Limits, Continuity & Differentiability
Continuous and differentiable functions, fixed point theorem
Grade 12
Question:
<p><strong>340.</strong> Let \(f'(x)\) be a continuous function which maps from \([0,1] \to [p(a),\ p(b)]\). If \(p(x)\) is a differentiable function on \([a, b]\) such that \(p(g(x)) = x\), \(g(0) = a\) and \(g(1) = b\), then which of the following is/are true?</p>
<p>(a) \(f(0) + 2 < f(1)\)</p>
<p>(b) \(f(1) \leq 1 + f(0)\)</p>
<p>(c) \(\dfrac{\displaystyle\int_0^1 f'(x)\, dx}{\displaystyle\int_0^1 g'(x)\, dx} \leq p'(c)\) for some \(c \in (a, b)\)</p>
<p>(d) There exists \(k \in [0, 1]\) such that \(f'(k) = k\)</p>
Step-by-Step Solution
Key Concept: Since f'(x) is continuous on [0,1] and p(g(x)) = x means g is the inverse function of p, we must recognize that g'(x) = 1/p'(g(x)). The relationship between the ranges and the inverse function composition reveals constraints on p'.
<p><strong>Step 1:</strong> Recognize the relationship p(g(x)) = x means g is the inverse function of p, so g = p⁻¹</p><p><strong>Step 2:</strong> Differentiate p(g(x)) = x with respect to x: p'(g(x)) · g'(x) = 1, which gives g'(x) = 1/p'(g(x))</p><p><strong>Step 3:</strong> Since f'(x) is continuous on [0,1] and maps to [p(a), p(b)], and g(0) = a, g(1) = b, we have the image of the continuous function g on [0,1] is [a,b]</p><p><strong>Step 4:</strong> The range of f' being [p(a), p(b)] combined with the inverse relationship means p is strictly monotonic on [a,b] and p' is either always positive or always negative (never zero)</p><p><strong>Step 5:</strong> This ensures g'(x) = 1/p'(g(x)) is well-defined and continuous on [0,1], making g differentiable on [0,1]</p><p><strong>Step 6:</strong> Therefore statements involving: (i) g being differentiable on [0,1], (ii) g' being continuous, (iii) p' being non-zero and monotonic, and (iv) the range relationships are all true</p><p>∴ Answer: D</p>
Correct Answer: D