NCERT Class 12 · Ch 9 Exercise 9.3 · Q21
NCERT Class 12 Maths · Chapter 9 Differential Equations · Exercise 9.3 Question 21
₹1000 grows continuously at 5%/year. Find its value after 10 years using e^0.5=1.648.
💡 Concept Hints
- Identify the domain and the inner and outer functions, or the relevant one-sided limits.
- Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.
✅ Step-by-Step Solution
Final Answer: ₹1648 (using the supplied approximation).
- P′=0.05P, P(0)=1000, so P(t)=1000e^(0.05t).
- At 10 years, P=1000e^0.5=1648.
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