Continuity and Differentiability — Practice

Question 1

Find the value of k that makes f(x) = {kx + 1 if x ≤ π; cos x if x > π} continuous at x = π.

Accepted Answer

For f to be continuous at x = π: LHL = RHL = f(π). f(π) = kπ + 1 (from the first piece, since π ≤ π). LHL: lim(x→π⁻) f(x) = lim(x→π⁻) (kx + 1) = kπ + 1. RHL: lim(x→π⁺) f(x) = lim(x→π⁺) cos x = cos π = −1. For continuity: LHL = RHL → kπ + 1 = −1 → kπ = −2 → k = −2/π. VERIFY: f(π) = (−2/π)π + 1 = −2 + 1 = −1 = RHL ✓.

Question 2

Differentiate y = xˢⁱⁿˣ + (sin x)ˣ with respect to x.

Accepted Answer

Let y = u + v where u = xˢⁱⁿˣ and v = (sin x)ˣ. DIFFERENTIATING u = xˢⁱⁿˣ: Take ln: ln u = sin x · ln x. Differentiate: (1/u)(du/dx) = cos x · ln x + sin x · (1/x). du/dx = u[cos x · ln x + sin x/x] = xˢⁱⁿˣ [cos x · ln x + sin x/x]. DIFFERENTIATING v = (sin x)ˣ: Take ln: ln v = x · ln(sin x). Differentiate: (1/v)(dv/dx) = ln(sin x) + x · (cos x/sin x) = ln(sin x) + x cot x. dv/dx = v[ln(sin x) + x cot x] = (sin x)ˣ [ln(sin x) + x cot x]. RESULT: dy/dx = du/dx + dv/dx = xˢⁱⁿˣ [cos x · ln x + sin x/x] + (sin x)ˣ [ln(sin x) + x cot x].

Question 3

Verify the Mean Value Theorem for f(x) = x(x−1)(x−2) on [0, 1/2].

Accepted Answer

VERIFY CONDITIONS: f(x) = x(x−1)(x−2) = x³ − 3x² + 2x is a polynomial → continuous on [0, 1/2] and differentiable on (0, 1/2). Conditions satisfied. FIND THE MVT SLOPE: f(0) = 0(−1)(−2) = 0. f(1/2) = (1/2)(−1/2)(−3/2) = 3/8. [f(1/2) − f(0)] / (1/2 − 0) = (3/8 − 0)/(1/2) = (3/8)(2) = 3/4. FIND c: f'(x) = 3x² − 6x + 2. Set f'(c) = 3/4: 3c² − 6c + 2 = 3/4. 3c² − 6c + 2 − 3/4 = 0. 3c² − 6c + 5/4 = 0. 12c² − 24c + 5 = 0. c = [24 ± √(576 − 240)] / 24 = [24 ± √336] / 24 = [24 ± 4√21] / 24 = (6 ± √21)/6. √21 ≈ 4.58. c₁ = (6 − 4.58)/6 ≈ 0.237. c₂ = (6 + 4.58)/6 ≈ 1.763. Only c₁ ≈ 0.237 ∈ (0, 0.5). Hence c = (6 − √21)/6 ∈ (0, 1/2). MVT VERIFIED.

Practice — Continuity and Differentiability

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