Question bank/JEE Main/Mathematics

The highest composite number satisfying the rational inequality
$1 < \frac{3x^{2} - 7x + 8}{x^{2} + 1} \leq 2$ is

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$1 < \frac{3x^{2} - 7x + 8}{x^{2} + 1}\ \leq \ 2$
$\frac{3x^{2} - 7x + 8}{x^{2} + 1} > 1$ and
$\frac{3n^{2} - 7x + 8}{n^{2} + 1} \leq 2$
(A) $\frac{3x^{2} - 7x + 8 - x^{2} - 1}{n^{2} + 1} > 0$
$\frac{2x^{2} - 7x + 7}{x^{2} + 1} > 0$
[IMAGE] (B) $\frac{3x^{2} - 7x + 8 - 2x^{2} - 2}{x^{2} + 1} \leq 0$
$\frac{x^{2} - 7x + 6}{x^{2} + 1} \leq 0$
Critical points: x = 1,6
Combining both (A) and (B)
$x \in (A) \cap (B)$
$\Rightarrow x \in \left( \frac{7 + \sqrt{7}}{4};6 \right\rbrack$
$\therefore$ highest composite number =

Source: Grass-G / M-05 Mathematics Paper 10 October.docx