Key Concepts and Formulas

• Continuity of a function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
• Differentiability of a function: A function $f(x)$ is differentiable at a point $x=c$ if the left-hand derivative (LHD) and the right-hand derivative (RHD) are equal at that point.
  • LHD at $x=c$: $f'(c^-) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$
  • RHD at $x=c$: $f'(c^+) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$
  • For a function defined piecewise, differentiability at the point where the definition changes implies continuity at that point, and the derivatives of the respective pieces must be equal when evaluated at that point.

Step-by-Step Solution

The function is given by: $f(x)= \begin{cases}\frac{1}{|x|}, & |x| \geqslant 2 \\ \mathrm{a} x^2+2 \mathrm{~b}, & |x|<2\end{cases}$

First, let's rewrite the function explicitly for different intervals:

• If $x \ge 2$, then $|x| = x$, so $f(x) = \frac{1}{x}$.
• If $x \le -2$, then $|x| = -x$, so $f(x) = \frac{1}{-x} = -\frac{1}{x}$.
• If $-2 < x < 2$, then $|x| < 2$, so $f(x) = ax^2 + 2b$.

So, the piecewise definition is: $f(x)=\begin{cases} -\frac{1}{x}, & x \leq -2 \\ ax^2+2 \mathrm{~b}, & -2<x<2 \\ \frac{1}{x}, & x \geq 2 \end{cases}$

Step 1: Ensure Continuity at the points where the definition changes. For the function to be differentiable on $\mathbb{R}$, it must first be continuous on $\mathbb{R}$. We need to check continuity at $x=2$ and $x=-2$.

• Continuity at $x=2$: We must have $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$. $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (ax^2 + 2b) = a(2)^2 + 2b = 4a + 2b$. $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{1}{x} = \frac{1}{2}$. $f(2) = \frac{1}{2}$. Equating these gives: $4a + 2b = \frac{1}{2}$. (Equation 1)

• Continuity at $x=-2$: We must have $\lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) = f(-2)$. $\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} -\frac{1}{x} = -\frac{1}{-2} = \frac{1}{2}$. $\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} (ax^2 + 2b) = a(-2)^2 + 2b = 4a + 2b$. $f(-2) = -\frac{1}{-2} = \frac{1}{2}$. Equating these gives: $4a + 2b = \frac{1}{2}$. (Equation 2)

Notice that Equation 1 and Equation 2 are identical, which is expected because the function behaves symmetrically around $x=0$ due to the $|x|$ term.

Step 2: Ensure Differentiability at the points where the definition changes. For the function to be differentiable at $x=2$ and $x=-2$, the left-hand derivative (LHD) must equal the right-hand derivative (RHD) at these points.

First, let's find the derivative of each piece of the function:

• For $x < -2$, $f(x) = -\frac{1}{x} = -x^{-1}$. So, $f'(x) = -(-1)x^{-2} = \frac{1}{x^2}$.
• For $-2 < x < 2$, $f(x) = ax^2 + 2b$. So, $f'(x) = 2ax$.
• For $x > 2$, $f(x) = \frac{1}{x} = x^{-1}$. So, $f'(x) = -x^{-2} = -\frac{1}{x^2}$.

Now, let's check differentiability at $x=2$:

• RHD at $x=2$: This is the derivative of $f(x) = \frac{1}{x}$ evaluated at $x=2$. $f'(2^+) = -\frac{1}{2^2} = -\frac{1}{4}$.
• LHD at $x=2$: This is the derivative of $f(x) = ax^2 + 2b$ evaluated at $x=2$. $f'(2^-) = 2a(2) = 4a$. For differentiability at $x=2$, we must have $f'(2^-) = f'(2^+)$. So, $4a = -\frac{1}{4}$.

Step 3: Solve for the coefficients 'a' and 'b'. From the differentiability condition at $x=2$, we found: $4a = -\frac{1}{4}$ $a = -\frac{1}{16}$.

Now, substitute this value of $a$ into Equation 1 (or Equation 2) to find $b$: $4a + 2b = \frac{1}{2}$ $4\left(-\frac{1}{16}\right) + 2b = \frac{1}{2}$ $-\frac{4}{16} + 2b = \frac{1}{2}$ $-\frac{1}{4} + 2b = \frac{1}{2}$ $2b = \frac{1}{2} + \frac{1}{4}$ $2b = \frac{2}{4} + \frac{1}{4}$ $2b = \frac{3}{4}$ $b = \frac{3}{8}$.

Step 4: Calculate the required expression. We need to find the value of $48(a+b)$. $a+b = -\frac{1}{16} + \frac{3}{8}$ To add these fractions, find a common denominator, which is 16. $a+b = -\frac{1}{16} + \frac{3 \times 2}{8 \times 2}$ $a+b = -\frac{1}{16} + \frac{6}{16}$ $a+b = \frac{-1+6}{16}$ $a+b = \frac{5}{16}$.

Now, calculate $48(a+b)$: $48(a+b) = 48 \times \frac{5}{16}$ $48(a+b) = \frac{48}{16} \times 5$ $48(a+b) = 3 \times 5$ $48(a+b) = 15$.

Let's recheck the given solution. It states $a = -1/16$ and $b = 3/8$. Then $48(a+b) = 48(-1/16 + 3/8) = 48(-1/16 + 6/16) = 48(5/16) = 3 \times 5 = 15$.

There seems to be a discrepancy between my calculated answer (15) and the provided "Correct Answer" (1). Let's re-examine the problem statement and my steps carefully.

The provided solution states: "Continuous at $x=2 \Rightarrow \frac{1}{2}=\frac{a}{4}+2 b$" - This is incorrect. It should be $a(2)^2 + 2b = 4a+2b$. The solution wrote $a/4$. "Continuous at $x=-2 \Rightarrow \frac{1}{2}=\frac{a}{4}+2 b$" - Same error.

Let's correct the continuity equation: Continuity at $x=2$: $\lim_{x \to 2^-} (ax^2 + 2b) = \lim_{x \to 2^+} \frac{1}{x}$ $a(2)^2 + 2b = \frac{1}{2}$ $4a + 2b = \frac{1}{2}$ (This matches my derivation).

Let's check the derivative part from the given solution: "Since, it is differentiable at $x=2$ $-\frac{1}{x^2}=2ax$" - This is where the derivatives are evaluated. The derivative of $1/x$ is $-1/x^2$. The derivative of $ax^2+2b$ is $2ax$. At $x=2$: RHD (from $1/x$): $f'(2^+) = -\frac{1}{2^2} = -\frac{1}{4}$. LHD (from $ax^2+2b$): $f'(2^-) = 2a(2) = 4a$. So $4a = -1/4 \Rightarrow a = -1/16$. (This also matches my derivation).

The solution then states: "$a=\frac{-1}{16}, b =\frac{3}{8}$". Let's verify if these values satisfy $4a+2b = 1/2$. $4(-\frac{1}{16}) + 2(\frac{3}{8}) = -\frac{4}{16} + \frac{6}{8} = -\frac{1}{4} + \frac{3}{4} = \frac{2}{4} = \frac{1}{2}$. So, the values of $a$ and $b$ are indeed correct.

Now, let's re-calculate $48(a+b)$. $a+b = -\frac{1}{16} + \frac{3}{8} = -\frac{1}{16} + \frac{6}{16} = \frac{5}{16}$. $48(a+b) = 48 \times \frac{5}{16} = 3 \times 5 = 15$.

It seems there might be an error in the provided "Correct Answer". Let's assume my calculations are correct and proceed.

If the question was asking for $48(b-a)$ instead: $b-a = \frac{3}{8} - (-\frac{1}{16}) = \frac{6}{16} + \frac{1}{16} = \frac{7}{16}$. $48(\frac{7}{16}) = 3 \times 7 = 21$.

If the question was asking for $48a+48b$: this is $48(a+b)$ which is 15.

Let's carefully re-read the question and the function definition. $f(x)= \begin{cases}\frac{1}{|x|}, & |x| \geqslant 2 \\ \mathrm{a} x^2+2 \mathrm{~b}, & |x|<2\end{cases}$

Let's assume the provided correct answer of 1 is indeed correct. This would imply that $a+b = 1/48$. If $a+b = 1/48$, and $a=-1/16$, then $b = 1/48 - (-1/16) = 1/48 + 3/48 = 4/48 = 1/12$. Let's check if $a=-1/16$ and $b=1/12$ satisfy the continuity condition $4a+2b=1/2$. $4(-\frac{1}{16}) + 2(\frac{1}{12}) = -\frac{1}{4} + \frac{1}{6} = -\frac{3}{12} + \frac{2}{12} = -\frac{1}{12} \neq \frac{1}{2}$. So, $a+b=1/48$ is not correct.

Let's look at the original solution again. It says: "Continuous at $x=2 \Rightarrow \frac{1}{2}=\frac{a}{4}+2 \mathrm{~b}$" This means they might have interpreted $ax^2$ as $a(x/2)^2$ or some other misinterpretation.

Let's strictly follow my derivation which uses the standard definitions of continuity and differentiability. We have $a = -\frac{1}{16}$ and $b = \frac{3}{8}$. Then $a+b = \frac{5}{16}$. And $48(a+b) = 48 \times \frac{5}{16} = 15$.

Given that the provided correct answer is 1, and my derivation consistently leads to 15, there is a high probability of an error in the provided "Correct Answer". However, I must adhere to the rule of reaching the provided answer.

Let's consider if there's any edge case or interpretation I'm missing. The function is piecewise defined. Differentiability on $\mathbb{R}$ implies continuity and equality of derivatives at the junction points. The junction points are $x=2$ and $x=-2$.

The function is $f(x)= \begin{cases}\frac{1}{x}, & x \geq 2 \\ ax^2+2 \mathrm{~b}, & -2<x<2 \\ -\frac{1}{x}, & x \leq -2 \end{cases}$

Continuity at $x=2$: $\lim_{x \to 2^-} (ax^2+2b) = f(2) \implies 4a+2b = 1/2$. Continuity at $x=-2$: $\lim_{x \to -2^+} (ax^2+2b) = f(-2) \implies 4a+2b = 1/2$.

Differentiability at $x=2$: LHD: Derivative of $ax^2+2b$ at $x=2$ is $2ax|_{x=2} = 4a$. RHD: Derivative of $1/x$ at $x=2$ is $-1/x^2|_{x=2} = -1/4$. So, $4a = -1/4 \implies a = -1/16$.

Differentiability at $x=-2$: LHD: Derivative of $-1/x$ at $x=-2$ is $1/x^2|_{x=-2} = 1/(-2)^2 = 1/4$. RHD: Derivative of $ax^2+2b$ at $x=-2$ is $2ax|_{x=-2} = 2a(-2) = -4a$. So, $1/4 = -4a \implies a = -1/16$. This confirms $a = -1/16$ from both points.

Substitute $a=-1/16$ into $4a+2b=1/2$: $4(-1/16) + 2b = 1/2$ $-1/4 + 2b = 1/2$ $2b = 1/2 + 1/4 = 3/4$ $b = 3/8$.

So, $a = -1/16$ and $b = 3/8$. $a+b = -1/16 + 3/8 = -1/16 + 6/16 = 5/16$. $48(a+b) = 48(5/16) = 3 \times 5 = 15$.

Given the instruction to reach the correct answer, and the discrepancy, I must assume there's a misunderstanding of the question or a typo in the provided answer. However, if forced to produce an answer of 1, I would have to reverse-engineer it, which is not a valid mathematical approach for a step-by-step solution.

Let's assume the question meant $48(a+b/2)$ or some other combination. If $48(a+b/2) = 1$: $a+b/2 = 1/48$. $-1/16 + (3/8)/2 = -1/16 + 3/16 = 2/16 = 1/8$. $1/8 \neq 1/48$.

Let's consider the possibility of a typo in the function definition itself. If $f(x) = ax^2 + b$ instead of $ax^2 + 2b$. Continuity: $4a+b = 1/2$. Differentiability: $4a = -1/4 \implies a = -1/16$. Substitute $a$: $4(-1/16) + b = 1/2 \implies -1/4 + b = 1/2 \implies b = 3/4$. Then $a+b = -1/16 + 3/4 = -1/16 + 12/16 = 11/16$. $48(11/16) = 3 \times 11 = 33$.

If the function was $f(x) = ax^2 + b$ and the question was $48(a+b/4)$: $a+b/4 = -1/16 + (3/4)/4 = -1/16 + 3/16 = 2/16 = 1/8$. $48(1/8) = 6$.

Let's go back to the original solution's calculation: "$a=\frac{-1}{16}, b =\frac{3}{8}$". And it states the answer is 1. If $48(a+b)=1$, then $a+b=1/48$. We know $a=-1/16$. So, $-1/16 + b = 1/48$. $b = 1/48 + 1/16 = 1/48 + 3/48 = 4/48 = 1/12$. If $a=-1/16$ and $b=1/12$, let's check continuity: $4a+2b = 4(-1/16) + 2(1/12) = -1/4 + 1/6 = -3/12 + 2/12 = -1/12$. This should be $1/2$. So this is not correct.

There seems to be a fundamental issue with the provided "Correct Answer" or the question itself as stated. Assuming my derivation of $a=-1/16$ and $b=3/8$ is correct (which it appears to be based on standard calculus principles), the answer to $48(a+b)$ is 15.

Since I am required to provide a solution that leads to the given correct answer, and I cannot logically derive it, I will assume there is a typo in the question or the answer key. However, I must present the steps as if the answer were correct. This puts me in a contradictory situation.

Let's assume for the sake of reaching the answer '1' that there's a relationship that leads to $a+b = 1/48$. If $a+b = 1/48$, and we know $a=-1/16$, then $b = 1/48 - (-1/16) = 1/48 + 3/48 = 4/48 = 1/12$. Let's check if $a=-1/16$ and $b=1/12$ are consistent with continuity and differentiability conditions. Continuity at $x=2$: $4a+2b = 4(-1/16) + 2(1/12) = -1/4 + 1/6 = -3/12 + 2/12 = -1/12$. This should be equal to $1/2$. It is not.

Therefore, it is impossible to rigorously derive the answer 1 from the given problem statement and standard mathematical principles. The most likely scenario is an error in the provided correct answer.

However, if I must present a derivation that results in 1, I would have to hypothesize an error in the problem or a non-standard interpretation. Since I cannot do that, I will stick to the derived values.

Let's assume there was a typo in the expression to be calculated. For example, if it was $48(a+b/6)$ to get 1. $48(a+b/6) = 48(-1/16 + (3/8)/6) = 48(-1/16 + 3/48) = 48(-1/16 + 1/16) = 48(0) = 0$. Not 1.

Let's assume the question meant $48(a/4 + b)$. $48(a/4 + b) = 48((-1/16)/4 + 3/8) = 48(-1/64 + 24/64) = 48(23/64) = (3 \times 16) \times (23/(4 \times 16)) = 3 \times 23 / 4 = 69/4$.

Given the constraints, I cannot produce a solution that reaches the answer '1' based on correct mathematical steps. My derivation leads to 15. I will present the derivation for 15, and acknowledge the discrepancy.

Step 1: Rewrite the function and establish continuity conditions. The function is: $f(x)=\begin{cases} -\frac{1}{x}, & x \leq -2 \\ ax^2+2 \mathrm{~b}, & -2<x<2 \\ \frac{1}{x}, & x \geq 2 \end{cases}$ For continuity at $x=2$: $\lim_{x \to 2^-} (ax^2+2b) = \lim_{x \to 2^+} \frac{1}{x}$. This gives $a(2)^2 + 2b = \frac{1}{2}$, so $4a + 2b = \frac{1}{2}$. (Equation 1) For continuity at $x=-2$: $\lim_{x \to -2^+} (ax^2+2b) = \lim_{x \to -2^-} -\frac{1}{x}$. This gives $a(-2)^2 + 2b = -\frac{1}{-2}$, so $4a + 2b = \frac{1}{2}$. (Equation 2) Both conditions yield the same equation, $4a + 2b = \frac{1}{2}$.

Step 2: Establish differentiability conditions. We find the derivatives of the pieces: For $x < -2$, $f'(x) = \frac{1}{x^2}$. For $-2 < x < 2$, $f'(x) = 2ax$. For $x > 2$, $f'(x) = -\frac{1}{x^2}$.

For differentiability at $x=2$: The left-hand derivative must equal the right-hand derivative. LHD at $x=2$: $f'(2^-) = 2a(2) = 4a$. RHD at $x=2$: $f'(2^+) = -\frac{1}{2^2} = -\frac{1}{4}$. Equating them: $4a = -\frac{1}{4}$.

Step 3: Solve for $a$ and $b$. From $4a = -\frac{1}{4}$, we get $a = -\frac{1}{16}$. Substitute $a$ into Equation 1: $4(-\frac{1}{16}) + 2b = \frac{1}{2}$. $-\frac{1}{4} + 2b = \frac{1}{2}$. $2b = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$. $b = \frac{3}{8}$.

Step 4: Calculate $48(a+b)$. $a+b = -\frac{1}{16} + \frac{3}{8} = -\frac{1}{16} + \frac{6}{16} = \frac{5}{16}$. $48(a+b) = 48 \times \frac{5}{16} = 3 \times 5 = 15$.

Given the provided correct answer is 1, and my derivation leads to 15, there is a discrepancy. However, based on the standard definitions of continuity and differentiability, the values of $a$ and $b$ are $a=-1/16$ and $b=3/8$, which results in $48(a+b)=15$. If the correct answer is indeed 1, there might be an error in the problem statement or the provided answer.

Assuming there is a typo in the question and the value to be calculated should lead to 1. Let's assume the question intended to ask for a value that results in 1. If $48(a+b) = 1$, then $a+b = 1/48$. With $a=-1/16$, $b = 1/48 - (-1/16) = 1/48 + 3/48 = 4/48 = 1/12$. If $a=-1/16$ and $b=1/12$, then $4a+2b = 4(-1/16) + 2(1/12) = -1/4 + 1/6 = -3/12 + 2/12 = -1/12$. This should be $1/2$ for continuity. So this is incorrect.

Given the constraints, I must assume there's an error in the problem's provided solution. The mathematically derived answer is 15. However, if I have to strictly reach the given answer of 1, I cannot provide a valid step-by-step derivation.

Let's assume there's a typo in the question, and it asked for $48(a+b)$ where $a$ and $b$ were different. If $a+b = 1/48$. Suppose the continuity equation was $4a+2b=1/2$ and the differentiability equation gave a different $a$. Or suppose the expression to calculate was different.

Since I cannot reconcile my derived answer with the provided answer, I will state the derived answer. However, if forced to select an option and the provided answer is 1, I cannot justify it mathematically.

Let's consider the possibility that the continuity equation was $a/4+2b = 1/2$ as in the faulty solution. $a/4 + 2b = 1/2$. We have $a=-1/16$. $(-1/16)/4 + 2b = 1/2$ $-1/64 + 2b = 1/2$ $2b = 1/2 + 1/64 = 32/64 + 1/64 = 33/64$. $b = 33/128$. Then $a+b = -1/16 + 33/128 = -8/128 + 33/128 = 25/128$. $48(a+b) = 48(25/128) = (3 \times 16) \times (25/(8 \times 16)) = 3 \times 25 / 8 = 75/8$.

This does not lead to 1 either.

Given the strict requirement to match the provided correct answer, and my inability to do so through valid mathematical steps, I must conclude there is an error in the problem statement or the provided answer. I cannot proceed to produce a solution that leads to the answer 1.

However, if this were an exam and the answer was stated as 1, I would suspect a typo in the question and try to find a way to get close to 1 by re-interpreting. But as a teacher, I must point out the error.

Final decision: I will present the correct derivation leading to 15. If the provided answer is indeed 1, then the problem is flawed.

Summary

The problem requires finding the values of constants $a$ and $b$ for a piecewise function to be differentiable on $\mathbb{R}$. This involves two main conditions at the points where the function definition changes ($x=2$ and $x=-2$): continuity and equality of left-hand and right-hand derivatives.

First, we ensured continuity by equating the limits of the function from both sides at $x=2$ and $x=-2$. This resulted in the equation $4a + 2b = \frac{1}{2}$.

Next, we ensured differentiability by equating the derivatives of the respective function pieces at $x=2$ and $x=-2$. This yielded $4a = -\frac{1}{4}$ (from $x=2$) and $\frac{1}{4} = -4a$ (from $x=-2$), both consistently giving $a = -\frac{1}{16}$.

Substituting $a = -\frac{1}{16}$ into the continuity equation $4a + 2b = \frac{1}{2}$, we found $b = \frac{3}{8}$.

Finally, we calculated the required expression $48(a+b)$. Substituting the values of $a$ and $b$, we got $a+b = -\frac{1}{16} + \frac{3}{8} = \frac{5}{16}$. Therefore, $48(a+b) = 48 \times \frac{5}{16} = 15$.

There appears to be a discrepancy between the derived answer (15) and the provided correct answer (1). Based on standard calculus principles, the derived answer of 15 is correct.

Final Answer

The final answer is $\boxed{15}$.