Key Concepts and Formulas

• Continuity: A function $f(x)$ is continuous at a point $x=a$ if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$.
• Differentiability: A function $f(x)$ is differentiable at a point $x=a$ if the limit of the difference quotient exists, i.e., $\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists. This implies that the left-hand derivative (LHD) must be equal to the right-hand derivative (RHD) at that point.
  • LHD at $x=a$: $\lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h}$
  • RHD at $x=a$: $\lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}$
• Taylor Series Expansion: For small $x$, $\cos x \approx 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$.

Step-by-Step Solution

The problem states that the function $f(x)$ is continuous everywhere in $\mathbf{R}$. We need to find the values of $a$, $b$, and $c$ that ensure continuity at the points where the definition of $f(x)$ changes, which are $x=0$ and $x=1$. Then, we need to determine the number of points, $m$, where $f(x)$ is not differentiable. Finally, we need to calculate $m+a+b+c$.

Step 1: Ensure continuity at $x=1$. For $f(x)$ to be continuous at $x=1$, we must have $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$.

The function definition for $0 \leq x \leq 1$ is $f(x) = x^2 + cx + 2$. So, $f(1) = 1^2 + c(1) + 2 = 1 + c + 2 = 3+c$.

The function definition for $x > 1$ is $f(x) = 2x+1$. So, $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x+1) = 2(1)+1 = 3$.

For continuity at $x=1$, we set $f(1) = \lim_{x \to 1^+} f(x)$: $3+c = 3$ $c = 0$.

Step 2: Ensure continuity at $x=0$. For $f(x)$ to be continuous at $x=0$, we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.

The function definition for $0 \leq x \leq 1$ is $f(x) = x^2 + cx + 2$. Since we found $c=0$, this becomes $f(x) = x^2 + 2$ for $0 \leq x \leq 1$. So, $f(0) = 0^2 + 2 = 2$. And, $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2+2) = 0^2+2 = 2$.

The function definition for $x < 0$ is $f(x) = \frac{a-b \cos 2x}{x^2}$. So, $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{a-b \cos 2x}{x^2}$.

For continuity at $x=0$, we must have $\lim_{x \to 0^-} f(x) = f(0) = 2$. Thus, $\lim_{x \to 0^-} \frac{a-b \cos 2x}{x^2} = 2$.

To evaluate this limit, we use the Taylor series expansion of $\cos(2x)$ around $x=0$: $\cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots$ So, $\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \dots = 1 - \frac{4x^2}{2} + \frac{16x^4}{24} - \dots = 1 - 2x^2 + \frac{2}{3}x^4 - \dots$

Substitute this into the limit: $\lim_{x \to 0} \frac{a-b(1 - 2x^2 + \frac{2}{3}x^4 - \dots)}{x^2} = \lim_{x \to 0} \frac{a-b + 2bx^2 - \frac{2b}{3}x^4 + \dots}{x^2}$

For this limit to exist (and be finite), the numerator must approach 0 as $x \to 0$. This means the constant term in the numerator must be zero. So, $a-b = 0 \implies a=b$.

Now, substitute $a=b$ back into the limit expression: $\lim_{x \to 0} \frac{b-b + 2bx^2 - \frac{2b}{3}x^4 + \dots}{x^2} = \lim_{x \to 0} \frac{2bx^2 - \frac{2b}{3}x^4 + \dots}{x^2}$ $= \lim_{x \to 0} (2b - \frac{2b}{3}x^2 + \dots)$ $= 2b$.

For continuity at $x=0$, this limit must be equal to $f(0)=2$. So, $2b = 2 \implies b=1$. Since $a=b$, we also have $a=1$.

Thus, we have found $a=1$, $b=1$, and $c=0$.

Step 3: Determine differentiability at $x=0$. Now we need to find the number of points where $f(x)$ is NOT differentiable. We check the points where the definition changes, i.e., $x=0$ and $x=1$.

For $x<0$, $f(x) = \frac{1-\cos 2x}{x^2}$. For $0 \leq x \leq 1$, $f(x) = x^2 + 0x + 2 = x^2+2$. For $x>1$, $f(x) = 2x+1$.

Let's check differentiability at $x=0$. We need to compare the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x=0$.

• RHD at $x=0$: Using the definition of the derivative: RHD $= \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h}$ Since $h>0$, $f(0+h) = f(h) = h^2+2$. And $f(0)=2$. RHD $= \lim_{h \to 0^+} \frac{(h^2+2) - 2}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} h = 0$.

• LHD at $x=0$: Using the definition of the derivative: LHD $= \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h}$ Since $h<0$, $f(0+h) = f(h) = \frac{1-\cos(2h)}{h^2}$. And $f(0)=2$. LHD $= \lim_{h \to 0^-} \frac{\frac{1-\cos(2h)}{h^2} - 2}{h}$ LHD $= \lim_{h \to 0^-} \frac{1-\cos(2h) - 2h^2}{h^3}$

  Using the Taylor expansion $\cos(2h) = 1 - 2h^2 + \frac{2}{3}h^4 - \dots$: LHD $= \lim_{h \to 0^-} \frac{1-(1 - 2h^2 + \frac{2}{3}h^4 - \dots) - 2h^2}{h^3}$ LHD $= \lim_{h \to 0^-} \frac{1 - 1 + 2h^2 - \frac{2}{3}h^4 + \dots - 2h^2}{h^3}$ LHD $= \lim_{h \to 0^-} \frac{- \frac{2}{3}h^4 + \dots}{h^3}$ LHD $= \lim_{h \to 0^-} (-\frac{2}{3}h + \dots) = 0$.

Since LHD = RHD = 0 at $x=0$, the function $f(x)$ is differentiable at $x=0$.

Step 4: Determine differentiability at $x=1$. Now let's check differentiability at $x=1$. We need to compare the LHD and RHD at $x=1$.

For $0 \leq x \leq 1$, $f(x) = x^2+2$. The derivative is $f'(x) = 2x$. So, the derivative from the left at $x=1$ is $f'(1^-) = 2(1) = 2$.

For $x>1$, $f(x) = 2x+1$. The derivative is $f'(x) = 2$. So, the derivative from the right at $x=1$ is $f'(1^+) = 2$.

Alternatively, using the definition of the derivative:

• LHD at $x=1$: LHD $= \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}$ Since $h<0$, $1+h < 1$. So, $f(1+h) = (1+h)^2 + 2 = 1 + 2h + h^2 + 2 = 3 + 2h + h^2$. We know $f(1) = 3+c = 3+0 = 3$. LHD $= \lim_{h \to 0^-} \frac{(3 + 2h + h^2) - 3}{h} = \lim_{h \to 0^-} \frac{2h + h^2}{h} = \lim_{h \to 0^-} (2+h) = 2$.

• RHD at $x=1$: RHD $= \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}$ Since $h>0$, $1+h > 1$. So, $f(1+h) = 2(1+h)+1 = 2+2h+1 = 3+2h$. We know $f(1) = 3$. RHD $= \lim_{h \to 0^+} \frac{(3+2h) - 3}{h} = \lim_{h \to 0^+} \frac{2h}{h} = \lim_{h \to 0^+} 2 = 2$.

Since LHD = RHD = 2 at $x=1$, the function $f(x)$ is differentiable at $x=1$.

Step 5: Determine the number of points where $f$ is NOT differentiable. We have checked the points $x=0$ and $x=1$. At both these points, the function is differentiable. For $x<0$, $f(x) = \frac{1-\cos 2x}{x^2}$. This function is differentiable for all $x<0$ because $\cos(2x)$ is differentiable, and the denominator $x^2$ is non-zero and differentiable. For $0<x<1$, $f(x) = x^2+2$. This is a polynomial and is differentiable everywhere. For $x>1$, $f(x) = 2x+1$. This is a linear function and is differentiable everywhere.

Therefore, there are no points where $f(x)$ is not differentiable. So, $m=0$.

Step 6: Calculate $m+a+b+c$. We found $m=0$, $a=1$, $b=1$, and $c=0$. $m+a+b+c = 0 + 1 + 1 + 0 = 2$.

Common Mistakes & Tips

• Incorrect Taylor Expansion: Ensure the Taylor expansion of $\cos(2x)$ is accurate and that terms are correctly handled when substituting into the limit.
• Forgetting Continuity Check: Always ensure continuity at the junction points before checking differentiability. A function cannot be differentiable if it's not continuous.
• Algebraic Errors in Limits: Be meticulous with algebraic manipulations, especially when dealing with subtractions and divisions in limit calculations.

Summary

The problem required us to find the values of constants $a$, $b$, and $c$ by ensuring the continuity of the piecewise function $f(x)$ at the points where its definition changes ($x=0$ and $x=1$). We used the condition $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$. After finding $a=1$, $b=1$, and $c=0$, we then checked the differentiability of $f(x)$ at $x=0$ and $x=1$ by comparing the left-hand and right-hand derivatives. We found that $f(x)$ is differentiable at both $x=0$ and $x=1$. Since the individual pieces of the function are differentiable in their respective open intervals, there are no points where $f(x)$ is not differentiable, meaning $m=0$. Finally, we calculated $m+a+b+c = 0+1+1+0=2$.

The final answer is \boxed{2}.