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

Answer 2

Given the piecewise function \( f(x) \) defined as: \[ f(x) = \begin{cases} \frac{a - b \cos 2x}{x^2}, & x < 0, \\ x^2 + cx + 2, & 0 \leq x \leq 1, \\ 2x + 1, & x > 1, \end{cases} \] and that \( f \) is continuous everywhere in \( \mathbb{R} \), find \( m + a + b + c \), where \( m \) is the number of points where \( f \) is NOT differentiable.

Concept Used:

For continuity at a point \( x_0 \), \( \lim_{x \to x_0^-} f(x) = f(x_0) = \lim_{x \to x_0^+} f(x) \).
For differentiability, the left-hand derivative must equal the right-hand derivative at the point.

Step-by-Step Solution:

Step 1: Ensure continuity at \( x = 0 \).

Left-hand limit as \( x \to 0^- \): \[ \lim_{x \to 0^-} \frac{a - b \cos 2x}{x^2} \] Using expansion \( \cos 2x = 1 - 2x^2 + \frac{(2x)^4}{4!} - \dots \), so: \[ a - b \cos 2x = a - b\left(1 - 2x^2 + O(x^4)\right) = (a - b) + 2b x^2 + O(x^4) \] Thus: \[ \frac{a - b \cos 2x}{x^2} = \frac{a - b}{x^2} + 2b + O(x^2) \] For the limit to exist as \( x \to 0^- \), we need \( a - b = 0 \Rightarrow a = b \). Then: \[ \lim_{x \to 0^-} f(x) = 2b \] Value at \( x = 0 \) from middle piece: \( f(0) = 0^2 + c\cdot 0 + 2 = 2 \).
So for continuity: \( 2b = 2 \Rightarrow b = 1 \), hence \( a = 1 \).

Step 2: Ensure continuity at \( x = 1 \).

Left-hand limit as \( x \to 1^- \): \( f(1) = 1^2 + c\cdot 1 + 2 = c + 3 \).
Right-hand limit as \( x \to 1^+ \): \( f(1^+) = 2(1) + 1 = 3 \).
For continuity: \( c + 3 = 3 \Rightarrow c = 0 \).

Step 3: Check differentiability at \( x = 0 \).

Left-hand derivative at 0: \[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} \] With \( a = b = 1 \), for \( x < 0 \): \[ f(x) = \frac{1 - \cos 2x}{x^2} = \frac{2 \sin^2 x}{x^2} = 2\left(\frac{\sin x}{x}\right)^2 \] So \( f(0^-) = 2 \). Then: \[ f'(0^-) = \lim_{h \to 0^-} \frac{2\left(\frac{\sin h}{h}\right)^2 - 2}{h} \] Let \( t = -h > 0 \), then \( h \to 0^- \Rightarrow t \to 0^+ \): \[ f'(0^-) = \lim_{t \to 0^+} \frac{2\left(\frac{\sin(-t)}{-t}\right)^2 - 2}{-t} = \lim_{t \to 0^+} \frac{2\left(\frac{\sin t}{t}\right)^2 - 2}{-t} \] As \( t \to 0 \), \( \frac{\sin t}{t} = 1 - \frac{t^2}{6} + O(t^4) \), so \( \left(\frac{\sin t}{t}\right)^2 = 1 - \frac{t^2}{3} + O(t^4) \).
Thus numerator \( 2\left(1 - \frac{t^2}{3} + O(t^4)\right) - 2 = -\frac{2}{3}t^2 + O(t^4) \).
So: \[ f'(0^-) = \lim_{t \to 0^+} \frac{-\frac{2}{3}t^2}{-t} = \lim_{t \to 0^+} \frac{2}{3}t = 0 \] Right-hand derivative at 0: \[ f'(0^+) = \lim_{h \to 0^+} \frac{(h^2 + 0\cdot h + 2) - 2}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = 0 \] So \( f'(0^-) = f'(0^+) = 0 \), hence differentiable at \( x = 0 \).

Step 4: Check differentiability at \( x = 1 \).

Left-hand derivative at 1: \[ f'(1^-) = \frac{d}{dx}(x^2 + 0\cdot x + 2)\big|_{x=1} = 2x\big|_{x=1} = 2 \] Right-hand derivative at 1: \[ f'(1^+) = \frac{d}{dx}(2x + 1)\big|_{x=1} = 2 \] So \( f'(1^-) = f'(1^+) = 2 \), hence differentiable at \( x = 1 \).

Step 5: Check differentiability in the interior of each piece.

For \( x < 0 \), \( f(x) = \frac{1 - \cos 2x}{x^2} \) is differentiable (smooth).
For \( 0 \le x \le 1 \), \( f(x) = x^2 + 2 \) is differentiable.
For \( x > 1 \), \( f(x) = 2x + 1 \) is differentiable.

Thus, \( f \) is differentiable everywhere, so \( m = 0 \).

Step 6: Compute \( m + a + b + c \).

We have \( m = 0 \), \( a = 1 \), \( b = 1 \), \( c = 0 \).
So \( m + a + b + c = 0 + 1 + 1 + 0 = 2 \).

Therefore, \( m + a + b + c = \mathbf{2} \).