IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

naman

f:R+----->R+.if [f(xy)]²=x(f(y))² for all positive no.s x such that f(2)=6.find f(50).

aforadi

ans is +-30 put
x as 25
y as 2
do rate

naman

there is another ques if u solve this i will definately rate u

Que.suppose f is a real func satisfying f(x+f(x))=4f(x) and f(1)=4.find f(21)

aysh

Hi Naman!!!
It's really easy...
f(x+f(x))=4f(x)...............(1)

Now,
f(1)=4
Thus, by eq. (1), we get
f(1+f(1))=4f(1)
f(1+4)=4.4
f(5)=16...........................(2)

Again,
f(5+f(5))=4f(5)
f(5+16)=4.16 {from eq. (2)}
f(21)=64.

PLEAZZZZZZ RATE IF U LIKE IT....

punterjack

[f(xy)]²=x(f(y))²

putting y=1, we get,

f(x)2=xf(1)2..................................................1

given that f(2)=6

=> substituting x=2, in 1

36=2f(1)2=> f(1)= root(18)----------------------------2

f(50)=f(5*10)=

5 *f(10)---------------------------------3

f(10)=f(2*5)=

2* f(5)-----------------------------------4

f(5)=f(5*1)=

5*f(1)----------------------------------------5

as from eq2 f(1)=root18, and substituting the values of f(5) and f(10) from 4 and 5 in eq 3 we get f(50) as 30

If my answer is correct then please be kind enough to rate me as ive spent quite a bit of time on this one. if not then please do correct me for i too being a student would like my wrong fundas to be corrected if i had any

phanindraramesh

[f(xy)]²=x.(f(y))²for the given conditions

let x=2 and y=1

[f(2)]²=2.[f(1)]²

36 =2.[f(1)]²

f(1) =

18

=3

2

now let x=50 and y=1

[f(50)]²=50.[f(1)]²

[f(50)]²=50.*18

[f(50)]²=900

f(50) =+_30

but range of f is R+

hence f(50)=30

funky

[f(25*2)]^2 =25*[f(2)]^2
where f(2)=6
hence f(50)= ( 25*36)^(1/2)
f(50)=30

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