Question

Let f: $(0, \infty) \rightarrow R$ be a twice differentiable function. If for some $a \neq 0, \int_{0}^{1} f(\lambda x) d \lambda = af(x)f(1)$, $f(1) = 1$ and $f(16) = \frac{1}{8}$, then $16-f'(\frac{1}{16})$ is equal to _______.

Answer 1

112

Answer 2

Solution Explanation:

1. Substitute λx = t (so t = λx ⇒ dλ = dt/x) in the given equation:   ∫₀¹ f(λx)dλ = (1/x)∫₀ˣ f(t)dt = af(x).
     Thus, (1/x)∫₀ˣ f(t)dt = af(x) ⟹ ∫₀ˣ f(t)dt = a x f(x).

2. Differentiate both sides with respect to x:   Left: d/dx[∫₀ˣ f(t)dt] = f(x).
     Right: d/dx[a x f(x)] = a f(x) + a x f′(x).
     So, f(x) = a f(x) + a x f′(x) ⟹ a x f′(x) = (1 – a) f(x).

3. Rearranging gives:   f′(x)/f(x) = (1 – a)/(a) ⋅ (1/x).
     Integrate: ln f(x) = [(1 – a)/(a)] ln x + constant.   Using f(1) = 1, constant = 0.
     Thus, f(x) = x^((1 – a)/a).

4. Use f(16) = 1/8:   16^((1 – a)/a) = 1/8
     Write 16 = 2⁴ and 1/8 = 2^(–3):
     (2⁴)^((1 – a)/a) = 2^(–3) ⟹ 2^(4(1 – a)/a) = 2^(–3).
     Equate exponents: 4(1 – a)/a = –3 ⟹ 4 – 4a = –3a ⟹ a = 4.

5. So, f(x) = x^((1 – 4)/4) = x^(–3/4).
     Differentiate: f′(x) = (–3/4) x^(–7/4).

6. Evaluate f′(1/16):   f′(1/16) = (–3/4) (1/16)^(–7/4).
     Since (1/16)^(–7/4) = 16^(7/4) = (16^(1/4))^7 = 2⁷ = 128,
     Thus, f′(1/16) = (–3/4) × 128 = –96.

7. Finally, compute:   16 – f′(1/16) = 16 – (–96) = 16 + 96 = 112.