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.
“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
MathsSequences & Series· Class 11
Geometric mean
“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)
MathsPermutations & Combinations· Class 11
Circular permutations
“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!
MathsPermutations & Combinations· Class 11
Combinations with a condition
“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
MathsBinomial Theorem· Class 11
Pascal's rule
“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.
MathsBinomial Theorem· Class 11
Number of terms in an expansion
“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.
MathsStraight Lines· Class 11
Locus
“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
MathsStraight Lines· Class 11
Collinear points
“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
MathsConic Sections· Class 11
General equation of a circle
“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)
MathsConic Sections· Class 11
Condition of tangency
“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.
“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
MathsConic Sections· Class 11
Foci of an ellipse
“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
MathsApplication of Derivatives· Class 12
Second derivative
“The curvature of a roller coaster tells you if you’re entering a dip (concave up) or climbing a hill (concave down).”
The second derivative d²y/dx² is the derivative of the slope. Positive d²y/dx² means the graph bends upward (concave up); negative means it bends downward (concave down). It measures how quickly the slope changes, not whether the function itself is increasing.
d²y/dx²
Memory trick: ↑ curvature = + second derivative
MathsApplication of Derivatives· Class 12
Tangent and normal
“A road’s tangent shows the car’s direction; the normal is the wall built exactly perpendicular to that road.”
At x₀ the tangent line has slope m = f'(x₀). The normal line is perpendicular, so its slope is –1/m (provided m≠0). If the tangent is vertical (undefined slope), the normal is horizontal, and vice-versa. Forgetting the negative reciprocal causes sign errors.
Memory trick: Normal slope = –1/(tangent slope)
MathsApplication of Derivatives· Class 12
Approximation using differentials
“When you turn the faucet knob a tiny bit, the extra water flow ≈ slope (flow rate) × knob rotation.”
For a differentiable function, a tiny change Δx produces a change Δy ≈ f'(x)·Δx. The linear term dominates because higher-order terms contain (Δx)², negligible for small Δx. In exams, ensure Δx is truly small; applying the formula to a large step leads to big error.
“To find the growth rate of a bacteria colony that multiplies as xˣ, you first take log, turning the exponent into a product.”
Take ln of both sides, use ln(ab)=ln a+ln b and ln(aᵇ)=b ln a, then differentiate. This turns products, quotients or variable powers into sums, simplifying the derivative. Trap: forgetting the factor 1/y when differentiating ln y, so the final dy/dx is off.
Memory trick: ln → bring exponent down.
MathsDifferentiation· Class 12
Parametric differentiation
“The path of a roller-coaster described by x=sin t, y=cos t gives its slope dy/dx = (dy/dt)/(dx/dt).”
For parametric curves x=f(t), y=g(t), dy/dx = (dy/dt)/(dx/dt) by the chain rule dy/dx = (dy/dt)*(dt/dx). Ensure dx/dt ≠ 0. Trap: swapping numerator and denominator or forgetting to check dx/dt zero.
Memory trick: dy/dx = (dy/dt) ÷ (dx/dt).
MathsIntegration· Class 12
Definite integral as a limit of a sum
“Pouring water into a tub cup-by-cup; as cups get infinitesimally small the total equals the tub’s volume.”
A definite integral is the limit of Riemann sums: Σ f(x_i)Δx as Δx→0. The sum of thin rectangles approaches the exact area. Assuming the sum equals the area without taking the limit is a common error.
Memory trick: Riemann → tiny cups → limit = area.
MathsIntegration· Class 12
Integral of even/odd functions
“A seesaw balanced at the centre: equal opposite weights cancel, leaving zero net torque.”
If f is odd, f(−x)=−f(x) so areas on [−a,0] and [0,a] cancel → integral zero. If f is even, f(−x)=f(x) so total area is twice the half-interval area. The symmetry must be exactly about 0; shifting the interval breaks it.
Memory trick: Odd → 0, Even → 2×half.
MathsIntegration· Class 12
Partial fractions
“Breaking a chocolate bar into smaller pieces lets you eat each piece separately.”
Partial fraction decomposition rewrites a rational function as a sum of simpler fractions with linear or irreducible quadratic denominators, each easy to integrate. Missing a repeated factor or quadratic term leads to an incorrect decomposition.
Memory trick: PF = split, then integrate each piece.
MathsDifferential Equations· Class 12
Linear differential equation
“A charging capacitor follows dy/dx + (1/RC) y = V/RC, solved by multiplying with e^(∫P dx).”
A first-order linear DE has form dy/dx + P(x) y = Q(x). Multiply by integrating factor μ = e^(∫P dx) to make left side a product derivative: d(μ y)/dx = μ Q. Then integrate. Common mistake: forgetting to simplify μ before integration.
Memory trick: Linear → μ = e^(∫P) → d(μy)=μQ.
MathsDifferential Equations· Class 12
Degree of a differential equation
“In kinetic energy ½ m (dx/dt)² the squared velocity shows degree 2 after removing the fraction ½ m.”
Degree is the exponent of the highest-order derivative after the equation is cleared of radicals, fractions and denominators, provided the DE is a polynomial in its derivatives. It is undefined if a derivative is inside a transcendental function or a denominator. The exam trap: assuming a degree exists for non-polynomial forms like sin(y') or 1/y'.
Memory trick: Degree = power of the top derivative after polynomialising.
MathsVectors & 3D Geometry· Class 12
Coplanarity of vectors
“Three sticks lying flat on a table are coplanar; mathematically their scalar triple product being zero tells the same.”
Three vectors a, b, c are coplanar iff their scalar triple product [a b c]=a·(b×c) equals zero, meaning the volume of the parallelepiped they form is zero. The trap: mixing up order; the product changes sign if the order is permuted, but zero stays zero.
Memory trick: [a b c]=0 → zero volume → coplanar
MathsVectors & 3D Geometry· Class 12
Shortest distance between skew lines
“Imagine two non-parallel train tracks that never meet; the shortest bridge between them is the common perpendicular.”
For skew lines L₁ and L₂ with direction vectors d₁, d₂ and a point A on L₁, the shortest distance is |(AB·(d₁×d₂))|/|d₁×d₂|, where B is any point on L₂. The numerator projects the connecting vector onto the common normal. A common mistake is to apply the intersecting-lines formula (set numerator to zero).
Memory trick: d₁×d₂ gives common normal; project AB onto it