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KEAM 2013 Math Paper

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Question : 103 of 120

Marks: +1, -0

If the point

(a, b)

y=\sqrt{x}

is closest to the point

(1, 0),

then the value of

ab

is

$\frac{1}{2}$
$\frac{\sqrt{2}}{2}$
$\frac{1}{4}$
$\frac{\sqrt{2}}{4}$
1

Solution:

Given curve is

y=\sqrt{x}

....(i)

since, the point

(a, b)

is on the curve (i).

Then,

b=\sqrt{a} \Rightarrow b^{2}=a

....(ii)

Let

D=

distance between

(a, b)

and (1,0) .

\therefore \;\; D^{2}=(a-1)^{2}+(b-0)^{2}=a^{2}+1-2a+b^{2}

\Rightarrow \;\; D^{2}=a^{2}+1-2a+a

[from Eq. (ii)]

\Rightarrow \;\; D^{2}=a^{2}-a+1

......(iii)

Let

P=a^{2}-a+1

Now,

P

is max\/min according as

D^{2}

is max\/min

\therefore \;\; \frac{dP}{da}=2a-1

and

\frac{d^{2}p}{da^{2}}=2>0

(min)

For maximum or minimum of

P

,

put

\frac{dP}{da}=0 \Rightarrow 2a-1=0

\Rightarrow \;\; a=\frac{1}{2}

\therefore \text{At } \left(a=\frac{1}{2}\right),

point

(a, b)

is closest to the point (1,0)

From Eq. (ii).

b^{2}=\frac{1}{2} \Rightarrow b=\frac{1}{\sqrt{2}}

\therefore \;\; ab=\frac{1}{2} \cdot \frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}

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