IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

neeraj_agarwal_1990

2f(x)=f(xy)+f(x/y)

Put y=1/x

2f(x) = f(1) +f(x^2)
since f(1)=1,
2f(x) =f(x^2)

differentiate..
2f'(x) = 2xf'(x^2)
=> xf'(x^2) = f'(x) ......(1)

now again consider 2f(x)=f(xy)+f(x/y)
differentiate..

2f'(x) = yf'(xy) + (1/y)f'(x/y)
put y=1/x..

2f'(x)=f'(1)/x + xf'(x^2)

using (1), and f'(1) = 1/ln6
2f'(x)=1/(xln6) + f'(x)

=> f'(x) = 1/xln6

=> dy/dx =1/xln6
=> yln6=lnx +c (on solving variable separable DE)
atx=1,y=0 (given)
putting these values, c=0

so y=f(x) = lnx / ln6

putting x=7776= 6^5
=> f(7776) = 5

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