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This post was updated on Apr 19, 2016; 12:37pm.
I found these choices to be true. Please let me know if mistakes are present.
Updated with Sir's answer key. 1. c 2. b 3. d 4. a 5. c 6. d 7. d 8. b 9. a 10. a 11. d 12. b 13. a 14. Question not correct 15. c 16. d 17. c 18. b 19. c 20. c 21. d 22. a 23. b 24. a 25. c 26. d 27. a 28. a 29. a 30. c |
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how di u got q. 16 answer as d)
plzz explain |
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In reply to this post by Dr. Strange
Question 2 Why is it sufficient?
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In reply to this post by Dr. Strange
can u plz tell how to do ques no.29
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@varnika
take f(x,y) = x^2-y^2 and solve |
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This post was updated on Apr 12, 2016; 8:46am.
In reply to this post by varnika1880
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thnx but my query is in the second eq of the question x,y{r2) . but the fn goes from r2to r1..won't this fact be considered while solving
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In reply to this post by samuraijack
Are you implying this because of the equality sign present?
Since it is mentioned that function is twice differentiable I have doubt whether this means higher derivatives don't exist or are equal to 0. In second case second derivative must be nonzero where first derivative is 0 and it will be necessary and sufficient condition.However if third or higher derivatives exist then it will only be a necessary condition as the point can be inflection point. My doubt lies on meaning of twice differentiable. |
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Yes, I think twice differentiable can be there to distinguish it from a straight line y=c; so we should consider inflection points too.
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So what is your opinion: can we call y=x^3 a twice as well as a thrice differentiable function or shall we call it only a thrice differentiable function?
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x^3 is both twice and thrice differentiable, we're only given it's a twice differentiable fuction nothing about possibilities of higher order derivatives, it may or may not exist. But again I am not sure.
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However I think you are right. Since it is not mentioned about higher derivatives I can't take it for granted. Changing my option and making it only a necessary condition.
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In reply to this post by Dr. Strange
@Dr Strange
Can you explain Q13,14,21,26. |
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13.System is homogenous.So either trivial soln(0,0,0,)exists or infinite soln exists.
So for non-zero soln to exist determinant should be 0. Making determinant zero we get the soln. 21.four cases are possible: (red,not red),(red,red),(not red,red),(not red,not red) P(red,red)=a/n* (a-1)/(n-1) P(not red,red)=(n-a)/n * a/(n-1) p(second ball is red)=P(red,red)+P(not red,red)= a/n 26.given constraint x=y put x=y in target equation and optimise |
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14.Since cumulative distribution function attains value 1 at x=infinity so it is only possible if a=0 and b=1
F(x)= 1 for x>=0 =x^2-x +1 for x<0 Since x is defined for [0,1] so second part becomes redundant. so the distribution is discrete having p(x)=1 at x=0 and p(x)=0 elsewhere problem is question mentions that 'a' belongs to (0,1) so a=0 is not a viable solution.I am stuck in here. If ' a ' is considered to be 0 then F becomes continuous in (0,1). |
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In reply to this post by Dr. Strange
I read it in a book which says it's both a sufficient and necessary condition
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In reply to this post by Dr. Strange
Can u please tell how u r getting d as an ans for 21
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I have displayed the solution of 21 in above post.
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In reply to this post by Dr. Strange
If we sum up the two value over two intervals and equate it to one ...and then in that eq replacing X by a ...we will get a=1 but 1 Doesn't belong to the given interval...so it might be the case that it's not continuous at a
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a can't be 1 because in that case in interval [0,1/2] F(x) is decreasing which is not possible.
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