Question

If $f(x) = \begin{cases} \alpha \sqrt{x+1}, & 0 \leq x \leq 3 \\ \beta x+2, & 3 < x \leq 5 \end{cases}$ is differentiable in (0,5) then $\{\alpha + \beta\}$ is

(Where $\{x\}$ represents fractional part of x)

• A. 0
• B. $\frac{1}{5}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

0

Answer 2

To ensure differentiability of the piecewise function $f(x)$ at $x=3$, we need to ensure both continuity and equality of derivatives at that point.

1. Continuity at $x=3$: We need to ensure that the function values match at $x=3$:

   $$\alpha \sqrt{3+1} = \beta(3) + 2$$ $$2\alpha = 3\beta + 2 \quad \text{(1)}$$

2. Differentiability at $x=3$: We need to ensure that the derivatives match at $x=3$. First, find the derivatives of both pieces: For $0 \leq x \leq 3$:

   $$f'(x) = \frac{\alpha}{2\sqrt{x+1}}$$ $$f'(3) = \frac{\alpha}{2\sqrt{3+1}} = \frac{\alpha}{4}$$

   For $3 < x \leq 5$:

   $$f'(x) = \beta$$

   Equating the derivatives at $x=3$:

   $$\frac{\alpha}{4} = \beta \quad \text{(2)}$$ $$\alpha = 4\beta$$

3. Solving for $\alpha$ and $\beta$: Substitute $\alpha = 4\beta$ into equation (1):

   $$2(4\beta) = 3\beta + 2$$ $$8\beta = 3\beta + 2$$ $$5\beta = 2$$ $$\beta = \frac{2}{5}$$

   Now, find $\alpha$:

   $$\alpha = 4\beta = 4\left(\frac{2}{5}\right) = \frac{8}{5}$$

4. Finding $\{\alpha + \beta\}$:

   $$\alpha + \beta = \frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2$$

   The fractional part of 2 is:

   $$\{2\} = 0$$

Therefore, $\{\alpha + \beta\} = 0$.