Question

Let $f(x)=\begin{cases}1-x,&x<1\\ (1-x)(2-x),& 1\le x\le2\end{cases}$. Which one of the following is not true?

• A. $f(x)$ is differentiable in (-1,1)
• B. $f(x)$ is differentiable in $(-\infty,1)$
• C. $f(x)$ is differentiable at $x=-2026$
• D. $f(x)$ is not differentiable at $x=1$
• E. $f(x)$ is continuous at $x=1.5$ and differentiable at $x=1.5$

Hint:

Math Tip: Always check continuity before checking differentiability! If a piecewise function is not continuous at a point, it is automatically not differentiable there, saving you from calculating limits of difference quotients.

Answer 1

The Correct Option is D

Concept:
Calculus - Continuity and Differentiability of Piecewise Functions.
A function is differentiable at $x=c$ if its Left Hand Derivative (LHD) equals its Right Hand Derivative (RHD), and it is continuous at that point.
Step 1: Check for continuity at the transition point $x=1$.
Calculate the Left Hand Limit (LHL) and Right Hand Limit (RHL) at $x=1$:

• $\text{LHL} = \lim_{x\to1^-} (1-x) = 1 - 1 = 0$
• $\text{RHL} = \lim_{x\to1^+} (1-x)(2-x) = (1-1)(2-1) = 0$
• $f(1) = (1-1)(2-1) = 0$

Since $\text{LHL} = \text{RHL} = f(1)$, the function is continuous at $x=1$.
Step 2: Find the derivatives of the two pieces.
Differentiate each piece of the function with respect to $x$:

• For $x<1$: $\frac{d}{dx}(1-x) = -1$
• For $1 \le x \le 2$: Expand the function first: $(1-x)(2-x) = x^2 - 3x + 2$.
  Then differentiate: $\frac{d}{dx}(x^2 - 3x + 2) = 2x - 3$

Step 3: Evaluate LHD and RHD at $x=1$.
Substitute $x=1$ into the respective derivative expressions:

• $\text{LHD at } x=1 \text{ is } -1$
• $\text{RHD at } x=1 \text{ is } 2(1) - 3 = -1$

Step 4: Determine the differentiability at $x=1$.
Since $\text{LHD} = \text{RHD} = -1$, the function $f(x)$ is differentiable at $x=1$.
Step 5: Analyze the given options.
We are looking for the statement that is not true (false):

• (A), (B), (C): The function is a simple polynomial $1-x$ for $x<1$, so it is continuous and differentiable everywhere in that region (including $-2026$ and $(-1, 1)$). These are true.
• (E): At $x=1.5$, it is a simple quadratic polynomial, which is continuous and differentiable. This is true.
• (D): States $f(x)$ is not differentiable at $x=1$. Our Step 4 proves this statement is false.