JEE Main 2025 Mathematics – Comprehensive Bullet Notes

Coordinate Geometry: Lines, Circles, Parabolas, Ellipses & Hyperbolas

• Shortest distance between skew lines

  • Formula: $d = \frac{|(\vec r<em>2-\vec r</em>1) \cdot (\vec a<em>1 \times \vec a</em>2)|}{|\vec a<em>1 \times \vec a</em>2|}$
  • Typical manipulation: equate given numerical distance to obtain constraints on unknown parameters (e.g. (\lambda,\,\mu)).
  • JEE‐Main Ex. – distance 3 between $\ell<em>1:\; \frac{x-5}{2}=\frac{y-3}{1}=\frac{z-4}{3}$ and $\ell</em>2:\; \frac{x-7}{3}=\frac{y-6}{2}=\frac{z-2}{4}$ delivered $d=\frac{5}{\sqrt{6}}$ (option 1).

• Image of a point in a line (reflection)

  • Given P and a line AB, the image Q is found by resolving the foot of the perpendicular & doubling the segment.
  • In Q.1, line through A(4,7,1),B(3,5,3); reflecting P(1,0,3) gave $Q(4,7,7)\Rightarrow \alpha+\beta+\gamma=18$ (option 3).

• Focal chords of a parabola

  • For (y^2=16x), any point (t) is ((4t^2,8t)). Product of slopes of lines from focus etc. used to split segment in ratio.
  • With given point (1,–4) in II quadrant, gcd(m,n)=1 led to $m^2+n^2=17$(Ans).

• Parabola axis inclined to axes

  • Axis y=x; distance of vertex & focus from origin known ⇒ standard form $((x-y)/\sqrt2)^2 = 4a((x+y)/\sqrt2)$.
  • Point (1,k) satisfying gave k∈{4,9,8,3}; allowed option 2 (k=9).

• Circle touching axes in 3rd quadrant

  • Circle centre (–3,–3), radius 3. Another circle through (1,3) externally tangent to the first ⇒ distance of centres = sum radii ⇒ centre at (1,3), radius $\sqrt{(1+3)^2+(3+3)^2}-3=\sqrt{52}-3$. Point of contact ((\alpha,\beta)). Expression $(\beta-\alpha)^2=\frac{22}{9}$ ⇒ m+n=22 (option 3).

• Ellipse with minor–axis = (\tfrac14) focal distance

  • Using $2b = \tfrac14\cdot 2ae \Rightarrow b=a e/4\; \Longrightarrow \,(1-e^2)=e^2/16 \Rightarrow e=\tfrac{4}{\sqrt{17}}$ (option 1).

• Hyperbola focus–directrix property

  • Given one focus (10,0) & directrix x=10/9, standard definition $\frac{SP}{PM}=e$ led to $9(e^2+l) = 16$ (option 3).

• Circle of minimum area enclosing ellipse

  • For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\;foci(±2,0),\;e=\sqrt{1-b^2/a^2}$. Minimum enclosing circle has radius = semi–major axis. Using latus–rectum length 29 parallel to axis; maximisation produced area 8 (option 4).

• Triangle line–axes area condition

  • Line L: x+by+c=0 cuts axes at (–c,0) and (0,–c/b). Area = $\frac12 |c^2/b|=48$ ⇒ $c^2=96b$. Perpendicular from O forms angle 45°, i.e. foot has equal coordinates ⇒ distance $|c|/\sqrt{1+b^2}=\sqrt2|c|/2$. Gives $b^2+c^2=97$ (option 3).

Differential & Integral Calculus

• Leibnitz type functional equation

  • $\int_{1}^{x}f(t)\,dt = \frac{5x f(x)}{x^2-1}$ uniformly ⇒ differentiate both sides to obtain Bernoulli ODE leading to explicit $f(x)=k(x+1)^5$. Evaluated at x=3 ⇒ f(3)=32 (option 2).

• Mean value property via inequality

  • Twice differentiable f with $\bigl(\int<em>1^x f(t)dt\bigr)^2\le9(x-1)\int</em>1^x f^2(t)dt$ implies C-S equality ⇒ f constant; boundary forces $f(3)=32.$

• Differential equations

  • Linear first-order: $(x^2+1) y'-2xy=(x^4+2x^2+1)\cos x$; integrating factor $(x^2+1)^{-1}$; yields $y = x^2\cos x + 1$ ⇒ $\int_1^3 y \,dx =24$ (Ans).
  • Second–order homogeneous with sec^2 forcing: $y''+2y\sec^2x=2\sec^2x+3\tan x\sec^2x$, IC y(0)=5/4 ⇒ y(\pi/4)=21.

• Improper limits → continuity constants

  • Piece‐wise $f(x)=(1+ax)^{1/x}, x<0;\, b^{x}, x=0;\, c^{1/x}, x>0$ continuous at 0 ⇒ $e^a=b=c^1\Rightarrow ea=bc$, numeric 48 (option 3).

• Limit manipulations:

  • $\lim_{x\to0}\frac{\tan^{-1}(2x)-\tan^{-1}(x)}{x}=1$ etc.

• Beta/Gamma usage

  • $\int<em>0^{\pi/2} x\sin(x)\,dx = \tfrac\pi2 -1$ and $\int</em>0^{\pi/2} x\sin^3x\,dx = \tfrac34\pi-1$ leading to required ratio 4.

Algebra: Sequences, Series & Numbers

• Highest power of 3 in $50!$

  • $\sum_{k\ge1}\left\lfloor\frac{50}{3^k}\right\rfloor=16+5+1=22$ (option 2).

• Even-length AP with separate odd/even sums

  • Sodd=24, Seven=30, last exceeds first by 5. Solving gives integer terms =4 (option 1).

• Term independent of x in $\bigl(x^2+1/x\bigr)^{10}$

  • General term $\binom{10}{r}x^{2r-10}$ ⇒ independent when $2r-10=0\Rightarrow r=5$ ⇒ coefficient $\binom{10}{5}=252\times?$ Provided answer 210.

• Infinite G.P. & telescoping sums

  • $1+3+11+25+…$ obtained by cubic polynomial difference method gave sum 7240.

• Binomial ratio (15th term from start vs end) leads to equation to solve for n, giving 2300.

Matrices & Determinants

• Cayley–Hamilton use: $A^2(A-2I)-4(A-I)=0$ ⇒ characteristic roots satisfy (\lambda^2(\lambda-2)-4(\lambda-1)=0). Reduce $A^5=\alpha A^2+\beta A+\gamma I$ then trace identity yields $\alpha+\beta+\gamma =12.$

• Properties of adjugate

  • $\det(\operatorname{adj}(\operatorname{adj}(kA))) = k^{3(3-1)} \det(A)^{3-2}$ etc. For A with |A|=–1 and k=3 ⇒ exponent counting gives answer 38.

• Singular 2×2 matrices with entries from {2,3,6,9}

  • Condition $ad-bc=0$ excluding simultaneous zero. Count pairs ⇒ 36.

Vectors & 3-D Geometry

• Coplanar perpendicular unit vector in plane of a and b

  • Use $\hat n = \frac{\vec a\times\vec b}{|\vec a\times\vec b|}$ then rotate 90° within plane ⇒ candidates show sign patterns; correct option (4).

• Tetrahedron with mutually perpendicular edges:

  • If areas of faces with common vertex A are p,q,r, opposite face area = $\sqrt{p^2+q^2+r^2}$ ⇒ with 5,6,7 gives $\sqrt{110}$ (option 3).

• Triangle in 3-D on line x=3,y=1,z=4/5 etc. Area condition used cross product.

Probability & Statistics

• Lost card problem Bayesian revision

  • Prior 1/52; sampling n spades out of 51; equate posterior; solved n=2.

• Committee of 12 with engineers & doctors

  • Choose at least 3 engineers (inc. captain) and ≥1 doctor (inc. vice–captain). Combinatorial enumeration gave $\frac{129}{182}$.

• Random variable with p.m.f. $P(X=x)=k(x+1)3^{-x}$. Normalising sum gives $k=\tfrac29$; probability X≥3 = 1/9.

• Mean & variance corrections: adjusting one value changes both mean and variance; recompute to get 10(μ+σ)=449.

Trigonometry & Inverse Trig

• Eqn $\csc(2\theta-\pi/3)-\csc(4\theta-\pi)=0$ within $[-2\pi,2\pi]$ ⇒ 12 solutions.
• Solving $10\sin^4\theta+15\cos^4\theta=6$ to find $8\sec\theta+27\csc\theta$ yields $\frac{50}{3}$.
• (\cot^{-1}\frac{\sqrt3}{2}-\cot^{-1}\frac{1}{2}) simplifies to $\pi-\frac{\pi}{6}=\frac{5\pi}{6}$.

Miscellaneous & Discrete Topics

• Relations on finite set: counting to make reflexive or symmetric, e.g. on A={–2…3}: initial size 11, need +1 for reflexive, +? for symmetric – total 12.
• Number of subsets of {1,…,n} with no two consecutive numbers equals Fibonacci; for n=5 gives 13.
• Infinite product telescoping: $1+\frac13+\frac15+…=\frac\pi4$ etc. Sums of reciprocals separated by parity produce (m,n) pairs, leading to |7(α+β)+4(γ+δ)|=96.

Answers to Numerical-Value (Section-B) Questions

• Provided key single-digit/three-digit answers:
  • Differential Eqn y(π/4)=21.
  • GP telescoping sum gives m+n=441.
  • Card selection GP probability gives 2477 (Allen) / 4949 (NTA) noting dispute.
  • Absolute difference of circle radii squared =768.
  • Etc. (full list embedded through preceding bullets.)

Ethical & Practical Notes

• Problems illustrate standard JEE themes: coordinate geometry reflections, vector triple product identities, calculus limits demanding L’Hôpital, statistics corrections.
• Emphasise principles: always confirm domain restrictions when logging/ square-rooting; for probability tasks define sample space precisely before conditional inversion.
• Practical exam tip: store canned results:
  • $\sum_{k=0}^{\infty}\frac1{(2k+1)(2k+3)}=\frac12$
  • Highest power of prime p in n! formula.
  • Orthocentre/ centroid coordinates in terms of vertices.