Question

Let $f: \mathbb{R} \to \mathbb{R}$ be defined as: f(x) = \begin{cases} \frac{a - b \cos 2x}{x^2}, & x < 0, \\\ x^2 + cx + 2, & 0 \leq x \leq 1, \\\ 2x + 1, & x > 1\. \end{cases}
If $f$ is continuous everywhere in $\mathbb{R}$ and $m$ is the number of points where $f$ is NOT differentiable, then $m + a + b + c$ equals:

• A. 1
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (D)

2

Answer 2

To ensure continuity at x = 0 and x = 1 :

At x = 0 : - For the limit from the left:

$\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^-}} \frac{{a - b \cos 2x}}{{x^2}} = \text{undefined unless } a = 0 \text{ and } b = 0 \text{ (to ensure a finite value)}$

- For the limit from the right:

$\lim_{{x \to 0^+}} f(x) = 0^2 + c \cdot 0 + 2 = 2$

- To ensure continuity at x = 0 , we must have:

$\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^+}} f(x) = 2$

Thus, a = 0 and b = 0.

At x = 1 : - For the limit from the left:

$\lim_{{x \to 1^-}} f(x) = 1^2 + c \cdot 1 + 2 = 3 + c$

- For the limit from the right:
$\lim_{{x \to 1^+}} f(x) = 2 \cdot 1 + 1 = 3$
To ensure continuity at x = 1 , we must have:
$3 + c = 3 \implies c = 0$
Now, we check differentiability at x = 0 and x = 1 : - At x = 0 , the left-hand derivative does not exist (due to division by x 2), so f is not differentiable at x = 0. - At x = 1 , the left-hand and right-hand derivatives are not equal, so f is not differentiable at x = 1.
Thus, m = 2.
Given a = 0 , b = 0 , and c = 0 , we find:
$m + a + b + c = 2 + 0 + 0 + 0 = 2$