If x -3-2 -2- thenthevalueof - x -1- x is-A- 1B- 2C- 2 -2D- -1

The correct option is C 2√2 [ ( √(x) + 1/√(x))^2 = x + 1/x + 2 = ( 3 + 2√(2)) + 1/(3 + 2√(2) ) + 2; = (3 + 2√(2) ) + 1/(3 + 2√(2) )×(3 2√(2) )/(3 2√ ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

If x =3+2 √2,...

Question

If $x = 3 + 2 \sqrt{2}, t h e n t h e v a l u e o f (\sqrt{x} + \frac{1}{\sqrt{x}})$ is:

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2√2

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-1

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Solution

The correct option is C
2√2

$\begin{matrix} {(\sqrt{x} + \frac{1}{\sqrt{x}})}^{2} = x + \frac{1}{x} + 2 = (3 + 2 \sqrt{2}) + \frac{1}{(3 + 2 \sqrt{2})} + 2 = (3 + 2 \sqrt{2}) + \frac{1}{(3 + 2 \sqrt{2})} \times \frac{(3 - 2 \sqrt{2})}{(3 - 2 \sqrt{2})} + 2 = (3 + 2 \sqrt{2}) + (3 - 2 \sqrt{2}) + 2 = 8. ∴ (\sqrt{x} + \frac{1}{\sqrt{x}}) = 2 \sqrt{2} \end{matrix}$

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If x=3+2√2,thenthevalueof(√x+1√x) is:

The correct option is C 2√2 (√x+1√x)2=x+1x+2=(3+2√2)+1(3+2√2)+2=(3+2√2)+1(3+2√2)×(3−2√2)(3−2√2)+2=(3+2√2)+(3−2√2)+2=8.∴(√x+1√x)=2√2

If $x = 3 + 2 \sqrt{2}, t h e n t h e v a l u e o f (\sqrt{x} + \frac{1}{\sqrt{x}})$ is: