AP EAPCET 04-Jul-2022 Shift 1 Paper

f(x⋅f(y))=f(xy)+x.......(i)
x↔y
⇒‌‌f(y⋅f(x))=f(xy)+y....(ii)
x↔f(x) in Eq. (i),
f(f(x)⋅f(y))=f(y⋅f(x))+f(x).........(iii)
⇒‌‌f(fx)⋅f(y))=f(xy)+y+f(x)
[from Eq. (ii)] ... (iv)
Again x↔y in Eq. (iv),
f(f(y)⋅f(x))=f(yx)+x+f(y).....(v)
Eq. (iv) =Eq⋅(v),
‌f(xy)+y+f(x)=f(xy)+x+f(y)
⇒f(x)−x=f(y)−y............(vi)
⇒‌‌f(x)−x=λ=f(y)−y
Substituting f(x)=λ+x in Eq. (i), we have
‌x⋅f(y)+λ=(xy+λ)+x
‌⇒xf(y)=xy+x
‌∴x(y+λ)=xy+x‌‌[∵f(y)=λ+y]
‌⇒λx=x
‌⇒λ=1
∴‌‌f(x)=x+λ=x+1
Now, ‌

lim

x→0

‌

(f(x)) 1 3 −1

(f(x))‌ 1 2 −1

=

lim

x→0

‌

(1+x) 1 3 −1

(1+x)‌ 1 2 −1

=‌

lim

x→0

(‌

(1+x) 1 4 −1

1+x−1

)(‌

1+x−1

(1+x)‌ 1 2 −1

)
=‌‌

1

3 2

=‌

2

3