TS Inter 1st Year Maths Model Paper 2027 — Full Paper With Solutions

TS Inter 1st Year Maths has two papers — Paper 1A (Algebra and Trigonometry) and Paper 1B (Coordinate Geometry). Each carries 75 marks in 3 hours.

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PAPER 1A — ALGEBRA AND TRIGONOMETRY (75 Marks)

Paper Structure

Section	Type	Marks
Section A	Very Short Answer — 10 questions (2 marks each)	20
Section B	Short Answer — 5 of 7 (4 marks each)	20
Section C	Long Answer — 2 of 3 (7 marks each)	14
Internal	Project/Assignment	21

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SECTION A — Very Short Answer (2 Marks Each)

Q1. If f(x) = 2x² − 3x + 1, find f(2).

Solution: f(2) = 2(4) − 3(2) + 1 = 8 − 6 + 1 = 3

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Q2. Find the domain of f(x) = √(4 − x²).

Solution: Need 4 − x² ≥ 0 → x² ≤ 4 → −2 ≤ x ≤ 2
Domain = [−2, 2]

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Q3. If A = and B = , find A∩B and A∪B.

Solution:
A∩B =
A∪B =

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Q4. Find the value of sin 15°.

Solution:
sin 15° = sin(45° − 30°) = sin45°cos30° − cos45°sin30°
= (1/√2)(√3/2) − (1/√2)(1/2)
= (√3 − 1)/(2√2) = (√6 − √2)/4

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Q5. If tan θ = 5/12, find sin θ and cos θ (θ in first quadrant).

Solution:
hypotenuse = √(25 + 144) = √169 = 13
sin θ = 5/13, cos θ = 12/13

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Q6. Show that sin²A + sin²B + sin²C = 2 when A + B + C = π.

Solution (outline):
Since C = π − A − B:
sin²C = sin²(A+B)
Expanding sin²A + sin²B + sin²(A+B) and using sum-to-product identities gives the result = 2 ✓

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Q7. Find the number of ways 5 boys and 3 girls can be seated in a row.

Solution: Total = 8 persons
Number of arrangements = 8! = 40320

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Q8. Find the 5th term in the expansion of (2x + 3y)⁸.

Solution:
T₅ = T₄₊₁ = ⁸C₄(2x)⁴(3y)⁴
= 70 × 16x⁴ × 81y⁴
= 90720 x⁴y⁴

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Q9. Find the sum of infinite GP: 1 + 1/3 + 1/9 + ...

Solution:
a = 1, r = 1/3 (|r| < 1)
S∞ = a/(1−r) = 1/(1−1/3) = 1/(2/3) = 3/2

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Q10. If nP₃ = 60, find n.

Solution:
n(n−1)(n−2) = 60 = 5×4×3
n = 5

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SECTION B — Short Answer (4 Marks Each — Attempt 5 of 7)

Q11. Prove that (sinA + cosA)(tanA + cotA) = secA + cosecA.

Solution:
LHS = (sinA + cosA)(sinA/cosA + cosA/sinA)
= (sinA + cosA)(sin²A + cos²A)/(sinAcosA)
= (sinA + cosA) × 1/(sinAcosA)
= sinA/(sinAcosA) + cosA/(sinAcosA)
= 1/cosA + 1/sinA
= secA + cosecA = RHS ✓

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Q12. Solve the quadratic equation: x² − 5x + 6 = 0. Verify using sum and product of roots.

Solution:
x² − 5x + 6 = (x−2)(x−3) = 0
Roots: α = 2, β = 3

Verification:
Sum: α + β = 5 = −(−5)/1 = −b/a ✓
Product: αβ = 6 = 6/1 = c/a ✓

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Q13. Find the angle between the lines y = 3x + 5 and y = 2x − 7.

Solution:
m₁ = 3, m₂ = 2
tan θ = |(m₁ − m₂)/(1 + m₁m₂)| = |(3−2)/(1+6)| = 1/7
θ = tan⁻¹(1/7)

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SECTION C — Long Answer (7 Marks Each — Attempt 2 of 3)

Q14. If A, B, C are angles of a triangle, prove that:
sin A + sin B + sin C = 4cos(A/2)cos(B/2)cos(C/2)

Solution:
Since A + B + C = π:
sin A + sin B = 2sin((A+B)/2)cos((A−B)/2) = 2cos(C/2)cos((A−B)/2)
sin C = 2sin(C/2)cos(C/2)

Also A+B = π − C → (A+B)/2 = π/2 − C/2

Adding:
= 2cos(C/2)cos((A−B)/2) + 2sin(C/2)cos(C/2)
= 2cos(C/2)[cos((A−B)/2) + sin(C/2)]
= 2cos(C/2)[cos((A−B)/2) + cos((A+B)/2)]
= 2cos(C/2) × 2cos(A/2)cos(B/2)
= 4cos(A/2)cos(B/2)cos(C/2) ✓

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PAPER 1B — COORDINATE GEOMETRY (75 Marks)

SECTION A — Very Short Answer

Q15. Find the slope of the line joining (2, 3) and (5, 7).

Solution: m = (7−3)/(5−2) = 4/3

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Q16. Find the equation of the line with slope 3 passing through (1, 2).

Solution: y − 2 = 3(x − 1) → y = 3x − 1

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Q17. Find the distance of point (3, 4) from the line 3x − 4y + 5 = 0.

Solution:
d = |3(3) − 4(4) + 5| / √(9+16) = |9 − 16 + 5| / 5 = |−2|/5 = 2/5

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Q18. Find the equation of circle with centre (2, 3) and radius 5.

Solution:
(x−2)² + (y−3)² = 25
x² + y² − 4x − 6y − 12 = 0

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SECTION B — Short Answer

Q19. Find the equation of the pair of lines through origin perpendicular to 3x² − 5xy + 2y² = 0.

Solution:
3x² − 5xy + 2y² = (3x−2y)(x−y) = 0
Lines: 3x = 2y and x = y
Perpendicular to these: 2x + 3y = 0 and x + y = 0 (through origin)
Combined equation: (2x+3y)(x+y) = 0
2x² + 5xy + 3y² = 0

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Q20. Find the locus of a point equidistant from (1, 2) and (3, 4).

Solution:
Let P(x, y). PA = PB
√((x−1)² + (y−2)²) = √((x−3)² + (y−4)²)
Squaring: x² − 2x + 1 + y² − 4y + 4 = x² − 6x + 9 + y² − 8y + 16
4x + 4y − 20 = 0
x + y − 5 = 0
(This is the perpendicular bisector of the segment joining the two points)

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Chapter-Wise Weightage

Chapter	Marks (1A)
Functions	8
Mathematical Induction	6
Matrices	10
Complex Numbers	8
Quadratic Equations	8
Theory of Equations	6
Binomial Theorem	8
Trigonometry (all chapters)	21

Chapter	Marks (1B)
Locus	6
Transformation of Axes	6
Straight Lines	18
Pair of Straight Lines	12
3D Geometry	8
Limits and Continuity	10
Differentiation	10
Errors and Approximations	5

Tip: TS Inter Maths Section A (Very Short Answer) is 20 marks in Paper 1A — 10 questions at 2 marks each. These are purely formula application. Never skip Section A — it is the most time-efficient section of the entire paper.