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AFCAT Model Paper 1

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Question : 52 of 100

Marks: +1, -0

If

x=\sqrt{3}+\sqrt{2}

, then find the value of

\left[x^{2}-\left(\frac{1}{x}\right)^{2}\right]

.

$4\sqrt{6}$
5
$5\sqrt{6}$
$8\sqrt{3}$

Solution:

Given:

x = \sqrt{3} + \sqrt{2}

Formula used:

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

Calculation:

x = \sqrt{3} + \sqrt{2}

...(i)

⇒

\frac{1}{x} = \frac{1}{\sqrt{3}+\sqrt{2}}

⇒

\left[\frac{1}{\sqrt{3}+\sqrt{2}}\right] \times \left[\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right]

⇒

(\sqrt{3}-\sqrt{2})

⇒

\frac{1}{x} = (\sqrt{3}-\sqrt{2})

...(ii)

On adding (i) and (ii), we get

\left[x+\frac{1}{x}\right] = \sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}

⇒

\left[x+\frac{1}{x}\right] = 2\sqrt{3}

...(iii)

On subtracting (ii) by (i), we get

x - \frac{1}{x} = \sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}

⇒

x - \frac{1}{x} = 2\sqrt{2}

Now, we have to find the value of

x^{2} - \frac{1}{x^{2}}

x^{2}-\frac{1}{x^{2}}=\left[x-\frac{1}{x}\right]\left[x+\frac{1}{x}\right]

From equation (ii) & (iii)

⇒

2\sqrt{2} \times 2\sqrt{3}

⇒

4\sqrt{6}

∴ The value of

x^{2}-\frac{1}{x^{2}}

is

4\sqrt{6}

.

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