If $x^{2} + y^{2} = 1$ then value of ${(3 x - {4 x}^{3})}^{2} + {(3 y - {4 y}^{3})}^{2}$ is equal to:

\frac{1}{2}

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\frac{3}{2}

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\frac{3}{4}

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Solution

The correct option is B 1
We have,
$x^{2} + y^{2} = 1$ $. . . . . . . . . . . . . (1)$

Since,
$(3 x - 4 x^{3})^{2} + (3 y - 4 y^{3})^{2}$
$= 9 x^{2} + 16 x^{6} - 24 x^{4} + 9 y^{2} + 16 y^{6} - 24 y^{4}$
$= 16 (x^{6} + y^{6}) - 24 (x^{4} + y^{4}) + 9 (x^{2} + y^{2})$
$= 16 (x^{6} + y^{6}) - 24 (x^{4} + y^{4}) + 9$ $x^{2} + y^{2} = 1$
$= 16 ((x^{2})^{3} + (y^{2})^{3}) - 24 (x^{4} + y^{4}) + 9$
$= 16 [(x^{2} + y^{2}) (x^{4} + y^{4} - x^{2} y^{2})] - 24 (x^{4} + y^{4}) + 9$
$= 16 (x^{4} + y^{4} - x^{2} y^{2}) - 24 (x^{4} + y^{4}) + 9$
$= 9 - 8 (x^{4} + y^{4}) - 16 x^{2} y^{2}$
$= 9 - 8 [(x^{2} + y^{2})^{2}]$
$= 9 - 8 [1^{2}]$
$= 9 - 8$
$= 1$

Hence, this is the answer.

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