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.
300 fundamentals
“The average winnings in a game of rolling a die and receiving the face value is the expected value.”
Expected value E(X)= Σ x_i·P(x_i) is the long-run average of a random variable. It works because each outcome contributes proportionally to its chance. A trap: mixing up sample mean with expected value; use probabilities, not observed frequencies.
Memory trick: E = Σ(value × chance).
“When you slide a set of parallel contour lines over a map, the first touch point on the feasible region is where the optimum lies.”
A linear objective reaches its extreme on the boundary of the feasible region. Usually the extreme occurs at a vertex, but if the objective line is parallel to a constraint edge, every point on that edge is optimal; a constant objective makes the whole region optimal.
Memory trick: Check vertices; if objective line aligns with an edge, the whole edge is optimal.
“Like climbing a ladder, proving the first rung and then showing each step leads to the next secures the whole climb.”
Induction shows P(n) holds for all natural n by (1) proving P(1) (base) and (2) assuming P(k) true to prove P(k+1) (inductive step). The hypothesis applies only to a single k, not all n. Skipping the base case or using an unjustified jump in the inductive step is a common exam trap.
Memory trick: Base = first rung; if you can move from any rung to the next, you can reach any height.
“|x|<a is like a person staying within a fence a meters from the house; |x|>a is stepping outside the fence.”
|x|<a (a>0) means –a<x<a, a single open interval; |x|>a means x<–a or x>a, two disjoint rays. The direction of inequality flips when dividing by a negative inside absolute value. In JEE, watch the strict vs. non-strict signs; ≤ gives closed intervals.
Memory trick: '<' gives interval, '>' gives two rays.
“A metronome’s tick repeats every beat, just like a periodic function repeats its values each period.”
A periodic function satisfies f(x+T)=f(x) for all x, where T>0 is called the period. The smallest such T is the fundamental period; any integer multiple also works. In JEE problems you must pick the least T, otherwise you may over-estimate the answer.
Memory trick: Fundamental period = smallest T with f(x+T)=f(x).
“A traffic light shows red, yellow, green – the signum function tells if a number is negative, zero, or positive.”
sgn(x) = -1 for x<0, 0 for x=0, and +1 for x>0. It is a piecewise constant function used to capture the sign of a number. A common trap is forgetting that sgn(0)=0, so sgn²(x) equals 1 only for x≠0.
Memory trick: sgn: negative → -1, zero → 0, positive → +1.
“When two speakers emit sound waves at angles A and B, the combined pressure can be rewritten as two sine waves with angles (A+B) and (A-B).”
2 sin A cos B = sin(A+B)+sin(A-B). It follows from the sine addition formulas by multiplying and rearranging. A common trap is dropping the factor 2 or ignoring that sin(A-B)=-sin(B-A) when B>A, which changes the sign.
Memory trick: 2 sin A cos B → two sines (A+B, A-B).
“A swinging pendulum with horizontal drive a and vertical drive b reaches its peak displacement equal to √(a²+b²).”
The expression a sinθ + b cosθ can be written as R sin(θ+φ) where R=√(a²+b²); thus its maximum magnitude is ±R. The trap is to think the max is a+b or |a|+|b| – only the resultant vector length matters.
√(a²+b²)
Memory trick: R is hypotenuse of a-b right triangle.
“A swing’s rope at the midpoint of its swing makes angle θ/2, and the height rise R(1-cosθ) follows from the half-angle identity.”
The half-angle identities are sin(θ/2)=±√[(1-cosθ)/2] and cos(θ/2)=±√[(1+cosθ)/2]; the sign is chosen from the quadrant of θ/2. They are derived from the double-angle formulas. A frequent error is to drop the ± and always take the positive root, giving wrong values outside the first quadrant.
Memory trick: ½θ: sin → √[(1-cosθ)/2]; cos → √[(1+cosθ)/2]; keep the ± sign.
“Like points of a pizza cut into equal slices, the n roots lie on a circle equally spaced.”
A non-zero complex number z = re^(iθ) has exactly n distinct nth roots: r^(1/n) e^(i(θ+2kπ)/n), k=0…n−1. They sit on a circle of radius r^(1/n) and are separated by 2π/n radians. Missing the 2kπ term loses some roots.
Memory trick: Root angles = (θ+2kπ)/n, step = 360°/n
“When two rotating vectors combine, the resulting vector’s direction is the sum of their angles.”
If z₁ = r₁e^(iα) and z₂ = r₂e^(iβ), then arg(z₁z₂)=α+β (mod 2π). Add the angles and, if the sum exceeds 2π, subtract 2π. A common trap is forgetting the modulo-2π reduction.
Memory trick: Multiply → add angles, wrap at 360°
“The brightness of a fading LED that dims linearly while its power supply decays exponentially follows an arithmetico-geometric pattern.”
An arithmetico-geometric progression has term Tₖ=(a+(k-1)d)·r^(k-1), product of an AP term and a GP term. Sum is found by multiplying the series by r and subtracting to cancel. Trap: using pure AP or GP sum formulas gives wrong result.
Memory trick: AGP = (AP term)×(GP term)
“When you peel an onion layer by layer, each outer layer cancels the inner, leaving only the core—like a telescoping sum.”
A telescoping series can be written as Σ[ f(k) – f(k+1) ]. Adjacent terms cancel, leaving only f(first) – f(last+1). Trap: missing a sign or the final term leaves extra pieces.
Memory trick: Write as Δ; only first and last survive
“The side of a square whose area equals a rectangle’s area is the geometric mean of the rectangle’s sides.”
The geometric mean of a and b is √(ab). It is the side of a square whose area equals the rectangle a×b, giving a balanced multiplicative average. Trap: applying GM when numbers have opposite signs gives a non-real result.
√(ab)
Memory trick: GM = √(product)
“Seating 5 friends around a round table yields (5-1)! = 24 distinct seatings.”
In a circle, rotations are identical, so linear permutations n! are reduced by a factor of n, giving (n−1)! distinct arrangements. Trap: treating the circle as a line and using n! overcounts by n.
(n−1)!
Memory trick: Circle → fix one, arrange rest!
“Finding ways to pick 3 cards from a deck without any ace = total ways minus ways that include at least one ace.”
For restricted selections, count all unrestricted ways and subtract the cases that violate the condition (the complement). This avoids direct enumeration of complex constraints. Trap: overlooking overlapping violations; apply inclusion-exclusion when needed.
Memory trick: Unwanted → total − bad
“Stacking oranges in a pyramid, each orange rests on two below – just like Pascal’s rule.”
nCr + nC(r−1) = (n+1)Cr for 1≤r≤n. The mistake is applying the rule at r=0 or r=n+1 where the terms outside the triangle are zero.
Memory trick: Above two add → next row entry.
“A ladder with n steps has n+1 rungs, just as (a+b)ⁿ yields n+1 terms.”
(a+b)ⁿ expands into n+1 distinct terms when a and b are unlike. If a=b, like terms combine, reducing the count – a frequent oversight.
Memory trick: Terms = exponent + 1.
“A kite on a string traces a perfect circle when the wind keeps it at a fixed distance from the hand.”
A locus is the set of all positions a moving point can occupy while satisfying a given condition. Express the condition as an equation to obtain the curve. Exam trap: assume a shape (e.g., circle) without deriving the equation.
Memory trick: L = Locations satisfying condition
“A ruler placed through three pins shows they lie on one straight line when no triangle can be formed.”
Three points are collinear when the area of the triangle they would form is zero, i.e., ½|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|=0. Use the determinant form to avoid sign errors. Trap: forgetting the absolute value or mixing point order.
Memory trick: Collinear → Area = 0
“When you draw a circle on graph paper, the coefficients g and f shift the centre like moving the compass point left/right and up/down.”
x²+y²+2gx+2fy+c=0 represents a circle with centre (−g,−f) and radius r=√(g²+f²−c). The quantity under the root must be positive for a real circle. Common slip: forgetting the minus sign in the centre coordinates.
Memory trick: Centre = (−g,−f)
“A basketball rolling straight hits a wall at just one point – the wall line is tangent to the ball’s circular path.”
A line is tangent to a circle if it meets the circle at exactly one point, i.e., the perpendicular distance from the centre (a,b) to the line y=mx+c equals the radius r. Hence (ma-b+c)² = r²(1+m²). The usual error is omitting the ‘a’ term and using (c-b)², which only holds when a=0.
Memory trick: Tangent ⇔ |ma-b+c| = r·√(1+m²) (include a!).
“A hyperbola’s branches chase straight lines like a runner forever approaching the finish line without crossing it.”
For hyperbola x²/a²−y²/b²=1 the asymptotes are the straight lines y=±(b/a)x passing through the centre. The curve gets arbitrarily close to these lines as |x|→∞. Remember for rectangular hyperbola a=b, asymptotes are y=±x.
Memory trick: Slope = b/a
“Imagine two lanterns at the foci; any point on the ellipse reflects light to both lanterns with equal total travel.”
Ellipse foci are at (±c,0) with c² = a²−b² (major axis along x). For any point P on the ellipse, PF₁+PF₂ = 2a, a constant. A frequent error is swapping a²−b² with b²−a²; keep c² = a²−b².
Memory trick: c² = a²−b²