Maths ! help !

[Anjali]

1.  The condition that the straight line y=mx+c may touch the hyperbola (x^2/a^2) - (y^2/b^2)=1  is
 C^2= a^2m^2-b^2
 C^2=a^2m^2+b^2
C^2=m^2-a^2b^2
C=b^2m^2-a^2

[Anjali]

[Anjali]

[Anjali]

[Amita_IITM]

I am trying to solve. Meanwhile can you tell me these are important for which exam?

[Granpa Simpson]

Q1) a.
Q2) 0.25 (since probability of accident in a particular year is 0.5, the probability of non occurance of accident in a particular year is 0.5..so probability of non occurance of accident in both the years is 0.5*0.5=0.25).
Q4) c

[Granpa Simpson]

Q3) I think you can use Euler's theorem and try...

[The Villain]

2b
 4 c

[Ashima]

(-infinity, 1/2)

[Ashima]

How did you take 0.5 as the prob of accident in a year? Its nowhere given that in a year either an accident will take place or won't take place.

[The Villain]

I think becaz there can be only two outcomes..accident or not accident...so we take each with equal prob 1/2..maybe

[Shefali]

Yes i agree with ron , prob should be half for each.
I forgot Euler's theorem.. nyone???

[Ashima]

In the same ques, had it been asked prob of two accidents the? Again it would be 0.5x0.5! Not a correct logic i feel :(

[Granpa Simpson]

assuming equally likely events probability of occurance of accident in a year is 0.5...

[Granpa Simpson]

Eulers Theorem states that for a homogeneous equation of degree n,
nf(x,y)= x*fx + y*fy
where fx and fy are the partial derivatives wrt x and y respectively..!!!!

[Dreyfus]

Its n(n-1)f...

[Granpa Simpson]

By using Euler’s Theorem, n*f(x,y)=x*fx+y*fy
Again differentiating partially wrt x and y respectively we get,
Or, n*fx = fx + x*fxx +  y*fyx...................(1)
Or, n*fy = fy + y*fyy +  x*fxy………………...(2)
By using Young’s Theorem fxy = fyx.
Multiplying (1) by x, (2) by y and adding and rearranging we get.
n*(x*fx + y*fy)= (x^2)*fxx+ (y)^2*fyy+ 2*x*y*fxy + (x*fx+y*fy)
Substituting x*fx+y*fy = nf we get,
n*n*f - n*f = (x^2)*fxx+ (y)^2*fyy+ 2*x*y*fxy .
or, (x^2)*fxx+ (y)^2*fyy+ 2*x*y*fxy = n*f*(n-1)

[Granpa Simpson]

Consider two equally likely events A & B s.t.
A= accident occurs in the first year.
B= accident occurs in the second year.
Also AUB= S
Now A and B are mutually exclusive as if accident occurs in year 1 then it cannot occur in year 2 and vice versa. Hence occurance of one event prevents occurance of the other hence mutually exclusive.
Now P(AUB) = P(A)+P(B)
again 1= P(A)+P(B),
if A and B are equally likely then P(A)=P(B)=1/2
Using this you will get the answer as 0.25.

[Anjali]

Subhayu can you explain 4 ?
For upwards , the 1st derivative must be greater than 0
Which concludes x<1
Now for concave second derivative must be less than 0
Which concludes x>1/2
Iam stuck here
Please help
And for that Euler's theorem I have taken a function (xy)^n and then I have substituted its partial derivatives in the given equation. Am I going right ?

[Anjali]

Awesome subhayu !