Monomath – CBSE | GSEB | IB | JEE Maths by Ajay Yadav (Math King of Katargam) – 15+ Years Experience. Study Notes, Video Lessons, Formula Sheets & PYQs.
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Applications of Derivatives

Previous Year Questions (2022-2024)

📅 Year 2022

  1. Find equation of tangent to curve y = x2 - 2x + 1 at x = 2.
    Show Solution
    y(2) = 4-4+1 = 1. Point: (2,1)\ndy/dx = 2x-2. At x=2: slope = 2\nEquation: y-1 = 2(x-2) => y = 2x-3
  2. Find maxima/minima of f(x) = x3 - 6x2 + 9x + 1.
    Show Solution
    f'(x) = 3x2 - 12x + 9 = 3(x2-4x+3) = 3(x-1)(x-3)\nCritical: x=1, x=3\nf''(x)=6x-12. f''(1)=-6 max at x=1, f=5\nf''(3)=6 > 0 => min at x=3, f=1

📅 Year 2023

  1. Find equation of tangent to curve y = x2 - 2x + 1 at x = 2.
    Show Solution
    y(2) = 4-4+1 = 1. Point: (2,1)\ndy/dx = 2x-2. At x=2: slope = 2\nEquation: y-1 = 2(x-2) => y = 2x-3
  2. Find maxima/minima of f(x) = x3 - 6x2 + 9x + 1.
    Show Solution
    f'(x) = 3x2 - 12x + 9 = 3(x2-4x+3) = 3(x-1)(x-3)\nCritical: x=1, x=3\nf''(x)=6x-12. f''(1)=-6 max at x=1, f=5\nf''(3)=6 > 0 => min at x=3, f=1

📅 Year 2024

  1. Find equation of tangent to curve y = x2 - 2x + 1 at x = 2.
    Show Solution
    y(2) = 4-4+1 = 1. Point: (2,1)\ndy/dx = 2x-2. At x=2: slope = 2\nEquation: y-1 = 2(x-2) => y = 2x-3
  2. Find maxima/minima of f(x) = x3 - 6x2 + 9x + 1.
    Show Solution
    f'(x) = 3x2 - 12x + 9 = 3(x2-4x+3) = 3(x-1)(x-3)\nCritical: x=1, x=3\nf''(x)=6x-12. f''(1)=-6 max at x=1, f=5\nf''(3)=6 > 0 => min at x=3, f=1