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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Continuity and Differentiability

NDA 2025 · Mathematics · Continuity and Differentiability

For the following two (02) items: Let the function f(x) = |x - 3| +...

For the following two (02) items:
Let the function f(x)=∣x−3∣+∣x−4∣f(x) = |x - 3| + |x - 4| be defined on the interval [0,5][0, 5].

What is dydx\dfrac{dy}{dx} at x=3.5x = 3.5 equal to?

Options

  1. A00
  2. B11
  3. C22
  4. D3.53.5

Correct answer

A. 00

Step-by-step solution

Given f(x)=∣x−3∣+∣x−4∣f(x) = |x - 3| + |x - 4| on the interval [0,5][0, 5].

At x=3.5x = 3.5, we have 3<x<43 < x < 4.

In the interval (3,4)(3, 4), x−3>0x - 3 > 0 and x−4<0x - 4 < 0.

Therefore, f(x)=(x−3)−(x−4)=1f(x) = (x - 3) - (x - 4) = 1.

Since f(x)f(x) is a constant in the neighborhood of x=3.5x = 3.5, its derivative is zero.

dydx=0\dfrac{dy}{dx} = 0 at x=3.5x = 3.5.
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