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Method of Intervals

The solution ...

Question

The solution set of $\frac{3 x^{2} - 7 x + 8}{x^{2} + 1} \leq 2$ is

A

[1, 4]

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B

[1, 2]

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C

[2, 6]

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D

[1, 6]

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Solution

The correct option is D $[1, 6]$
$\frac{3 x^{2} - 7 x + 8}{x^{2} + 1} \leq 2 \Rightarrow \frac{3 x^{2} - 7 x + 8}{x^{2} + 1} - 2 \leq 0 \Rightarrow \frac{x^{2} - 7 x + 6}{x^{2} + 1} \leq 0 \Rightarrow \frac{(x - 6) (x - 1)}{x^{2} + 1} \leq 0$

$∴ x \in [1, 6]$

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Method of Intervals

Standard XII Mathematics

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