<p>Let \( d(x, [a, b]) = \min\{|x - y| : a \le y \le b\} \). A function \( f : R \to [0,1] \) is defined by \( f(x) = \dfrac{d(x, [0,1])}{d(x,[0,1]) + d(x,[2,3])} \), 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: Analyze f(x) by computing d(x,[0,1]) and d(x,[2,3]) in three regions: x<0, 0≤x≤1, 1<x<2, 2≤x≤3, and x>3. The function's behavior depends critically on the distance to each interval, which changes piecewise.
<p><strong>Step 1: Compute d(x,[0,1]) and d(x,[2,3]) in each region</strong></p><p>• For x < 0: d(x,[0,1]) = -x, d(x,[2,3]) = 2-x</p><p>• For 0 ≤ x ≤ 1: d(x,[0,1]) = 0, d(x,[2,3]) = 2-x</p><p>• For 1 < x < 2: d(x,[0,1]) = x-1, d(x,[2,3]) = 2-x</p><p>• For 2 ≤ x ≤ 3: d(x,[0,1]) = x-1, d(x,[2,3]) = 0</p><p>• For x > 3: d(x,[0,1]) = x-1, d(x,[2,3]) = x-3</p></p><p><strong>Step 2: Evaluate f(x) in each region</strong></p><p>• x ∈ [0,1]: f(x) = 0/(0 + 2-x) = 0</p><p>• x ∈ [2,3]: f(x) = (x-1)/(x-1 + 0) = 1</p><p>• x ∈ (1,2): f(x) = (x-1)/(x-1 + 2-x) = (x-1)/1 = x-1 ∈ (0,1)</p><p>• x < 0: f(x) = (-x)/(−x + 2-x) = -x/(2-2x) = 1/2 (constant)</p><p>• x > 3: f(x) = (x-1)/(x-1 + x-3) = (x-1)/(2x-4) = 1/2 (constant)</p></p><p><strong>Step 3: Identify properties</strong></p><p>✓ f is continuous on ℝ</p><p>✓ Range of f is [0,1]</p><p>✓ f(x) = 1/2 for x ∈ (-∞,0) ∪ (3,∞)</p><p>✓ f is strictly increasing on (1,2)</p><p>✓ f(0) = 0, f(3) = 1</p><p><strong>Identify the incorrect statement(s) among the options to determine answer A</strong></p>
Correct Answer: A