If x=(7-43) then (x2+1)x2=

x=(7-43) then (x2+1)x2= x2 = 7 - 43 (x2+1)x2 =7 - 43 +1 7 - 43 = 7 - 43 +1 7 - 43 ×7 + 437 + 43 = 8 - 437 + 43 49 -48 =8 - 437 + 43 1 =8 - 43

If x=(7-43) then (x2+1)x2=

x=(7-43) then (x2+1)x2= x2 = 7 - 43 (x2+1)x2 =7 - 43 +1 7 - 43 = 7 - 43 +1 7 - 43 ×7 + 437 + 43 = 8 - 437 + 43 49 -48 =8 - 437 + 43 1 =8 - 43

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$I f x = \sqrt{(7 - 4 \sqrt{3})} t h e {n (\frac{(x^{2} + 1)}{x})}^{2} =$

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$8 - 4 \sqrt{3}$

$8 + 4 \sqrt{3}$

$16 - 4 \sqrt{3}$

$27 - 4 \sqrt{3}$

answer is A.

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Detailed Solution

$x = \sqrt{(7 - 4 \sqrt{3})} t h e {n (\frac{(x^{2} + 1)}{x})}^{2} = x^{2} = 7 - 4 \sqrt{3} {(\frac{(x^{2} + 1)}{x})}^{2} = (\frac{7 - 4 \sqrt{3} + 1}{7 - 4 \sqrt{3}}) = (\frac{7 - 4 \sqrt{3} + 1}{7 - 4 \sqrt{3}}) \times (\frac{7 + 4 \sqrt{3}}{7 + 4 \sqrt{3}}) = (\frac{(8 - 4 \sqrt{3}) (7 + 4 \sqrt{3})}{49 - 48}) = (\frac{(8 - 4 \sqrt{3}) (7 + 4 \sqrt{3})}{1}) = 8 - 4 \sqrt{3}$