Let f be a differentiable function in the interval (0, ∞) such that f(1)=1 and lim t → x (t² f(x)-x² f(t))/(t-x)=1 for each x>0. Then 2 f(2)+3 f(3) is equal to _________.

Solution

Maths sir · 4 steps

Add and subtract x² f(x) in the numerator, and the limit turns into a linear differential equation.

1. Open the limit. Write the numerator as (t² - x²) f(x) - x²(f(t) - f(x)) and divide by t - x. As t → x, the first part gives 2x f(x) and the second gives x² f'(x).

   2x f(x) - x² f'(x) = 1 ⇒ f'(x) - 2/x f(x) = -1/x²

2. Integrating factor. The integrating factor is e^(-∫ 2/x dx) = 1/x². So d/dx(f(x)/x²) = -1/x⁴, which gives f(x)/x² = 1/3x³ + C.
3. Use f(1) = 1. 1 = 1/3 + C, so C = 2/3. Hence f(x) = 1/3x + 2x²/3.
4. So. 2f(2) = 1/3 + 16/3 = 17/3 and 3f(3) = 1/3 + 18 = 55/3. Their sum is 72/3 = 24.

The common slip: 46. Losing the minus sign on the right side gives f'(x) - 2/xf(x) = 1/x², then f(x) = -1/3x + 4x²/3 and a sum of 46, not 24.

The idea to keep. When a limit looks like a difference quotient, add and subtract a term to split it into a derivative. What is left is usually a linear equation you solve with an integrating factor.

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