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
“A detective notes: “If the window is broken, the alarm rang”; the contrapositive is “If the alarm didn’t ring, the window isn’t broken”.”
The statement ‘If p then q’ is logically identical to ‘If not q then not p’. Both have the same truth table, so they are always true together. In exams, forgetting to negate both parts gives the converse, not the contrapositive, and loses marks.
Memory trick: Flip + Negate both → contrapositive
“A recipe says, “If you add sugar, the cake is sweet”; the converse would be “If the cake is sweet, you added sugar”, which isn’t guaranteed.”
The converse of ‘If p then q’ is ‘If q then p’; the inverse is ‘If not p then not q’. Neither is logically equivalent to the original statement. Students often mistake the inverse for the converse or assume they’re true like the original.
Memory trick: Converse swaps, inverse negates both
“A rubber band stretched around two sticks can’t reach a third longer stick unless you add the third side.”
For any triangle, the sum of any two sides must be greater than the third side; equality occurs only when the points are collinear. This ensures a non-degenerate triangle. Trap: writing ≥ instead of >, which would admit a straight line.
Memory trick: Two sides always outstretch the third
“Three cars travel the same distance at 60, 80 and 100 km/h; their AM speed is highest, HM (based on times) is lowest, and GM lies between.”
For any set of positive numbers, AM ≥ GM ≥ HM, with equality only when all numbers are equal. GM is defined for non-negative numbers, but the chain needs all numbers strictly positive because HM uses reciprocals. A common trap is applying the order when a value is zero or negative, which makes HM undefined or breaks the inequality.
Memory trick: A > G > H for positives – think ‘A-G-Higher’ (HM smallest).
“The number of ways 4 runners can finish a race is 4! = 24.”
n! is the product of all positive integers up to n, defined recursively as n! = n·(n−1)! with 0! = 1. It counts permutations of n distinct items. Trap: using factorial for non-integers or forgetting 0! = 1 in calculations.
n! = n×(n−1)!
Memory trick: 0! = 1, start the product
“Counting steps on a staircase where each step is 2 cm higher, the total height after n steps equals n/2 × (first + last).”
The sum of n terms of an AP is Sₙ = n/2 (first + last) = n/2[2a+(n-1)d]. Remember to use the actual first and last terms, not just a and d. A frequent mistake is mixing up a+(n-1)d with a+nd.
Sₙ=n/2(a+l)
Memory trick: Pair first with last: S = n/2 × (first+last).
“Choosing 3 toppings out of 8 is the same as leaving out the 5 you don’t want.”
nCr equals nC(n−r) because picking r objects is equivalent to deciding which n−r objects to leave out. This symmetry cuts work, especially when r>n/2. Forgetting it can double calculations or cause overflow.
nCr=nC(n−r)
Memory trick: r or n−r, same pick.
“Measuring the gap between two parallel railway tracks with a ruler uses |c1-c2|/√(a²+b²).”
For parallel lines ax+by+c1=0 and ax+by+c2=0, the perpendicular distance is |c1−c2| divided by √(a²+b²). Derive by picking a point on one line and measuring to the other. Trap: drop the absolute value or use √(a²−b²).
Memory trick: Dist = |Δc| / √(a²+b²)
“The steepness of a road that curves like tan x grows as sec² x, just like a shadow lengthening under a moving sun.”
d/dx(tan x)=sec² x because tan x = sin x/cos x and applying the quotient rule yields sec² x. Remember sec² x = 1+tan² x. Trap: writing derivative as sec x or forgetting the square.
d/dx(tanx)=sec²x
Memory trick: tan → sec².
“A chef tasting a soup adds a tiny fraction of spice; the flavor change follows (1+δ)ⁿ and the series works only while the pinch is less than the whole dish (|δ|<1).”
For any real n, (1+x)ⁿ = Σ_(r=0)^∞ C(n,r)x^r, an infinite series that converges for |x|<1. At the endpoints x=±1 convergence is not automatic: at x=1 it converges only if n>-1, at x=-1 only if n>0. In JEE you must state the radius = 1 and check the two endpoints, otherwise the answer is marked wrong.
Memory trick: Radius = 1 → ‘unit circle’; then test x=+1 (need n>-1) and x=-1 (need n>0).
“A piano key can be silent yet still present; D<0 means the equation has two invisible (complex) roots, not zero roots.”
D=b²−4ac < 0 does not mean ‘no solution’; it means the quadratic has two non-real complex conjugate roots. Exams expect you to state ‘complex roots’ rather than ‘no root’. Forgetting this leads to loss of marks.
Memory trick: D<0 → two complex roots, not zero.
“A seesaw balances only when the total torque follows the rule sum = –b/a, just like the two kids on each side.”
For ax²+bx+c=0 the roots α and β satisfy α+β = –b/a and α·β = c/a (Vieta’s relations). The common slip is to forget the minus sign on b, especially when a≠1. Remember the sign when forming equations from given sums or products.
Memory trick: Sum = –b/a, Product = c/a – keep the minus with b.
“A parabolic bridge arch meets the river surface: two contact points if D>0, just touches if D=0, floats above if D<0.”
The discriminant D = b²-4ac decides root nature: D>0 gives two distinct real roots, D=0 gives one repeated real root, D<0 gives two complex conjugate roots. This follows directly from solving ax²+bx+c=0. In JEE, compute D first; a zero discriminant still yields a real root, not “no solution”.
Memory trick: D>0 → 2 real; D=0 → 1 real; D<0 → 0 real (complex)
“A doctor checks temperature to know fever, not run full tests; similarly you check D to know root nature.”
Only the sign of D matters for the nature of roots; you need not find the roots themselves. Many waste time solving the quadratic when the question asks merely whether the roots are real, equal or complex. Check D and stop.
Memory trick: Ask D, not x.
“Your GPS computes the quickest turn by solving a quadratic with x = [–b ± √(b²−4ac)]/2a, just like the formula.”
The quadratic formula x = [–b ± √(b²−4ac)]/2a gives the two solutions of any quadratic ax²+bx+c=0 (a≠0). It is derived from completing the square and works for any coefficients. A frequent error is to miss the ‘2a’ denominator or change the sign of b.
x = [−b ± √(b²−4ac)] / 2a
Memory trick: Minus b ± √(b²−4ac) over 2a.
“When two friends split a pizza, the total slices they claim follow sum = –b/a and each gets c/a of the whole.”
Vieta’s relations state α+β = –b/a and α·β = c/a for roots α, β of ax²+bx+c=0. They let you construct a quadratic when the sum or product of roots is known, and vice-versa. Watch out for sign mistakes when b or c are negative.
sum = −b/a, product = c/a
Memory trick: Sum –b/a, Prod c/a – signs matter.
“A ball thrown on a flat floor lands (real roots) only if D≥0; if D<0 it never touches the ground, giving complex roots.”
If D≥0 the parabola y = ax²+bx+c meets the x-axis, giving real roots (touching when D=0). If D<0 the curve stays above or below the axis, so the solutions are complex conjugates. Remember D=0 still yields a single real root, not two.
Memory trick: D≥0 hits axis, D<0 flies.
“Factoring x²-5x+6 as (x-2)(x-3) shows 2 and 3 are the x-values that make the quadratic zero.”
Roots are the x-values that satisfy the equation =0. Each linear factor (x-root) corresponds to one root, so a quadratic = a(x-α)(x-β). Beware: include the leading coefficient a when writing factors.
Memory trick: Root → zero; Factor = (x-root)
“The equation x²+1=0 never touches the x-axis, yet its solutions are i and –i.”
A negative discriminant (D<0) gives two complex conjugate roots, not “no roots”. The quadratic always has two roots in the complex plane. The trap is to equate “no real root” with “no root”.
Memory trick: D<0 → imaginary pair
“Multiplying a complex number by i turns it 90° counter-clockwise on the Argand diagram, like rotating a vector on a compass.”
The imaginary unit i is defined by i² = –1, giving a way to take square roots of negative real numbers. In the complex plane, multiplying by i corresponds to a 90° counter-clockwise rotation, so applying it twice flips the direction (i² = –1). On exams never use √a·√b = √(ab) when a or b is negative; that rule holds only for non-negative reals.
Memory trick: i = 90° CCW turn; i² = flip (-1)
“Treating √(–1)·√(–1) as √1 is like assuming two left-hand turns make a right turn; they actually give a left turn.”
The identity √a·√b = √(ab) holds only for a,b ≥ 0. For a = b = –1, √(–1)·√(–1)=i·i = –1, not +1. Hence √(–1)·√(–1)= –1, confirming i² = –1.
Memory trick: √(–1)·√(–1)= –1 because i²=–1
“|z| is the straight-line distance from the origin to the point (a,b) on the Argand plane.”
For z = a+bi, modulus |z| = √(a²+b²) is its distance from the origin. Its conjugate ̅z = a–bi reflects across the real axis. Multiplying z by ̅z gives |z|², a real number, useful for rationalising denominators.
Memory trick: z·̅z = |z|²
“Skipping the conjugate when dividing is like forgetting to cancel a common factor; the result stays complex.”
A frequent slip is dividing by a complex number without using its conjugate, leaving an imaginary denominator. Use z/ w = (z·̅w)/|w|² to make the denominator real. Forgetting this step yields an incorrect answer.
Memory trick: Use conjugate → real denominator
“When you plot the voltage phasor of a household AC outlet, its real and imaginary parts become x-y coordinates on the Argand plane.”
A complex number a+bi is represented by the point (a,b) on the Argand plane: real part on the horizontal axis, imaginary coefficient on the vertical. Swapping the axes leaves the distance √(a²+b²) unchanged but gives a different angle. Exam trap: mixing axes changes the argument (quadrant) while the modulus stays the same.
Memory trick: Re → right, Im → up