Differential Equations · Question 36

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 d = dtx

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

f(x) = k x + c, where k = 1 - aa

f(x) = Cxk

f(1) = 1 C = 1

f(16) = 18 24k = 2-3 k = - 34 (then a = 4 0)

f(x) = x-3/4, f'(x) = - 34x-7/4

f' ( 116 ) = - 34 167/4 = - 34 27 = -96

16 - f' ( 116 ) = 16 + 96 = 112

Answer: 112