Exams › JEE Advanced › Maths

Let f(x) be a twice-differentiable function with f(0) = 0, f''(0) = 0, and f''(0) = 2 (here the last condition means the second derivative at 0 equals 2 and f'(0) = 0 is implied for a finite limit). Find: lim (x -> 0) [f(x) + 2f(2x) + 3f(3x)] / x².

1. 3
2. 36
3. 72
4. 18

Correct answer: 36

Solution

With f(0)=0, f'(0)=0, f''(0)=2: f(x) approx x² near 0. So f(kx) approx k²*x². Numerator: f(x) + 2f(2x) + 3f(3x) approx x² + 2*(4x²) + 3*(9x²) = (1 + 8 + 27)*x² = 36*x². Dividing by x² gives 36.

Related JEE Advanced Maths questions

• Evaluate f(x) = lim x→∞ (x²n − 1) / (x²n + 1). Which of the following is true?
• Determine the value of L if L = lim x→0 [(sin x + ae^x + be^−x + c ln(1 + x)) / x³], given that L is finite and not infinite.
• Consider the function f(x) defined as f(x) = lim n→∞ {n(x + n)(x + n/2)...(x + n/n) / n!(x² + n²)(x² + n²/4)...(x² + n²/n²)}^x/n, where x > 0. Which of the following is true?
• For a positive integer n, let f(n) be defined as n plus the sum of terms in the form (a + bn − cn²) / (dn + en²), where the coefficients vary across terms. Specifically, f(n) = n + (16 + 5n − 3n²) / (4n + 3n²) + (32 + n − 3n²) / (8n + 3n²) + (48 − 3n − 3n²) / (12n + 3n²) +... + (25n − 7n²) / (7n²). Determine the value of lim n→∞ f(n).
• Consider the function f: (0,1) → R defined by f(x) = √n whenever x lies in the interval [1/(n+1), 1/n), where n belongs to the set of natural numbers. Let g: (0,1) → R be a function satisfying the inequality ∫x¹ (1−t)/t dt ≤ g(x) < 2√x for all x in (0,1). Determine the value of lim x→0 f(x)g(x).
• Let h(x) = h(2x) for every real x, where h is a continuous function and h(2012) = pi/2. Define the limit M = lim_(x->0) [cos²(h(x)) + 1 - sin³(h(x))] / sin²(x). What is the value of 4M?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →