CONTINUITY AND DIFFERENTIABILITY — Online MCQ Test
MATHS · Grade 12 · CBSE(NCERT)
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CONTINUITY AND DIFFERENTIABILITY — Important Questions & Answers
A function f(x) is continuous at x = a if which of the following conditions is satisfied?
- A. lim(x→a) f(x) = f(a) and the limit exists
- B. f(a) is defined
- C. lim(x→a⁺) f(x) exists
- D. f(x) is differentiable at x = a
Answer: A. lim(x→a) f(x) = f(a) and the limit exists
Continuity at x = a requires that the limit exists and equals the function value at that point: lim(x→a) f(x) = f(a).
Continuity at x = a requires that the limit exists and equals the function value at that point: lim(x→a) f(x) = f(a).
What is the derivative of f(x) = 5x³ + 2x² - 7 with respect to x?
- A. 15x² + 4x
- B. 15x² + 4x - 7
- C. 5x² + 2x
- D. 3x² + 2x
Answer: A. 15x² + 4x
Using power rule: d/dx(5x³) = 15x², d/dx(2x²) = 4x, and d/dx(-7) = 0. Therefore, f'(x) = 15x² + 4x.
Using power rule: d/dx(5x³) = 15x², d/dx(2x²) = 4x, and d/dx(-7) = 0. Therefore, f'(x) = 15x² + 4x.
Find the derivative of f(x) = eˣ sin(x) using the product rule.
- A. eˣ(sin(x) + cos(x))
- B. eˣ cos(x)
- C. eˣ sin(x) + eˣ cos(x)
- D. sin(x) + eˣ
Answer: A. eˣ(sin(x) + cos(x))
Using product rule: f'(x) = (eˣ)'·sin(x) + eˣ·(sin(x))' = eˣ sin(x) + eˣ cos(x) = eˣ(sin(x) + cos(x)).
Using product rule: f'(x) = (eˣ)'·sin(x) + eˣ·(sin(x))' = eˣ sin(x) + eˣ cos(x) = eˣ(sin(x) + cos(x)).
Find the left derivative and right derivative of f(x) = |x| at x = 0. Which of the following is correct?
- A. Left derivative = -1, Right derivative = 1
- B. Left derivative = 1, Right derivative = -1
- C. Left derivative = 0, Right derivative = 0
- D. Left derivative = 1, Right derivative = 1
Answer: A. Left derivative = -1, Right derivative = 1
For x < 0: f(x) = -x, so left derivative = d/dx(-x)|ₓ₌₀ = -1. For x > 0: f(x) = x, so right derivative = d/dx(x)|ₓ₌₀ = 1.
For x < 0: f(x) = -x, so left derivative = d/dx(-x)|ₓ₌₀ = -1. For x > 0: f(x) = x, so right derivative = d/dx(x)|ₓ₌₀ = 1.
A function f(x) is defined as f(x) = {(x² - 9)/(x - 3), x ≠ 3; k, x = 3}. What value of k makes f continuous at x = 3?
- A. k = 3
- B. k = 6
- C. k = 0
- D. k = 9
Answer: B. k = 6
Simplify: (x² - 9)/(x - 3) = (x-3)(x+3)/(x-3) = x + 3 for x ≠ 3. So lim(x→3) f(x) = 3 + 3 = 6. For continuity, k = 6.
Simplify: (x² - 9)/(x - 3) = (x-3)(x+3)/(x-3) = x + 3 for x ≠ 3. So lim(x→3) f(x) = 3 + 3 = 6. For continuity, k = 6.