A variable Y is a function of t. Given that Y (t = 0) = 1 and Y (t = 1) = 2.
dt
— in
the interval t = [0, 1] can be approximated as _
Q26 - Q55 carry two marks each.
Official Answer Key:
1 to 1
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2020 Q.35)**
**1. Governing Concept & Definition**:
- **Topic**: Calculus & Mean Value Theorem
- By the Mean Value Theorem for derivatives, if $Y(t)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, there exists at least one $c \in (a, b)$ such that:
$\left. \frac{dY}{dt} \right|_{t=c} = \frac{Y(b) - Y(a)}{b - a}$
**2. Detailed Step-by-Step Calculation**:
- **Given**:
- Interval: $[0, 1] \implies a = 0, b = 1$
- $Y(0) = 1$
- $Y(1) = 2$
- **Average Rate of Change**:
$\frac{Y(1) - Y(0)}{1 - 0} = \frac{2 - 1}{1} = 1$
**3. Final Answer**:
- **Calculated Value**: **1**
- **Official GATE Accepted Range**: **1 to 1**
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