Class 12 Maths Chapter 5: Continuity and Differentiability worksheet with answer key

Answer: (d) discontinuous, since the limit at 3 exists but is not equal to f(3)

Left hand limit = 2(3) + 1 = 7; right hand limit = 3 + 4 = 7; so lim_(x→3) f(x) = 7, but f(3) = 4 ≠ 7. The limit exists and differs from the value, so f is discontinuous at 3.

Answer: (a) −3x² sin(x³)

With t = x³: d/dx cos t = −sin t · dt/dx = −sin(x³) · 3x².

Answer: (d) log 5

d/dx (a^x) = a^x log a, so the derivative is 5^x log 5, which at x = 0 is 5⁰ log 5 = log 5.

Answer: (c) 12

For x ≠ 2, (x³ − 8)/(x − 2) = x² + 2x + 4, so lim_(x→2) f(x) = 4 + 4 + 4 = 12. Continuity at 2 needs f(2) = 12, so k = 12.

Answer: (a) f is a continuous function, since it is continuous at every point of its domain

0 is not in the domain, so continuity at 0 is not asked. At every c ≠ 0, lim_(x→c) 1/x = 1/c = f(c), so f is a continuous function. +∞ is not a real number, so (d) is false.

dx/dt = e^t − e^−t = y and dy/dt = e^t + e^−t = x, so dy/dx = (dy/dt)/(dx/dt) = x/y (for t ≠ 0, where y ≠ 0).

• dx/dt = y and dy/dt = x (1)
• dy/dx = x/y (1)

dy/dx = 2x/(x² + 1). By the quotient rule, d²y/dx² = [2(x² + 1) − 2x · 2x]/(x² + 1)² = 2(1 − x²)/(x² + 1)².

• dy/dx = 2x/(x² + 1) (1)
• d²y/dx² = 2(1 − x²)/(x² + 1)² (1)

Left hand limit and f(2): 2k + 3. Right hand limit: 4 − k. Continuity needs 2k + 3 = 4 − k, so 3k = 1 and k = 1/3.

• writes 2k + 3 = 4 − k from the two sides (1)
• k = 1/3 (1)

Continuity: f(1) = 1, left hand limit 1, right hand limit 2 − 1 = 1. Left hand derivative: lim_(h→0⁻) [(1 + h)² − 1]/h = lim (2 + h) = 2. Right hand derivative: lim_(h→0⁺) [2(1 + h) − 1 − 1]/h = 2. They are finite and equal, so f is differentiable at 1 with f′(1) = 2.

• continuity at x = 1 checked (1)
• left hand derivative = 2 (1)
• right hand derivative = 2, so differentiable (1)

Each piece is a polynomial, so f is continuous except possibly at the joins. At x = 1: f(1) = a + 5 and the right hand limit is 1 + b, so a + 5 = 1 + b. At x = 3: the left hand limit is 9 + b and f(3) = 6b − 1, so 9 + b = 6b − 1, giving b = 2. Then a = 1 + 2 − 5 = −2.

• condition at x = 1: a + 5 = 1 + b (1)
• condition at x = 3: 9 + b = 6b − 1, b = 2 (1)
• a = −2 (1)

(i) With x = tan θ, θ ∈ (−π/2, π/4): (1 + tan θ)/(1 − tan θ) = tan(π/4 + θ), and π/4 + θ ∈ (−π/4, π/2), so y = π/4 + θ = π/4 + tan⁻¹ x and dy/dx = 1/(1 + x²). (ii) Directly, dy/dx = [1/(1 + u²)] du/dx with u = (1 + x)/(1 − x), du/dx = 2/(1 − x)²; 1 + u² = [(1 − x)² + (1 + x)²]/(1 − x)² = 2(1 + x²)/(1 − x)². So dy/dx = 2/[2(1 + x²)] = 1/(1 + x²), the same.

• substitution gives y = π/4 + tan⁻¹ x, dy/dx = 1/(1 + x²) (1)
• chain rule with du/dx = 2/(1 − x)² (1)
• simplifies to 1/(1 + x²): agrees (1)