Functions Paragraph

From the given function

f(0) = 2f(0) + f(0)²

=> f(0) = -1

Then φ(0) = 0

But ans is wrong

What's the mistake???

f(0)=0 or f(0)=-1..u missed that

Ok Thanks both of u :)

Then for 2nd one

Suppose f(0) = -1

Then x=1 & y=0

f(1+0) = -1

=> φ(1) = 0

=> Discontinuity at x= 1 ???

Can't v consider f(0) = -1 ???

But ans is given as (4)

Hiiiiiiiiiiiii

I am solving this question by a general method....try to get it.
Since function is of two variables (x,y) hence x and y are independent on each other...means for puting x=2....doesn't bind us to put y=some value...I mean change in y doesn't depend on change in x.
So dy/dx = 0 ....{Because dy/dx means change in y per unit change in x}

SO now question is
f(x+y) = f(x) + f(y) + f(x).f(y)
differentiate both side...
f'(x+y){1+dy/dx} = f'(x) + f'(y)*dy/dx + f'(x)*f(y) + f(x)*f'(y)*dy/dx
put dy/dx = 0
we have

f'(x+y) = f'(x) + f'(x)*f(y) ----------------(1)
Now put y=0
f'(x) = f'(x) + f'(x)*f(0)
f'(x) = f'(x) [ 1 + f(0) ]
so f(0) = 0 ----------------------(2)

Now we have \phi(x) = 1+f(x)
so \phi(0) = 1+f(0)
hence,\phi(0) = 1 + 0 ....{from (2) }
so \phi(0) = 1.
Therefore correct option of first question is (c)

Question no:2-----

Given\omega(x) = ln(1+f(x))
now from eq(1)
put x=0
f'(y) = f'(0) + f'(0) * f(y)
Let f'(0) =k
so f'(y) = k(1+f(y))
Hence this can be written as
f'(x) = k(1+f(x)) {by replacing y by x}
also manupulated as
[f'(x)/(1+f(x)].dx = k.dx

integrate it we have
ln(1+f(x)) = k.x + c
Now put x=0 to find out the value of c.
Hence ln(1+f(0)) = c = ln(1) = 0
So ln(1+f(x)) = \omega(x) = k.x

And k.x is always continuos everywhere...

\omega(x) = k.x ....means
\omega(x) is continuos everywhere..

Hence option is (4)

If you will understand this ......i will feel very happy...