Let f : (0, ) R be a twice differentiable function. If for some a 0, 01 f( x) d = af(x), f(1) = 1 and f(16) = 18, then 16 - f' ( 116 ) is equal to .
Answer
112
Solution
Put x = t, so d = dtx and the given relation becomes
1x 0x f(t) dt = af(x) 0x f(t) dt = a x f(x)
Differentiating w.r.t. x:
f(x) = af(x) + a x f'(x) f'(x)f(x) = 1 - aa 1x
Integrating, f(x) = k x + c, where k = 1 - aa, so f(x) = Cxk.
f(1) = 1 gives C = 1, and f(16) = 18 gives
24k = 2-3 k = - 34 (then a = 4 0)
Hence f(x) = x-3/4 and f'(x) = - 34x-7/4, so
f' ( 116 ) = - 34 167/4 = - 34 27 = -96
16 - f' ( 116 ) = 16 + 96 = 112
Answer: 112