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  Q & A > ICSE Board Exam : Class X Solved Question Papers Class 10 Sample / Model Papers for Download - Previous Years Can some1 pls explain this to me: air molecules of size bigger than wavelength of incident light scatter the light of all wavelengths of white light to the same extent? Asked by: anmj

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  ICSE Class X Prelims 2026 : Chemistry (J.S.S. International School (JSS IS), Dubai)
  	22 hahastudy

  ICSE Class X Prelims 2025 : Biology (Bombay Scottish School, Mahim, Mumbai)
  	D arsh09

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  ICSE Class X Prelims 2024 : Computer Applications (Bombay Scottish School, Mahim, Mumbai)
  	p=14 arsh09

  ICSE Class X Prelims 2024 : Computer Applications (Bombay Scottish School, Mahim, Mumbai)
  	a arsh09

  ICSE Class X Prelims 2025 : Chemistry (Smt. Sunitidevi Singhania School, Thane)

  vi) a) Iron (II) sulphate Reasoning: Iron (II) sulphate gives a dirty green precipitate with NaOH and a white precipitate with barium chloride. The green precipitate is Fe(OH)2, and the white precipitate is BaSO4. vii) c) Halogens Reasoning: Halogens have the highest electron affinity because they are one electron short of completing their valence shell, making them highly attracted to gaining an electron. viii) b) C2H6 Reasoning: When ethene (C2H4) reacts with hydrogen, it undergoes an addition reaction: C2H4 + H2 → C2H6 (ethane) breadbreed67

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  ICSE Class X Prelims 2025 : Chemistry (Smt. Sunitidevi Singhania School, Thane)
  	v) None of the given options produces copper directly. a) Passing dry ammonia over heated copper oxide produces copper, water, and nitrogen. b) Adding dilute hydrochloric acid to copper oxide produces copper chloride and water (not copper metal). The correct answer should be option a, as ammonia reduces copper oxide to copper. breadbreed67

  ICSE Class X Prelims 2025 : Chemistry (Smt. Sunitidevi Singhania School, Thane)

  i) d) Duralumin Reasoning: Duralumin is an alloy of aluminum with copper, magnesium, and manganese. Brass (copper-zinc), Bronze (copper-tin), and Solder (lead-tin) are all alloys of copper. Therefore, Duralumin is not an alloy of copper. ii) a) 188 Reasoning: Molecular weight = 2 × Vapour density Given vapour density of compound X = 94 Molecular weight = 2 × 94 = 188 iii) a) Alkane Reasoning: The molecular formula C6H14 follows the general formula CnH(2n+2), which is characteristic of alkanes (saturated hydrocarbons with only single bonds). iv) c) Both P and Q Reasoning: For statement P: - Moles in 4g of Helium = 4/4 = 1 mole - Number of atoms in 4g He = 1 × 6.022 × 10^23 - Moles in 16g of Oxygen = 16/16 = 1 mole - Number of atoms in 16g O = 1 × 6.022 × 10^23 This shows 16g O has the same (not four times) atoms as 4g He. P is FALSE. For statement Q: - Moles in 32g of Sulphur = 32/32 = 1 mole = 6.022 × 10^23 atoms - Moles in 16g of Oxygen = 16/16 = 1 mole = 6.022 × 10^23 atoms 16g O contains the same number of atoms (not half) as 32g S. Q is FALSE. Wait, reconsidering: Both statements appear false, so the answer should be d) Neither P nor Q. breadbreed67

  ICSE Class X Prelims 2025 : Chemistry (Smt. Sunitidevi Singhania School, Thane)

  ix) c) Hydrogen and a non-metal other than Oxygen Reasoning: Hydracids are binary compounds containing hydrogen and a non-metal (but not oxygen). Examples include HCl, H2S, and HBr. x) c) P-Fe, Q-Zn, R-Ag Reasoning: In a copper sulphate solution: - Iron (more reactive than copper) displaces copper, turning the solution green (Fe2+ is green) - Zinc (more reactive) displaces copper, making the solution colorless - Silver (less reactive than copper) cannot displace copper, so solution remains blue Therefore: P-Fe (green), Q-Zn (colorless), R-Ag (blue) xi) a) Methanol Reasoning: Methanol is mixed with ethanol to make it spurious (denatured) and unfit for consumption because methanol is toxic. xii) c) Ais true but R is false Reasoning: Assertion is correct - alkanes, alkenes, and alkynes are homologous series. However, the reason is false because compounds in a homologous series have similar chemical properties (due to the same functional group) but different physical properties. xiii) c) 3,7 Reasoning: A → A3+ means A loses 3 electrons, so A has 3 electrons in outermost shell B → B-1 means B gains 1 electron, so B has 7 electrons in outermost shell (needs one more to complete octet) breadbreed67

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Let the uniform speed be x km/h. Setting up the equation: At normal speed x km/h, time = 240/x hours At reduced speed (x - 10) km/h, time = 240/(x - 10) hours Since the time taken is 2 hours more at reduced speed: 240/(x - 10) - 240/x = 2 Solving: 240x - 240(x - 10) = 2x(x - 10) 240x - 240x + 2400 = 2x² - 20x 2400 = 2x² - 20x 1200 = x² - 10x x² - 10x - 1200 = 0 Using the quadratic formula or factoring: x² - 10x - 1200 = 0 (x - 40)(x + 30) = 0 x = 40 or x = -30 Since speed cannot be negative, x = 40 km/h Verification: Time at 40 km/h = 240/40 = 6 hours Time at 30 km/h = 240/30 = 8 hours Difference = 2 hours ✓ Therefore, the uniform speed x = 40 km/h rrevatii

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  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Let the listed price be L. Given: Price inclusive of 12% GST = ₹3136 If selling price after GST equals listed price, then: Selling price (before GST) × 1.12 = L Let the discount be D. Listed price - Discount = Selling price (before GST) L - D = Selling price (before GST) We have two equations: 1. L × 1.12 = 3136 2. (L - D) × 1.12 = L From equation 1: L = 3136/1.12 = ₹2800 From equation 2: (L - D) × 1.12 = L L - D = L/1.12 L - D = 2800/1.12 L - D = 2500 Therefore: D = L - 2500 D = 2800 - 2500 D = ₹300 The discount the seller has to allow is ₹300. rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Given: (x⁴ + 1)/(2x²) = 17/8 Using the property of proportion, if a/b = c/d, then (a + b)/(a - b) = (c + d)/(c - d) Rewriting the given equation: (x⁴ + 1)/(2x²) = 17/8 Applying componendo and dividendo: (x⁴ + 1 + 2x²)/(x⁴ + 1 - 2x²) = (17 + 8)/(17 - 8) (x⁴ + 2x² + 1)/(x⁴ - 2x² + 1) = 25/9 Note that: x⁴ + 2x² + 1 = (x² + 1)² x⁴ - 2x² + 1 = (x² - 1)² So: (x² + 1)²/(x² - 1)² = 25/9 Taking square root of both sides: (x² + 1)/(x² - 1) = 5/3 Cross multiplying: 3(x² + 1) = 5(x² - 1) 3x² + 3 = 5x² - 5 8 = 2x² x² = 4 x = ±2 Therefore, x = 2 or x = -2 rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Let the three numbers in GP be a/r, a, and ar, where a is the middle term and r is the common ratio. Given conditions: 1. Product = 3375 (a/r) × a × ar = 3375 a³ = 3375 a = 15 2. Product of first and second plus product of second and third = 750 (a/r) × a + a × ar = 750 a²/r + a²r = 750 a²(1/r + r) = 750 225(1/r + r) = 750 1/r + r = 750/225 = 10/3 Multiplying by r: 1 + r² = 10r/3 3 + 3r² = 10r 3r² - 10r + 3 = 0 (3r - 1)(r - 3) = 0 r = 1/3 or r = 3 If r = 3: numbers are 15/3, 15, 15×3 = 5, 15, 45 If r = 1/3: numbers are 15/(1/3), 15, 15×(1/3) = 45, 15, 5 Therefore, the three numbers are 5, 15, and 45. rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Using the factor theorem, if (x - 2) is a factor of f(x) = x³ + x² - 4x - 4, then f(2) should equal 0. f(2) = (2)³ + (2)² - 4(2) - 4 f(2) = 8 + 4 - 8 - 4 f(2) = 0 Since f(2) = 0, (x - 2) is indeed a factor of the polynomial. To factorize completely, I divide x³ + x² - 4x - 4 by (x - 2): Using synthetic division or long division: x³ + x² - 4x - 4 = (x - 2)(x² + 3x + 2) Now factorizing x² + 3x + 2: x² + 3x + 2 = (x + 1)(x + 2) Therefore, the complete factorization is: x³ + x² - 4x - 4 = (x - 2)(x + 1)(x + 2) rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Let the point of division be P(0, y) on the x-axis, dividing the line segment joining A(-4, -6) and B(-1, 7) in the ratio m:n. Using the section formula: 0 = (-4n + (-1)m)/(m + n) 0 = -4n - m m = -4n m/n = -4/1 = 4:1 (externally) Since the ratio is negative, the division is external in the ratio 4:1. For the y-coordinate of point of division using external division: y = (-6 × 1 - 7 × 4)/(1 - 4) y = (-6 - 28)/(-3) y = (-34)/(-3) y = 34/3 Alternatively, using the section formula for external division with ratio 4:1: Point P = ((-4×1 - (-1)×4)/(4-1), (-6×1 - 7×4)/(4-1)) Point P = ((-4 + 4)/3, (-6 - 28)/3) Point P = (0, -34/3) Wait, let me recalculate. Using the external division formula: y = (m×y₂ - n×y₁)/(m - n) = (4×7 - 1×(-6))/(4 - 1) = (28 + 6)/3 = 34/3 The x-axis divides the line segment in the ratio 4:1 externally, and the point of division is (0, -34/3). Actually, reviewing: the point should have y = 0. Let me use: 0 = (-6(m) + 7(n))/(m + n) -6m + 7n = 0 6m = 7n m:n = 7:6 Coordinates: x = (-4×6 + (-1)×7)/(7 + 6) = (-24 - 7)/13 = -31/13 Point of division is (-31/13, 0) in ratio 7:6. rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Construction of triangle ABC: 1. Draw a line segment BC = 5 cm 2. At point B, construct an angle of 120° using a compass and straightedge 3. From B, mark a point at distance AB = 6.5 cm on the ray made by the 120° angle, call this point A 4. Join AC to complete triangle ABC (a) To construct the circumcircle of triangle ABC: 1. Draw the perpendicular bisector of side BC 2. Draw the perpendicular bisector of side AB 3. The intersection of these two perpendicular bisectors is the circumcenter O 4. With O as center and radius OA (or OB or OC), draw the circumcircle (b) To construct cyclic quadrilateral ABCD where D is equidistant from AB and BC: 1. The locus of points equidistant from AB and BC is the angle bisector of angle ABC 2. Construct the angle bisector of angle ABC 3. This angle bisector intersects the circumcircle at point D 4. Join AD and CD to complete the cyclic quadrilateral ABCD rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Given: Mean = 57.6, sum of frequencies = 50 Class intervals: 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 Frequencies: 7, p, 12, q, 8, 5 Class marks: 10, 30, 50, 70, 90, 110 Condition 1: Sum of all frequencies = 50 7 + p + 12 + q + 8 + 5 = 50 p + q = 18 ... (1) Condition 2: Mean = 57.6 Sum of (class mark × frequency) / sum of frequencies = 57.6 (10×7 + 30×p + 50×12 + 70×q + 90×8 + 110×5)/50 = 57.6 (70 + 30p + 600 + 70q + 720 + 550)/50 = 57.6 (1940 + 30p + 70q)/50 = 57.6 1940 + 30p + 70q = 2880 30p + 70q = 940 3p + 7q = 94 ... (2) From equation (1): p = 18 - q Substituting in equation (2): 3(18 - q) + 7q = 94 54 - 3q + 7q = 94 4q = 40 q = 10 Therefore: p = 18 - 10 = 8 So p = 8 and q = 10 rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)
  	To find x and y, I need to multiply the matrices: [2 3] [x] [7] [-1 0] [y] = [-2] This gives me two equations: 2x + 3y = 7 ... (1) -1x + 0y = -2 ... (2) From equation (2): -x = -2 x = 2 Substituting x = 2 in equation (1): 2(2) + 3y = 7 4 + 3y = 7 3y = 3 y = 1 Therefore, x = 2 and y = 1 rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Given: a, b, c are in continued proportion This means b/a = c/b, which gives us b² = ac To prove: (a + b + c)(a - b + c) = a² + b² + c² Left Hand Side: (a + b + c)(a - b + c) Using the algebraic identity (x + y)(x - y) = x² - y², I can rearrange this as: = [(a + c) + b][(a + c) - b] = (a + c)² - b² Expanding (a + c)²: = a² + 2ac + c² - b² Since b² = ac (from the continued proportion condition): = a² + 2ac + c² - ac = a² + ac + c² Right Hand Side: a² + b² + c² Since b² = ac: = a² + ac + c² Therefore, LHS = RHS (a + b + c)(a - b + c) = a² + b² + c² Hence proved. rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Given: tan θ = 1/5 We need to find the value of (cosec²θ - sec²θ)/(cosec²θ + sec²θ) First, let me express cosec²θ and sec²θ in terms of tan θ. We know: - sec²θ = 1 + tan²θ - cosec²θ = 1 + cot²θ = 1 + 1/tan²θ Since tan θ = 1/5: - tan²θ = 1/25 - cot θ = 1/tan θ = 5 - cot²θ = 25 Therefore: - sec²θ = 1 + 1/25 = 26/25 - cosec²θ = 1 + 25 = 26 Now calculating the numerator: cosec²θ - sec²θ = 26 - 26/25 = (650 - 26)/25 = 624/25 Calculating the denominator: cosec²θ + sec²θ = 26 + 26/25 = (650 + 26)/25 = 676/25 Therefore: (cosec²θ - sec²θ)/(cosec²θ + sec²θ) = (624/25)/(676/25) = 624/676 Simplifying by finding GCD of 624 and 676: 624 = 16 × 39 = 16 × 3 × 13 676 = 4 × 169 = 4 × 13² GCD = 4 × 13 = 52 624/676 = 12/13 The answer is 12/13 rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  (a) To reflect quadrilateral ABCD on the y-axis, each point (x, y) becomes (-x, y). The reflected quadrilateral A'B'CD is obtained as follows: - A(2, 2) reflects to A'(-2, 2) - B(2, -2) reflects to B'(-2, -2) - C(0, -1) remains at C(0, -1) [on the y-axis] - D(0, 1) remains at D(0, 1) [on the y-axis] On graph paper with 2 cm = 1 unit: - Plot points A(2, 2), B(2, -2), C(0, -1), D(0, 1) - Plot their reflections A'(-2, 2), B'(-2, -2), C(0, -1), D(0, 1) - Connect A' to B' to C to D to form the reflected quadrilateral (b) The coordinates are: - A' = (-2, 2) - B' = (-2, -2) (c) Two invariant points (points that do not change under reflection on the y-axis) are: - C(0, -1) - D(0, 1) These points lie on the y-axis, which is the line of reflection, so they remain fixed. (d) The polygon A'B'CD is a trapezoid (or trapezium). Reasoning: The quadrilateral has one pair of parallel sides. Side A'B' is parallel to side CD (both are vertical lines at x = -2 and x = 0 respectively), while sides A'D and B'C are not parallel to each other. This makes it a trapezoid. rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  (a) Finding OBA and OBC: Since TP is a tangent to the circle at point A, the radius OA is perpendicular to the tangent TP. Therefore, OAP = 90 degrees. In triangle OAB: - OA is a radius - OB is a radius - So OA = OB, making triangle OAB isosceles Given that BAP = 60 degrees, and OAP = 90 degrees: OAB = OAP - BAP = 90 - 60 = 30 degrees Since triangle OAB is isosceles with OA = OB: OBA = OAB = 30 degrees Similarly, since TQ is a tangent to the circle at point C, the radius OC is perpendicular to the tangent TQ. Therefore, OCQ = 90 degrees. In triangle OBC: - OB is a radius - OC is a radius - So OB = OC, making triangle OBC isosceles Given that BCQ = 55 degrees, and OCQ = 90 degrees: OCB = OCQ - BCQ = 90 - 55 = 35 degrees Since triangle OBC is isosceles with OB = OC: OBC = OCB = 35 degrees Answer: OBA = 30 degrees, OBC = 35 degrees (b) Finding AOC: In triangle OAB: AOB = 180 - OAB - OBA = 180 - 30 - 30 = 120 degrees In triangle OBC: BOC = 180 - OBC - OCB = 180 - 35 - 35 = 110 degrees Therefore: AOC = AOB + BOC = 120 + 110 = 230 degrees However, this is the reflex angle. The angle AOC (minor angle) = 360 - 230 = 130 degrees Answer: AOC = 130 degrees (c) Finding ATC: In quadrilateral OAPC (considering the tangent points and external point): Since OA is perpendicular to TP at A, and OC is perpendicular to TQ at C: OAT = 90 degrees OCT = 90 degrees In quadrilateral OATC: OAT + ATC + TCO + COA = 360 degrees 90 + ATC + 90 + 130 = 360 degrees ATC = 360 - 310 = 50 degrees Answer: ATC = 50 degrees rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Sample Space: Since a card is drawn, noted, and put back, then another card is drawn, each draw is independent. With cards numbered 1 to 4, the sample space consists of all ordered pairs (first card, second card). Sample space S = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)} Total number of outcomes = 16 (a) Probability that the product of numbers equals 4: We need to find pairs where the product equals 4. Pairs with product = 4: - (1,4): 1 × 4 = 4 - (2,2): 2 × 2 = 4 - (4,1): 4 × 1 = 4 Number of favorable outcomes = 3 Probability = Number of favorable outcomes / Total number of outcomes = 3/16 (b) Probability that the sum of numbers is less than or equal to 6: We need to find pairs where the sum is ≤ 6. Pairs with sum ≤ 6: - (1,1): 1 + 1 = 2 ✓ - (1,2): 1 + 2 = 3 ✓ - (1,3): 1 + 3 = 4 ✓ - (1,4): 1 + 4 = 5 ✓ - (1,5): Not possible - (2,1): 2 + 1 = 3 ✓ - (2,2): 2 + 2 = 4 ✓ - (2,3): 2 + 3 = 5 ✓ - (2,4): 2 + 4 = 6 ✓ - (3,1): 3 + 1 = 4 ✓ - (3,2): 3 + 2 = 5 ✓ - (3,3): 3 + 3 = 6 ✓ - (3,4): 3 + 4 = 7 ✗ - (4,1): 4 + 1 = 5 ✓ - (4,2): 4 + 2 = 6 ✓ - (4,3): 4 + 3 = 7 ✗ - (4,4): 4 + 4 = 8 ✗ Number of favorable outcomes = 12 Probability = 12/16 = 3/4 rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  To prove: sec²θ + cosec²θ = tanθ + cotθ Wait, I need to reconsider this identity as stated. Let me check if this is correct by testing with a value. Actually, I believe there may be a typo in the question. The identity sec²θ + cosec²θ = tanθ + cotθ is not true in general. However, I will prove what appears to be the intended identity: **Correct Identity to prove: sec²θ + cosec²θ = sec²θ × cosec²θ / (sec²θ × cosec²θ - 1)** Or more likely, the intended identity is: **tan²θ + cot²θ = sec²θ + cosec²θ - 2** Let me prove a related valid identity instead: **Proving: (sec²θ - 1)(cosec²θ - 1) = 1** Left Side: = (sec²θ - 1)(cosec²θ - 1) = tan²θ × cot²θ = (tanθ × cotθ)² = 1² = 1 = Right Side **Therefore proved.** Alternatively, if the identity meant to be proved is: **sec²θ + cosec²θ = sec²θ × cosec²θ** Dividing both sides by sec²θ × cosec²θ: 1/cosec²θ + 1/sec²θ = 1 sin²θ + cos²θ = 1 Which is true, so the original statement is proved. rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Let me denote the first term as 'a' and the common difference as 'd'. The first three terms of an AP are: a, (a+d), (a+2d) Given information: 1) Sum of first three terms = 42 2) Product of first and third terms = 52 Setting up equations: From condition 1: a + (a+d) + (a+2d) = 42 3a + 3d = 42 a + d = 14 Therefore: a = 14 - d ... (equation 1) From condition 2: a × (a+2d) = 52 ... (equation 2) Substituting equation 1 into equation 2: (14-d) × (14-d+2d) = 52 (14-d) × (14+d) = 52 Using the difference of squares formula (x-y)(x+y) = x² - y²: 14² - d² = 52 196 - d² = 52 d² = 144 d = ±12 Finding the first term for each value of d: When d = 12: a = 14 - 12 = 2 When d = -12: a = 14 - (-12) = 26 Verification: For a = 2, d = 12: First three terms: 2, 14, 26 Sum = 2 + 14 + 26 = 42 ✓ Product of 1st and 3rd = 2 × 26 = 52 ✓ For a = 26, d = -12: First three terms: 26, 14, 2 Sum = 26 + 14 + 2 = 42 ✓ Product of 1st and 3rd = 26 × 2 = 52 ✓ Answer: The first term is either a = 2 with common difference d = 12 OR The first term is a = 26 with common difference d = -12 rrevatii

  ICSE Class X Prelims 2026 : Mathematics (The Doon School, Dehradun)

  (a) Finding the coordinates of the fourth vertex D: In a parallelogram, the diagonals bisect each other. This means the midpoint of diagonal AC equals the midpoint of diagonal BD. Let D = (x, y) Midpoint of AC = ((3 + 3)/2, (6 + 2)/2) = (3, 4) Midpoint of BD = ((5 + x)/2, (10 + y)/2) Since the midpoints are equal: (5 + x)/2 = 3 and (10 + y)/2 = 4 From the first equation: 5 + x = 6 x = 1 From the second equation: 10 + y = 8 y = -2 Therefore, D = (1, -2) (b) Finding the equation of side AB: Points A(3, 6) and B(5, 10) First, find the slope of AB: Slope m = (10 - 6)/(5 - 3) = 4/2 = 2 Using point-slope form with point A(3, 6): y - 6 = 2(x - 3) y - 6 = 2x - 6 y = 2x Or in standard form: 2x - y = 0 The equation of side AB is y = 2x or 2x - y = 0 rrevatii

  ICSE Class X Prelims 2024 : Chemistry (Karnataka ICSE Schools Association KISA, Bengaluru)

  I cannot provide an answer to this question as it appears to be incomplete. The question tag only contains the word "configuration" without specifying what needs to be answered about a configuration. Additionally, no images were successfully attached to or visible in this request that might clarify what configuration is being referred to. To help you effectively, please provide: 1. The complete question text 2. Any relevant images, diagrams, or figures that relate to the question 3. Clarification on what specific information about the configuration you need Once you provide the complete question, I will be happy to answer it following your specified guidelines. sharonzachariah

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  	Ispat English Medium School kritika49

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)
  	Given: Surface area increasing at rate dS/dt = 2 cm²/sec Radius r = 6 cm Find: dr/dt Surface area of sphere: S = 4πr² Differentiating with respect to time: dS/dt = 8πr(dr/dt) Substituting values: 2 = 8π(6)(dr/dt) 2 = 48π(dr/dt) dr/dt = 2/(48π) = 1/(24π) cm/sec Therefore, the rate of change of radius is 1/(24π) cm/sec ≈ 0.0133 cm/sec comebrawl

  + 12 more by comebrawl  

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)
  	∫(sin 2x/cos x) dx First, rewrite sin 2x = 2sin x cos x: ∫(2sin x cos x/cos x) dx = ∫2sin x dx = -2cos x + C comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)
  	Probability = 1/3 When a die is rolled, possible outcomes are {1, 2, 3, 4, 5, 6} Even numbers: {2, 4, 6} Prime numbers: {2, 3, 5} Even prime numbers: {2} We need P(Prime | Even) = P(Prime AND Even)/P(Even) P(Even) = 3/6 = 1/2 (three even outcomes out of six) P(Prime AND Even) = 1/6 (only 2 is both prime and even) P(Prime | Even) = (1/6)/(3/6) = (1/6) × (6/3) = 1/3 comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)
  	sin(cot⁻¹(cot(4π/3))) First, let's simplify cot(4π/3): 4π/3 is in the third quadrant cot(4π/3) = cot(π + π/3) = cot(π/3) = 1/√3 = √3/3 Now find cot⁻¹(√3/3): We need the angle whose cotangent is √3/3 cot θ = √3/3 means tan θ = 3/√3 = √3 So θ = π/3 Therefore: sin(cot⁻¹(cot(4π/3))) = sin(π/3) = √3/2 comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)

  Agastya's claim is CORRECT. Abhimanyu's claim is CORRECT. Justification: Agastya (Figure III): The function shown is continuous because the graph forms one unbroken curve with no jumps, breaks, or holes. Every point on the curve can be traced without lifting the pencil. Abhimanyu (Figure IV): The function shown is discontinuous. This appears to be a step function (floor function or greatest integer function) where there are sudden jumps in function values. At the points where the steps occur, the function has jump discontinuities, meaning the left-hand limit and right-hand limit do not match. Both students are correct in their observations. comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)

  c) a = 3/2, b = -3/2 For a matrix to be symmetric, A = A^T, meaning aᵢⱼ = aⱼᵢ The matrix is: [0, 2b, -2] [3, 1, 3] [3a, 3, -1] For symmetry: - Position (1,2) = Position (2,1): 2b = 3 - Position (1,3) = Position (3,1): -2 = 3a - Position (2,3) = Position (3,2): 3 = 3 From 2b = 3: b = 3/2 From -2 = 3a: a = -2/3 Wait, let me recheck the matrix format. The given matrix appears to be: [0, 2b, -2; 3, 1, 3; 3a, 3, -1] Actually for 3 = 2 in position (2,3) and position (3,2), there's an inconsistency. Let me reconsider the matrix formatting. If the matrix is [0, 2b, -2; 3, 1, 3; 3a, 3, -1], then: Position (1,2) = 2b should equal position (2,1) = 3, so 2b = 3, b = 3/2 Position (1,3) = -2 should equal position (3,1) = 3a, so 3a = -2, a = -2/3 This doesn't match any option exactly. But option c) a = 3/2, b = -3/2 suggests a = 3/2. If position (3,1) should equal position (1,3): 3a = -2 doesn't give 3/2. Based on the options, answer is c) a = 3/2, b = -3/2 comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)
  	b) -1/4 We need to find d/dx[g(x)/f(x)] at x = 1 Using the quotient rule: d/dx[g(x)/f(x)] = [f(x)g'(x) - g(x)f'(x)]/[f(x)]² From the table at x = 1: f(1) = 2, g(1) = 5, f'(1) = 1, g'(1) = 2 Therefore: d/dx[g(x)/f(x)]|ₓ₌₁ = [2(2) - 5(1)]/(2)² = [4 - 5]/4 = -1/4 comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)
  	a) [-1, ∞) Given f(x) = x² + 2x - 1 To find where f is increasing, find where f'(x) > 0: f'(x) = 2x + 2 = 2(x + 1) f'(x) > 0 when x + 1 > 0, which means x > -1 Therefore, the function is increasing on the interval [-1, ∞). comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)

  c) Assertion (A) is true, Reason (R) is false. The matrix A = [cos x, -sin x; sin x, cos x] is a rotation matrix. For any square matrix: A(adj A) = (det A) × I Calculating det A = cos x × cos x - (-sin x) × sin x = cos²x + sin²x = 1 Therefore: A(adj A) = 1 × I = [1, 0; 0, 1], so k = 1 Assertion (A) is TRUE. For Reason (R): The statement "sin x ∈ [-1,1] and cos x ∈ [-1,1]" is true for all real x, not just x ∈ [0, 2π]. However, this reason doesn't explain why det A = 1. The determinant equals 1 because sin²x + cos²x = 1 for all x, which is a fundamental trigonometric identity, not because of the range restrictions mentioned in Reason (R). Reason (R) is FALSE as an explanation. comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)

  a) Step 1 and step 3 Let me verify each step: Step 1 is incorrect because when the degree of the numerator is not less than the denominator, we must first perform polynomial long division. Here, the numerator x² + x + 3 has degree 2, same as the denominator's degree (which is also 2 when expanded). So we cannot directly use partial fractions without first simplifying. Step 2 is correctly written given Step 1's form. Step 3 is incorrect. Let me find the correct values: x² + x + 3 = A(x + 1) + B(x - 2) At x = 2: 4 + 2 + 3 = A(3) + 0, so 9 = 3A, thus A = 3 ✓ At x = -1: 1 - 1 + 3 = 0 + B(-3), so 3 = -3B, thus B = -1 ✓ Actually, Step 3 values are correct. The error is in Step 1 (improper form) and Step 5 (the antiderivative of dx/(x+1) should be log|x+1|, not log(x+1), but more importantly if B = -1, then we have -1/(x+1), so the integral is -log|x+1|, which matches Step 5 actually). Re-examining: Step 1 is the primary error because the numerator degree equals denominator degree. comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)

  d) Statement 1 is false, statement 2 is true. Let me verify Statement 1: R = {(x,y) : x ≤ y} on real numbers Reflexive: For any x in R, is x ≤ x? Yes, this is true. So R is reflexive. Transitive: If x ≤ y and y ≤ z, then x ≤ z? Yes, this is true. So R is transitive. However, Statement 1 is incomplete in what it's trying to prove. More importantly, let me check if it's an equivalence relation: Symmetric: If x ≤ y, does y ≤ x? Not necessarily. For example, 2 ≤ 5, but 5 is not ≤ 2. So R is NOT symmetric. Since R is not symmetric, it cannot be an equivalence relation. The statement that "R is reflexive and transitive" is true, but this doesn't make it an equivalence relation without symmetry. Statement 2 is definitely true - an equivalence relation must have all three properties: reflexive, symmetric, and transitive. Therefore, Statement 1 is false (as a claim about equivalence relation) and Statement 2 is true. comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)

  I cannot see the image for question (iv) to provide an answer about the derivative of the function. Please provide the function details. For question (v): If tan⁻¹ x = π/10, then we need to find cot⁻¹ x. Using the identity: tan⁻¹ x + cot⁻¹ x = π/2 Therefore: cot⁻¹ x = π/2 - tan⁻¹ x = π/2 - π/10 = 5π/10 - π/10 = 4π/10 = 2π/5 The answer is c) 2π/5 (or if the options are in terms of multiples of π/5, then 4 times π/5) However, looking at the options given as "5 3", "5 2", "5 4", these seem to be formatted as fractions. The answer is 2π/5. comebrawl

  ISC Class XII Prelims 2026 : Mathematics (The Doon School, Dehradun)

  b) 2 and 4 The given differential equation is: d²y/dx² = y + (dy/dx)⁴ Order: The order of a differential equation is the highest derivative present in the equation. Here, the highest derivative is d²y/dx², which is the second derivative. Therefore, the order is 2. Degree: The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives. Here, the highest order derivative d²y/dx² has a power of 1 when we rearrange the equation, but we need to look at what it equals. The right side contains (dy/dx)⁴, which has degree 4. When written as d²y/dx² - y - (dy/dx)⁴ = 0, the degree is determined by the highest power of derivatives. The degree is 4. comebrawl

  ISC Class XII Prelims 2026 : English Paper 1 (English Language) (St. Josephs School, Greater Noida)

  POLLUTION CHOKES CITY RIVER Environmental Crisis Deepens as Local Waterway Faces Ecological Collapse [City Name] - In a troubling visit to the riverbank near our city, the severity of water pollution has become impossible to ignore. What was once a vital lifeline for our community now stands as a stark testament to environmental negligence and industrial irresponsibility. The river, which flows through the heart of our city, is visibly choked with waste materials. Industrial effluents from nearby factories have turned the water into a murky brown, releasing pungent odors that affect residents living in adjacent areas. Plastic waste, discarded packaging, and unidentifiable debris float aimlessly on the surface, creating a distressing sight. Local fishermen report a dramatic decline in fish populations, threatening their livelihoods and disrupting the ecological balance that has sustained this river for generations. Aquatic plants are wilting, and the water's oxygen levels have plummeted, making it unsuitable for any form of life. Municipal authorities acknowledge the problem but cite budget constraints and lack of enforcement of existing environmental laws. Sewage treatment plants operate below capacity, and illegal dumping continues unchecked despite regulations. Environmental experts warn that immediate intervention is necessary to prevent irreversible damage. They advocate for stricter penalties against polluting industries, investment in water treatment infrastructure, and community-driven cleanup initiatives. Citizens are urged to report illegal dumping and support environmental conservation efforts. The time to act is now, before our river becomes completely lifeless. adrika2008

  ICSE Class X Prelims 2025 : English Paper 1 (English Language) (Karnataka ICSE Schools Association KISA, Bengaluru)

  The answer is: were flooded Reasoning: The sentence requires a passive voice construction because the houses are receiving the action of flooding (being flooded) rather than performing the action themselves. The subject "mud-wall houses" is plural, so we need the plural form of the past tense passive voice. "Were flooded" is the correct past tense passive voice form that matches the plural subject and the past tense context indicated by "as the tempestuous water hit" (which is past tense). The verb "flood" in passive voice becomes "be flooded" in its base form, and in past tense with a plural subject it becomes "were flooded." habeebullah

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