Limits, Continuity & Differentiability
Intermediate Value Theorem
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 atleast 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 atleast 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 atleast 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 atleast one root in \((a, b)\).</p>
Step-by-Step Solution
Key Concept: Use the Intermediate Value Theorem (IVT) carefully: IVT requires continuity and opposite signs at endpoints to guarantee a root exists. Also, verify algebraic claims about continuous functions by checking if they follow from IVT.
<p><strong>Analyzing each statement:</strong></p><p><strong>Statement (a):</strong> For f(x) = sin x - x on [π/4, π/2]:<br/>f(π/4) = sin(π/4) - π/4 = (√2/2) - π/4 ≈ 0.707 - 0.785 = -0.078 < 0<br/>f(π/2) = sin(π/2) - π/2 = 1 - π/2 ≈ 1 - 1.571 = -0.571 < 0<br/>Both values negative, so IVT doesn't guarantee a root. <strong>INCORRECT</strong> ✓</p><p><strong>Statement (b):</strong> For g(x) = tan x - x on [π/6, π/3]:<br/>g(π/6) = tan(π/6) - π/6 = (1/√3) - π/6 ≈ 0.577 - 0.524 = 0.053 > 0<br/>g(π/3) = tan(π/3) - π/3 = √3 - π/3 ≈ 1.732 - 1.047 = 0.685 > 0<br/>Both values positive, so IVT doesn't guarantee a root. <strong>INCORRECT</strong> ✓</p><p><strong>Statement (c):</strong> For any continuous f on [a,b], by IVT, f takes all values between f(a) and f(b). The value (2f(a) + 3f(b))/5 is a weighted average strictly between f(a) and f(b) (or equals them). Since f is continuous, ∃c ∈ [a,b] where f(c) equals this value. <strong>CORRECT</strong> ✓</p><p><strong>Statement (d):</strong> If f(a) and f(b) have opposite signs AND f is continuous on [a,b], then by IVT, ∃c where f(c) = 0. However, the statement doesn't explicitly state continuity—this is a necessary condition. If f is not continuous, the root may not exist. <strong>INCORRECT</strong> (missing continuity hypothesis) ✓</p><p>∴ Answer: A, B, D</p>
Correct Answer: A, B, D