Suppose #f'(x)=p sin(1/2x)# where #p# is a constant, #f(0)=1# and #f(2pi)=0#. Find p and hence find f(x).?

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Answer 1 · 2018-02-11T19:03:15

# p=-1/4, and, f(x)=1/2{cos(x/2)+1}#.

#f'(x)=p*sin(x/2), (p" const.)"#

By the Definition of Integral, #f(x)=intf'(x)dx+c#,

#=intp*sin(x/2)dx+c#,

#=p*intsin(x/2)dx+c#,

#=p{-cos(x/2)/(1/2)}+c#.

#rArr f(x)=-2pcos(x/2)+c.............(ast)#.

Given that, #f(0)=1, (ast) rArr 1=-2pcos0+c#.

#:. c-2p=1..................................................................(ast_1)#.

Next, #f(2pi)=0, (ast) rArr 0=-2pcos(2pi/2)+c#.

#:. c+2p=0.................................................................(ast_2)#.

Solving #(ast_1) and (ast_2), c=1/2, p=-1/4#.

# rArr f(x)=1/2{cos(x/2)+1}#.