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) can be -1 or 0

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...

@Uttara....
It is not lengthy.......i have solved and write every steps...so that it is clear to you how to solve....If you have to learn any new method the....it must that you learn it clearly....shortcuts or available in markets...but shortcut with concept are not available everywhere....
So If i have to solve i can solve it in 2 minutes...because i am familiar with this method....after all if u also try to understand it more clearly...then u will not face any problem in such type of function