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AS & A Level Further Mathematics (9231)

Hyperbolic functions

5 questions · about 1% of the marks we indexed · examined in 5 of 9 sessions.

Questions by year

2025

  • 2025 May/June · Paper 22 · Q6 (a)3 marksDefinitions & identities

2024

  • 2024 May/June · Paper 22 · Q6 (a)2 marksDefinitions & identities

2023

  • 2023 Oct/Nov · Paper 22 · Q7 (a)3 marksDefinitions & identities
  • 2023 May/June · Paper 22 · Q8 (a)3 marksDefinitions & identities

2022

  • 2022 May/June · Paper 22 · Q2 (a, b)8 marksDefinitions & identities · Hyperbolic equations

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Definitions & identities · Core · Paper 2 style · 5 marks
  1. (a) Using the definitions of sinh x and cosh x in terms of exponentials, prove that cosh²x − sinh²x = 1. [3]
  2. (b) Given that sinh x = 3/4, find the exact value of cosh x and of tanh x. [2]
Show answer
  1. (a) Proof (see working).
    M1: cosh²x = (e^(2x) + 2 + e^(−2x))/4, sinh²x = (e^(2x) − 2 + e^(−2x))/4.M1: subtract.A1: = 4/4 = 1.
  2. (b) cosh x = 5/4, tanh x = 3/5
    M1: cosh²x = 1 + 9/16 = 25/16, cosh x > 0.A1: cosh x = 5/4; tanh x = (3/4)/(5/4) = 3/5.
Q2Definitions & identities · Core · Paper 2 style · 3 marks
  1. (a) Find the exact values of sinh(ln 4) and tanh(ln 4), giving each as a fraction. [3]
Show answer
  1. (a) sinh(ln 4) = 15/8, tanh(ln 4) = 15/17
    M1: e^(ln 4) = 4, e^(−ln 4) = 1/4.A1: sinh(ln 4) = (4 − 1/4)/2 = 15/8.A1: tanh(ln 4) = (4 − 1/4)/(4 + 1/4) = (16 − 1)/(16 + 1) = 15/17.
Q3Definitions & identities · Core · Paper 2 style · 5 marks
  1. (a) Use the definitions sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2. Prove that sinh 2x = 2 sinh x cosh x. [2]
  2. (b) Given that cosh x = 13/12 and x > 0, find the exact value of sinh 2x. [3]
Show answer
  1. (a) Proof (see working).
    M1: 2 × (eˣ − e⁻ˣ)/2 × (eˣ + e⁻ˣ)/2 = (e^(2x) − e^(−2x))/2.A1: = sinh 2x.
  2. (b) sinh 2x = 65/72
    M1: sinh²x = 169/144 − 1 = 25/144.A1: sinh x = 5/12 (x > 0).A1: sinh 2x = 2 × 5/12 × 13/12 = 130/144 = 65/72.
Q4Definitions & identities · Core · Paper 2 style · 5 marks
  1. (a) Use the definitions sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2. Prove that cosh 2x = 2cosh²x − 1. [3]
  2. (b) Given that cosh x = 3, find the value of cosh 2x and of cosh 4x. [2]
Show answer
  1. (a) Proof (see working).
    M1: 2cosh²x = 2(e^(2x) + 2 + e^(−2x))/4 = (e^(2x) + e^(−2x))/2 + 1.M1: subtract 1.A1: = (e^(2x) + e^(−2x))/2 = cosh 2x.
  2. (b) cosh 2x = 17, cosh 4x = 577
    M1: cosh 2x = 2(9) − 1 = 17.A1: cosh 4x = 2(17²) − 1 = 577.
Q5Definitions & identities · Core · Paper 2 style · 5 marks
  1. (a) Express 4 cosh x − 2 sinh x in terms of eˣ, simplifying your answer. [2]
  2. (b) Hence find the minimum value of 4 cosh x − 2 sinh x and the exact value of x at which it occurs. [3]
Show answer
  1. (a) eˣ + 3e⁻ˣ
    M1: 2(eˣ + e⁻ˣ) − (eˣ − e⁻ˣ).A1: = eˣ + 3e⁻ˣ.
  2. (b) Minimum 2√3 at x = (1/2) ln 3
    M1: differentiate: eˣ − 3e⁻ˣ = 0 ⇒ e^(2x) = 3.A1: x = (1/2) ln 3 (second derivative eˣ + 3e⁻ˣ > 0, so a minimum).A1: value √3 + 3/√3 = 2√3.
Q6Definitions & identities · Core · Paper 2 style · 4 marks
  1. (a) Starting from cosh²x − sinh²x = 1, prove that coth²x − 1 = cosech²x for x ≠ 0. [2]
  2. (b) Given that coth x = −5/3, find the exact value of cosech x. [2]
Show answer
  1. (a) Proof (see working).
    M1: divide every term by sinh²x (sinh x ≠ 0 for x ≠ 0).A1: cosh²x/sinh²x − 1 = 1/sinh²x, i.e. coth²x − 1 = cosech²x.
  2. (b) cosech x = −4/3
    M1: cosech²x = 25/9 − 1 = 16/9.A1: coth x < 0 ⇒ x < 0 ⇒ sinh x < 0, so cosech x = −4/3.
Q7Definitions & identities · Core · Paper 2 style · 4 marks
  1. (a) Show that (cosh x + sinh x)ⁿ = cosh nx + sinh nx for any positive integer n. [2]
  2. (b) Hence, or otherwise, find the exact value of (cosh(ln 2) + sinh(ln 2))⁵ − (cosh(ln 2) − sinh(ln 2))⁵. [2]
Show answer
  1. (a) Proof (see working).
    M1: cosh x + sinh x = eˣ, so (cosh x + sinh x)ⁿ = e^(nx).A1: e^(nx) = cosh nx + sinh nx.
  2. (b) 1023/32
    M1: = e^(5 ln 2) − e^(−5 ln 2) = 32 − 1/32.A1: = 1023/32.
Q8Definitions & identities · Standard · Paper 2 style · 6 marks
  1. (a) Prove that arsinh x = ln(x + √(x² + 1)). [4]
  2. (b) Hence find the exact value of arsinh(3/4) − arsinh(−4/3). [2]
Show answer
  1. (a) Proof (see working).
    M1: let y = arsinh x, so x = sinh y = (e^y − e^(−y))/2.M1: e^(2y) − 2x e^y − 1 = 0, a quadratic in e^y.A1: e^y = x ± √(x² + 1).A1: e^y > 0 rejects the minus sign, so y = ln(x + √(x² + 1)).
  2. (b) ln 6
    M1: arsinh(3/4) = ln(3/4 + 5/4) = ln 2; arsinh(−4/3) = ln(−4/3 + 5/3) = ln(1/3).A1: ln 2 − ln(1/3) = ln 6.
Q9Definitions & identities · Standard · Paper 2 style · 6 marks
  1. (a) Prove that arcosh x = ln(x + √(x² − 1)) for x ≥ 1. [4]
  2. (b) Find the exact value of arcosh(5/3) + arcosh(17/8), simplifying your answer to a single logarithm. [2]
Show answer
  1. (a) Proof (see working).
    M1: y = arcosh x (y ≥ 0), x = (e^y + e^(−y))/2.M1: e^(2y) − 2x e^y + 1 = 0.A1: e^y = x ± √(x² − 1).A1: y ≥ 0 needs e^y ≥ 1; the product of the roots is 1, so the larger root x + √(x² − 1) is taken.
  2. (b) ln 12
    M1: arcosh(5/3) = ln(5/3 + 4/3) = ln 3; arcosh(17/8) = ln(17/8 + 15/8) = ln 4.A1: ln 12.
Q10Definitions & identities · Standard · Paper 2 style · 6 marks
  1. (a) Prove that artanh x = (1/2) ln((1 + x)/(1 − x)) for −1 < x < 1. [4]
  2. (b) Express artanh(1/3) + artanh(1/5) as a single logarithm. [2]
Show answer
  1. (a) Proof (see working).
    M1: y = artanh x, x = (e^y − e^(−y))/(e^y + e^(−y)) = (e^(2y) − 1)/(e^(2y) + 1).M1: x e^(2y) + x = e^(2y) − 1.A1: e^(2y) = (1 + x)/(1 − x).A1: y = (1/2) ln((1 + x)/(1 − x)).
  2. (b) (1/2) ln 3 (= ln √3)
    M1: (1/2) ln 2 + (1/2) ln(3/2).A1: = (1/2) ln 3.
Q11Definitions & identities · Standard · Paper 2 style · 5 marks
  1. (a) Using exponentials, prove that sinh 3x = 3 sinh x + 4 sinh³x. [4]
  2. (b) Given that sinh x = 1/2, find the exact value of sinh 3x. [1]
Show answer
  1. (a) Proof (see working).
    M1: 4 sinh³x = 4(eˣ − e⁻ˣ)³/8 = (e^(3x) − 3eˣ + 3e⁻ˣ − e^(−3x))/2.M1: 3 sinh x = (3eˣ − 3e⁻ˣ)/2.M1: add.A1: = (e^(3x) − e^(−3x))/2 = sinh 3x.
  2. (b) 2
    B1: 3/2 + 4/8 = 2.
Q12Definitions & identities · Standard · Paper 2 style · 6 marks
  1. (a) Use the definitions sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2. Prove that cosh(x + y) = cosh x cosh y + sinh x sinh y. [4]
  2. (b) Given that sinh a = 3/4 and sinh b = 5/12, with a, b > 0, find the exact value of cosh(a + b). [2]
Show answer
  1. (a) Proof (see working).
    M1: cosh x cosh y = (e^(x+y) + e^(x−y) + e^(−x+y) + e^(−x−y))/4.M1: sinh x sinh y = (e^(x+y) − e^(x−y) − e^(−x+y) + e^(−x−y))/4.M1: add: (2e^(x+y) + 2e^(−x−y))/4.A1: = cosh(x + y).
  2. (b) 5/3
    M1: cosh a = 5/4, cosh b = 13/12.A1: 65/48 + 15/48 = 80/48 = 5/3.
Q13Definitions & identities · Standard · Paper 2 style · 5 marks
  1. (a) Prove that 1 − tanh²x = sech²x. [2]
  2. (b) Given that tanh x = −5/13, find the exact values of cosh x and sinh x. [3]
Show answer
  1. (a) Proof (see working).
    M1: divide cosh²x − sinh²x = 1 by cosh²x.A1: 1 − tanh²x = sech²x.
  2. (b) cosh x = 13/12, sinh x = −5/12
    M1: sech²x = 1 − 25/169 = 144/169.A1: cosh x = 13/12 (cosh x ≥ 1).A1: sinh x = tanh x cosh x = −5/12.
Q14Definitions & identities · Standard · Paper 2 style · 4 marks
  1. (a) Given that x = ln(2 + √3), find the exact values of cosh x, sinh x and tanh x. [4]
Show answer
  1. (a) cosh x = 2, sinh x = √3, tanh x = √3/2
    M1: e⁻ˣ = 1/(2 + √3) = 2 − √3.A1: cosh x = ((2 + √3) + (2 − √3))/2 = 2.A1: sinh x = ((2 + √3) − (2 − √3))/2 = √3.A1: tanh x = √3/2.
Q15Definitions & identities · Standard · Paper 2 style · 5 marks
  1. (a) Show that cosh 2t = (1 + tanh²t)/(1 − tanh²t). [3]
  2. (b) Hence find the exact value of cosh(2 artanh(1/2)). [2]
Show answer
  1. (a) Proof (see working).
    M1: cosh 2t = cosh²t + sinh²t.M1: divide numerator and denominator by cosh²t, using cosh²t − sinh²t = 1.A1: = (1 + tanh²t)/(1 − tanh²t).
  2. (b) 5/3
    M1: tanh t = 1/2 ⇒ (1 + 1/4)/(1 − 1/4).A1: = 5/3.
Q16Definitions & identities · Challenge · Paper 2 style · 7 marks
  1. (a) Using exponentials, prove that cosh 3x = 4cosh³x − 3 cosh x. [4]
  2. (b) Given that cosh x = 5/4, find the exact value of cosh 3x, and verify your answer by first finding x as a logarithm. [3]
Show answer
  1. (a) Proof (see working).
    M1: 4cosh³x = (eˣ + e⁻ˣ)³/2 = (e^(3x) + 3eˣ + 3e⁻ˣ + e^(−3x))/2.M1: 3 cosh x = (3eˣ + 3e⁻ˣ)/2.M1: subtract.A1: = (e^(3x) + e^(−3x))/2 = cosh 3x.
  2. (b) cosh 3x = 65/16 (x = ±ln 2)
    M1: 4(125/64) − 15/4 = 500/64 − 240/64.A1: = 260/64 = 65/16.B1: cosh x = 5/4 ⇒ x = ±ln 2; cosh(3 ln 2) = (8 + 1/8)/2 = 65/16 ✓.
Q17Definitions & identities · Challenge · Paper 2 style · 7 marks
  1. (a) Using the result for tanh(x + y) in terms of tanh x and tanh y, which you should first prove from the addition formulae for sinh and cosh, show that tanh 2x = 2 tanh x/(1 + tanh²x). [4]
  2. (b) Given that tanh 2x = 4/5, find the exact value of tanh x. [3]
Show answer
  1. (a) Proof (see working).
    M1: tanh(x + y) = (sinh x cosh y + cosh x sinh y)/(cosh x cosh y + sinh x sinh y).M1: divide top and bottom by cosh x cosh y.A1: = (tanh x + tanh y)/(1 + tanh x tanh y).A1: put y = x: tanh 2x = 2 tanh x/(1 + tanh²x).
  2. (b) tanh x = 1/2
    M1: 4(1 + t²) = 10t ⇒ 2t² − 5t + 2 = 0.A1: t = 1/2 or 2.A1: |tanh x| < 1, so tanh x = 1/2.
Q18Definitions & identities · Challenge · Paper 2 style · 6 marks
  1. (a) Express 5 cosh x + 3 sinh x in the form R cosh(x + α), where R > 0, giving R exactly and α as a logarithm. [4]
  2. (b) State the minimum value of 5 cosh x + 3 sinh x and the value of x at which it occurs. [2]
Show answer
  1. (a) 4 cosh(x + ln 2)
    M1: R cosh x cosh α + R sinh x sinh α: R cosh α = 5, R sinh α = 3.A1: R² = 25 − 9 = 16, R = 4.M1: tanh α = 3/5 ⇒ α = (1/2) ln((1 + 3/5)/(1 − 3/5)).A1: α = (1/2) ln 4 = ln 2.
  2. (b) Minimum 4 at x = −ln 2
    B1: minimum of cosh is 1 when x + ln 2 = 0.B1: value 4 at x = −ln 2.
Q19Definitions & identities · Challenge · Paper 2 style · 6 marks
  1. (a) Prove that sinh x + sinh y = 2 sinh((x + y)/2) cosh((x − y)/2). [4]
  2. (b) Hence show that sinh 3x + sinh x = 0 has only one real solution. [2]
Show answer
  1. (a) Proof (see working).
    M1: let p = (x + y)/2, q = (x − y)/2, so x = p + q, y = p − q.M1: RHS = 2 × (e^p − e^(−p))/2 × (e^q + e^(−q))/2.M1: = (e^(p+q) + e^(p−q) − e^(−p+q) − e^(−p−q))/2.A1: = (eˣ − e⁻ˣ)/2 + (e^y − e^(−y))/2 = sinh x + sinh y.
  2. (b) x = 0 is the only solution
    M1: 2 sinh 2x cosh x = 0; cosh x ≥ 1 never zero.A1: sinh 2x = 0 ⇒ x = 0 only.
Q20Definitions & identities · Challenge · Paper 2 style · 7 marks
  1. (a) By writing each side in logarithmic form, show that artanh x = arsinh(x/√(1 − x²)) for −1 < x < 1. [5]
  2. (b) Use this result to find the exact value of x for which artanh x = arsinh(1). [2]
Show answer
  1. (a) Proof (see working).
    M1: arsinh(u) = ln(u + √(u² + 1)) with u = x/√(1 − x²).M1: u² + 1 = 1/(1 − x²), so √(u² + 1) = 1/√(1 − x²).A1: arsinh u = ln((1 + x)/√(1 − x²)).M1: (1 + x)/√(1 − x²) = √((1 + x)²/((1 − x)(1 + x))) = √((1 + x)/(1 − x)).A1: so arsinh u = (1/2) ln((1 + x)/(1 − x)) = artanh x.
  2. (b) x = 1/√2 (= √2/2)
    M1: x/√(1 − x²) = 1 ⇒ x² = 1 − x².A1: x = 1/√2 (positive since arsinh 1 > 0).
Q21Definitions & identities · Challenge · Paper 2 style · 7 marks
  1. (a) Show that cosh 2y = 2cosh²y − 1 and hence prove that arcosh(2x² − 1) = 2 arcosh x for x ≥ 1. [4]
  2. (b) Hence find the exact value of arcosh 7 in terms of a single logarithm of a number of the form a + b√3, where a and b are integers. [3]
Show answer
  1. (a) Proof (see working).
    B1: cosh 2y = 2cosh²y − 1 (from exponentials or cosh 2y = cosh²y + sinh²y).M1: let y = arcosh x ≥ 0, so cosh y = x.M1: cosh 2y = 2x² − 1, and 2y ≥ 0.A1: so 2y = arcosh(2x² − 1), i.e. arcosh(2x² − 1) = 2 arcosh x.
  2. (b) arcosh 7 = 2 ln(2 + √3) = ln(7 + 4√3)
    M1: 2x² − 1 = 7 ⇒ x = 2.A1: arcosh 7 = 2 arcosh 2 = 2 ln(2 + √3).A1: = ln((2 + √3)²) = ln(7 + 4√3).
Q22Definitions & identities · Challenge · Paper 2 style · 6 marks
  1. (a) Given that sinh x = −2, find the exact values of cosh x and tanh x. [2]
  2. (b) Find the exact value of x, and show that sinh 2x + cosh 2x = 9 − 4√5. [4]
Show answer
  1. (a) cosh x = √5, tanh x = −2/√5
    M1: cosh²x = 1 + 4 = 5, cosh x > 0.A1: cosh x = √5, tanh x = −2/√5 = −2√5/5.
  2. (b) x = ln(√5 − 2); sinh 2x + cosh 2x = e^(2x) = 9 − 4√5
    M1: x = arsinh(−2) = ln(−2 + √5).A1: x = ln(√5 − 2).M1: sinh 2x + cosh 2x = e^(2x) = (√5 − 2)².A1: = 5 − 4√5 + 4 = 9 − 4√5.
Q23Hyperbolic equations · Core · Paper 2 style · 4 marks
  1. (a) Solve the equation 3 cosh x − sinh x = 3, giving your answers in exact form. [4]
Show answer
  1. (a) x = 0 or x = ln 2
    M1: 3(eˣ + e⁻ˣ)/2 − (eˣ − e⁻ˣ)/2 = eˣ + 2e⁻ˣ = 3.M1: e^(2x) − 3eˣ + 2 = 0.A1: (eˣ − 1)(eˣ − 2) = 0.A1: x = 0 or x = ln 2.
Q24Hyperbolic equations · Core · Paper 2 style · 4 marks
  1. (a) Solve the equation 2 cosh x = 5, giving your answers as logarithms in exact form. [3]
  2. (b) Give the positive solution correct to 3 significant figures. [1]
Show answer
  1. (a) x = ± ln((5 + √21)/2)
    M1: eˣ + e⁻ˣ = 5 ⇒ e^(2x) − 5eˣ + 1 = 0.A1: eˣ = (5 ± √21)/2.A1: x = ln((5 + √21)/2) or ln((5 − √21)/2) = −ln((5 + √21)/2).
  2. (b) 1.57
    B1: ln(4.79128…) = 1.5668 → 1.57.
Q25Hyperbolic equations · Standard · Paper 2 style · 5 marks
  1. (a) Solve the equation cosh 2x − 5 sinh x = 4, giving your answers as logarithms in exact form. [5]
Show answer
  1. (a) x = ln(3 + √10) or x = ln((√5 − 1)/2)
    M1: use cosh 2x = 1 + 2sinh²x.M1: 2sinh²x − 5 sinh x − 3 = 0.A1: (2 sinh x + 1)(sinh x − 3) = 0, sinh x = 3 or −1/2.A1: x = ln(3 + √10).A1: x = ln(−1/2 + √(5/4)) = ln((√5 − 1)/2).
Q26Hyperbolic equations · Standard · Paper 2 style · 5 marks
  1. (a) Solve the equation 2 cosh 2x − 9 cosh x + 7 = 0, giving your answers in exact logarithmic form where appropriate. [5]
Show answer
  1. (a) x = 0, x = ln 2, x = −ln 2
    M1: cosh 2x = 2cosh²x − 1.M1: 4cosh²x − 9 cosh x + 5 = 0.A1: (cosh x − 1)(4 cosh x − 5) = 0.A1: cosh x = 1 ⇒ x = 0.A1: cosh x = 5/4 ⇒ x = ±ln 2.
Q27Hyperbolic equations · Standard · Paper 2 style · 5 marks
  1. (a) Solve the equation 3 sech²x + 4 tanh x + 1 = 0, giving your answer in the form a ln b. [5]
Show answer
  1. (a) x = −(1/2) ln 5
    M1: sech²x = 1 − tanh²x.M1: 3tanh²x − 4 tanh x − 4 = 0.A1: (3 tanh x + 2)(tanh x − 2) = 0.A1: tanh x = 2 rejected (|tanh x| < 1); tanh x = −2/3.A1: x = (1/2) ln((1/3)/(5/3)) = −(1/2) ln 5.
Q28Hyperbolic equations · Challenge · Paper 2 style · 6 marks
  1. (a) Express 13 cosh x − 5 sinh x in the form R cosh(x − α), where R > 0 and α is given as a logarithm. [3]
  2. (b) Hence solve 13 cosh x − 5 sinh x = 13. [3]
Show answer
  1. (a) 12 cosh(x − ln(3/2))
    M1: R cosh α = 13, R sinh α = 5 ⇒ R = √(169 − 25) = 12.M1: tanh α = 5/13 ⇒ α = (1/2) ln((18/13)/(8/13)).A1: α = (1/2) ln(9/4) = ln(3/2).
  2. (b) x = 0 or x = 2 ln(3/2) = ln(9/4)
    M1: cosh(x − ln(3/2)) = 13/12.A1: arcosh(13/12) = ln(13/12 + 5/12) = ln(3/2), so x − ln(3/2) = ±ln(3/2).A1: x = 0 or x = ln(9/4).
Q29Hyperbolic equations · Challenge · Paper 2 style · 5 marks
  1. (a) Given that sinh 3x = 3 sinh x + 4 sinh³x, solve the equation sinh 3x = 7 sinh x, giving the non-zero solutions as logarithms in exact form. [5]
Show answer
  1. (a) x = 0, x = ±ln(1 + √2)
    M1: 4sinh³x − 4 sinh x = 0.M1: 4 sinh x (sinh x − 1)(sinh x + 1) = 0.A1: sinh x = 0 ⇒ x = 0.A1: sinh x = 1 ⇒ x = ln(1 + √2).A1: sinh x = −1 ⇒ x = ln(√2 − 1) = −ln(1 + √2).
Q30Hyperbolic equations · Challenge · Paper 2 style · 6 marks
  1. (a) Show that the equation tanh x + sech x = 7/5 can be written as e^(2x) − 5eˣ + 6 = 0. [4]
  2. (b) Hence solve tanh x + sech x = 7/5. [2]
Show answer
  1. (a) Shown (see working).
    M1: multiply by cosh x: sinh x + 1 = (7/5) cosh x.M1: (eˣ − e⁻ˣ)/2 + 1 = (7/10)(eˣ + e⁻ˣ); multiply by 10eˣ.M1: 5e^(2x) − 5 + 10eˣ = 7e^(2x) + 7.A1: 2e^(2x) − 10eˣ + 12 = 0 ⇒ e^(2x) − 5eˣ + 6 = 0.
  2. (b) x = ln 2 or x = ln 3
    M1: (eˣ − 2)(eˣ − 3) = 0.A1: x = ln 2, ln 3.

Similar questions

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Definitions & identities 5 questions · 5/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Definitions & identities 145/9
Hyperbolic equations Rare51/9

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