NDA
Mathematics
Continuity and Differentiability
2025
NDA 2025 (Phase 2)
NDA Mathematics Question (2025) — Solution
Question
For the following two (02) items: Let the function f(x) = |x - 3| + |x - 4| be defined on the interval [0, 5]. Consider the following statements: I. The function is differentiable at x = 3. II. The function is differentiable at x = 4. Which of the statements given above is/are correct?
Options
- A. I only
- B. II only
- C. Both I and II
- D. Neither I nor II
Answer
D. Neither I nor II
Step-by-step solution
The given function is f(x) = |x - 3| + |x - 4| defined on [0, 5]. We can redefine the function by removing the absolute value signs in different intervals: f(x) = cases -(x - 3) - (x - 4), & 0 x Simplifying this, we get: f(x) = cases 7 - 2x, & 0 x Differentiating f(x) with respect to x in the open intervals: f'(x) = cases -2, & 0 At x = 3: Left Hand Derivative (LHD) = -2 Right Hand Derivative (RHD) = 0 Since LHD RHD, f(x) is not differentiable at x = 3. At x = 4: Left Hand Derivative (LHD) = 0 Right Hand Derivative (RHD) = 2 Since LHD RHD, f(x) is not differentiable at x = 4. Therefore, neither statement I nor statement II is correct. Answer: Neither I nor II
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