Question

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Answer 1

Answer:

Required relation between a and b is a > b.        

Option ( 2 ) is correct.

Step-by-step explanation:

Given numeric value of  a is $\dfrac{9}{\sqrt{11}-\sqrt2}$ and b is $\dfrac{6}{3\sqrt3}$.

Using Rationalization : Multiply denominator & numerator, by the original denominator with an opposite sign between the irrational numbers, so that the form condition( in denominator ) can satisfy this formula : ( a + b )( a - b ) = a^2 - b^2.

It means, here, in the numeric value of a, we have to multiply the numerator and denominator by √11 + √2.  

Thus,  

$a=\dfrac{9}{\sqrt{11}-\sqrt2}\times\dfrac{\sqrt{11}+\sqrt2}{\sqrt{11}+\sqrt2}\\\\\\\implies a=\dfrac{9(\sqrt{11}+\sqrt2)}{(\sqrt{11}-\sqrt{2})(\sqrt{11}+\sqrt{2})}\\\\\\\implies a=\dfrac{9(\sqrt{11}+\sqrt{2})}{(\sqrt{11})^2-(\sqrt2)^2}\\\\\\\implies a=\dfrac{9(\sqrt{11}+\sqrt2)}{11-2}\\\\\\\implies a=\dfrac{9(\sqrt{11}+\sqrt{2}}{9}$

= > a = √11 + √2                

= > a^2 = ( √11 +  √2 )^2                  { Square on both sides }

= > a^2 = ( √11 )^2 + ( √2 )^2 + 2( √11 x  √2 )                    ( a + b )^2 = a^2 + b^2 + 2ab

= > a^2 = 11 + 2 + 2 √22

= > a^2 = 13 + 2 √22                  

Given,

        $\implies b=\dfrac{6}{3\sqrt3}$

= > b = 2 /  √3

= > b = ( 2 x √3 ) / ( √3 x  √3 )                  { Multiply & Divide by  √3 }

= > b = 2 √3 / 3

= > b^2 = ( 2 √3 / 3 )^2

= > b^2 = ( 4 x 3 / 9 )

= > b^2 = 4 / 3  

= > b^2 » 1.4

= > b^2 » 1.4 < 13 < 13 + 2√22    

= > b^2 < 13 + 2√22                  

= > b^2  < a^2  

We know that the square of any negative number can also be greater than the square of any positive number, but here, numeric value of a is √11 + √2, which can't be negative, since it is without any negative sign. Therefore,

= > b < a    or a > b  

Hence the required relation between a and b is a > b.