Exams › JEE Advanced › Maths

The normal to the curve y = f(x) at x = 0 is given by 3x - y + 3 = 0. Find the value of lim_(x->1) (x-1)² / [f((x-1)²) - 5*f(4*(x-1)²) + 4*f(7*(x-1)²)].

1. -1/3
2. 1/3
3. 3
4. -3

Correct answer: -1/3

Solution

Normal 3x-y+3=0 gives f(0)=3 and slope of tangent f'(0)=-1/3. With t=(x-1)²->0, the denominator ~ (-3t), so the limit = t/(-3t) = -1/3.

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 →