Question

★ STANDARD QUESTION 3 ★

★ CONTENT QUALITY SUPPORT REQUIRED ★

If
$\frac{1}{ \sqrt{2011 + \sqrt{ {2011}^{2} - 1} } } = \sqrt{m} - \sqrt{n}$
( m , n ) ∈ I⁺
Then evaluate the expression ( m + n )

★ LEVEL - 100 ★

Answer 1

Final result :
m = 1006 and n = 1005.

Steps involved:
1) Conjugate Multiplication
2)Solving expression .

For Calculation ,see pic

Hope, you understand my answer!

Answer 2

Hi friend, Harish here.

Given that,

$\frac{1}{ \sqrt{2011+ \sqrt{2011^{2}- 1 } } } = \sqrt{m} - \sqrt{n}$

To find,

The value of m+n.

Solution,

$\frac{1}{ \sqrt{2011+ \sqrt{2011^{2}- 1 } } } = \sqrt{m} - \sqrt{n}$

Now squaring on both sides we get.

⇒ $\frac{1}{2011+ \sqrt{2011^{2}- 1 } } =m+n-2\sqrt{mn}$

Now by rationalizing the denominator of LHS. We get,

$\frac{1}{2011+ \sqrt{2011^{2}- 1 } } \times \frac{2011- \sqrt{2011^{2}- 1 }}{2011- \sqrt{2011^{2}- 1 }} = \frac{2011- \sqrt{2011^{2}- 1 }}{1}$

⇒  ${2011- \sqrt{2011^{2}- 1 } = m+n-2 \sqrt{mn}$

⇒  $2011- \sqrt{(2010)(2012)} = m+n- \sqrt{4mn}$

Now let us compare both the sides.

Then,

$\sqrt{4mn} = \sqrt{(2010)(2012)}$

⇒  $mn = \frac{2010}{2} \times \frac{2012}{2}$

⇒ $mn = (1005) (1006)$

Then, We also know that.

m + n = 2011. ( By comparing).

So, m & n values are 1005 ,1006 . 

Therefore m+n is 2011.
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