graphs
1. |||x-1|+|x-2|-2|-3|
2. |x+1|+|x+2|+|x+3|+|x+4|
how to draw graphs for such mod questions. would be helpful if u suggest some vids, ive watched nv sir for mod but he didnt discuss it in that.
20 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.I would suggest playing with graphs.
bro jese x ki koi value put kro uske corresponding wali y ki value ko jese 10 th mia plot krte the wese kro
first mai
x=0 pe y ki value 2 aa rhi hai
ese hi alg alg value pe y ki value find out kro aur plot kro
1. First off, starts with |x-1|+|x-2|. Draw it graph, now '-2' implies graph is shifted by 2 units below. Applying the mod implies x-axis ke neeche wale part ki mirror image lo w.r.t x-axis and y<0 wale part ko mita do. Then shift graph by 3 below and again axis ke about mirror image lelo
2. Iss wale mein it might be useful to see that for x<-4, slope=-4, for x in (4,3), slope=-2 and so on. Now Only find values at critical points. Now, kyunki tumhe line ka slope pata hai aur ek point pata hai line aa jayegi.
make cases
then open the equation
cases toh bahut bann jayega ig?
cases se kam mei ye value daalne wala kaam will help na??
hn
Any vid or smthing to get some grip on this?
2nd one is quite easy.
you know that fn is gonna be linear in x in any case right? so just get the values at zeroes of the mods and draw straight lines accordingly.
1st one:
basic graphical transformations.
boat shaped graph first -> 2 units down.
mod of that again.
3 units down.
mod of that thing again.
y=|x-1|+|x-3| if you know how to plot functions like these then the rest is easy for 1st one
Yea but that’s the method for subjective questions
i should consider three cases here right
x>=1 x>=3 1<x<3 and then see where is the func increasing and decreasing
Yeah a boat shaped graph will form
that's the basic approach but when dealing with 2 mods you can just get rid of the intermediate steps.
Example:
y=|x-a|+|x-b|; b>a;
for all x: a<x<b
y=|b-a|
Ok so I get |x-2|+|x-5| ka graph would be a boat shaped graph
What bout |x-2|-|x-5| and
| |x-2|-|x-5| |
Bro see the only legitimate and classic method is to make split function (cases) and plot the straight lines in bounded intervals
Ok so experiment and practice , I'll follow that ig . Thanx everyone
Yeah practice with desmos
Will do. Marking it solved now
+solved @Priyam Alok @@Ameya² @SirLancelotDuLac @petrushka @Slembash @Tlop
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