Let f:(-∞, ∞)-{0} → ℝ be a differentiable function such that f′(1)=lim a → ∞ a² f(1/a). Then lim a → ∞ a(a+1)/2 tan ⁻¹(1/a)+a²-2 log ₑ a is equal to

Solution

Maths sir · 4 steps

Put a = 1/x, pull out the factor a², and the expression becomes a²f(1/a). Then the given rule hands you f'(1).

1. Substitute. Let a = 1/x. Then a(a+1)/2 = (1+x)/2x², a² = 1/x², tan⁻¹ 1/a = tan⁻¹ x and -2 logₑ a = 2 logₑ x.
2. Find f. Take 1/x² = a² out of every term. The expression becomes a² f(1/a), where f(x) = (1+x)/2 tan⁻¹ x + 1 + 2x²logₑ x. By the given rule, the limit is f'(1).

   f'(x) = 1/2 tan⁻¹ x + (1+x)/2(1+x²) + 4x logₑ x + 2x

3. Put x = 1. 1/2·π/4 = π/8, then 2/(2 × 2) = 1/2, then 4 logₑ 1 = 0, and finally 2x = 2.
4. So. The limit is f'(1) = π/8 + 1/2 + 2 = 5/2 + π/8, which is option (a).

The idea to keep. When a question gives you a rule like lim a² f(1/a) = f'(1), rewrite the expression in that exact shape. Put a = 1/x, read off f(x), and then just differentiate.

Answer from the NTA final answer key. Solution by PrepWiser.