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

$[0, 2]$

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$[0, \frac{3}{2}]$

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$[0, \frac{7}{2}]$

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(- $\infty$ , 0 ]

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Solution

The correct option is B
$[0, \frac{3}{2}]$

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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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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\neq

\neq

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