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NCHMCT JEE 2024 Solved Paper

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Question : 11 of 200

Marks: +1, -0

If

\mathrm{a}^{4}+\frac{1}{\mathrm{a}^{4}}=322

, then the value of

\mathrm{a}-\frac{1}{\mathrm{a}}

is :

$\frac{3}{2}$
2
$\frac{8}{27}$
4

Solution:

Concept:

Use the identity

\left(a-\frac{1}{a}\right)^2 = a^2+\frac{1}{a^2}-2

and relate it to

a^4+\frac{1}{a^4}

.

Explanation:

Given

a^{4}+\frac{1}{a^{4}}=322

.

\left(a^{2}+\frac{1}{a^{2}}\right)^{2}=a^{4}+\frac{1}{a^{4}}+2=322+2=324

So,

a^{2}+\frac{1}{a^{2}}=18

.

Now,

\left(a-\frac{1}{a}\right)^{2}=a^{2}+\frac{1}{a^{2}}-2=18-2=16

Hence,

a-\frac{1}{a}=\pm 4

.

Among the given options, the positive value is

4

.

Answer:

Option D:

4

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