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

The Correct Option is D

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 x²), 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:
\[ m + a + b + c = 2 + 0 + 0 + 0 = 2 \]