The solution set of the inequality -6- x 5x2-7-2 x -3-9-25 -2 - 0 can be given by - a - b - c - then c - b -

X ϵ ( ∞, 1/5 ) First of all we must have 6 x ≥ 0 ∴ x ≤ 6 Also 5^x^2 7.2x+3.9 5^2^1/2≥ 0 ⇒ x^2 72/10x+39/10≥5/2 i.e., 25/10 ⇒ 5x^2 36x+7 ≥ 0 ⇒ (x 7)(5x 1)≥ 0 ...

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Standard X

Mathematics

Logarithmic inequalities

The solution ...

Question

The solution set of the inequality
$\sqrt{6 - x} (5^{x^{2} - 7.2 x + 3.9} - 25 \sqrt{2}) \geq 0$ can be given by $(- a, \frac{b}{c})$ , then c - b = ___

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Solution

$x ϵ (- \infty, \frac{1}{5})$
First of all we must have
$6 - x \geq 0 ∴ x \leq 6$
Also $5^{x^{2} - 7.2 x + 3.9} - 5^{2^{\frac{1}{2}}} \geq 0$
$\Rightarrow x^{2} - \frac{72}{10} x + \frac{39}{10} \geq \frac{5}{2} i . e ., \frac{25}{10}$
$\Rightarrow 5 x^{2} - 36 x + 7 \geq 0 \Rightarrow (x - 7) (5 x - 1) \geq 0$
$\Rightarrow x \leq \frac{1}{5}, x \geq 7 (R e j e c t e d a s x \leq 6)$
$∴ x \leq \frac{1}{5} \Rightarrow x ϵ (- \infty, \frac{1}{5})$

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The solution set of the inequality √6−x(5x2−7.2x+3.9−25√2)≥0 can be given by (−a,bc), then c - b = ___

xϵ(−∞,15) First of all we must have 6−x≥0 ∴x≤6 Also 5x2−7.2x+3.9−5212≥0 ⇒x2−7210x+3910≥52i.e.,2510 ⇒5x2−36x+7≥0⇒(x−7)(5x−1)≥0 ⇒x≤15,x≥7(Rejected as x≤6) ∴x≤15⇒xϵ(−∞,15)

The solution set of the inequality
$\sqrt{6 - x} (5^{x^{2} - 7.2 x + 3.9} - 25 \sqrt{2}) \geq 0$ can be given by $(- a, \frac{b}{c})$ , then c - b = ___