@albiesriram said:
n=2
11*14*13=2002
N= 2*log(base2002)2002 =2
11*14*13=2002
N= 2*log(base2002)2002 =2
albiesriram
at least 2 real roots
Q) what is the reminder when 56! is divided by 57?
For first the OA says C 
.
For second OA is B 
@kunjesh said:Q) what is the reminder when 56! is divided by 57?
57 is 19*3
56! has both 3 and 19 in it so 0
@albiesriram said:For first the OA says C .For second OA is B
how c
let roots be x and y
x+y = 1/x^2 + 1/y^2
x+y = (x+y)^2 -xy/x^2y^2
x+y=-b/a
xy = c/a
so
-b/a = [b^2/a^2 - c/a] /(c/a)^2
now a, b ,c if in hp should satisfy this
put a=1 b = 1/2 c = 1/3
it is not getting equal ???????
@albiesriram said:
1. a + b = 1/a^2 + 1/b^2
(-b/a) = a^2 + b^2/(ab)^2
(-b/a) = (-b/a)^2 - 2(c/a)/(c/a)^2
2a^2c = ab^2 + bc^2
now ... b/a = 2(a/c)(c/b)/(a/c + c/b)
solving this will get ....2ca^2 = ab^2 + bc^2
so...a/c,b/a,c/b are in HP..
@Asfakul said:A shopkeeper cheats by 10% while purchasing and cheats by 12% while selling the goods. if he professes to sell the goods at cost price then find his profit percentage.?
it will be
1100 -- for rs x
880 for 0.8x
x-0.8x/0.8x = 0.25*100=25%
@abhishek.2011 said:how clet roots be x and yx+y = 1/x^2 + 1/y^2x+y = (x+y)^2 -xy/x^2y^2x+y=-b/axy = c/aso-b/a = [b^2/a^2 - c/a] /(c/a)^2now a, b ,c if in hp should satisfy thisput a=1 b = 1/2 c = 1/3it is not getting equal ???????
-2xy na?
also they have asked the relation between a/c, b/a, c/b and not a,b,c
i am sorry i misread the question i was thinking that we need to prove that a b c are in HP :(
sorry
