CAT 2023 Slot 3 Solved Paper

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Question : 45 of 66

Marks: +1, -0

If

{\x}

is a positive real number such that

x^8+ (\;{1}/{x}) ^8=47

, then the value of

x^9+ (\;{1}/{x}) ^9

is

$40 {√{5}}$
$30 {√{5}}$
$36 {√{5}}$
$34 {√{5}}$

Solution:

It is given that

x^8+ (\;{1}/{x}) ^8=47

, which can be written as:

⇒ (x^4) ^2+ (\;{1}/{x^4}) ^2=47

\;⇒ (x^4+\;{1}/{x^4}) ^2-2 ⋅ x^4 ⋅ \;{1}/{x^4}=47

\;⇒ (x^4+\;{1}/{x^4}) ^2=49

\;⇒x^4+\;{1}/{x^4}=7

Similarly,

x^4+\;{1}/{x^4}=7

can be expressed as:

\;⇒ (x^2) ^2+ (\;{1}/{x^2}) ^2=7

\;⇒ (x^2+\;{1}/{x^2}) ^2-2 ⋅ x^2 ⋅ \;{1}/{x^2}=7

\;= (x^2+\;{1}/{x^2}) ^2=9

\;⇒ x^2+\;{1}/{x^2}=3

By the same logic, we get

x+\;{1}/{x}={√{5}}

Now,

x^3+\;{1}/{x^3}= (x+\;{1}/{x}) ^3-3 ⋅ x ⋅ \;{1}/{x} (x+\;{1}/{x})

⇒ x^3+\;{1}/{x^3}=({√{5}})^3-3 {√{5}}=2 {√{5}}

By the same logic, we can say that

\;⇒x^9+\;{1}/{x^9}= (x^3+\;{1}/{x^3}) ^3-3 ⋅ x^3 ⋅ \;{1}/{x^3} (x^3+\;{1}/{x^3})

\;⇒x^9+\;{1}/{x^9}=(2 {√{5}})^3-3(2 {√{5}})

\;⇒x^9+\;{1}/{x^9}=40 {√{5}}-6 {√{5}}=34 {√{5}}

The correct option is D

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