If `f(x) = sum_(r=1)^(n)a_(r)|x|^(r)`, where `a_(i)` s are real constants, then f(x) is

Options

A. continuous at x = 0, for all `a_(i)`
B. differentiable at x = 0, for all `a_(i) in R`
C. differentiable at x = 0, for all `a_(2k+1) = 0`
D. None of the above

Answer

To determine the nature of the function \( f(x) = \sum_{r=1}^{n} a_r |x|^r \), where \( a_r \) are real constants, we will analyze the function step by step.

### Step 1: Understanding the Function
The function \( f(x) \) is defined as a sum of terms involving \( |x|^r \). Since \( |x| \) is always non-negative, each term \( |x|^r \) is also non-negative for any real \( x \).

### Step 2: Evaluating \( f(x) \) at \( x = 0 \)
Let's evaluate \( f(0) \):
\[ f(0) = \sum_{r=1}^{n} a_r |0|^r = \sum_{r=1}^{n} a_r \cdot 0 = 0 \]
Thus, \( f(0) = 0 \).

### Step 3: Evaluating the Left-Hand Limit as \( x \to 0^- \)
Next, we evaluate the limit as \( x \) approaches 0 from the left:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sum_{r=1}^{n} a_r |x|^r = \lim_{x \to 0^-} \sum_{r=1}^{n} a_r (-x)^r \]
As \( x \to 0^- \), \( |x| = -x \), so:
\[ \lim_{x \to 0^-} f(x) = \sum_{r=1}^{n} a_r (0) = 0 \]

### Step 4: Evaluating the Right-Hand Limit as \( x \to 0^+ \)
Now, we evaluate the limit as \( x \) approaches 0 from the right:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \sum_{r=1}^{n} a_r |x|^r = \lim_{x \to 0^+} \sum_{r=1}^{n} a_r x^r \]
As \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \sum_{r=1}^{n} a_r (0) = 0 \]

### Step 5: Continuity at \( x = 0 \)
Since both the left-hand limit and the right-hand limit as \( x \to 0 \) equal \( f(0) = 0 \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 6: Differentiability at \( x = 0 \)
To check differentiability, we need to compute the derivatives from both sides.

#### Left-Hand Derivative:
For \( x < 0 \):
\[ f'(x) = \sum_{r=1}^{n} a_r r (-x)^{r-1} \]
Taking the limit as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f'(x) = \sum_{r=1}^{n} a_r r (0)^{r-1} = 0 \]

#### Right-Hand Derivative:
For \( x > 0 \):
\[ f'(x) = \sum_{r=1}^{n} a_r r x^{r-1} \]
Taking the limit as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f'(x) = \sum_{r=1}^{n} a_r r (0)^{r-1} = 0 \]

### Conclusion
Since the left-hand and right-hand derivatives at \( x = 0 \) are equal, \( f(x) \) is differentiable at \( x = 0 \).

Thus, we conclude that \( f(x) \) is continuous and differentiable at \( x = 0 \).

### Final Answer
The function \( f(x) \) is continuous and differentiable at \( x = 0 \).

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