Question

Let \( f : [0,3] \to \mathbb{R} \) be defined by \[ f(x) = \begin{cases} 0, & 0 \leq x < 1, \\ e^{x^2} - e, & 1 \leq x < 2, \\ e^{x^2} + 1, & 2 \leq x \leq 3. \end{cases} \] Now, define \( F : [0, 3] \to \mathbb{R} \) by \[ F(0) = 0 \,\,\text{and}\,\, F(x) = \int_0^x f(t) \, dt, \text{ for } 0 < x \leq 3. \] Then 
 

• A. \( F \) is differentiable at \( x = 1 \) and \( F'(1) = 0 \)
• B. \( F \) is differentiable at \( x = 2 \) and \( F'(2) = 0 \)
• C. \( F \) is not differentiable at \( x = 1 \)
• D. \( F \) is differentiable at \( x = 2 \) and \( F'(2) = 1 \)

Hint:

When working with piecewise functions, always check for differentiability at the boundaries where the function changes form.

Answer 1

The Correct Option is A

Step 1: Find $F(x)$ for each interval

For $0 \leq x < 1$: $$F(x) = \int_0^x 0 , dt = 0$$

For $1 \leq x < 2$: $$F(x) = \int_0^1 0 , dt + \int_1^x (e^{t^2} - e) dt = \int_1^x (e^{t^2} - e) dt$$

For $2 \leq x \leq 3$: $$F(x) = \int_0^1 0 , dt + \int_1^2 (e^{t^2} - e) dt + \int_2^x (e^{t^2} + 1) dt$$

Step 2: Check differentiability at $x = 1$

Left derivative: $F'(1^-) = f(1^-) = 0$

Right derivative: $F'(1^+) = f(1^+) = e^1 - e = e - e = 0$

Since $F'(1^-) = F'(1^+) = 0$, $F$ is differentiable at $x = 1$ with $F'(1) = 0$.

Option (A) is TRUE 

Step 3: Check differentiability at $x = 2$

Left derivative: $F'(2^-) = f(2^-) = e^4 - e$

Right derivative: $F'(2^+) = f(2^+) = e^4 + 1$

Since $F'(2^-) = e^4 - e \neq e^4 + 1 = F'(2^+)$, $F$ is not differentiable at $x = 2$.

Options (B) and (D) are FALSE 

Option (C) is FALSE (since $F$ is differentiable at $x = 1$) 

Answer: (A) $F$ is differentiable at $x = 1$ and $F'(1) = 0$