<p>Let \( d(x,[a,b]) = \min\{|x-y| : a \le y \le b\} \). A function \( f: R \to [0,1] \) is defined by \[ f(x) = \frac{d(x,[0,1])}{d(x,[0,1]) + d(x,[2,3])} \] then which of the following is(are) <strong>incorrect</strong>?</p>
<p>(a) \( f(x) \) is decreasing in \( (-\infty, 0) \) and increasing in \( (3, \infty) \).</p>
<p>(b) The function \( f \) is bijective.</p>
<p>(c) Number of points where \( f(x) \) is non-derivable is 4.</p>
<p>(d) Number of solution of the equation \( f(x) = \dfrac{1}{2} \) is 2.</p>
Step-by-Step Solution
Key Concept: The function f(x) is defined piecewise based on the distance from x to the intervals [0,1] and [2,3]. For x in different regions (x<0, 0≤x≤1, 1<x<2, 2≤x≤3, x>3), the distances change, creating different formulas for f(x). Understanding continuity and boundary behavior is crucial.
<p><strong>Step 1:</strong> Identify d(x,[0,1]) and d(x,[2,3]) for different regions.</p><p>For <strong>x < 0:</strong> d(x,[0,1]) = -x, d(x,[2,3]) = 2-x, so f(x) = -x/(2-2x) = x/(2x-2) = 1/2 - 1/(2(x-1))</p><p>For <strong>0 ≤ x ≤ 1:</strong> d(x,[0,1]) = 0, d(x,[2,3]) = 2-x, so f(x) = 0/(2-x) = 0</p><p>For <strong>1 < x < 2:</strong> d(x,[0,1]) = x-1, d(x,[2,3]) = 2-x, so f(x) = (x-1)/((x-1)+(2-x)) = (x-1)/1 = x-1</p><p>For <strong>2 ≤ x ≤ 3:</strong> d(x,[0,1]) = x-1, d(x,[2,3]) = 0, so f(x) = (x-1)/(x-1) = 1</p><p>For <strong>x > 3:</strong> d(x,[0,1]) = x-1, d(x,[2,3]) = x-3, so f(x) = (x-1)/(x-1+x-3) = (x-1)/(2x-4)</p><p><strong>Step 2:</strong> Check properties:</p><p>• f is continuous on (0,1) and (2,3) ✓, but discontinuous at x=1⁻ (limit = 0) and x=2⁺ (limit = 1) ✗</p><p>• f(0) = 0, f(1) = 0, f(2) = 1, f(3) = 1 ✓</p><p>• f is strictly increasing on (1,2) ✓</p><p>• Range = [0,1] ✓</p><p>∴ Any statement claiming f is continuous everywhere or has other global properties contradicting the piecewise analysis is <strong>incorrect</strong></p>
Correct Answer: A