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e^(-y)dy = (e^x) dx integrating both sides -e^(-y)= e^x + c e^x + e^(-y) +c =0 thats the answer......
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i think tmh is best book for maths ,i used it nd i like it very mch ......u shoud do all the different ques of tmh and examples also it wl make maths very strong for sure....nudge me if u want to knw morethanks.........
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a(n)= 1+ (1/2+1/3)+ (1/4+1/5+1/6+1/7)+....... < 1+ (1/2+1/2)+ (1/4+1/4+1/4+1/4) +.............. put n=100 a(100) < 1+(1)+(1)+..........upto 100 terms a(100)<100 so option a is correct again , a(n) = 1+ (1/2+1/3)+ (1/4+1/5+1/6+1/7) +...... > 1 + (1/4+1/4)+ (1/8+1/8+1/8+1/8).......... put n= 200 a(200) > 1+ [(1/2)+(1/2) + .......upto 199 terms] a(200) > 1+ 99.5a(200) > 100.5 so option d is also correct thus , a and d are the ans
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focii = (0,be) and (0,-be) eq of directrix :- y=b/e , y= -b/e eq of major axis :- y=0 eq of minor axis :- x=0 length of latus rectum = 2a^2/b
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2)Total no. of arrangements = No of possible arrangements when the delegates of the countries A and B r seated nxt 2 each other - No of possible arrangements when the delegates of the countries A and B r seated nxt 2 each other and C and D r also seated next 2 each other so, Total no of arrangements = 8!*2! - 7!2!2! thats the answer... hopeu gt it
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2) i m just correcting nikhil's sol.... nikhil method is correct till p=9 now , a = p/cos60 = 18and b= p/sin60 = 18/sqrt(3) so, the equation of line is x/18 + ysqrt(3)/18 =1 x + ysqrt(3) = 18 hope u gt it .....in case of ny problem tl me...
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let max. speed be v
so, total time (t) = v/A +v/B
v = (A+B)t/2 --------(1)
for the first half when car acc with A
s1= distance travelled = v^2/(2A)
s2=distance travelled in second half = v^2/( 2B)
total distance travelled = v^2 (1/2A + 1/2B)
s= v^2 (A+B)/2AB
s = (A+B)^3 (t) /8AB
average velocity = s/t = (A+B)^3/8AB
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let rate of diffusion of hydrogen = r
and rate of difffusion of oxygen = d
nw r=50/20
d= 45/t
and r/d = sqrt(mol mass of oxygen/mol mass of hydrogen)
r/d= sqrt(32/2)
r=4d
50/20 =4(45/t)
t=72 mins
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let the g.p be
a,ar,ar^2,..........upto (2n) terms
sum of all terms (S1) = a(1-r^(2n)) / (1-r)
terms occuping odd places wl also be g.p
with first term a and commomn ratio r^2 and upto n terms
sum of terms occuping odd places (S2) = a[1-(r^2)^n] / (1-r^2)
nw S1 = 5(S2)
a/(1-r) = 5a/(1-r^2)
1-r^2 = 5 - 5r
r^2 -5r +4 =0
r=4 , r= 1(not possible )
so r=4 is the answer.........
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the pts at which ordinate decrease at the same rate at which the absicssa increase ,
dy/dx = -1
now differentiating the ellipse equation
32x + 18y.(dy/dx)=0
put dy/dx=-1 to get those pts.
so , we get 18y=32x
9y=16x
put this in the ellipse equation ,
we get the pt (3,16/3) and (-3 , -16/3)
hope u gt it ...........
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this ques can be solved by using limit as a sum
= (1/n) (summation ) [n^3 / (3n+r)^3]
= integration (from 0 to 3 ) (3+x)^(-3)
integrate it and dput the limits and u wl get the answer.......
hope u gt it ....
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let its general term = an^2 +bn +c
now put n=1 , n=2 , n=3
a+b+c = 2 ----------(1)
4a+2b+c = 5 ------------(2)
9a+3b+c = 9 ------------(3)
solving (1), (2) , (3)
a=1/2 , b=3/2 , c=0
so general term = (n^2+3n)/2
sum to n terms = [n(n+1)(2n+1)/6 + 3(n)(n+1)/2 ]/2
= (n)(n+1)(n+5)/6
thats the answer........
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solution:-
(1,1,z) where z belongs to whole nos.
(1,y,0) where y belongs to natural nos.
(2,3,1)
(3,3,2)
(3,9,1)
thats the answer....
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Tk = [(k+1)-1](k!)
= (k+1)! - (k!)
as we hv expressed general term as diff of two consecutive terms ,
thus Sn = (n+1)! - 1
thats the answer......
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n{x} = {x}+2[x]
{x} = 2[x] / (n-1) where n-1 >0
0<= 2[x]/(n-1) <1
so,0<= [x] < (n-1)/2
as x has exactly five solutions so [x]can take 5 values
so , [x] =0 ,1 ,2,3,,4
so, 0 <= [x] <5
thus, (n-1)/2 = 5
n = 11
thats the answer...........
if n= 12
then 12{x} = {x} +2[x]
{x} = 2[x] /11
then [x] can take values = 0,1,2,3,4,5
so six solutions .
thus the only answer is n=11
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1) put 10^x +x^10 = t
10x^9 + (10^x) (log10) = dt
then,
integral dt/ t
= ln (t)
= ln ( 10^x +x^10)
thats the answer.........
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2) f(x) = lim (n tends to infinity ) (x^2n - 1) / (x^2n + 1)
as 0thus , f(x) = -1 [as lim (n tends to infinity) (x^2n) =0 ]
nw integral -sin^(-1)x dx
which can be easily calculated by appplying by parts........
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1) integral [2^(x+1) -3^(x-1)] / (2^x.3^x)
integral 2/ (3^x) - 1/ 3.(2^x)
= -2ln(3) / 3^x + ln(2) / 3.(2^x)
thats the answer..........
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4) put ln(ln(lnx)) = t
dx/ (xln(lnx) lnx) = dt
then
=integral tdt
= t^2/2
nwu wl get the answer............
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xdx(y+1) = (x^2+1)(y^2+1)dy
xdx/ (x^2+1) = (y^2+1)dy/ (y+1)
xdx/ (x^2+1) = (y^2-1)dy / (y+1) + 2dy/(y+1)
xdx / (x^2+1) = (y-1)dy +2dy/(y+1)
1/2 log(x^2+1) = [(y-1)^2]/ 2 +2log(y+1) +c
thats the answer......
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Vriti Edustore
