Question

If function defined by f(x) = $\begin{cases} \frac{(x-m)}{|x-m|}, & x \leq 0 \\ 2x^2 + 3ax + b, & 0 < x < 1, \text{ is continuous \& differentiable} \\ m^2x + b - 2, & x \geq 1 \end{cases}$

everywhere, then:

• A. b + m = -1
• B. b + m = 1
• C. b + m = -3
• D. m² + a + b = 3

Answer 1

Answer: (B), (D)

b + m = 1, m² + a + b = 3

Answer 2

The given function is: $f(x) = \begin{cases} \frac{(x-m)}{|x-m|}, & x \leq 0 \\ 2x^2 + 3ax + b, & 0 < x < 1 \\ m^2x + b - 2, & x \geq 1 \end{cases}$

The function is continuous and differentiable everywhere.

Step 1: Analyze the first piece of the function for $x \leq 0$.

For $f(x)$ to be defined for all $x \leq 0$, the denominator $|x-m|$ must not be zero. This means $x \neq m$ for any $x \leq 0$.
Therefore, $m$ cannot be in the interval $(-\infty, 0]$. This implies $m > 0$.
If $m > 0$, then for any $x \leq 0$, $x-m$ will always be negative.
So, $|x-m| = -(x-m)$.
Thus, $f(x) = \frac{x-m}{-(x-m)} = -1$ for all $x \leq 0$.
This part of the function is a constant, so it is continuous and differentiable for $x < 0$, and its derivative is $0$.

Now the function can be rewritten as: $f(x) = \begin{cases} -1, & x \leq 0 \\ 2x^2 + 3ax + b, & 0 < x < 1 \\ m^2x + b - 2, & x \geq 1 \end{cases}$ And we have the condition $m > 0$.

Step 2: Apply continuity and differentiability conditions at $x=0$.

For continuity at $x=0$:
$\lim_{x \to 0^-} f(x) = f(0) = -1$
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x^2 + 3ax + b) = 2(0)^2 + 3a(0) + b = b$
For continuity, the left-hand limit, right-hand limit, and function value must be equal:
$b = -1$.

For differentiability at $x=0$:
First, find the derivatives of the respective pieces:
$f'(x) = \begin{cases} 0, & x < 0 \\ 4x + 3a, & 0 < x < 1 \\ m^2, & x > 1 \end{cases}$
Left-hand derivative at $x=0$: $f'(0^-) = 0$.
Right-hand derivative at $x=0$: $f'(0^+) = \lim_{x \to 0^+} (4x + 3a) = 4(0) + 3a = 3a$.
For differentiability, $f'(0^-) = f'(0^+)$:
$0 = 3a \implies a = 0$.

So far, we have $a=0$ and $b=-1$. The function becomes: $f(x) = \begin{cases} -1, & x \leq 0 \\ 2x^2 - 1, & 0 < x < 1 \\ m^2x - 1 - 2, & x \geq 1 \end{cases}$
$f(x) = \begin{cases} -1, & x \leq 0 \\ 2x^2 - 1, & 0 < x < 1 \\ m^2x - 3, & x \geq 1 \end{cases}$

Step 3: Apply continuity and differentiability conditions at $x=1$.

For continuity at $x=1$:
$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x^2 - 1) = 2(1)^2 - 1 = 1$.
$\lim_{x \to 1^+} f(x) = f(1) = m^2(1) - 3 = m^2 - 3$.
For continuity:
$1 = m^2 - 3 \implies m^2 = 4$.

For differentiability at $x=1$:
Left-hand derivative at $x=1$: $f'(1^-) = \lim_{x \to 1^-} (4x) = 4(1) = 4$.
Right-hand derivative at $x=1$: $f'(1^+) = \lim_{x \to 1^+} (m^2) = m^2$.
For differentiability:
$4 = m^2$.

Both continuity and differentiability conditions at $x=1$ give $m^2 = 4$.
Since we established $m > 0$, we take the positive root: $m = 2$.

Step 4: Summarize the values of constants and check the options.

The determined values are:
$a = 0$
$b = -1$
$m = 2$

Now, let's check the given options:
(A) $b + m = (-1) + 2 = 1$.
(B) $b + m = 1$. (This is true)
(C) $b + m = -3$. (This is false)
(D) $m^2 + a + b = (2)^2 + 0 + (-1) = 4 + 0 - 1 = 3$. (This is true)

Both options (B) and (D) are correct based on the derived values of $a, b, m$.