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

01 15 PM IST - 17 Mar'26

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!

01 00 PM IST - 17 Mar'26

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.

12 00 PM IST - 17 Mar'26

Midway Through the Maths Exam

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

11 00 AM IST - 17 Mar'26

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.

10 00 AM IST - 17 Mar'26

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. 

09 00 AM IST - 17 Mar'26

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.

08 00 AM IST - 17 Mar'26

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.

07 00 AM IST - 17 Mar'26

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.

06 00 AM IST - 17 Mar'26

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!

05 00 AM IST - 17 Mar'26

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).

04 00 AM IST - 17 Mar'26

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

03 00 AM IST - 17 Mar'26

Parametric Forms of Conics

Parabola

x = at²
y = 2at

Ellipse

x = a cosθ
y = b sinθ

Hyperbola

x = a secθ
y = b tanθ

02 00 AM IST - 17 Mar'26

Tamil Nadu 12th Grading System 2026

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

01 00 AM IST - 17 Mar'26

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!

12 00 AM IST - 17 Mar'26

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

11 30 PM IST - 16 Mar'26

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

11 00 PM IST - 16 Mar'26

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

10 30 PM IST - 16 Mar'26

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

10 00 PM IST - 16 Mar'26

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.

09 30 PM IST - 16 Mar'26

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.

09 00 PM IST - 16 Mar'26

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)]

08 45 PM IST - 16 Mar'26

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

08 30 PM IST - 16 Mar'26

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

08 15 PM IST - 16 Mar'26

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 +

08 00 PM IST - 16 Mar'26

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 − ...

07 45 PM IST - 16 Mar'26

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

07 30 PM IST - 16 Mar'26

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.

07 15 PM IST - 16 Mar'26

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 ]

07 00 PM IST - 16 Mar'26

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

06 30 PM IST - 16 Mar'26

1-Mark Important Questions with Answers - 3

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

06 15 PM IST - 16 Mar'26

1-Mark Important Questions with Answers - 2

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

06 00 PM IST - 16 Mar'26

1-Mark Important Questions with Answers - 1

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

05 45 PM IST - 16 Mar'26

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

05 30 PM IST - 16 Mar'26

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

05 15 PM IST - 16 Mar'26

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)

05 00 PM IST - 16 Mar'26

Conics (General Second-Degree Equation)

General equation:

ax² + 2hxy + by² + 2gx + 2fy + c = 0

Classification condition:

04 45 PM IST - 16 Mar'26

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

04 30 PM IST - 16 Mar'26

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)

04 15 PM IST - 16 Mar'26

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

04 00 PM IST - 16 Mar'26

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

03 45 PM IST - 16 Mar'26

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)

03 30 PM IST - 16 Mar'26

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

03 15 PM IST - 16 Mar'26

Lagrange Identity

Lagrange identity:

|a × b|² = |a|² |b|² − (a · b)²

03 00 PM IST - 16 Mar'26

Jacobi Identity

Jacobi identity for vectors:

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

02 45 PM IST - 16 Mar'26

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)

02 30 PM IST - 16 Mar'26

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.

02 15 PM IST - 16 Mar'26

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.

02 00 PM IST - 16 Mar'26

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.

01 45 PM IST - 16 Mar'26

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!

01 30 PM IST - 16 Mar'26

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.

01 15 PM IST - 16 Mar'26

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.

01 00 PM IST - 16 Mar'26

Topper's Presentation Strategy

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

12 45 PM IST - 16 Mar'26

Maths Chapter-Wise Weightage

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

12 30 PM IST - 16 Mar'26

Tamil Nadu 12th Maths Blueprint 2026

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

12 15 PM IST - 16 Mar'26

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.

12 00 PM IST - 16 Mar'26

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.