Limits, Continuity & Differentiability
Continuity and Differentiability of functions
Grade 12
Question:
<p>Let <em>f</em>(<em>x</em>) = min {|tan <em>x</em>|, |cot <em>x</em>|}. Then <em>f</em>(<em>x</em>) is</p>
<p>(a) everywhere continuous</p>
<p>(b) everywhere differentiable</p>
<p>(c) not differentiable at <em>x</em> = 0</p>
<p>(d) not differentiable when \(x = \dfrac{\pi}{4}\)</p>
Step-by-Step Solution
Key Concept: Analyze the minimum of two periodic functions by finding where |tan x| = |cot x| and comparing their values in each region. The function's continuity and differentiability depend on whether sharp corners exist at intersection points.
<p><strong>Step 1:</strong> Find where |tan x| = |cot x|. This occurs when |tan x| = 1/|tan x|, so |tan x|² = 1, giving |tan x| = 1. Thus x = π/4 + nπ/2 (excluding asymptotes).</p><p><strong>Step 2:</strong> At these intersection points, f(x) = 1. Between intersections, either |tan x| < |cot x| or vice versa. Near x = π/4: for x slightly less than π/4, |tan x| < 1 < |cot x|, so f(x) = |tan x|. For x slightly more than π/4, |tan x| > 1 > |cot x|, so f(x) = |cot x|.</p><p><strong>Step 3:</strong> Check continuity: As x → π/4⁻, f(x) = |tan x| → 1. As x → π/4⁺, f(x) = |cot x| → 1. And f(π/4) = 1. So f is <strong>continuous everywhere</strong> on its domain.</p><p><strong>Step 4:</strong> Check differentiability: The left derivative at x = π/4 uses |tan x| (derivative = sec²x > 1), while right derivative uses |cot x| (derivative = -csc²x has different magnitude). The function has a corner at each intersection point, so f is <strong>not differentiable at x = π/4 + nπ/2</strong>.</p><p>∴ Answer: a (continuous everywhere on domain), d (not differentiable at specific points)</p>
Correct Answer: a, d