1,200 fundamentals
“A palindrome’s centre letters are the middle terms of a binomial expansion.”
If n is even, there is one middle term at r = n/2. If n is odd, there are two middle terms at r = (n−1)/2 and r = (n+1)/2. Students often swap the even-odd cases.
Memory trick: Even n → r=n/2; odd n → r=(n±1)/2.
“Flipping a fair coin n times gives 2ⁿ outcomes, exactly the sum of all nCr.”
Σ_(r=0)ⁿ nCr = (1+1)ⁿ = 2ⁿ. The exam trick is to set x=1 in (1+x)ⁿ. Forgetting the r=0 term loses the +1.
ΣnCr = 2ⁿ
Memory trick: Set x=1 in (1+x)ⁿ → sum = 2ⁿ.
“When you cut a rope so the left piece is m parts and the right piece n parts, the cut point’s coordinates follow the same rule.”
For A(x1,y1) and B(x2,y2), if P divides AB such that AP:PB = m:n, then P = ((n·x1 + m·x2)/(m+n), (n·y1 + m·y2)/(m+n)). The coefficient of each endpoint is the opposite segment length. A common trap is swapping m and n, which yields the wrong point.
Memory trick: n with first point, m with second – ‘n-first, m-second’.
“A laser beam hits a wall at the shortest possible path, which is the perpendicular distance given by the formula.”
The perpendicular distance from (x₀,y₀) to ax+by+c=0 is d = |ax₀+by₀+c| / √(a²+b²). Absolute value removes sign; the denominator comes from the line’s normal vector length. Forgetting the absolute value gives a signed distance, not the required length.
d=|ax₀+by₀+c|/√(a²+b²)
Memory trick: |ax₀+by₀+c| over √(a²+b²).
“The angle where two intersecting roads meet is found from their slopes using tanθ = |(m₁−m₂)/(1+m₁m₂)|.”
For lines with slopes m₁ and m₂, tan θ = |(m₁−m₂)/(1+m₁m₂)| gives the acute angle between them. The denominator must be 1 + m₁m₂, not a sum of slopes. Using the wrong sign or omitting the absolute value yields the wrong angle.
Memory trick: tanθ = |Δm|/(1+m₁m₂).
“A pencil touching a coin at one point draws the tangent, which is perpendicular to the radius through that point.”
A tangent touches a circle at exactly one point and the radius drawn to that point is perpendicular to the tangent line. This follows from the shortest-distance property of a radius. In problems, any line through the point is not automatically a tangent; the ⟂ condition must be checked.
Memory trick: Tangent ⟂ radius at point of contact
“A planet’s orbit around the Sun traces an ellipse described by x²/a² + y²/b² = 1.”
The equation x²/a² + y²/b² = 1 represents an ellipse centered at the origin with semi-major axis a along the x-axis and semi-minor axis b along the y-axis. Any point (x,y) on it satisfies that the sum of distances to the two foci equals 2a. Remember: a and b are semi-axes, not full lengths; if a<b the major axis is vertical.
x²/a²+y²/b²=1
Memory trick: a = half the longest diameter, b = half the shortest.
“The shape of a cooling tower’s cross-section matches the hyperbola x²/a²−y²/b²=1, opening left-right.”
For the hyperbola x²/a²−y²/b²=1 the transverse axis lies along the x-direction (the term with +). The difference of distances to the foci is constant =2a. A sign mistake (using + for both terms) turns the equation into an ellipse, a frequent pitfall.
x²/a²−y²/b²=1
Memory trick: Hyperbola: one plus, one minus → opens along plus sign
“The latus-rectum is like the width of a satellite dish at the focus, measured perpendicular to the dish’s axis.”
In any conic the latus rectum is the chord through the focus and perpendicular to the major (or transverse) axis. For a parabola y²=4ax its length is 4a; for ellipse/hyperbola it is 2b²/a. Remember: it must be perpendicular—do not mix with any focal chord.
Memory trick: L = line through focus ⟂ axis
“A pendulum’s angle swings, and its velocity follows cosine, while the acceleration follows –sine.”
d/dx(sin x)=cos x and d/dx(cos x)=−sin x. The derivative of sine leads the function by 90°, cosine leads with a minus sign. Mistake: mixing up the sign for cosine’s derivative.
Memory trick: sin→cos, cos→−sin.
“A bank account with continuous compounding grows at a rate equal to its balance—just like eˣ.”
d/dx(eˣ)=eˣ; the exponential is its own derivative. For ln x, derivative is 1/x, reflecting the inverse relationship with eˣ. Trap: applying ln rule at x≤0 where it’s undefined.
Memory trick: eˣ self-derivative; ln' = 1/x.
“Dividing the fuel flow (u) by the wheel radius (v) while both change, the bike’s speed follows (u'v-uv')/v².”
The quotient rule gives the derivative of u/v as (u'v-uv')/v². It comes from writing u·v⁻¹ and applying the product rule. Common trap: forgetting to square v or reversing the sign of the second term.
(u/v)'=(u'v−uv')/v²
Memory trick: Low·dHigh − High·dLow over Low².
“Finding the slope of a circle x²+y²=25 is like measuring the tilt of a spinning coin; you differentiate both x and y terms.”
Treat y as y(x) and differentiate each term; for y-terms multiply by dy/dx (chain rule). Then collect dy/dx on one side and solve. Trap: omitting dy/dx on the differentiated y-terms, leading to a wrong slope.
Memory trick: Differentiate, then factor dy/dx.
“A roller coaster that starts and ends at the same height must have at least one flat spot in between.”
Rolle’s theorem: for a function continuous on [a,b], differentiable on (a,b) and with f(a)=f(b), there exists c∈(a,b) where f'(c)=0. The differentiability inside the interval is essential; equal ends alone are insufficient.
Memory trick: Equal ends → flat point
“While driving from city A to B, at some moment your speedometer reads exactly the trip’s average speed.”
Mean Value Theorem states that for a continuous f on [a,b] and differentiable on (a,b) there is c∈(a,b) with f'(c)= (f(b)-f(a))/(b-a). Both continuity and differentiability are required; missing either invalidates the claim.
Memory trick: Some instant = overall slope
“Peeling a banana: you first hold the peel (u) and then pull the fruit (dv); the total effort equals hold×pull minus the small slip when the peel slides (∫v du).”
Integration by parts is the reverse of the product rule: d(uv)=u dv+v du ⇒ ∫u dv = uv−∫v du. Pick u so du simplifies and dv is easy to integrate. A frequent mistake is dropping the minus sign or choosing u that makes du harder than the original integral.
∫u dv = uv − ∫v du
Memory trick: LIATE order: Log → Inverse trig → Algebraic → Trig → Exp for choosing u.
“Switching a recipe from cups to milliliters simplifies the math, just like substitution changes variables in an integral.”
In substitution, set t = g(x) so that dt = g′(x)dx, turning the integral into ∫f(t)dt. This works when the integrand contains a function and its derivative. In definite integrals, you must also replace the limits; forgetting this gives an error.
Memory trick: Change variable, also change limits.
“When chemists plot pH vs concentration, the curve’s slope 1/[H+] integrates to ln|[H+]|, the natural log.”
∫(1/x)dx = ln|x| + C for x≠0; the absolute value ensures the result is valid for negative x as well. Ignoring the bars yields ln x, which is undefined for x<0 and leads to sign mistakes.
∫1/x dx=ln|x|+C
Memory trick: 1/x integrates to ln|x|, keep the bars.
“Running a 400 m track from start to finish or from finish back to start covers the same distance.”
Using substitution u = a+b−x gives du = −dx; swapping limits restores the sign, so the area under f from a to b equals that under its mirror f(a+b−x). Forgetting to reverse limits after the substitution flips the sign and yields a wrong result.
Memory trick: Mirror swap, same integral.
“Imagine rain collecting between two sloping roofs; at any x the water depth equals the upper roof minus the lower roof.”
The area between y=f(x) and y=g(x) is ∫ₐᵇ[ f(x)−g(x) ]dx, where f is the upper curve on [a,b]. First locate their intersection points to set limits; if the curves cross, split the integral so the upper-lower order stays correct. A frequent mistake is integrating without confirming which curve is on top, which can produce a negative result.
Memory trick: Intersections → limits → (upper−lower) in each piece
“When you adjust a thermostat, the highest rate of temperature change it can sense is the order, while the exponent of that rate is the degree.”
The order of a differential equation is the highest order derivative appearing in it. After clearing radicals or fractions involving derivatives, the degree is the highest power of that highest-order derivative. Trap: degree is defined only for polynomial form; if a derivative is inside a root or a transcendental function, degree is not defined.
Memory trick: Order = top derivative; Degree = its power after clearing roots.
“Separating ingredients in a recipe mirrors moving all x-terms to one side and y-terms to the other before mixing.”
If dy/dx = f(x)·g(y), the equation is separable: rewrite as dy/g(y) = f(x)dx and integrate both sides. The trap: after integration, always add the constant only once, not to each side separately.
Memory trick: Separate → integrate → C once.
“Walk 3 km east then 4 km north; the straight line from start to finish is the resultant vector of the two moves.”
Vector addition follows the head-to-tail rule: place the tail of the second vector at the head of the first; the vector from the free tail to the free head is the resultant. The order of addition does not affect the result.
Memory trick: Resultant = start→end; triangle closes
“A GPS arrow points only direction, not distance – it’s a unit vector.”
A unit vector ˆa = a/|a| has length 1 and points in the same direction as a. It is used to express pure direction without magnitude. Forgetting to normalize leads to wrong direction cosines in 3-D problems.
â = a/|a|
Memory trick: ̂ = vector ÷ its length.