GOPAL JI MEMORIAL SCHOOL

Reoti, Ballia (U.P.) · Affiliated to CBSE · www.gjms.edu.in

SAMPLE QUESTION PAPER · Session 2026-27
Mathematics at Advanced Level (Optional) — Class IX

Time Allowed: 1 HourMaximum Marks: 25

General Instructions:

This is the optional Advanced-level paper, attempted in addition to the compulsory 80-mark Mathematics paper. It carries no internal assessment.
The paper has 13 questions across three Sections and is compulsory in full. All questions are Higher Order Thinking Skills (HOTS) type.
Section A has 5 objective/very-short questions of 1 mark each. Section B has 4 questions of 2 marks each. Section C has 4 questions of 3 marks each.
There is no overall choice. However, internal choice (OR) is provided in one question of 2 marks and one question of 3 marks.
Content is drawn only from the six Advanced chapters: Sets, Logarithms, Relations & Functions, Coordinate Geometry, Combinatorics and Progressions.
Use of calculators is not permitted. Take log values in base 10 unless a base is stated.
Scoring 50% or more (≥ 13 marks) earns the marksheet remark "Advanced Level successfully cleared". Marks are not added to the aggregate; there is no downside.

SECTION A Objective / Very Short (HOTS) — 5 × 1 = 5 marks

1.1 A set A has 4 elements. A student claims the number of proper subsets of A equals the number of elements in its power set. Is the student correct? Justify with the two counts.

2.1 Without a calculator, evaluate log₂ 32 + log₃(1/27).

3.1 A quiz team of 2 members is to be picked from 8 students. Reasoning by counting, how many different teams are possible, and why is this ⁸C₂ rather than ⁸P₂?

4.1 A line has slope 2/3. A second line is perpendicular to it. State the slope of the second line and the product of the two slopes.

5.1 For the infinite GP 1 + 1/3 + 1/9 + 1/27 + …, state whether it converges and, if so, find its sum. Justify the convergence condition.

SECTION B Short Answer (HOTS) — 4 × 2 = 8 marks

6.2 In a class of 100 students surveyed about drinks, 40 like tea, 30 like coffee and 15 like both. Using the inclusion–exclusion principle, find how many like at least one of the two drinks and how many like neither. Represent your reasoning with a Venn diagram description.

7.2 Solve for x: log(x) + log(x − 3) = 1 (base 10). Clearly show the check that rejects any extraneous root.

Solve for x: log₂(x) + log₂(x − 2) = 3, rejecting any invalid root with reasoning.

8.2 A function is defined by f(x) = x² − 2x on the set of integers. Find f(3) and f(−1), and hence explain why the pairs (3, 3) and (−1, 3) do not stop f from being a function.

9.2 In a GP, the 3rd term is 12 and the 6th term is 96. Find the common ratio r and the first term a. Verify your r by re-deriving the 6th term.

SECTION C Long Answer (HOTS) — 4 × 3 = 12 marks

10.3 A survey of 100 students records subjects opted:

Mathematics: 45, Physics: 38, Chemistry: 42. Both Maths & Physics: 15, both Physics & Chemistry: 18, both Maths & Chemistry: 12, all three: 8.

Using inclusion–exclusion for three sets, find (i) how many opted at least one subject, (ii) how many opted none, and (iii) how many opted exactly one subject.

11.3 A committee of 4 is to be formed from 7 men and 5 women, with the condition that it contains at least 2 women. Using the block/case method, find the number of such committees. Show each case count.

Eight points lie on a plane, no three collinear, except that a particular set of 4 of them are collinear. How many triangles can be formed by joining these points? Explain the subtraction you make.

12.3 A straight line passes through the points (2, 3) and (4, 7).

Find its equation in the form Ax + By + C = 0.
Find its x-intercept and y-intercept.
Find the area of the triangle the line makes with the two coordinate axes.

13.3 [Indian Knowledge Systems] The 14th-century Kerala mathematician Mādhava of Sangamagrāma discovered the infinite series π/4 = 1 − 1/3 + 1/5 − 1/7 + … , a landmark in the method of successive summation.

Using the method of differences (telescoping), prove that 1/(1·2) + 1/(2·3) + … + 1/(n(n+1)) = n/(n+1), and evaluate it for n = 10.
Using the first four terms of Mādhava's series, obtain an approximate value of π and comment on why the approximation is rough with so few terms.

— END OF PAPER —

MARKING SCHEMEfor teachers

SECTION A — 1 mark each

1. Power set of A has 2⁴ = 16 elements (all subsets). Number of proper subsets = 2⁴ − 1 = 15 (every subset except A itself). Since 15 ≠ 16, the student is incorrect. (Correct counts 16 and 15 with conclusion — 1)

2. log₂ 32 = log₂ 2⁵ = 5; log₃(1/27) = log₃ 3⁻³ = −3. Sum = 5 + (−3) = 2. (Both values ½ + ½; answer 2 — 1)

3. A team is an unordered selection, so order does not matter → use combinations. ⁸C₂ = (8×7)/(2×1) = 28. (⁸P₂ = 56 would double-count each team as two orderings.) (Reason + answer 28 — 1)

4. For perpendicular lines m₁·m₂ = −1, so slope = −3/2; product of slopes = (2/3)(−3/2) = −1. (Slope + product — 1)

5. Here a = 1, r = 1/3. Since |r| = 1/3 < 1, the series converges. Sum = a/(1 − r) = 1/(1 − 1/3) = 1/(2/3) = 3/2. (Convergence condition + sum 3/2 — 1)

SECTION B — 2 marks each

6. n(T∪C) = n(T) + n(C) − n(T∩C) = 40 + 30 − 15 = 55 like at least one. (1)
Neither = 100 − 55 = 45. Venn: two overlapping circles, overlap 15, tea-only 25, coffee-only 15, outside both 45. (Neither = 45 with Venn description — 1)

7. log(x) + log(x−3) = log[x(x−3)] = 1 ⇒ x(x−3) = 10¹ = 10 ⇒ x² − 3x − 10 = 0 ⇒ (x − 5)(x + 2) = 0 ⇒ x = 5 or x = −2. (Set-up + roots — 1)
Check: x must make both logs defined (x > 0 and x − 3 > 0 ⇒ x > 3). x = −2 is rejected; x = 5 is valid. x = 5. (Extraneous-root rejection — 1)
OR: log₂[x(x−2)] = 3 ⇒ x(x−2) = 2³ = 8 ⇒ x² − 2x − 8 = 0 ⇒ (x−4)(x+2) = 0 ⇒ x = 4 or x = −2. Need x > 2, so reject x = −2 ⇒ x = 4. (Set-up 1 + rejection 1)

8. f(3) = 3² − 2·3 = 9 − 6 = 3; f(−1) = (−1)² − 2(−1) = 1 + 2 = 3. (Both values — 1)
A function requires each input to have exactly one output. Here 3 and −1 are different inputs giving the same output 3 — that is allowed (many-to-one). It would fail only if one input gave two outputs. So f is still a valid function. (Correct reasoning — 1)

9. ar² = 12 and ar⁵ = 96. Dividing: r³ = 96/12 = 8 ⇒ r = 2. (1)
Then a·2² = 12 ⇒ a = 12/4 = 3. Check 6th term: ar⁵ = 3·2⁵ = 3·32 = 96 ✓. (a = 3 with verification — 1)

SECTION C — 3 marks each

10. Let M, P, C be the sets. n(M∪P∪C) = 45 + 38 + 42 − 15 − 18 − 12 + 8. (Correct formula set-up — 1)
= 125 − 45 + 8 = 88 opted at least one.
(ii) None = 100 − 88 = 12. (at least one + none — 1)
(iii) Exactly one: only M = 45 − 15 − 12 + 8 = 26; only P = 38 − 15 − 18 + 8 = 13; only C = 42 − 18 − 12 + 8 = 20. Exactly one = 26 + 13 + 20 = 59. (Exactly-one computation — 1)

11. "At least 2 women" ⇒ cases (women, men) = (2,2), (3,1), (4,0):
(2W,2M): ⁵C₂·⁷C₂ = 10·21 = 210. (½)
(3W,1M): ⁵C₃·⁷C₁ = 10·7 = 70. (½)
(4W,0M): ⁵C₄·⁷C₀ = 5·1 = 5. (½)
Total = 210 + 70 + 5 = 285 committees. (Sum + cases — total 3)
OR: Triangles need 3 non-collinear points. From all 8 points: ⁸C₃ = 56. (1) But the 4 collinear points give ⁴C₃ = 4 "triangles" that are actually degenerate (a straight line), so subtract them. (reason — 1) Triangles = 56 − 4 = 52. (answer — 1)

12. (a) Slope m = (7 − 3)/(4 − 2) = 4/2 = 2. Line: y − 3 = 2(x − 2) ⇒ y = 2x − 1 ⇒ 2x − y − 1 = 0. (1)
(b) x-intercept (y = 0): 2x − 1 = 0 ⇒ x = 1/2. y-intercept (x = 0): −y − 1 = 0 ⇒ y = −1. (1)
(c) Triangle with axes has legs |1/2| and |−1|. Area = ½ · (1/2) · 1 = 1/4 sq. unit (= 0.25). (1)

13. (a) Method of differences: 1/(k(k+1)) = 1/k − 1/(k+1). Summing k = 1…n telescopes:
(1 − 1/2) + (1/2 − 1/3) + … + (1/n − 1/(n+1)) = 1 − 1/(n+1) = n/(n+1). (telescoping proof — 1½)
For n = 10: 10/11 ≈ 0.909. (½)
(b) π ≈ 4(1 − 1/3 + 1/5 − 1/7) = 4(1 − 0.3333 + 0.2 − 0.1429) = 4(0.7238) ≈ 2.895. (computation — ½) It is rough because Mādhava's series is very slowly (alternating) convergent — each extra term shrinks the error only by roughly 1/(next odd number), so four terms leave a large gap from the true π ≈ 3.1416. (reasoning — ½)