Inequalities, Intervals: Very Difficult Problems with Solutions

Solution:
Cube: [tex]S_1=6a^2[/tex]. Box with dimensions [tex]a+4[/tex], [tex]a-2[/tex], [tex]a-1[/tex]:
[tex]S_2=2\big[(a+4)(a-2)+(a+4)(a-1)+(a-2)(a-1)\big][/tex]
[tex]=2(a^2+2a-8+a^2+3a-4+a^2-3a+2)=6a^2+4a-20[/tex]
[tex]6a^2+4a-20\lt 6a^2 \Rightarrow 4a\lt 20 \Rightarrow a\lt 5[/tex]
The width must be positive: [tex]a-2>0 \Rightarrow a>2[/tex]. So [tex]2\lt a\lt 5[/tex].

Solution:
The speeds are [tex]65[/tex] and [tex]75[/tex] mph, so the trains approach each other at [tex]140[/tex] mph. After [tex]t[/tex] hours the distance between them is [tex]|420-140t|[/tex] (after they meet, they move apart).
[tex]|420-140t|\lt 28 \Rightarrow -28\lt 420-140t\lt 28[/tex]
[tex]392\lt 140t\lt 448 \Rightarrow 2. 8\lt t\lt 3. 2[/tex]
That is, from [tex]2[/tex] h [tex]48[/tex] min to [tex]3[/tex] h [tex]12[/tex] min after departure.

Solution:
Case 1: [tex]x\le -2[/tex]. Then [tex]f(x)=x+3\le 1[/tex], so [tex]f(f(x))=x+6>1 \Rightarrow x>-5[/tex]. Solutions: [tex]-5\lt x\le -2[/tex].
Case 2: [tex]-2\lt x\le 1[/tex]. Then [tex]f(x)=x+3>1[/tex], so [tex]f(f(x))=1-2(x+3)=-2x-5>1 \Rightarrow x\lt -3[/tex] – no solutions in this case.
Case 3: [tex]x>1[/tex]. Then [tex]f(x)=1-2x\lt -1\le 1[/tex], so [tex]f(f(x))=(1-2x)+3=4-2x>1 \Rightarrow x\lt \frac{3}{2}[/tex]. Solutions: [tex]1\lt x\lt \frac{3}{2}[/tex].
Answer: [tex]x\in(-5, -2]\cup\left(1, \frac{3}{2}\right)[/tex]

Solution:
Let the buses carry [tex]a[/tex] and [tex]b[/tex] people ([tex]a\ge 5[/tex], [tex]b\ge 15[/tex] are whole numbers):
[tex]a+b>24[/tex], [tex]2(b-15)\lt a[/tex], [tex]7(a-5)\lt b[/tex]
From the third inequality [tex]b>7a-35[/tex]; substituting into the second: [tex]a>2b-30>2(7a-35)-30=14a-100[/tex], so [tex]13a\lt 100[/tex], i.e. [tex]a\le 7[/tex].
From the second inequality [tex]b\lt \frac{a+30}{2}[/tex], and from the first [tex]b>24-a[/tex]. We check [tex]a\in\{5, 6, 7\}[/tex]:
[tex]a=5:\ 19\lt b\lt 17. 5[/tex] – impossible
[tex]a=6:\ 18\lt b\lt 18[/tex] – impossible
[tex]a=7:\ 17\lt b\lt 18. 5 \Rightarrow b=18[/tex]
Check: [tex]7+18=25>24[/tex], [tex]2\times 3=6\lt 7[/tex], [tex]7\times 2=14\lt 18[/tex]