Tamil Nadu 12th Maths Answer Key 2026 (OUT) LIVE: Detailed Unofficial Key Solutions; Paper Difficulty Level

Tamil Nadu 12th Maths Answer Key 2026 LIVE:The Tamil Nadu Class 12 Maths exam 2026 is concluded today,March 17, 2026 at 1:15 PM. Our team of subject-matter experts is preparing a comprehensive, step-by-step solution set covering all sections, with a specific focus on the 20 compulsory one-mark multiple-choice questions in Section A, as soon as the exam concludes. We will also provide a comparative analysis of the difficulty level, highlighting whether the problems from high-weightage topics like Vector Algebra and Differential Calculus were direct from the textbook or required lateral thinking. We will also examine the balance between theory-based questions and calculation-intensive problems to give a clear picture of the overall performance trends for 2026.

Also read |Tamil Nadu 12th Result 2026 Release Date Prediction

Tamil Nadu 12th Maths Answer Key 2026 Available

What After Tamil Nadu 12th Maths Answer Key 2026?

Following the release of the Tamil Nadu 12th Maths Answer Key 2026, students must quickly shift their focus to the next major examination scheduled for Monday, March 23, 2026. Depending on their stream, students will face crucial papers such as Biology, Botany, History, or Business Mathematics and Statistics, leaving a nearly week-long gap for intensive preparation. To maximize this time, students are encouraged to prioritize high-weightage chapters, such as Genetics and Physiology for Biology, or Financial Mathematics for Business Maths, while solving at least three previous years' question papers to master time management.

Tamil Nadu 12th Maths Exam 2026 LIVE

• Mar 17, 2026 01:15 PM IST

  Tamil Nadu 12th Maths Exam 2026 Ends!

  The exam has officially concluded. Students are now stepping out of the halls. Initial reactions from students about the difficulty level and unofficial answer key are expected shortly. Stay tuned to our LIVE blog for all updates!

  
   

• Mar 17, 2026 01:00 PM IST

  Last 15 Minutes Remaining

  Only 15 minutes remain for the Tamil Nadu Class 12 Maths exam to conclude. Students are likely using this time to finish final steps, verify answers, and ensure all questions are attempted wherever possible.

  
   

• Mar 17, 2026 12:00 PM IST

  Midway Through the Maths Exam

  The exam has now crossed the halfway mark. Students are likely working on long-answer and application-based questions.

  
   

• Mar 17, 2026 11:00 AM IST

  One Hour Into the Exam

  Students are now about an hour into the Mathematics exam. This is usually the stage when many complete the first set of questions and move on to longer problems.

  
   

• Mar 17, 2026 10:00 AM IST

  Tamil Nadu 12th Maths Exam 2026 Begins!

  The TN Class 12 Mathematics board exam has officially begun. Students are now inside examination halls and working on their question papers. 

  
   

• Mar 17, 2026 09:00 AM IST

  Students Reaching Exam Centres

  Students have started arriving at their respective exam centres. Many are revising formulas, discussing last-minute doubts with friends, or quietly going through short notes before entering the examination hall.

• Mar 17, 2026 08:00 AM IST

  Students Leaving Homes

  Across Tamil Nadu, thousands of Class 12 students are now leaving their homes to appear for the Mathematics board exam. Parents are accompanying many students to exam centres.

  
   

• Mar 17, 2026 07:00 AM IST

  Morning Traffic Alert!

  Students and parents heading to exam centres are advised to start early to avoid the morning rush. Traffic is expected to increase near schools and main roads as exam time approaches.

• Mar 17, 2026 06:00 AM IST

  Good Morning, Students! Exam Day is Here

  Good morning to all Class 12 students appearing for the Tamil Nadu Maths exam today! Take a deep breath and start your day calmly. Have a healthy breakfast, stay hydrated, and keep your exam essentials ready. Try reaching 30-40 minutes before the exam timings. All the best!

  
   

• Mar 17, 2026 05:00 AM IST

  Descartes’ Rule of Signs

  Rule for positive real roots:

  Number of positive real roots =
  number of sign changes in f(x)
  or less than it by an even number.

  Rule for negative real roots:

  Replace x with −x.

  Number of negative real roots =
  number of sign changes in f(−x).

  
   

• Mar 17, 2026 04:00 AM IST

  Tangent Forms (Parametric)

  Parabola

  Equation of tangent:

  ty = x + at²

  Ellipse

  Equation of tangent:

  x cosθ / a + y sinθ / b = 1

  Hyperbola

  Equation of tangent:

  x secθ / a − y tanθ / b = 1

  
   

• Mar 17, 2026 03:00 AM IST

  Parametric Forms of Conics

  Parabola

  x = at²
  y = 2at

  Ellipse

  x = a cosθ
  y = b sinθ

  Hyperbola

  x = a secθ
  y = b tanθ

  
   

• Mar 17, 2026 02:00 AM IST

  Tamil Nadu 12th Grading System 2026

  Tamil Nadu 12th Grading System 2026 can be checked from the table below:

  Grade	Marks Range	Grade Point	Performance
  A1	91 - 100	10	Outstanding
  A2	81 - 90	9	Excellent
  B1	71 - 80	8	Very Good
  B2	61 - 70	7	Good
  C1	51 - 60	6	Above Average
  C2	41 - 50	5	Average
  D	35 - 40	4	Pass
  E1	21 - 34	—	Fail/Disqualified

• Mar 17, 2026 01:00 AM IST

  Essentials Checklist for Maths Exam

  Before you sleep, make sure everything is ready for tomorrow:

  1. Hall Ticket / Admit Card
  2. At least 2-3 Blue or Black Pens
  3. Pencil, Eraser, Sharpener
  4. Geometry Box (Scale, Compass, Protractor)
  5. Transparent Water Bottle
  6. Watch to manage exam time
  7. Quick Formula Sheet for morning revision

  Keep your bag packed tonight so you can leave home calmly. Good rest is also part of exam preparation!

  
   

• Mar 17, 2026 12:00 AM IST

  What not to do right now?

  With just a few hours left for the Tamil Nadu Class 12 Maths exam, avoid these mistakes:

  • Don’t start learning completely new chapters now
  • Don’t panic if you can’t recall a formula instantly
  • Don’t stay awake the whole night solving tough problems
  • Don’t scroll endlessly on social media or random discussions
  • Don’t compare your preparation with friends at this moment

• Mar 16, 2026 11:30 PM IST

  Rational Root Theorem

  If

  p/q is a rational root of polynomial

  a₀xⁿ + a₁xⁿ⁻¹ + ... + aₙ = 0

  Then

  p divides constant term

  q divides leading coefficient

  
   

• Mar 16, 2026 11:00 PM IST

  Formation of Polynomial Equation

  If roots are α and β

  Equation is

  x² − (α + β)x + αβ = 0

  If roots are α, β, γ

  Equation is

  x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ = 0

  
   

• Mar 16, 2026 10:30 PM IST

  Fundamental Theorem of Algebra

  Every polynomial equation of degree n has exactly n roots (real or complex), counting multiplicity.

  Example:

  Degree 4 polynomial → 4 roots

  
   

• Mar 16, 2026 10:00 PM IST

  Conjugate Root Theorem

  If a polynomial has real coefficients, and

  a + ib is a root,

  then

  a − ib is also a root.

  Thus complex roots occur in conjugate pairs.

  
   

• Mar 16, 2026 09:30 PM IST

  nᵗʰ Roots of a Complex Number

  If

  z = r(cosθ + i sinθ)

  Then the nᵗʰ roots are:

  zₖ = r^(1/n) [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]

  Where

  k = 0,1,2,…,n−1

  Important result:

  There are exactly n distinct nᵗʰ roots.

  
   

• Mar 16, 2026 09:00 PM IST

  De Moivre’s Theorem

  If a complex number is in polar form:

  z = r(cosθ + i sinθ)

  Then

  zⁿ = rⁿ[cos(nθ) + i sin(nθ)]

  
   

• Mar 16, 2026 08:45 PM IST

  Asymptotes

  Vertical asymptote:

  x = a when f(x) → ∞

  Horizontal asymptote:

  If

  lim x→∞ f(x) = L

  then

  y = L

  Oblique Asymptote

  Slope:

  m = lim (x→∞) f(x)/x

  Intercept:

  c = lim (x→∞) [f(x) − mx]

  Equation:

  y = mx + c

  
   

• Mar 16, 2026 08:30 PM IST

  Applications of Second Derivative

  Second derivative:

  f″(x)

  Concavity:

  f″(x) > 0 → concave upward

  f″(x) < 0 → concave downward

  Second derivative test:

  Local maximum → f′(x)=0 and f″(x)<0
  Local minimum → f′(x)=0 and f″(x)>0

  
   

• Mar 16, 2026 08:15 PM IST

  Applications of First Derivative

  Critical points:

  f′(x) = 0

  Increasing function:

  f′(x) > 0

  Decreasing function:

  f′(x) < 0

  Local maximum condition:

  f′(x) changes from + to −

  Local minimum condition:

  f′(x) changes from − to +

  
   

• Mar 16, 2026 08:00 PM IST

  Maclaurin Series Expansion

  Maclaurin series:

  f(x) = f(0) + x f′(0) + x²/2! f″(0) + x³/3! f‴(0) + ...

  Important expansions:

  eˣ = 1 + x + x²/2! + x³/3! + ...

  sin x = x − x³/3! + x⁵/5! − ...

  cos x = 1 − x²/2! + x⁴/4! − ...

  log(1+x) = x − x²/2 + x³/3 − ...

  
   

• Mar 16, 2026 07:45 PM IST

  Differential Calculus Solved Important Question - 4

  Find the Taylor series expansion about x = 2 for

  f(x) = x³ + 2x + 1.

  Taylor expansion about x = a:

  f(x) = f(a) + f'(a)(x − a) + f''(a)/2! (x − a)² + f'''(a)/3! (x − a)³

  Given:

  f(x) = x³ + 2x + 1

  First derivatives:

  f'(x) = 3x² + 2
  f''(x) = 6x
  f'''(x) = 6

  Now evaluate at x = 2

  f(2) = 8 + 4 + 1 = 13

  f'(2) = 3(4) + 2 = 14

  f''(2) = 12

  f'''(2) = 6

  Substitute:

  f(x) = 13 + 14(x − 2) + (12/2)(x − 2)² + (6/6)(x − 2)³

  Simplify:

  f(x) = 13 + 14(x − 2) + 6(x − 2)² + (x − 2)³

  Answer

  Taylor expansion about x = 2:

  f(x) = 13 + 14(x − 2) + 6(x − 2)² + (x − 2)³

  
   

• Mar 16, 2026 07:30 PM IST

  Differential Calculus Solved Important Question - 3

  Find two positive numbers whose sum is 12 and whose product is maximum.

  Let the two numbers be

  x and 12 − x

  Product:

  P = x(12 − x)

  P = 12x − x²

  Differentiate:

  dP/dx = 12 − 2x

  For maximum product:

  12 − 2x = 0

  2x = 12

  x = 6

  Second number:

  12 − 6 = 6

  Maximum product:

  P = 6 × 6 = 36

  Answer

  The two numbers are 6 and 6.
  Maximum product = 36.

  
   

• Mar 16, 2026 07:15 PM IST

  Differential Calculus Solved Important Question - 2

  Solve the differential equation

  d²y/dx² − dy/dx = 1/x.

  Let

  p = dy/dx

  Then

  dp/dx − p = 1/x

  This is a linear differential equation.

  Integrating factor (I.F.):

  I.F. = e^(−∫1 dx)

  I.F. = e^(−x)

  Multiply both sides by e^(−x)

  e^(−x) dp/dx − pe^(−x) = e^(−x)/x

  Left side becomes derivative:

  d/dx (pe^(−x)) = e^(−x)/x

  Integrate:

  pe^(−x) = ∫(e^(−x)/x) dx + C

  Since p = dy/dx

  dy/dx = e^x [ ∫(e^(−x)/x) dx + C ]

  This gives the general solution.

  Answer

  dy/dx = e^x [ ∫(e^(−x)/x) dx + C ]

  
   

• Mar 16, 2026 07:00 PM IST

  Differential Calculus Solved Important Question - 1

  Find df for f(x) = x² + 3x and evaluate it for x = 3 and dx = 0.02.

  Given:

  f(x) = x² + 3x

  df = f'(x) dx

  First find derivative:

  f'(x) = 2x + 3

  So,

  df = (2x + 3) dx

  Now substitute x = 3 and dx = 0.02

  df = (2 × 3 + 3)(0.02)

  df = (6 + 3)(0.02)

  df = 9 × 0.02

  df = 0.18

  Answer

  df = (2x + 3) dx
  At x = 3, df = 0.18

• Mar 16, 2026 06:30 PM IST

  1-Mark Important Questions with Answers - 3

  Given below are some more objective-type important questions with their correct answers:

  
  

• Mar 16, 2026 06:15 PM IST

  1-Mark Important Questions with Answers - 2

  You can check some more objective-type important questions below, along with their correct answers:

  
  

• Mar 16, 2026 06:00 PM IST

  1-Mark Important Questions with Answers - 1

  You can check some objective-type important questions below, along with their correct answers:

  
   

• Mar 16, 2026 05:45 PM IST

  Hyperbola

  Standard equation:

  Focus:

  (±c , 0)

  Relation:

  c² = a² + b²

  Eccentricity:

  e = c/a

  Tangent to Hyperbola

  Equation of tangent at point (x₁,y₁):

  (xx₁ / a²) − (yy₁ / b²) = 1

  Asymptotes of Hyperbola

  Equations:

  y = ± (b/a)x

  
  

• Mar 16, 2026 05:30 PM IST

  Ellipse

  Standard equation:

  Where a > b

  Focus:

  (±c , 0)

  Relationship:

  c² = a² − b²

  Eccentricity:

  e = c/a

  Length of latus rectum:

  2b² / a

  Tangent to Ellipse

  Equation of tangent at point (x₁,y₁):

  (xx₁ / a²) + (yy₁ / b²) = 1

  Normal to Ellipse

  Equation of normal:

  (ax / x₁) − (by / y₁) = a² − b²

  
  

• Mar 16, 2026 05:15 PM IST

  Parabola

  Standard equation:

  y^2 = 4ax

  Vertex:

  (0,0)

  Focus:

  (a,0)

  Directrix:

  x = −a

  Length of latus rectum:

  4a

  Tangent to Parabola

  Equation of tangent at point (x₁,y₁):

  yy₁ = 2a(x + x₁)

  Normal to Parabola

  Equation of normal:

  y − y₁ = −(y₁ / 2a)(x − x₁)

  
   

• Mar 16, 2026 05:00 PM IST

  Conics (General Second-Degree Equation)

  General equation:

  ax² + 2hxy + by² + 2gx + 2fy + c = 0

  Classification condition:

  Ellipse → h² < ab
  Parabola → h² = ab
  Hyperbola → h² > ab

• Mar 16, 2026 04:45 PM IST

  Analytical Geometry Solved Important Question - 4

  Find the equation of the ellipse whose foci are (2, 1) and (−2, 1) and the length of the latus rectum is 6.

  Foci:

  (2,1) and (−2,1)

  Centre = midpoint

  = (0,1)

  Distance of focus from centre:

  c = 2

  Latus rectum formula:

  Length = 2b² / a

  Given:

  2b² / a = 6

  b² = 3a

  Ellipse relation:

  c² = a² − b²

  4 = a² − b²

  Substitute b² = 3a

  4 = a² − 3a

  a² − 3a − 4 = 0

  (a − 4)(a + 1) = 0

  a = 4

  Then

  b² = 3a = 12

  Ellipse equation (centre shifted to (0,1)):

  x²/16 + (y − 1)²/12 = 1

  Answer

  Equation of ellipse:

  x²/16 + (y − 1)²/12 = 1

  
   

• Mar 16, 2026 04:30 PM IST

  Analytical Geometry Solved Important Question - 3

  Find the equation of the tangent and normal to the curve

  x = 7cos t

  y = 2sin t.

  Differentiate:

  dx/dt = −7sin t
  dy/dt = 2cos t

  Slope:

  dy/dx = (dy/dt) / (dx/dt)

  dy/dx = (2cos t) / (−7sin t)

  dy/dx = −(2/7) cot t

  Point on curve:

  (7cos t, 2sin t)

  Equation of tangent

  y − 2sin t = −(2/7) cot t (x − 7cos t)

  Equation of normal

  Slope of normal = reciprocal negative

  = 7/2 tan t

  y − 2sin t = (7/2) tan t (x − 7cos t)

  Answer

  Tangent:

  y − 2sin t = −(2/7) cot t (x − 7cos t)

  Normal:

  y − 2sin t = (7/2) tan t (x − 7cos t)

  
   

• Mar 16, 2026 04:15 PM IST

  Analytical Geometry Solved Important Question - 2

  Find the equation of the tangent and normal to the parabola x² + 6x + 4y − 5 = 0 at (1, −3).

  Solution

  Given curve:

  x² + 6x + 4y − 5 = 0

  Differentiate implicitly:

  2x + 6 + 4(dy/dx) = 0

  4(dy/dx) = −2x − 6

  dy/dx = −(2x + 6)/4

  dy/dx = −(x + 3)/2

  At point (1, −3):

  Slope of tangent

  m = −(1 + 3)/2
  m = −4/2
  m = −2

  Equation of tangent

  y − y₁ = m(x − x₁)

  y + 3 = −2(x − 1)

  y + 3 = −2x + 2

  y = −2x − 1

  Equation of normal

  Slope of normal = −1/m

  = −1/(−2)

  = 1/2

  y + 3 = 1/2(x − 1)

  2y + 6 = x − 1

  x − 2y − 7 = 0

  Answer

  Tangent: y = −2x − 1

  Normal: x − 2y − 7 = 0

• Mar 16, 2026 04:00 PM IST

  Analytical Geometry Solved Important Question - 1

  If the line y = 4x + c is a tangent to the circle x² + y² = 9, find c.

  Circle:
  x² + y² = 9

  Centre = (0, 0)
  Radius = 3

  Given line:
  y = 4x + c

  Write in standard form:

  4x − y + c = 0

  For a tangent, the distance from centre to line = radius

  Distance formula:

  |4(0) − 1(0) + c| / √(4² + (−1)²) = 3

  |c| / √17 = 3

  |c| = 3√17

  Therefore,

  c = ±3√17

  Answer

  c = 3√17 or c = −3√17

  
   

• Mar 16, 2026 03:45 PM IST

  Image of a Point in a Plane

  If point P(x₁,y₁,z₁) is reflected in plane ax + by + cz + d = 0

  Image point P′:

  x′ = x₁ − 2a(ax₁ + by₁ + cz₁ + d)/(a²+b²+c²)

  y′ = y₁ − 2b(ax₁ + by₁ + cz₁ + d)/(a²+b²+c²)

  z′ = z₁ − 2c(ax₁ + by₁ + cz₁ + d)/(a²+b²+c²)

  
   

• Mar 16, 2026 03:30 PM IST

  Equation of a Plane (Vector Form)

  General vector equation:

  r · n = d

  Where
  r = position vector of any point on plane
  n = normal vector

  Cartesian form:

  ax + by + cz + d = 0

  Plane through point (x₁,y₁,z₁):

  a(x − x₁) + b(y − y₁) + c(z − z₁) = 0

  
   

• Mar 16, 2026 03:15 PM IST

  Lagrange Identity

  Lagrange identity:

  |a × b|² = |a|² |b|² − (a · b)²

  
   

• Mar 16, 2026 03:00 PM IST

  Jacobi Identity

  Jacobi identity for vectors:

  a × (b × c) + b × (c × a) + c × (a × b) = 0

  
   

• Mar 16, 2026 02:45 PM IST

  Vector Algebra Solved Important Question - 4

  Find the vector equation and Cartesian equation of the plane passing through (0, 1, −5) and parallel to the straight lines

  r = (2i − 4j + k) + s(2i + 3j + 6k)
  r = (3i + 5j + k) + t(i + j − k)

  Direction vectors of the lines are

  d₁ = 2i + 3j + 6k
  d₂ = i + j − k

  A plane parallel to both lines contains these direction vectors.

  Normal vector of the plane

  n = d₁ × d₂

  | i j k |
  | 2 3 6 |
  | 1 1 −1 |

  = i(3×−1 − 6×1)
  − j(2×−1 − 6×1)

  • k(2×1 − 3×1)
     

  = i(−3 − 6) − j(−2 − 6) + k(2 − 3)

  = −9i + 8j − k

  Normal vector = (−9, 8, −1)

  Point on plane = (0, 1, −5)

  Plane equation

  −9(x − 0) + 8(y − 1) − (z + 5) = 0

  −9x + 8y − 8 − z − 5 = 0

  −9x + 8y − z − 13 = 0

  Cartesian Equation of Plane

  −9x + 8y − z − 13 = 0

  Vector Equation of Plane

  r = (0i + 1j − 5k) + λ(2i + 3j + 6k) + μ(i + j − k)

• Mar 16, 2026 02:30 PM IST

  Vector Algebra Solved Important Question - 3

  Prove that

  (a − b) + (b − c) + (c − a) = 0.

  (a − b) + (b − c) + (c − a)

  = a − b + b − c + c − a

  = a − a − b + b − c + c

  = 0

  Therefore,

  (a − b) + (b − c) + (c − a) = 0.

  
   

• Mar 16, 2026 02:15 PM IST

  Vector Algebra Solved Important Question - 2

  Show that the vectors

  2i + 3j − k, i − j, and 3i + 6j − k are coplanar.

  Let

  a = 2i + 3j − k
  b = i − j
  c = 3i + 6j − k

  Vectors are coplanar if the scalar triple product is zero.

  | 2 3 −1 |
  | 1 −1 0 | = 0
  | 3 6 −1 |

  Expanding,

  = 2[(-1)(-1) − (0)(6)]
  − 3[(1)(-1) − (0)(3)]
  − 1[(1)(6) − (-1)(3)]

  = 2(1) − 3(-1) − (6 + 3)

  = 2 + 3 − 9

  = 0

  Since the scalar triple product is 0, the vectors are coplanar.

• Mar 16, 2026 02:00 PM IST

  Vector Algebra Solved Important Question - 1

  If a, b, c are three vectors, prove that

  (a × c) + (a × b) + (a × b × c) = 0.

  (a − b) + (b − c) + (c − a)

  = a − b + b − c + c − a

  = a − a − b + b − c + c

  = 0

  Hence proved that

  (a − b) + (b − c) + (c − a) = 0.

  
   

• Mar 16, 2026 01:45 PM IST

  Important Formulas, Short Notes, & Important Questions Coming Up

  In the upcoming updates, we will share important formulas, short notes from important chapters, and a few important questions that are often asked in the exam. These quick revision points will help you refresh important concepts and feel more confident before the paper. Stay tuned to this live blog!

  
   

• Mar 16, 2026 01:30 PM IST

  TN Maths exam 2026 Prep Tip

  Experts suggest that since the paper can be lengthy, your primary goal should be speed and accuracy. You have to master the three highest weightage chapters: Vector Algebra, Analytical Geometry, and Differential Calculus, as these are heavy hitters for the 5-mark sections. Toppers often recommend solving every single example and exercise sum in the official textbook, as board questions rarely ignore questions from this source.

  
   

• Mar 16, 2026 01:15 PM IST

  Checklist for Extra Marks in Maths Exam

  1. In chapters like Applications of Differential Calculus (Rate of Change), always include units like cm2/sec.
  2. In Part D (5-mark questions), read both options (a) and (b) carefully. Choose the one that involves less calculation to save time.
  3. When using properties of definite integrals or determinants, write the property name or the property itself in brackets next to the step.

• Mar 16, 2026 01:00 PM IST

  Topper's Presentation Strategy

  When writing the exam, presentation is your best friend for earning "bonus" marks from the examiner's perspective. 

  Feature	How to do it	Why it works
  Formula Boxing	Write the main formula used in a box before starting the calculation.	Shows the examiner you know the concept clearly.
  Margin Work	Keep all rough calculations in a neat right-side margin.	Keeps the main answer sheet clean and professional.
  Final Answer	End with "Result: [Value] [Units]" and draw a box around it.	Makes it easy for the corrector to give you full marks instantly.
  Diagrams	Use a sharp pencil and ruler for all graphs and vector diagrams.	Neatness often influences the mindset of the person grading.

• Mar 16, 2026 12:45 PM IST

  Maths Chapter-Wise Weightage

  The following table shows the approximate marks for each chapter based on the standard blueprint.

  Chapter No.	Chapter Name	Approx. Marks
  1	Applications of Matrices and Determinants	12
  2	Complex Numbers	10
  3	Theory of Equations	9
  4	Inverse Trigonometric Functions	8
  5	Two-Dimensional Analytical Geometry-II	14
  6	Applications of Vector Algebra	14
  7	Applications of Differential Calculus	14
  8	Differentials and Partial Derivatives	9
  9	Applications of Integration	12
  10	Ordinary Differential Equations	10
  11	Probability Distributions	10
  12	Discrete Mathematics	10

• Mar 16, 2026 12:30 PM IST

  Tamil Nadu 12th Maths Blueprint 2026

  To know about the prescribed blueprint, students can check out the table below.

  Sections	No. of Questions	No. of Questions to Attempt	Total Marks
  I	20	20	20
  II	10	7	14
  III	10	7	21
  IV	7	7	35
  Total	47	41	90

• Mar 16, 2026 12:15 PM IST

  Hall Ticket Reminder

  The hall ticket for the Tamil Nadu 12th Maths Exam 2026 is a mandatory document that you must carry to your exam centre tomorrow, March 17. For regular school-going students, the hall tickets were released through the school login, and you should have already collected your physical copy, signed and stamped by your school principal.

  
   

• Mar 16, 2026 12:00 PM IST

  Tamil Nadu 12th Maths Exam 2026 Date & Timings

  The Tamil Nadu Class 12 Mathematics exam for 2026 is officially scheduled for Tuesday, March 17, 2026. The exam timings are fixed from 10:00 AM to 1:15 PM. 

  When you arrive at the hall, the first 10 minutes (10:00 AM to 10:10 AM) are strictly for reading the question paper so you can plan which sections to tackle first. 

  From 10:10 AM to 10:15 AM, you will have five minutes to verify your particulars on the answer sheet. The actual writing time starts at 10:15 AM and goes on until 1:15 PM.