Application of Derivatives
Karnataka Board · Class 12 · Mathematics
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Find the derivative of f(x) = 3x² + 2x - 5 and evaluate f'(2).
A particle moves along a line such that its position is given by s(t) = t³ - 6t² + 9t + 2. Find the velocity at t = 1.
For what values of x is the function f(x) = x³ - 3x + 1 increasing?
Find the critical points of f(x) = x³ - 6x² + 9x - 2.
Sample Questions
Which of the following are true about the function f(x) = 2x³ - 6x² + 6x - 1?
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f'(x) = 6x² - 12x + 6, f has a critical point at x = 1, f'(1) = 0, f''(x) = 12x - 12
f'(x) = 6x² - 12x + 6 = 6(x² - 2x + 1) = 6(x - 1)². So f'(1) = 0, making x = 1 a critical point. f''(x) = 12x - 12. Since f'(x) = 6(x - 1)² ≥ 0 for all x, f is always increasing except at x = 1 where it has zero slope.
The area of a square is increasing at 8 cm²/sec. Find the rate at which the side is increasing when the side is 4 cm.
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1 cm/sec
Let s = side length, A = area = s². Given: dA/dt = 8 cm²/sec, s = 4 cm. Find: ds/dt. Since A = s², we have dA/dt = 2s(ds/dt). Substituting: 8 = 2(4)(ds/dt), so 8 = 8(ds/dt), therefore ds/dt = 1 cm/sec.
Find the local maximum value of f(x) = -x² + 4x - 3.
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1
f'(x) = -2x + 4. Setting f'(x) = 0: -2x + 4 = 0, so x = 2. Since f''(x) = -2 < 0, x = 2 gives a local maximum. The maximum value is f(2) = -(2)² + 4(2) - 3 = -4 + 8 - 3 = 1.
A ladder 10 m long leans against a vertical wall. If the bottom slides away at 2 m/s, find the rate at which the top is sliding down when the bottom is 6 m from the wall.
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1.5 m/s
Let x = distance from wall to bottom, y = height up wall. Given: x² + y² = 100, dx/dt = 2 m/s, x = 6 m. Find dy/dt. When x = 6: y = √(100 - 36) = 8 m. Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0. So 2(6)(2) + 2(8)(dy/dt) = 0, giving 24 + 16(dy/dt) = 0, so dy/dt = -1.5 m/s. The top slides down at 1.5 m/s.
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Sources & Official References
- Karnataka SSLC — kseeb.kar.nic.in
- Dept of Pre-University Education, Karnataka
- National Education Policy 2020 — education.gov.in
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