@vbhvgupta said:Q6 plz explain
Neither of the statement is sufficient :neutral:
[1] x could be anything. ~ R
[2] x^4=1=> x= 1 or -1
@YouMadFellow said:If the statement were "The box contains only 10 blue balls" , it would have been fine. But it is incomplete I think. Waiting for the OA ..
Bro....I dont have OA....
@vijay_chandola said:kahan hai I don't see any Determine the first negative term in the expansion of sqrt{(1+2x)^7}
6th term?
value of 6th term={7/2*5/2*3/2*1/2*(-1/2)*(2x)^5}/5!=(-105/120)x^5
value of 6th term={7/2*5/2*3/2*1/2*(-1/2)*(2x)^5}/5!=(-105/120)x^5
@vijay_chandola said:Can you please explain it?I've got confused b/w 5th or 6th.
bhai upar thoda expand kiya hai maine .......actually
1st term =1
2nd term =7/2*2x
3rd term=7/2*5/2*(2x)^2/2!
4th term=7/2*5/2*3/2*(2x)^3/3!
5th term=7/2*5/2*3/2*1/2*(2x)^4/4!
6th term=7/2*5/2*3/2*1/2*(-1/2)*(2x)^5/5!
hope it helps......
1st term =1
2nd term =7/2*2x
3rd term=7/2*5/2*(2x)^2/2!
4th term=7/2*5/2*3/2*(2x)^3/3!
5th term=7/2*5/2*3/2*1/2*(2x)^4/4!
6th term=7/2*5/2*3/2*1/2*(-1/2)*(2x)^5/5!
hope it helps......
@vijay_chandola said:kahan hai I don't see any Determine the first negative term in the expansion of sqrt{(1+2x)^7}
Replies 11540/41/42 are the same 
(1 + 2x)^(7/2)
C(n) = (7/2) C n * (2^n)
Seems like 6th term 😐 ... :splat:
@19rsb said:bhai upar thoda expand kiya hai maine .......actually1st term =12nd term =7/2*2x3rd term=7/2*5/2*(2x)^2/2!4th term=7/2*5/2*3/2*(2x)^3/3!5th term=7/2*5/2*3/2*1/2*(2x)^4/4!6th term=7/2*5/2*3/2*1/2*(-1/2)*(2x)^5/5!hope it helps......
@YouMadFellow said:Replies 11540/41/42 are the same (1 + 2x)^(7/2)C(n) = (7/2) C n * (2^n) Seems like 5th term ...
OA is 6.
Acutally I did it like this:
(n-r+1)
=> r > n+1
value of r=5==> T5+1= T6 :banghead:
@19rsb said:6th term?
value of 6th term={7/2*5/2*3/2*1/2*(-1/2)*(2x)^5}/5!=(-105/120)x^5
Hi, sorry to interupt
can u please explain how you guys r getting 7/2 *5/2 ..........
is this some binomial expansion in some way !! @YouMadFellow said:
Replies 11540/41/42 are the same(1 + 2x)^(7/2)C(n) = (7/2) C n * (2^n)Seems like 6th term ...
@vijay_chandola said:Can you please explain it?I've got confused b/w 5th or 6th.
@techsurge said:Hi, sorry to interupt can u please explain how you guys r getting 7/2 *5/2 ..........is this some binomila expansion in some way !! :embarassed:
7/2 is the power.
So, the coefficient term would go like this:
n*(n-1)*(n-2)*(n-r+1)/r!
Q7
@techsurge said:Hi, sorry to interupt can u please explain how you guys r getting 7/2 *5/2 ..........is this some binomila expansion in some way !! :embarassed:
(1+x)^p=1+px+p(p-1)x^2/2!+p(p-1)(p-2)x^3/3!...........
@vijay_chandola said:7/2 is the power.So, the coefficient term would go like this:n*(n-1)*(n-2)*(n-r+1)/r!
thanks , i forgot binomial is possible for non integral powers also :banghead:
@vijay_chandola said:OA is 6.Acutally I did it like this:(n-r+1) => r > n+1value of r=5==> T5+1= T6
Arey, I edited the post, check above. I forgot the nC1 term initially 😐 ...
@vbhvgupta said:Q7
neither of the statement is sufficient.
[1] equation would be like this: y=x+k
[2] x=k
If R = {(1, 1), (2, 2), (1, 2), (2, 1), (3, 3)} and S = {(1, 1), (2, 2), (2, 3), (3, 2), (3, 3)} are two relations in the set X = {1, 2, 3}, the incorrect statement is:
a)R and S are both equivalence relations
b)R ∩ S is an equivalence relations
c)R^−1 ∩ S^−1 is an equivalence relations
a)R and S are both equivalence relations
b)R ∩ S is an equivalence relations
c)R^−1 ∩ S^−1 is an equivalence relations
d)R ∪ S is an equivalence relations
@falcao said:If R = {(1, 1), (2, 2), (1, 2), (2, 1), (3, 3)} and S = {(1, 1), (2, 2), (2, 3), (3, 2), (3, 3)} are two relations in the set X = {1, 2, 3}, the incorrect statement is:a)R and S are both equivalence relationsb)R ∩ S is an equivalence relationsc)R^−1 ∩ S^−1 is an equivalence relationsd)R ∪ S is an equivalence relations
Is it option D) .. Not much knowledge about equivalence, but seems so 😃 ..
.. Not much knowledge about equivalence, but seems so 😃 ..
Let S be the set of real numbers that can be represented as repeating decimals of the form 0.(abc) where a,b,c are distinct digits. Find the sum of the elements of S. Here () indicates recurring.