36 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.@hardcoreisdead @Prasan
krta thodi der mei
help how do i this
okk
confusing ha thoda
ye
lcm ( n^2-5n+8 ,1/n+m) = 2
since n belongs to N , n^2-5n+8 belong to Integer
lcm is 2 , matlab n^2-5n+8 can be either 1 or 2
yes n is 2 or 3
then
yep
1/2+m , 1/3+m aisi values ho jo lcm 2 de
ohhh
wait
so
woh condition kaise satisfy hogi wahan atak rha
hm
1/2+m can be 1 or 2?
or any fraction less than 1
1/2, 1/4... should also suffice
wat bout 0.35
hm yeah
so numerator should be 1
and denom should be an integer
can 1/2+m be 0.35?
idts
then how would lcm be 2
1/7 = 1/2+m will satisfy conditions?
it should
kaise
cuz 1/7 and 2 ka lcm will be 2
1/7 * 14 = 2
why cant 2+m = 1/2
same doubt
ans given is c...this means m cant be -1.5
but
it should be
whi
ping me if u get ans
this ques be bugging me for days ðŸ˜
k
LCM(n^2-5n+8, 1/(n+m))=2
Now, we analyse n^2-5n+8. It achieves its minimum at n=2.5. 6.25-12.5+8=1.75. For there to be a period of 2, the quadratic must acquire some value 2/k (k being a natural number). From the above the only value k can acquire is 1. So n^2-5n+8=2 which means n=3 or 2.
So, lcm(2, 1/(n+m))=2; which means 1/(n+m)=2/l for l being a non-zero integer.
So n+m=l/2 or m=l/2-n
Where note that l is a non-zero integer and n belongs to {2,3}
This gives the answer as (d)? I'm too sleep deprived rn.
@Imine
Given ans is c
Let m=-2.5 and n=2, then tan^2(pi.x/2)+cot(-0.5pi.x), then the period of this is 2 though?
https://www.desmos.com/calculator/y9vdbuuyhh Desmos settle it ig :/
so wrong ans ig?
galat hi diya hoga phir
tyyy
yea 😠why does arihant have so many errors
bhai bohot ans galat hai toh leave the book
ull end up wasting a lot of time
true true