10TH SLIP TEST 2 PRINCIPLE OF EVALUATION

💡 Note: Clean valuation key formatted according to Telangana SCERT standards. Fits fully on a single A4 page.

Class: X | Subject: Mathematics

Slip Test – II

Max Marks: 20

SECTION I (1 × 6 = 6 Marks)

Q1. Construct Similar Triangle (ΔA'BC' ~ ΔABC with scale factor 3/5) [AS 5 - Visualization & Representation] [ Marks: 6 ]

Rough Diagram & Given Data (AB = 5.6 cm, BC = 7.2 cm, CA = 4.8 cm)	1.0 M
Constructing base BC = 7.2 cm, arcs for BA = 5.6 cm and CA = 4.8 cm to complete ΔABC	1.5 M
Drawing acute ray BX and marking 5 equal divisions B₁, B₂, B₃, B₄, B₅	1.5 M
Joining B₅C and drawing B₃C' ∥ B₅C (C' on BC); then C'A' ∥ CA (A' on AB)	1.5 M
Concluding ΔA'BC' is the required similar triangle	0.5 M

Q2. Prove sin θ = (P² − 1) / (P² + 1) given sec θ + tan θ = P [AS 2 - Reasoning & Proof] [ Marks: 6 ]

Using sec²θ − tan²θ = 1 ⇨ (sec θ + tan θ)(sec θ − tan θ) = 1 ⇨ sec θ − tan θ = 1/P	1.5 M
Adding equations: 2 sec θ = P + (1/P) = (P² + 1)/P ⇨ sec θ = (P² + 1)/(2P)	1.5 M
Subtracting equations: 2 tan θ = P − (1/P) = (P² − 1)/P ⇨ tan θ = (P² − 1)/(2P)	1.5 M
Dividing tan θ by sec θ: sin θ = tan θ / sec θ = (P² − 1) / (P² + 1) (Hence Proved)	1.5 M

SECTION II (2 × 4 = 8 Marks)

Q3. Graph of p(x) = x² + 2x − 3 and its zeros [AS 5 - Visualization & Representation] [ Marks: 4 ]

Table of values: x = [-4, -3, -2, -1, 0, 1, 2] ⇨ y = [5, 0, -3, -4, -3, 0, 5]	1.0 M
Plotting points accurately on Cartesian plane with Scale (X: 1 cm = 1 unit, Y: 1 cm = 1 unit)	1.5 M
Drawing parabolic curve intersecting X-axis at (-3, 0) and (1, 0)	1.0 M
Conclusion: Zeros of the polynomial are x = -3 and x = 1	0.5 M

Q4. In ΔABC, DE ∥ BC (AB = 14 cm, AD = 3.5 cm, AE = 2.5 cm). Find AC [AS 1 - Problem Solving] [ Marks: 4 ]

Applying Basic Proportionality Theorem (BPT): AD / AB = AE / AC	1.0 M
Substituting given values: 3.5 / 14 = 2.5 / AC	1.0 M
Simplifying ratio: 1 / 4 = 2.5 / AC ⇨ AC = 4 × 2.5	1.0 M
Final Answer: AC = 10 cm	1.0 M

SECTION III (3 × 2 = 6 Marks)

Q5. If p(x) = x⁴ + 1, find p(2) − p(-2) [AS 1 - Problem Solving] [ Marks: 2 ]

p(2) = (2)⁴ + 1 = 17 and p(-2) = (-2)⁴ + 1 = 17	1.0 M
p(2) − p(-2) = 17 − 17 = 0	1.0 M

Q6. Express tan θ in terms of sin θ [AS 4 - Connections] [ Marks: 2 ]

Formulas: tan θ = sin θ / cos θ and cos θ = √(1 − sin²θ)	1.0 M
Final Expression: tan θ = sin θ √(1 − sin²θ)	1.0 M

Q7. Pole height diagram (Distance = 15 m, Angle of elevation = 45°) [AS 5 - Visualization & Representation] [ Marks: 2 ]

Drawing right triangle ABC (AB = vertical pole, BC = ground = 15 m)	1.0 M
Marking right angle at B and angle of elevation at C (∠ACB = 45°)	1.0 M

eTeacherDesk