Question

Let$f(x) = |2x^2 + 5|x - 3|, x \in \mathbb{R}$. If $m$ and $n$ denote the number of points were $f$is not continuous and not differentiable respectively, then$m + n$is equal to:

• A. 5
• B. 2
• C. 0
• D. 3

Answer 1

Answer: (D)

3

Answer 2

We analyze the function $f(x) = |2x^2 + 5|x| - 3|$ in two steps: checking continuity and differentiability.

Step 1: Continuity

The function $f(x)$ is a composition of absolute values and polynomials, which are continuous everywhere. Hence, $f(x)$ is continuous for all $x \in \mathbb{R}$.
$m = 0 \quad (\text{Number of points where } f(x) \text{ is not continuous})$

Step 2: Differentiability

The function $f(x)$ involves absolute values, which may cause non-differentiability at specific points:

1. First, the outermost absolute value $|2x^2 + 5|x| - 3|$ is non-differentiable where $2x^2 + 5|x| - 3 = 0$. Solving: $2x^2 + 5|x| - 3 = 0 \implies \text{Critical points are } x = -\frac{3}{2}, 0, \frac{3}{2}.$
2. Additionally, the inner term $|x|$ is non-differentiable at $x = 0$.

Hence, the total number of points of non-differentiability is:
$n = 3 \quad (\text{at } x = -\frac{3}{2}, 0, \frac{3}{2}).$

Final Calculation
$m + n = 0 + 3 = 3.$