Application of Derivatives — Class 12 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Application of Derivatives" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Application of Derivatives" — 7 important questions with detailed answers for CBSE…
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Key Questions Covered:
- Find the rate of change of the area of a circle with respect to its radius.
- A ladder of length 10 m leans against a wall. If the base moves away from the…
- Find the maximum and minimum values of f(x) = x³ - 3x on the interval [-2, 2].
- + 4 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Find the rate of change of the area of a circle with resp… | ✓ Solved |
| A ladder of length 10 m leans against a wall. If the base… | ✓ Solved |
| Find the maximum and minimum values of f(x) = x³ - 3x on … | ✓ Solved |
Showing 3 of 7 questions
Q1: Find the rate of change of the area of a circle with respect to its radius.
Let A be the area of a circle and r be its radius.
We know: A = πr²
The rate of change of area with respect to radius is dA/dr.
dA/dr = d/dr(πr²)
= π × d/dr(r²)
= π × 2r
= 2πr
CONCLUSION: dA/dr = 2πr
Interpretation: When the radius is r, a small increase dr in the radius cause...
Q2: A ladder of length 10 m leans against a wall. If the base moves away from the wall at 2 m/s, find the rate at which the top of the ladder slides down the wall when the base is 6 m from the wall.
Let x = distance of base from wall, y = height of top of ladder on the wall.
Given: ladder length = 10 m (constant), dx/dt = 2 m/s
By Pythagoras: x² + y² = 10²
x² + y² = 100
Differentiating both sides with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0
x(dx/dt) + y(dy/dt) = 0
dy/dt = -x(dx/dt)/y
Wh...
Q3: Find the maximum and minimum values of f(x) = x³ - 3x on the interval [-2, 2].
Given: f(x) = x³ - 3x on [-2, 2]
Step 1: Find critical points
f'(x) = 3x² - 3 = 3(x² - 1) = 3(x-1)(x+1)
Setting f'(x) = 0:
3(x-1)(x+1) = 0
Critical points: x = 1 and x = -1
Both critical points lie in [-2, 2].
Step 2: Evaluate f at critical points and endpoints
f(-2) = (-2)³ - 3(-2) = -8 + 6 = -...
Showing 3 of 7 questions. Visit the full page for complete solutions.