SSC CGL 16 Aug 2017 Shift 1 Paper

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Question : 62 of 100

Marks: +1, -0

If

x=(√{2}+1) /(√{2}-1),

then what is the value of

(x^{5}+x^{4}+x^{2}+x ) / x^{3}\ ?

40
37.5
38
$20√2$

Solution:

x=(√{2}+1) /(√{2}-1) = {√{2}+1}/{√{2}-1}

×{√{2}+1}/{√ {2}+1}

= {2+1+2√{2}}/{2-1}=3+2 √ {2}

and

{1}/{x}= {1}/{3+2√{2}}= {1}/{3+2 √ {2}}

×{3-2 √{2}}/{3-2 √{2}}

= {3-2√{2}}/{9-8}=3-2 √ {2}

x+ {1}/{x}=3+2√{2}+3-2 √ {2}=6

On squaring both sides,

x^{2}+ {1}/{x^{2}}+2=36

⇒

x^{2}+ {1}/{x^{2}}=34

∴

Expression

= {x^{5}+x^{4}+x^{2}+x}/{x^{3}}

=x^{2}+x+ {1}/{x}+ {1}/{x^{2}}

=x^{2}+ {1}/{x^{2}}+ {1}/{x}+x =34+6=40

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