jnuQ

[kavya]

[kavya]

[kavya]

[Sinistral]

10) d

dA/dt = uA
solving gives: A=exp(ut+c), c is a constant

similarly, dB/dt=(u+1)B
B= exp[(u+1)t+k]

at t=0, A=B ==> c=k

t=1: A=exp(u+c)
       B=exp(u+1+c)=exp(u+c)*exp(1)=Ae, ie A times e

[Sinistral]

equation of a circle with centre at (a,b): (x-a)^2  + (y-b)^2  = r^2

putting (8,0): (8-a)^2  + b^2 = r^2
putting (0,6):  a^2       + (6-b)^2  = r^2   ----- (I)

equating the above 2 equations (since RHS equal) gives a =(7+3b)/4

putting this value of a  in equation (I) gives a quadratic in b. making its discriminant >= 0 will give r^2  >= 25
hence (b)

[Sinistral]

(64) coefficient of x^2 are of different sign implies one parabola faces upwards and the other faces downwards. ( -ve sign ==> downward and vice versa)

since discriminant is negative in both the cases implies that both parabolas do not touch or cut x axis.
hence one parabola will always remain above x axis (whose coefficient of x^2 >0) and the other will always remain below x axis (whose coefficient of x^2 < 0). therefore they will never intersect.

option (c)

[laracroft]

Hey sinistral, i didnt get it... how did u calculate the exp. ??
plz explain !

[laracroft]

hi sinistral
you have a very good way of approaching these questions !!
it would be really kind of you if u could share from where have u prepared fr eco entrances?


I find mathematics really tough to solve and decipher... any suggestion would be of great help !

[Sinistral]

dA/dt = uA
dA/A = udt
integrating both sides:
lnA = ut + c , where c is integration constant

OR A = e^(ut+c)
     A = exp(ut+c)

similarly integrating dB/dt = (u+1)B

[kavya]

thank u sinistral :)