Let f:(0, ∞) → R be a twice differentiable function. If for some a≠ 0, ∫₀¹ f(λ x) d λ=a f(x), f(1)=1 and f(16)=1/8, then 16-f′(1/16) is equal to __________.

Solution

Maths sir · 5 steps

Put t = λ x to turn it into an integral up to x, then differentiate to get a power function.

1. Change the variable. Let t = λ x, so dλ = dt/x, and t runs from 0 to x. The condition becomes 1/x∫₀ˣ f(t) dt = a f(x), that is, ∫₀ˣ f(t) dt = a x f(x).
2. Differentiate both sides. By the Fundamental Theorem of Calculus, the left side gives f(x). The right side gives a f(x) + a x f'(x). So (1-a) f(x) = a x f'(x).

   f'(x)/f(x) = (1-a)/a·1/x ⇒ f(x) = C xᵏ, k = (1-a)/a

3. Use the given values. f(1) = 1 gives C = 1. Then f(16) = 16ᵏ = 2⁴ᵏ = 1/8 = 2⁻³, so k = -3/4 and f(x) = x^(-3/4) (this needs a = 4, which is non-zero).
4. Find the derivative. f'(x) = -3/4x^(-7/4), so f'(1/16) = -3/4× 16^(7/4) = -3/4× 2⁷ = -96.
5. So. 16 - f'(1/16) = 16 - (-96) = 112.

The common slip: -80. Dropping the minus sign of f'(1/16) = -96 gives 16 - 96 = -80, not 112.

The idea to keep. When the integrand is f(λ x) with limits 0 to 1, substitute t = λ x to get ∫₀ˣ f(t) dt, then differentiate. A relation like f = c x f' always means f is a power of x.

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