For $x > 1$ , $y = log x - (x - 1)$ satisfies the inequality

x - 1 > y

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x^{2} - 1 > y

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y > x - 1

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\frac{x - 1}{x} < y

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Solution

The correct options are
A $x - 1 > y$
B $\frac{x - 1}{x} < y$
C $x^{2} - 1 > y$
Differentiating the function $y = log x - (x - 1)$ , we get
$f^{'} (x) = \frac{1}{x} - 1 = \frac{1 - x}{x} < 0$ for $x \in (1, \infty)$
Hence $f (x)$ is a decreasing function, on $[1, \infty)$ .
Thus, for $x > 1$ , we have $f (x) < f (1) \Rightarrow log x < x - 1$ . Also $x^{2} - 1 > x - 1$ , so that $x^{2} - 1 > log x$ .
Similarly, $(x - 1) / x < log x$ .

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