Question

Let \[ f(x)= \begin{cases} |x|, & -\infty\lt x\lt 2 \\ |2x-4|, & 2\leq x\leq 20 \end{cases} \] \(x=a\) is a point where \(f(x)\) is continuous but not differentiable and \(x=b\) is a point where \(f(x)\) is not differentiable \((a\neq b)\). Then \(a+b=\)

• A. \(1\)
• B. \(2\)
• C. \(-2\)
• D. \(0\)

Hint:

Modulus functions are generally not differentiable where the expression inside modulus becomes zero. Always check continuity separately at piecewise junction points.

Answer 1

The Correct Option is B

Step 1: Write out the piecewise function.
\[f(x) = \begin{cases} |x|, & -\infty < x < 2 \\ |2x-4|, & 2 \leq x \leq 20 \end{cases}\] We need $x = a$ where $f$ is continuous but not differentiable, and $x = b$ where $f$ is not differentiable, with $a \neq b$.
Step 2: Identify the non-differentiable point of $|x|$ in $(-\infty, 2)$.
The function $|x|$ has a sharp corner at $x = 0$. Left derivative: $\lim_{h\to 0^-}\dfrac{|h|-0}{h} = -1$. Right derivative: $\lim_{h\to 0^+}\dfrac{|h|-0}{h} = +1$. These are unequal, so $f$ is not differentiable at $x = 0$. But $f$ IS continuous at $x = 0$ (since $|x| \to 0$ from both sides). Hence $a = 0$.
Step 3: Identify the corner of $|2x-4|$.
The expression $|2x-4|$ equals zero when $x = 2$. For $x > 2$: $|2x-4| = 2x-4$. For $x < 2$: $|2x-4| = 4-2x$. There is a corner at $x = 2$.
Step 4: Check continuity of $f$ at $x = 2$.
Left-hand limit (using $|x|$ piece): $\lim_{x\to 2^-}|x| = 2$. Right-hand value (using $|2x-4|$ piece): $f(2) = |4-4| = 0$. Since the left-hand limit $(2) \neq f(2) = 0$, the function $f$ is discontinuous at $x = 2$. A function that is discontinuous at a point is definitely not differentiable there. So $b = 2$.
Step 5: Confirm $a \neq b$.
$a = 0$ and $b = 2$, which are distinct. The roles are different: at $x=0$ the function is continuous but not differentiable; at $x=2$ the function is not even continuous.
Step 6: Compute $a + b$.
\[a + b = 0 + 2 = 2\]
Step 7: State the final answer.
\[ \boxed{2} \]