The least integral value of x- which satisfy the inequality x-2-3x-4x-2-6x-8-0 is

X^2 3x+4x^2 6x+8≤0 ⇒x^2 3x+4(x 4)(x 2)≤0 We know that, x^2 3x+4 0 ∵ D = 9 120 The critical points are 2, 4 ∴ x ∈ (2, 4) Hence, the least integral value ...

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Standard XI

Mathematics

Points of Discontinuity

The least int...

Question

The least integral value of $x$ , which satisfy the inequality $\frac{x^{2} - 3 x + 4}{x^{2} - 6 x + 8} \leq 0$ is

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Solution

$\frac{x^{2} - 3 x + 4}{x^{2} - 6 x + 8} \leq 0$
$\Rightarrow \frac{x^{2} - 3 x + 4}{(x - 4) (x - 2)} \leq 0$

We know that,
$x^{2} - 3 x + 4 > 0 ∵ D = 9 - 12 < 0$
The critical points are $2, 4$
$∴ x \in (2, 4)$

Hence, the least integral value of $x$ is $3$ .

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Points of Discontinuity

Standard XI Mathematics

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The least integral value of x, which satisfy the inequality x2−3x+4x2−6x+8≤0 is

x2−3x+4x2−6x+8≤0 ⇒x2−3x+4(x−4)(x−2)≤0 We know that, x2−3x+4>0 ∵D=9−12<0 The critical points are 2,4 ∴x∈(2,4) Hence, the least integral value of x is 3.

The least integral value of $x$ , which satisfy the inequality $\frac{x^{2} - 3 x + 4}{x^{2} - 6 x + 8} \leq 0$ is

$\frac{x^{2} - 3 x + 4}{x^{2} - 6 x + 8} \leq 0$
$\Rightarrow \frac{x^{2} - 3 x + 4}{(x - 4) (x - 2)} \leq 0$

We know that,
$x^{2} - 3 x + 4 > 0 ∵ D = 9 - 12 < 0$
The critical points are $2, 4$
$∴ x \in (2, 4)$

Hence, the least integral value of $x$ is $3$ .