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Manipal Entrance Test 2014 Math Paper

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Question : 41 of 80

Marks: +1, -0

The largest term in the sequence

a_{n}=\frac{n^{2}}{n^{3}+200}

is given by

$\frac{529}{49}$
$\frac{8}{89}$
$\frac{49}{543}$
None of these

Solution:

Consider the function

f(x)=\frac{x^{2}}{x^{3}+200}

f'(x)=x\frac{400-x^{3}}{(x^{3}+200)^{2}}=0

⇒

x^{3}=400

⇒

x=(400)^{1/3}

When

x<0, f'(x)>0

x>0, f'(x)<0

∴

f(x)

has maximum at

x=(400)^{1/3}

Since,

7<(400)^{1/3}<8

, either

a_{7}

or

a_{8}

is the greatest term of the sequence.

∵

a_{7}=\frac{49}{543}, a_{8}=\frac{8}{89}

and

\frac{49}{543} > \frac{8}{89}

a_{7}=\frac{49}{543}

is the greatest.

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