Question

प्रथम n-विषम संख्याओं का योगफल है​

Answer 1

Answer:

The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.Suppose that X and R∖X are both open. Pick a point a∈X and a point b∈R∖X. There’s no harm in assuming that a<b. Let A={x∈[a,b]:[a,x]⊆X}. A is bounded, so it has a least upper bound u. Clearly u≤b.

Now let ϵ>0, and consider the interval J=(u−ϵ,u+ϵ). Since u is the least upper bound of A, A∩(u−ϵ,u]≠∅, and therefore certainly J∩A≠∅. This shows that u can’t be in R∖X, since no open nbhd of u is a subset of R∖X, Thus, u∈A, and therefore u<b (since b∉A). Let v=min{u+ϵ,b}. Clearly v∉A, so there is some x∈[a,v)∖X. since [a,u]⊆X, we must have x∈[u,v)⊆[u,u+ϵ). Thus, J∖X≠∅, and J⊈X. Thus, u can’t be in X, either. Since u has to be somewhere, this is a contradiction, showing that X and R∖X can’t both be open.The possible subsets are ϕ,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}.The given equations are ab+bc=44 ..........(i)

And, ac+bc=23 .........(ii)

⇒c(a+b)=23, which is a prime number

So, the two factors must be 1 & 23.

c=1 and $$a + b = 23$$

⇒b=23−a

Put these values in (i), b(a+c)=44

⇒(23−a)(a+1)=44

⇒23a−a

2

+23−a=44

⇒a

2

−22a+21=0

⇒a

2

−21a−a+21=0

⇒a(a−21)−1(a−21)=0

⇒(a−21)(a−1)=0

⇒a=1,21

For, a=1,b=22

For a=21,b=2

∴b=22,2

The solution sets are (1,22,1);(21,2,1).

Hence, the number of solutions sets =2.