The inequality $(x - 1) ln (2 - x) < 0 < 0$ holds, if $x$ satisfies

1 < x < 2

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x > 0

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0 < x < 1

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N o n e o f t h e s e

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Solution

The correct option is A $1 < x < 2$
Given,

$(x - 1) ln (2 - x) < 0 < 0$

$(x - 1) ln (2 - x) < 0$ and $0 < 0$

$0 < 0$ is false

$∴ (x - 1) ln (2 - x) < 0$

$ln (2 - x) < 0$

$ln (2 - x) < ln (e^{0})$

$ln (2 - x) < ln (1)$

$2 - x < 1$ and $2 - x > 0$

$1 < x < 2$

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