Limits, Continuity & Differentiability
Intermediate Value Theorem and properties of continuous functions
Grade 12
Question:
<p><strong>367.</strong> Which of the following statement(s) is(are) <strong>incorrect</strong>?</p><p>(a) The equation \(\sin x - x = 0\) has a real root in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\).</p><p>(b) The equation \(\tan x - x = 0\) has a real root in \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\).</p><p>(c) If \(f\) is continuous function in \([a, b]\), then there exists at least one \(c \in [a, b]\) such that \(f(c) = \frac{2f(a) + 3f(b)}{5}\).</p><p>(d) If \(f(a)\) and \(f(b)\) are of opposite signs then equation \(f(x) = 0\) has necessarily at least one root in \((a, b)\).</p>
<p>(a) The equation \(\sin x - x = 0\) has a real root in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\).</p>
<p>(b) The equation \(\tan x - x = 0\) has a real root in \(\left(\frac{\pi}{6}, \frac{\pi}{3}\right)\).</p>
<p>(c) If \(f\) is continuous function in \([a, b]\), then there exists at least one \(c \in [a, b]\) such that \(f(c) = \frac{2f(a) + 3f(b)}{5}\).</p>
<p>(d) If \(f(a)\) and \(f(b)\) are of opposite signs then equation \(f(x) = 0\) has necessarily at least one root in \((a, b)\).</p>
Step-by-Step Solution
Key Concept: We need to identify which statements are INCORRECT by analyzing the Intermediate Value Theorem and its applicability conditions. A statement is incorrect if it violates IVT conditions or makes a claim that cannot be guaranteed.
<p><strong>Step 1: Analyze Statement (a) - sin x - x = 0 in (π/4, π/2)</strong></p><p>Let f(x) = sin x - x. We need to check if this has a root in (π/4, π/2).</p><p>f(π/4) = sin(π/4) - π/4 = (√2/2) - π/4 ≈ 0.707 - 0.785 ≈ -0.078 < 0</p><p>f(π/2) = sin(π/2) - π/2 = 1 - π/2 ≈ 1 - 1.571 ≈ -0.571 < 0</p><p>Since both f(π/4) and f(π/2) are negative, IVT cannot guarantee a root in this interval. <strong>Statement (a) is INCORRECT.</strong></p><p><strong>Step 2: Analyze Statement (b) - tan x - x = 0 in (π/6, π/3)</strong></p><p>Let g(x) = tan x - x. Check continuity: g(x) is continuous on (π/6, π/3) since π/6 and π/3 are both less than π/2.</p><p>g(π/6) = tan(π/6) - π/6 = (1/√3) - π/6 ≈ 0.577 - 0.524 ≈ 0.053 > 0</p><p>g(π/3) = tan(π/3) - π/3 = √3 - π/3 ≈ 1.732 - 1.047 ≈ 0.685 > 0</p><p>Both values are positive, so IVT does not guarantee a root. <strong>Statement (b) is INCORRECT.</strong></p><p><strong>Step 3: Analyze Statement (c) - Mean Value Property for Weighted Average</strong></p><p>The value [2f(a) + 3f(b)]/5 is a weighted average of f(a) and f(b) with weights 2/5 and 3/5 (sum = 1).</p><p>By the Intermediate Value Theorem, for any value k between min{f(a), f(b)} and max{f(a), f(b)}, there exists c ∈ [a,b] with f(c) = k.</p><p>The weighted average [2f(a) + 3f(b)]/5 always lies between min and max of {f(a), f(b)}, so there must exist c ∈ [a,b] with f(c) = [2f(a) + 3f(b)]/5. <strong>Statement (c) is CORRECT.</strong></p><p><strong>Step 4: Analyze Statement (d) - Opposite Signs Guarantee a Root</strong></p><p>The statement claims: if f(a) and f(b) have opposite signs, then f(x) = 0 has a root in (a,b). This is ONLY true if f is continuous on [a,b].</p><p>The statement does NOT require f to be continuous. Counterexample: f(x) = {-1 if x ∈ [0, 1), 1 if x = 1}. Here f(0) = -1 and f(1) = 1 (opposite signs), but f is discontinuous and has no root in [0,1].</p><p><strong>Statement (d) is INCORRECT because it doesn't require continuity.</strong></p><p><strong>∴ Answer: A, B, D</strong></p>
Correct Answer: A, B, D