x and y are the sides of two squares such that y=x−x2 T

x and y are the sides of two squares such that $y = x - x^{2}$ . The rate of change of area of the second square with respect to that of the first square is

$2 x^{2} + 3 x^{2} + 1$

$3 x^{2} + 2 x^{2} - 1$

$2 x^{2} - 3 x + 1$

$3 x^{2} + 2 x + 1$

Please scroll down to see the correct answer and solution guide.

Right Answer is: C

SOLUTION

Let $A = x^{2}$ and $B = y^{2} = (x - x^{2})^{2}$
$∴ \frac{d B}{d A} = \frac{\frac{d B}{d x}}{\frac{d A}{d x}} = \frac{2 (x - x^{2}) (1 - 2 x)}{2 x} = (1 - x) (1 - 2 x) = 2 x^{2} - 3 x + 1$

x and y are the sides of two squares such that y=x−x2 T

Right Answer is: C

SOLUTION

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