Question

\[ \text{Assertion (A): If } y = f(x) = (|x| - |x - 1|)^2, \text{ then } \left.\frac{dy}{dx}\right|_{x = 1} = 1 \] \[ \text{Reason (R): If } \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \text{ exists, then it is called the derivative of } f(x) \text{ at } x = a. \] Then:

• (A) is true, (R) is true, (R) is correct explanation to (A)
• (A) is true, (R) is true, (R) is not the correct explanation to (A)
• (A) is true, (R) is false
• (A) is false, (R) is true

Hint:

In assertion-reason questions, test each statement independently first. Then assess if the reason logically explains the assertion only if both are true.

Answer 1

The Correct Option is C

First, analyze \( y = (|x| - |x - 1|)^2 \). We break it into piecewise expressions.
Consider the intervals:
1. For \( x \ge 1 \):
\[ |x| = x, \quad |x - 1| = x - 1 \Rightarrow y = (x - (x - 1))^2 = 1^2 = 1 \] 2. For \( 0 \le x < 1 \):
\[ |x| = x, \quad |x - 1| = 1 - x \Rightarrow y = (x - (1 - x))^2 = (2x - 1)^2 \] 3. For \( x < 0 \):
\[ |x| = -x, \quad |x - 1| = 1 - x \Rightarrow y = (-x - (1 - x))^2 = (-1)^2 = 1 \] So, piecewise:
\[ f(x) = \begin{cases} 1, & x < 0 \quad \text{or } x > 1 \\ (2x - 1)^2, & 0 \le x < 1 \\ 1, & x = 1 \end{cases} \] Now compute derivative at \( x = 1 \) using one-sided limits:
From left:
\[ \lim_{x \to 1^-} f(x) = (2x - 1)^2 \Rightarrow \frac{d}{dx}[(2x - 1)^2] = 2 \cdot (2x - 1) \cdot 2 = 8x - 4 \Rightarrow \left. \frac{dy}{dx} \right|_{x \to 1^-} = 8(1) - 4 = 4 \] From right:
\[ f(x) = 1 \Rightarrow \left. \frac{dy}{dx} \right|_{x \to 1^+} = 0 \] Since left and right derivatives are not equal, derivative at \( x = 1 \) does not exist.
So:
- Assertion is false.
- The reason is a correct mathematical statement, but the assertion itself is wrong.
\[ \boxed{(4) (A) is false, (R) is true} \]