Question

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Q17. If $\sf{\dfrac{1}{\sqrt{2018+\sqrt{2018^2-1}}}=\sqrt{p}-\sqrt{q}}$ where p and q are positive integer then p – q is :

(A) 1
(B) 2
(C) 3
(D) 4​

Answer 1

Appropriate Question

If

$\rm :\longmapsto\:\sf{\dfrac{1}{\sqrt{2018+\sqrt{2018^2-1}}}=\sqrt{p}-\sqrt{q}}$

where p and q are positive real numbers, then the value of p - q is

(A) 1

(B) 2

(C) 3

(D) 4

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:\sf{\dfrac{1}{\sqrt{2018+\sqrt{2018^2-1}}}=\sqrt{p}-\sqrt{q}}$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:p - q$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf{(x + y)(x - y) \: = \: {x}^{2} - {y}^{2} }}$

$\boxed{ \bf{ {(x - y)}^{2} \: = \: {x}^{2} + {y}^{2} - 2xy }}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\sf{\dfrac{1}{\sqrt{2018+\sqrt{2018^2-1}}}=\sqrt{p}-\sqrt{q}}$

Consider,

$\rm :\longmapsto\:\sf{\dfrac{1}{\sqrt{2018+\sqrt{2018^2-1}}}}$

On rationalizing the denominator, we get

$\rm = \:\sf{\dfrac{1}{\sqrt{2018+\sqrt{2018^2-1}}}} \times \dfrac{\sqrt{2018 - \sqrt{2018^2-1}}}{\sqrt{2018 - \sqrt{2018^2-1}}}$

$\rm \: \: = \:\dfrac{\sqrt{2018 - \sqrt{2018^2-1}}}{\sqrt{(2018) ^{2} - (\sqrt{2018^2-1})^{2} }}$

$\rm \: \: = \:\dfrac{\sqrt{2018 - \sqrt{(2018-1)(2018 + 1)}}}{\sqrt{(2018)^{2} - (2018^2-1)}}$

$\rm \: \: = \:\dfrac{\sqrt{2018 - \sqrt{(2017)(2019)}}}{\sqrt{(2018)^{2} - 2018^2 + 1}}$

$\rm \: \: = \:\dfrac{\sqrt{2018 - \sqrt{(2017)(2019)}}}{\sqrt{1}}$

$\rm \: \: = \: \sqrt{2018 - \sqrt{(2019)(2017)} }$

$\rm \: \: = \: \sqrt{\dfrac{4036}{2} - \sqrt{(2019)(2017)} }$

$\rm \: \: = \: \sqrt{\dfrac{2019 + 2017}{2} - \sqrt{(2019)(2017)} }$

$\rm \: \: = \: \sqrt{\dfrac{2019}{2} + \dfrac{2017}{2} - 2 \sqrt{ \dfrac{(2019)(2017)}{4} } }$

$\rm \: \: = \: \sqrt{\dfrac{2019}{2} + \dfrac{2017}{2} - 2 \sqrt{ \dfrac{(2019)}{2} \dfrac{(2017)}{2} } }$

$\rm \: \: = \: \sqrt{ {\bigg( \sqrt{\dfrac{2019}{2}} - \sqrt{\dfrac{2017}{2} } \bigg) }^{2} }$

Hence,

$\rm :\longmapsto\:\sf{\dfrac{1}{\sqrt{2018+\sqrt{2018^2-1}}}}= \sqrt{\dfrac{2019}{2} } - \sqrt{\dfrac{2017}{2} }$

But it is given that

$\rm :\longmapsto\:\sf{\dfrac{1}{\sqrt{2018+\sqrt{2018^2-1}}}=\sqrt{p}-\sqrt{q}}$

So, on comparing we get

$\bf :\longmapsto\:p = \dfrac{2019}{2}$

and

$\bf :\longmapsto\:q = \dfrac{2017}{2}$

Thus,

$\bf :\longmapsto\:p - q$

$\rm \: \: = \:\dfrac{2019}{2} - \dfrac{2017}{2}$

$\rm \: \: = \:\dfrac{2019 - 2017}{2}$

$\rm \: \: = \:\dfrac{2}{2}$

$\rm \: \: = \:1$

$\bf\implies \:p - q = 1$

Therefore,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{\boxed{\bf{Option \: (A) \: is \: correct }}}$

Answer 2

Answer:

Use (a+b) * (a-b) = a^2 - b^2

The key is to represent the two numbers in the product as X-1 and X+1 and then use the above identity.

Starting from innermost root we write 2017 = 2018-1

And 2019 = 2018+1

Hence in the root 1+ 2017*2019 is equivalent to

1+(2018-1)*(2018+1) = 1+ 2018^2 - 1^2

As 1^2 =1

Therefore we have 1+2018^2 -1 = 2018^2

The innermost root is solved and we now have 1+2016*2018

We can use the same approach to solve this root and we will get 2017 after which we will solve using the same approach

The answer will come out to be 2015.

P.s. Comment if you still have any problem