22 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.solve |x-3| + |x-2|= 1
this is the sum, ik x=3 and x=2, but how is x-3 <= 0 and x-2>= 0
bro check interval of (2,3
(2,3)
[2,3] is the ans which i got and its right but idu how i got the ans
modulus function hai to 2 cases to banne padege na jese | x-2| me jo x hai wo 2 se bada bhi ho skta hai or chota bhi
So take 2.5 and sub in x-3 and x-2 to check
;-; im not hindi
ey i know that too xd but i want to know why x-3<=0 and x-2>= 0
and i feel im lackin behind while everyone is able to do better than me + i think im getting all sums on fluke.
Modulus fuction has 2 cases , like in |x-2| , x can be greater than 2 and less then @Deleted User [translation, did my best am not native Hindi either]
Yah this true too, it depends on which case
x-2>=0 and x-2<=0.. right
Idk the exact reason too but..
me myself check it by putting values, like anything greater than 3 will not give answer --> hence x<=3 , so x-3<=0
similarly anything smaller than 2 also won't give answer so x>=2 , so x-2>=0
This is my thought process, just putting values and checking btw in 11th grd, so it would be better if a senior helps out in thinking for a easier way (if thr is any)
same bro im in 11th too
I feel im getting all problems by fluke and i wont be able to do as these problems pass on, is it a starter problem?
yes i agree, major Q of mod and other functions, sometimes if lucky, easily chose the ans
like jahan cases wagera banane padhte hain
quite normal ig
for such Q, getting ans is fine but asking others too is beneficial, u get to know shorter or other ways to think upon it
yea cengage says another way but im getting it using only trial and error
sometimes trial n error is better, anyways JEE just wants ans
a good part about the exam, it doesn't require the process
i believe jee advanced does, im want to get into IIT.
same wanna get into IIT bom
idk about jee adv
Always do option checking imo
Some annoying questions require sub and checking even tho u get the answer , there is some exceptions
Same , I learnt abt mod lietrally tday
Oh nice
all IITians damn, well i wish yall the best on your journey, hope whatever path yall choose bring you success. Thank you for helping me.
Firstly, solve for (x-3) + (x-2)=1 and remember this is when x>=3, hence the solution must be in that interval
Then, assume x<2, solve (3-x) + (2-x) =1
and so on (2 more cases)
Combine all solutions
yoo i got it. |x-y| = |x|+|y|
so xy<= 0
thanks btw, gotta close channel.
+solved @stormycloud @Slembash @Real potato @RisingPhoenix @Comrade Rock Astley
Post locked and archived successfully!
Archived by
<@1141371202380763276> (1141371202380763276)
Time
<t:1718642257:R>
Solved by
<@821702877721657374> (821702877721657374), <@1165653012996444253> (1165653012996444253), <@1088352651567173632> (1088352651567173632), <@945760989209579540> (945760989209579540), <@769492044837552139> (769492044837552139)