Question

What are the minimum and maximum values of x, for which x3 + x2 – 6x ≥ 0? minimum = –3, maximum = 0 minimum = 0, maximum = 2 minimum = –3, no maximum minimum = 2, no maximum Please answer quickly

Answer 1

Answer

1. d

2. a

3. d

4. b

5. c

6. b

7. c

8. d

9. d

10. a

Step-by-step explanation:

for the edge2020 test, all the answers

Answer 2

Answer: Option C.

Step-by-step explanation:

The given inequality is

$x^3+x^2-6x\geq 0$

Taking x common, we get

$x(x^2+x-6)\geq 0$

Splitting the middle term, we get

$x(x^2+3x-2x-6)\geq 0$

$x(x(x+3)-2(x+3))\geq 0$

$x(x+3)(x-2)\geq 0$

Related equation is

$x(x+3)(x-2)=0$

$x=-3,0,2$

These three points divide the number line in 4 parts.

Interval       Sign of $x(x+3)(x-2)\geq 0$      Statement

(-∞,-3)              $(-)(-)(-)=(-)$                           False      

(-3,0)              $(-)(+)(-)=(+)$                            True  

(0,2)              $(+)(+)(-)=(-)$                              False      

(2,∞)              $(+)(+)(+)=(+)$                              True  

So, the solution set is (-3,0) ∪ (2,∞).

Thus, minimum value is –3 and no maximum value.

Therefore, the correct option is C.