UPSEE 2017 Solved Paper

Given,
y=√2x−x2
By differentiating both side w.r.t. x′, we get

dy

dx

=

1

2√2x−x2

×(2−2x)
=

1−x

√2x−x2

Here,

dy

dx

is defined for
2x−x2>0
i.e. 0<x<2
Now,

dy

dx

>0
When, 1−x>0
[∵√2x−x2 is always positive]
⇒x<1
∴ Given function increases in 0<x<1
And

dy

dx

<0
When, ∫−x<0
[∵√2x−x2 is always positive ]
x>1
∴ Given function decreases in 1<x<2.