Question

Let \( f(x) = |x - 3| + |x + 5| \) and \( A = \left\{ a \in \mathbb{R} / \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \text{ exists} \right\} \). Then the number of real numbers which are in \( [-8, 3] \cup [5, \infty) \) but not in \( A \) is

• A. \( 2 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( 3 \)

Hint:

The derivative of \( |x - c| \) is \( \text{sgn}(x - c) \) for \( x \neq c \), and it does not exist at \( x = c \). For a function involving sums of absolute values, check for non-differentiability at the points where each absolute value term changes its sign. The set \( A \) contains all points where the derivative exists.

Answer 1

The Correct Option is C

The function \( f(x) = |x - 3| + |x + 5| \) is piecewise defined as: $$ f(x) = \begin{cases} -2x - 2, & x<-5
8, & -5 \le x<3
2x + 2, & x \ge 3 \end{cases} $$ The limit \( \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \) exists if \( f(x) \) is differentiable at \( x = a \).
Absolute value functions are not differentiable at the points where the expression inside the absolute value is zero.
Here, these points are \( x = 3 \) and \( x = -5 \).
Thus, the set \( A = \mathbb{R} \setminus \{ -5, 3 \} \).
We are interested in the number of elements in the set \( ([-8, 3] \cup [5, \infty)) \setminus A \).
The set \( [-8, 3] \cup [5, \infty) \) includes the interval from -8 to 3 (inclusive) and the interval from 5 to infinity (inclusive).
The points not in \( A \) are \( -5 \) and \( 3 \).
We check if these points belong to \( [-8, 3] \cup [5, \infty) \).
\( -5 \in [-8, 3] \).
\( 3 \in [-8, 3] \).
Therefore, the real numbers in \( [-8, 3] \cup [5, \infty) \) but not in \( A \) are \( -5 \) and \( 3 \).
The number of such real numbers is 2.