The function y={2x-x²}

Correct Answer is : increase in (0,1)(0,1)(0,1) but decreases in (1,2)(1,2)(1,2)

increase in (0,1) but decreases in (1,2)

increases in (1,2) but decreases {in}(0,1)

Correct Answer is : increase in (0,1)(0,1)(0,1) but decreases in (1,2)(1,2)(1,2)

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UPSEE 2017 Solved Paper

Question : 115 of 150

Marks: +1, -0

The function

y=\sqrt{2x-x^{2}}

Solution:

Given,

y=\sqrt{2x-x^{2}}

By differentiating both side w.r.t.

x'

, we get

\frac{dy}{dx} = \frac{1}{2\sqrt{2x-x^{2}}} \times (2-2x)

= \frac{1-x}{\sqrt{2x-x^{2}}}

Here,

\frac{dy}{dx}

is defined for

2x-x^{2} > 0

i.e.

0 < x < 2

Now,

\frac{dy}{dx} > 0

When,

1-x > 0

[\because \sqrt{2x-x^{2}}

is always positive]

\Rightarrow x < 1

\therefore

Given function increases in

0 < x < 1

And

\frac{dy}{dx} < 0

When,

\int -x < 0

[\because \sqrt{2x-x^{2}}

is always positive

]

x > 1

\therefore

Given function decreases in

1 < x < 2

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