Sol. We have,

f = for all x, y ∈ R such that y ≠ 0

and f (y) ≠ 0

∴ f(1) =

⇒ {f(1)} 2 –f(1) = 0

⇒ f(1) = 0 or, f(1) = 1

⇒ f(1) = 1 [  f(1) ≠ 0]

Now,

f ′ (x) =

=

= f(x).

= . [  f(1) = 1]

= f ′ (1)

⇒ = [  f ′ (1) = 2]

⇒ = dx

Integrating both sides, we get

log f(x) = 2 log x + log C

⇒ f(x) = Cx 2 ... (i)

But, f(1) = 1 Therefore, C = 1

Putting C = 1 in (i), we obtain

f(x) = x 2 Thus, we have to find the area enclosed by the curves y = x

2 , x 2 + y 2 =2 and x- axis. Graphs of these curves and region enclosed are shown in fig. Clearly, y =x 2 and

x 2 + y 2 = 2 intersect at (1, 1) and (–1, 1). The shaded region in fig. is sliced into horizontal strips. The approximating rectangle shown in fig. is of length = (x 2 –x 1 ), width = Δ y and it can more vertically between y = 0 and y = 1. So,

Required area

=

=

=

= + sin –1 – = sq. units.