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WBJEE 2012 Math Solved Paper

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Question : 28 of 80

Marks: +1, -0

Maximum value of the function

f(x)=\frac{x}{8}+\frac{2}{x}

on the interval

[1, 6]

is

1
$\frac{9}{8}$
$\frac{13}{12}$
$\frac{17}{8}$

Solution:

f(x)=\;\frac{x}{8}+\;\frac{2}{x}

\therefore f'(x)=\;\frac{1}{8}-\;\frac{2}{x^{2}}=\;\frac{x^{2}-16}{8x^{2}}

For maximum or minimum

f'(x)

must be vanish,

\Rightarrow\frac{x^{2}-16}{8x^{2}}=0 \Rightarrow x=4,-4

x\in[1,6]

\therefore x \neq -4

Also, in

[1, 4],f'(x)<0 \Rightarrow f(x)

is decreasing.

In

[4, 6],f'(x)>0 \Rightarrow f(x)

is increasing

f(1)=\;\frac{1}{8}+\;\frac{2}{1}=\;\frac{17}{8}

f(6)=\;\frac{6}{8}+\;\frac{2}{6}=\;\frac{3}{4}+\;\frac{1}{3}=\;\frac{13}{12}

Hence, maximum value of

f(x)

in

[1, 6]

is

\frac{17}{8}

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