The set of all integral values of $x$ for which $5 x - 1 < (x + 1)^{2} < 7 x - 3$ , is

ϕ

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{1}

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Solution

The correct option is D $ϕ$
$(5 x - 1) < (x + 1)^{2} < (7 x - 3)$
$(5 x - 1) < x^{2} + 1 + 2 x < (7 x - 3)$
$x^{2} - 3 x + 2 > 0$ & $x^{2} - 5 x + 4 < 0$
Factorizing we get
$(x - 1) (x - 2) > 0$ & $(x - 1) (x - 4) < 0$
$x \in (- \infty, 1) \cup (2, \infty)$ & $x \in (1, 4)$
Now common integral value is nothing so there are no integral values.
Option A is correct.

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