If x is real, find the range of y from the equation $x^{2} (y - 1) - 2 x + (2 y - 1)$ = 0

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Solution

The correct option is B

If x is real,

D $\geq$ 0

$b^{2} - 4 a c \geq 0$

$4 - 4 (y - 1) (2 y - 1) \geq 0$

$1 - (y - 1) (2 y - 1) \geq 0$

(y - 1) (2y - 1) $\leq$ 1

$2 y^{2} - 3 y \leq 0$

$y (2 y - 3) \leq 0$

(y - 0)( $y - (\frac{3}{2})) \leq 0$

$0 \leq y \leq$ $\frac{3}{2}$

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Similar questions

x^{2} + y^{2} - 2 x + 4 y = 0

The differential equation of the family of curves represented by the equation $x^{2} y = a$ , is

\frac{1}{2 x} - \frac{1}{y} = - 1

\frac{1}{x} + \frac{1}{2 y} = 8

\neq

\neq

x y + 3 y^{2} - x + 4 y - 7 = 0

2 x y + y^{2} - 2 x - 2 y + 1 = 0

\frac{1}{2 x} - \frac{1}{y} = - 1; \frac{1}{x} + \frac{1}{2 y} = 8; x \neq 0, y \neq 0

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