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Step-by-Step Solution
Step 1: Understand the piecewise definition of the function
The function is given by:
$ f(x) =
\begin{cases}
\frac{a - b \cos(2x)}{x^2}, & x < 0,\\[6pt]
x^2 + c\,x + 2, & 0 \le x \le 1,\\[6pt]
2x + 1, & x > 1
\end{cases}
$
Step 2: Impose continuity at $x = 0$
For continuity at $x=0$, the limit as $x \to 0^{-}$ must match the value at $x=0$,
which in turn must match the limit as $x \to 0^{+}$ (although typically the limit
from the right should just equal the value at $0$ if the function is continuous at $0$).
(a) Value at $x=0$ (from $0 \le x \le 1$ part):
$ f(0) = 0^2 + c \cdot 0 + 2 = 2. $
(b) Left-hand limit as $x \to 0^{-}$ (from $x
(a) Value from the left ($0 \le x \le 1$ part):
$ f(1^-) = 1^2 + c \cdot 1 + 2 = 1 + c + 2 = 3 + c. $
(b) Value from the right ($x>1$ part):
$ f(1^+) = 2 \cdot 1 + 1 = 3. $
Equating for continuity: $3 + c = 3 \,\Longrightarrow\, c = 0.$
Step 4: Check differentiability at $x = 0$
We compare the left-hand derivative and the right-hand derivative at $x=0$.
(a) Left-hand derivative (for $x
Step 5: Check differentiability at $x = 1$
Next, compare the left-hand and right-hand derivatives at $x=1$.
(a) Left-hand derivative (for $0 \le x \le 1$):
With $c=0$, $f(x) = x^2 + 2$, so
$ f'(x) = 2x. $
At $ x=1 $,
$ f'(1) = 2 \cdot 1 = 2. $
(b) Right-hand derivative (for $x>1$):
$ f(x) = 2x + 1, $
so
$ f'(x) = 2. $
At $ x=1 $,
$ f'(1) = 2. $
Both one-sided derivatives at $x=1$ match (each is $2$), so $f$ is differentiable at $x=1$.
Step 6: Determine the number of non-differentiable points ($m$)
We have checked $x=0$ and $x=1$, and $f$ is differentiable at both points.
There are no other transition points since the function is otherwise defined
smoothly in each region. Hence,
$ m = 0. $
Step 7: Find $m + a + b + c$
From our deductions, $a = 1$, $b = 1$, $c = 0$, and $m = 0$. Thus,
$ m + a + b + c = 0 + 1 + 1 + 0 = 2. $
Therefore, the required value is $ \boxed{2}. $
