Let f: R → R be defined as f(x) = |x| + |x2|. The total number of points at which f attains either a local maximum or a local minimum is

Correct option is A. 5

Let's analyze the function f(x) = |x| + |x2|. Since |x2| = x2 for all x ∈ R, the function simplifies to f(x) = |x| + x2.

For x ≥ 0, f(x) = x + x2. The derivative is f'(x) = 1 + 2x. Setting f'(x) = 0 gives x = -1/2, which is not in the domain x ≥ 0. Thus, f(x) is strictly increasing for x ≥ 0.

For x < 0, f(x) = -x + x2. The derivative is f'(x) = -1 + 2x. Setting f'(x) = 0 gives x = 1/2, which is not in the domain x < 0. Let's check the behavior around x = 0:

As x approaches 0 from the left (x → 0-), f'(x) approaches -1 (negative), and as x approaches 0 from the right (x → 0+), f'(x) approaches 1 (positive).

Therefore, f(x) has a local minimum at x = 0.

Now let's consider the second derivative:

For x > 0, f''(x) = 2 > 0 (concave up)
For x < 0, f''(x) = 2 > 0 (concave up)

Since the second derivative is always positive, the function is always concave up. The only critical point is at x = 0, which is a local minimum.
However, the function f(x) = |x| + x² has a sharp point at x=0. The derivative is not defined at x=0, but the function is continuous. There's a minimum at x=0. The function is differentiable everywhere except at x=0. Hence there is a minimum at x = 0. The function is always increasing for x > 0 and decreasing for x < 0, so there is no other local minimum. This means we have one local minimum at x = 0.

Considering the absolute value, we have a minimum at x = 0. The derivative doesn't exist at x = 0 because of the |x| term. However, the function is continuous, and the limit of the derivative from both sides suggests a local minimum.

The function has a local minimum at x = 0. There are no local maxima. Hence, there is only 1 point (x=0) where the function attains a local extremum.