17 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.c>0 a<0 b^2-4ac>0 thats for sure
main fuss is b
-b/2a<0 a is -ve so b must be too ( this gives A , the correct ans in module)
a<0
-b/2a<0
-b<2a
b>-2a
a is -ve so -2a is +ve so b is +ve
so what should be the correct ans /
@SirLancelotDuLac sorry for the ping but pls help
uh
I haven't done any quadratic yet but from my parabola knowledge I know that y=ax^2 should make the curve be like a valley instead of a hill
so I would mark D without even thinking since it's not given a is negative which kind of invalidates the question tbh
in this case this graph satisfies all the statement 2 options but it's clearly nothing like the graph shown
basically what I'm saying is that the question is wrong and you should move on (don't listen to me)
what book are you using btw
I need to send a bad review
A is -ve
-b/2a is also -ve
Hence b is +ve
Look a<0 (Opening downward). Now look at the derivative at x=0, it is -ve. So b<0. The f(0)=c>0 from graph
actually a<0
fiitjee package ðŸ˜
but agar b +ve ho gya toh -b/2a would become overall positive thus invalidating -b/2a<0
hmm that makes sense
can you also tell why my approach is wrong
this one
@BlindSniper (BS)
ew
(Line 3:) -b/2a<0 Doesn't imply -b<2a
oh. mb
however that does imply -b<0 for which b nust be >0
2a is negative, so inequality will reverse
oh rightttt, my bad
+solved @SirLancelotDuLac @idk either
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