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
“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.
“The volume of a skewed box built from three edge vectors equals their scalar triple product.”
The scalar triple product a·(b×c) equals the signed volume of the parallelepiped formed by a, b, c. Cyclic permutations give the same value; swapping two vectors changes the sign. Zero indicates the three vectors lie in one plane.
[a b c]
Memory trick: a·(b×c)=volume; zero→coplanar.
“A flashlight beam hitting a wall shows the beam’s shadow length – that shadow is the projection of the beam onto the wall.”
The scalar projection of a on b equals (a·b)/|b|, giving the length of a along b’s direction. From a·b=|a||b|cosθ, dividing by |b| leaves |a|cosθ. In JEE, remember the denominator is |b|, not |b|² (that belongs to the vector projection).
(a·b)/|b|
Memory trick: Scalar proj = dot ÷ magnitude → ‘dot over mag’
“A ladder leaning against a wall forms angles with floor and wall; its direction cosines describe its 3-D tilt.”
Direction cosines l, m, n are cos α, cos β, cos γ where α,β,γ are the angles a line makes with the x, y, z axes. They always obey l²+m²+n²=1. Mistake: treating them as simple ratios without the cosine sign or forgetting the unit-vector condition.
Memory trick: L²+M²+N²=1
“When two book covers intersect, the angle you feel between them is the same as the angle between their normal arrows.”
The angle between two planes equals the angle between their normal vectors. Since normals are perpendicular to the planes, rotating one plane to the other is equivalent to rotating its normal to the other’s. Remember to take the acute angle; using the supplement gives a wrong answer.
Memory trick: Use normals; take acute θ
“Flipping a sheet of music notes over its diagonal swaps the staff lines, just like transposing swaps rows and columns.”
The transpose Aᵀ of a matrix A is obtained by interchanging its rows with columns; element aᵢⱼ becomes aⱼᵢ. This operation is useful for dot products and symmetry checks. Remember that (Aᵀ)ᵀ = A.
Memory trick: Rows ↔ columns, so (i,j)→(j,i)
“A perfectly still lake reflects the sky unchanged (symmetric), whereas a rotating fan blade gives a reversed image (skew-symmetric).”
A matrix is symmetric if Aᵀ = A, meaning it equals its mirror image across the main diagonal. It is skew-symmetric if Aᵀ = –A, which forces all diagonal entries to be zero. In JEE, verify symmetry by checking aᵢⱼ=aⱼᵢ and skewness by aᵢⱼ=–aⱼᵢ.
Memory trick: Sym: aij = aji; Skew: aij = –aji, diag 0
“A GPS map corrects coordinates only if its transformation matrix has a non-zero determinant.”
For a square matrix A, A⁻¹ = adj(A)/det(A). The inverse exists only when det(A) ≠ 0. Compute adjugate by cofactors and transpose. A common trap is to forget that a negative determinant still yields an inverse, only the sign of each entry changes.
A⁻¹=adjA/detA
Memory trick: A⁻¹ = adj/det, det≠0.
“In a 3-wire circuit, Cramer’s rule gives each current by swapping the voltage column into the conductance matrix.”
Cramer’s rule solves Ax = b by x_i = det(A_i)/det(A), where A_i replaces column i of A with vector b. It works only if det(A) ≠ 0; otherwise the system is either dependent or inconsistent. The exam trap is using the rule when the main determinant vanishes.
Memory trick: x_i = det(replace i)/det(main).
“The area of a triangular garden plot equals half the absolute value of the 3×3 determinant of its corner coordinates.”
For vertices (x1,y1), (x2,y2), (x3,y3), area = ½ | x1(y2−y3)+x2(y3−y1)+x3(y1−y2) | = ½ |det| of the 3×3 matrix with a column of 1’s. Forgetting the absolute value can give a negative area; ordering of points only changes sign.
Memory trick: Area = |det|/2 for 3 points.
“Rolling a die, the chance of getting a 2 or any even number uses the addition rule.”
For any two events, P(A∪B)=P(A)+P(B)−P(A∩B). Adding the separate probabilities counts the overlap twice, so subtract it once. Forgetting the subtraction can produce a probability larger than 1, which is impossible.
P(A∪B)=P(A)+P(B)−P(A∩B)
Memory trick: Add A and B, then subtract their overlap
“Two independent coin tosses both landing heads have probability (1/2)*(1/2)=1/4.”
When events are independent, the joint probability equals the product of the individual probabilities: P(A∩B)=P(A)P(B). If dependence exists, the product rule does not hold and you must use P(A|B)P(B). Applying multiplication blindly is a frequent error.
Memory trick: Indep events → product of probabilities
“Counting heads when flipping a fair coin 10 times is a binomial experiment.”
The binomial distribution gives the probability of exactly k successes in n independent trials with constant success probability p. It follows from multiplying p^k (1-p)^(n-k) and the combination nCk. Remember the trial count must be fixed; using it for dependent events is a common error.
Memory trick: B-n-p: Binomial needs fixed n and constant p.