Question

If $f(x) = \min \{2x^2 + 3, 6x\} + |x-1| \cos(x^2 - \frac{1}{4})$, then the number of points of non derivability of $f(x)$ is/are

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Non-derivability occurs where:
1. Functions in a $\min$ or $\max$ set intersect with different slopes.
2. The argument of an absolute value becomes zero (unless multiplied by another function that is zero at that point).
Step 2: Detailed Explanation:
1. Analyze $g(x) = \min \{2x^2 + 3, 6x\}$:
Set $2x^2 + 3 = 6x \implies 2x^2 - 6x + 3 = 0$.
Roots $x = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$.
At these two points, the quadratic and linear functions intersect. Since their slopes are different ($4x$ vs $6$), the function is non-differentiable at these 2 points.
2. Analyze $h(x) = |x-1| \cos(x^2 - 1/4)$:
The term $|x-1|$ is non-differentiable at $x=1$.
Check if $\cos(x^2 - 1/4)$ is zero at $x=1$:
$\cos(1 - 1/4) = \cos(3/4) \neq 0$.
Since the multiplying factor is non-zero, $x=1$ is a third point of non-derivability.
3. Total points = 2 (from $\min$) + 1 (from $|x-1|$) = 3.
Step 4: Final Answer:
The number of points of non-derivability is 3.