If x = 8 + 3 7 and x y = 1 , then the value of 1 x 2 + 1 y 2 is

Explanation: ∵ x = 8 + 3 7 y = 1 8 + 3 7 = 8 − 3 7 Now, 1 x 2 + 1 y 2 = x 2 + y 2 ( x y ) 2 = ( x + y ) 2 − 2 x y ( x y ) 2 = ( x + y ) 2 − 2 = ( 8 + 3 7 + 8 − 3 7 ) 2 − 2 = 16 2 − 2 = 254

Question:

If $x = 8 + 3 \sqrt{7}$ and $x y = 1,$ then the value of $\frac{1}{x^{2}} + \frac{1}{y^{2}}$ is

WBJEE

Updated On: Aug 9, 2024

(A) 254
(B) 192
(C) 292
(D) 66

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The Correct Option is A

Solution and Explanation

Explanation:

∵ x = 8 + 3 \sqrt{7}

y = \frac{1}{8 + 3 \sqrt{7}} = 8 - 3 \sqrt{7}

Now,

\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{x^{2} + y^{2}}{(x y)^{2}} = \frac{(x + y)^{2} - 2 x y}{(x y)^{2}}

= (x + y)^{2} - 2

= (8 + 3 \sqrt{7} + 8 - 3 \sqrt{7})^{2} - 2

= 16^{2} - 2 = 254

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