Question

Let $f(x) = [x]$ and

$g(x) = \begin{cases} x, & x \in [0, 1) \\ x-1, & x \in [1, 2) \\ x-2, & x \in [2, 3) \\ 0, & x = 3 \end{cases}$

Then $f(x) + g(x)$ is

[Note: [k] denotes the greatest integer function less than or equal to k.]

• A. discontinuous at x = 1 and x = 2.
• B. continuous in [0, 3] but non-derivable in [0,3].
• C. not twice differentiable in [0, 3].
• D. twice differentiable in [0, 3]

Answer 1

Answer: (D)

twice differentiable in [0, 3]

Answer 2

To determine the properties of $h(x) = f(x) + g(x)$, we first define $f(x)$ and $g(x)$ explicitly in the given intervals.

The function $f(x) = [x]$ is defined as: $f(x) = 0$ for $x \in [0, 1)$ $f(x) = 1$ for $x \in [1, 2)$ $f(x) = 2$ for $x \in [2, 3)$ $f(x) = 3$ for $x = 3$ (since $[3]=3$)

The function $g(x)$ is given as: $g(x) = x$ for $x \in [0, 1)$ $g(x) = x-1$ for $x \in [1, 2)$ $g(x) = x-2$ for $x \in [2, 3)$ $g(x) = 0$ for $x = 3$

Now, let's find $h(x) = f(x) + g(x)$ for each interval:

1. For $x \in [0, 1)$: $h(x) = f(x) + g(x) = 0 + x = x$

2. For $x \in [1, 2)$: $h(x) = f(x) + g(x) = 1 + (x-1) = x$

3. For $x \in [2, 3)$: $h(x) = f(x) + g(x) = 2 + (x-2) = x$

4. For $x = 3$: $h(3) = f(3) + g(3) = 3 + 0 = 3$

Combining these results, we can see that $h(x) = x$ for all $x \in [0, 3)$. Also, at $x=3$, $h(3)=3$. Therefore, the function $h(x)$ simplifies to: $h(x) = x$ for all $x \in [0, 3]$.

Now, we analyze the continuity and differentiability of $h(x) = x$ in the interval $[0, 3]$.

Continuity: The function $h(x) = x$ is a polynomial function. Polynomial functions are continuous everywhere. Thus, $h(x)$ is continuous in $[0, 3]$.

Differentiability: The first derivative of $h(x) = x$ is $h'(x) = \frac{d}{dx}(x) = 1$. Since $h'(x) = 1$ exists and is finite for all $x \in [0, 3]$, $h(x)$ is differentiable in $[0, 3]$.

Twice Differentiability: The second derivative of $h(x) = x$ is $h''(x) = \frac{d}{dx}(h'(x)) = \frac{d}{dx}(1) = 0$. Since $h''(x) = 0$ exists and is finite for all $x \in [0, 3]$, $h(x)$ is twice differentiable in $[0, 3]$.

Therefore, $h(x)$ is twice differentiable in $[0, 3]$.