2o13 Q10 (1 Viewer)

[mreditor16]

does anyone have a neat/quick way of getting this out?

thanks!

btw carret's 2013 3u solutions is dead, so I couldn't take a look at that....

 

[Kurosaki]

does anyone have a neat/quick way of getting this out?

thanks!

btw carret's 2013 3u solutions is dead, so I couldn't take a look at that....

Kuro is here =).

Fastest way I can think of is take cases as to whether .

By inspection, solution to the original equation is
You can actually eliminate A and B straight off using this, and then you just need to choose between C and D; D can be eliminated because between -2 and 3, |2x-1| is less than or equal to 5.

 

[mreditor16]

Kuro is here =).

Fastest way I can think of is take cases as to whether .

surely there is a quicker way than that? :/

btw GL for chem tomorrow you do chem right?

 

[Kurosaki]

surely there is a quicker way than that? :/

btw GL for chem tomorrow you do chem right?

Well, it eliminates A and B straight off if you notice the solution to the equation given is . And yeah I do, thank you =)

 

[Axio]

Imo graphing is the fastest way. If you do a 5sec sketch, you can tell by inspection that it is -2 <= x <=3. Then by looking at the answers, the very first one I tried out was c) simply because it had the right boundaries.

 

[Carrotsticks]

Kuro is here =).

Fastest way I can think of is take cases as to whether .

By inspection, solution to the original equation is
You can actually eliminate A and B straight off using this, and then you just need to choose between C and D; D can be eliminated because between -2 and 3, |2x-1| is less than or equal to 5.

This, unfortunately.

 

[integral95]

I remember drawing up a graph, inspected the values then test them in the answers given lol.

 

[WhoStanLeee]

I first solved the equation, which got me -2≤x≤3. As this solution indicates equality at x = -2 and x = 3, this automatically eliminates (A) and (B) because they are undefined at one of/both of these points. A quick sketch or subbing in a value -2≤x≤3 for (D) will indicate that it is also incorrect. This leave (C) as the correct response.

 

[photastic]

Draw the two graphs, and you clearly see that its between -2 and 3

 

[mreditor16]

okay so i'll just stick with critical points and go from there...

 

[photastic]

okay so i'll just stick with critical points and go from there...

Umm, I just drew |x+2| and 5-|x-3| Scaled graphs and the answer is when they overlap.

 

[faisalabdul16]

Umm, I just drew |x+2| and 5-|x-3| Scaled graphs and the answer is when they overlap.

yeah i also did it graphically!

 

[IR]

that reminds me of argand diagram. See if you can draw an ellipse from that and that prove that the inequality holds true. bahahaha. na jks. do what fais said ^^^^^^^^^ or carrot >>>>>

 

[RealiseNothing]

Let x=3 to eliminate A/B, then x=0 to eliminate D.