Key Concepts and Formulas

• Continuity: 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: A function $f(x)$ is differentiable at a point $x=c$ if the left-hand derivative (LHD) and the right-hand derivative (RHD) exist and are equal at $x=c$.
  • LHD at $x=c$: $f'(c^-) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}$
  • RHD at $x=c$: $f'(c^+) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$
• Relationship between Differentiability and Continuity: If a function is differentiable at a point, it must be continuous at that point.

Step-by-Step Solution

Step 1: Rewrite the function definition piecewise. The given function is f(x) = \left\{ {\matrix{ {{1 \over {|x|}}} & {;\,|x|\, \ge 1} \cr {a{x^2} + b} & {;\,|x|\, < 1} \cr } } \right.. We need to consider the absolute value $|x|$.

• If $|x| \ge 1$, then $x \le -1$ or $x \ge 1$. In this case, $|x| = x$ for $x \ge 1$ and $|x| = -x$ for $x \le -1$. So, $f(x) = \frac{1}{x}$ for $x \ge 1$ and $f(x) = \frac{1}{-x} = -\frac{1}{x}$ for $x \le -1$.
• If $|x| < 1$, then $-1 < x < 1$. In this case, $f(x) = ax^2 + b$.

Thus, the function can be rewritten as: f(x) = \left\{ {\matrix{ { - {1 \over x};} & {x \le - 1} \cr {a{x^2} + b;} & { - 1 < x < 1} \cr {{1 \over x};} & {x \ge 1} \cr } } \right.

Step 2: Use the continuity condition at x = 1. Since the function is differentiable at every point, it must be continuous at $x=1$. For continuity at $x=1$, the left-hand limit, the right-hand limit, and the function value must be equal.

• Left-hand limit at $x=1$: $\mathop {\lim }\limits_{x \to {1^- }} f(x) = \mathop {\lim }\limits_{x \to {1^- }} (ax^2 + b) = a(1)^2 + b = a+b$.
• Right-hand limit at $x=1$: $\mathop {\lim }\limits_{x \to {1^+ }} f(x) = \mathop {\lim }\limits_{x \to {1^+ }} \frac{1}{x} = \frac{1}{1} = 1$.
• Function value at $x=1$: $f(1) = \frac{1}{1} = 1$.

Equating the left-hand limit and the right-hand limit (or function value): $a+b = 1$ (Equation 1)

Step 3: Use the continuity condition at x = -1. The function must also be continuous at $x=-1$.

• Left-hand limit at $x=-1$: $\mathop {\lim }\limits_{x \to {-1^- }} f(x) = \mathop {\lim }\limits_{x \to {-1^- }} (-\frac{1}{x}) = -\frac{1}{-1} = 1$.
• Right-hand limit at $x=-1$: $\mathop {\lim }\limits_{x \to {-1^+ }} f(x) = \mathop {\lim }\limits_{x \to {-1^+ }} (ax^2 + b) = a(-1)^2 + b = a+b$.
• Function value at $x=-1$: $f(-1) = -\frac{1}{-1} = 1$.

Equating the limits: $1 = a+b$ This gives us the same equation as Equation 1, so it doesn't provide new information about $a$ and $b$ individually, but confirms consistency.

Step 4: Use the differentiability condition at x = 1. Since the function is differentiable at $x=1$, the left-hand derivative (LHD) must equal the right-hand derivative (RHD) at $x=1$.

First, let's find the derivatives of the respective pieces:

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

Now, we evaluate the derivatives at $x=1$:

• Right-hand derivative (RHD) at $x=1$: $f'(1^+) = \mathop {\lim }\limits_{x \to {1^+ }} f'(x) = \mathop {\lim }\limits_{x \to {1^+ }} \left(-\frac{1}{x^2}\right) = -\frac{1}{(1)^2} = -1$.

• Left-hand derivative (LHD) at $x=1$: $f'(1^-) = \mathop {\lim }\limits_{x \to {1^- }} f'(x) = \mathop {\lim }\limits_{x \to {1^- }} (2ax) = 2a(1) = 2a$.

Equating LHD and RHD at $x=1$: $2a = -1$ $a = -\frac{1}{2}$

Step 5: Use the differentiability condition at x = -1. The function must also be differentiable at $x=-1$.

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

Now, we evaluate the derivatives at $x=-1$:

• Left-hand derivative (LHD) at $x=-1$: $f'(-1^-) = \mathop {\lim }\limits_{x \to {-1^- }} f'(x) = \mathop {\lim }\limits_{x \to {-1^- }} \left(\frac{1}{x^2}\right) = \frac{1}{(-1)^2} = 1$.

• Right-hand derivative (RHD) at $x=-1$: $f'(-1^+) = \mathop {\lim }\limits_{x \to {-1^+ }} f'(x) = \mathop {\lim }\limits_{x \to {-1^+ }} (2ax) = 2a(-1) = -2a$.

Equating LHD and RHD at $x=-1$: $1 = -2a$ $a = -\frac{1}{2}$ This confirms the value of $a$ we found from the condition at $x=1$.

Step 6: Solve for b using Equation 1. We found $a = -\frac{1}{2}$ and from Equation 1, we have $a+b=1$. Substitute the value of $a$: $-\frac{1}{2} + b = 1$ $b = 1 + \frac{1}{2}$ $b = \frac{3}{2}$

Therefore, the values of $a$ and $b$ are $-\frac{1}{2}$ and $\frac{3}{2}$ respectively.

Common Mistakes & Tips

• Forgetting the absolute value: Properly defining the function for $x \le -1$, $-1 < x < 1$, and $x \ge 1$ is crucial.
• Confusing continuity and differentiability conditions: Ensure you use the limit equality for continuity and the derivative equality for differentiability.
• Errors in differentiation: Double-check the derivatives of the piecewise functions, especially with negative signs and powers.
• Algebraic errors: Be meticulous with arithmetic, especially when dealing with fractions and negative numbers.

Summary

To ensure the function is differentiable everywhere, we first enforced continuity at the points where the definition changes ($x=1$ and $x=-1$). This gave us an equation relating $a$ and $b$. Then, we used the condition that the left-hand derivative must equal the right-hand derivative at these points. By calculating the derivatives of the respective pieces and evaluating them at $x=1$ (or $x=-1$), we obtained the value of $a$. Finally, we substituted the value of $a$ back into the continuity equation to find the value of $b$.

The final answer is $a = -\frac{1}{2}$ and $b = \frac{3}{2}$.

The final answer is $\boxed{\text{-1/2, 3/2}}$.