Messages posted by tibu

The diff. Eqn of the family of circles having the centres on the Y-axis:-

(a) xd²y/dx²-(dy/dx)³-dy/dx=0

(b) xd²y/dx²+(dy/dx)³+dy/dx=0

(c) d²y/dx²-(dy/dx)³-xdy/dx=0

(d) d²y/dx²+(dy/dx)³+xdy/dx=0

The diff. eqn of all parabolas whose axis is the Y-axis

(a) xd²y/dx²=dy/dx

(b) xdy/dx=y

(c) y=dy/dx

(d) xdy/dx=y

y²=m(a²-x²) is a soln of :-

(a) xy dy/dx + yd²y/dx²=0

(b)xyd²y/dx²+x(dy/dx)²-ydy/dx=0

(c)d²y/dx²+yx+(dy/dx)²=0

(d)yd²y/dx²+x(dydx)²+y=0

sorry...i put it now.

Please solve this one

diff. eqn of (x+y)²=2cx:

(a)y-x=2xdy/dx
(b)x+y=xdy/dx
(c)y=xdy/dx
(d)y-x=2ydy/dx

Come on Yaar,People like me who live in small towns dont get a good choice of books for entrance prep.Only local guides that to just for the state engg. exam is available.I have downloaded and gone through some of the books you have put and have found it very useful.Please try and find some indian author's books.I have heard that the arihant series is good from other goiit members.i depend mainly on the materials i download from the net.

actually no....could you explain more clearly.

thats what i felt but the same question and answer has been put in HC verma as well as Arihant.It has come in Jee 85 for 3 marks

Q.Two Identical trains are moving on rails along the equator on the Earth in opposite directions with same speed.They will exert the same pressure on the rails . True/False

A.false because the angular speed and hence the required centripetal force fo both the trains will be different..

Why? Cant get the logic....

tibu

Its best to conduct the test before march as in march the possibly good days would be 11th and 12th coz' its after english for the cbse students.

wheras in feb,all sundays are perfect for the test.
Tibu

The no of possibly correct answers= 4C1+4C2+4C3+4C4

=4+6+4+1=15.

The number of chances given to student=3.

Therefore the probablility that the student gets the right answer in

one of the goes=1/15 + 14/15*1/14 + 13/14*1/13

= 43/210

This should be the answer.

Tibu

it is x^2

_{[1/3 ]}

^{[1 ]} {(x-x3)^1/3 /x⁴]

give me the method not the final answer.

_{[ 0]}

^{[ ]} {x²sin2x sin([pi/2 cosx) /2x-pi }dx from 0 to pi

Give me the method of solving the problem and not just the final answer.

f '(x)=3cosx f(x)

f'(x)/f(x)=3cosx

integrating bothe sides

_{[ ]}

^{[ ]} f'(x)/f(x) dx =_{[ ]}

^{[ ]} 3 cosx

f(x)=t f'(x)dx=dt

_{[ ]}

[ ] dt/t =_{[ ]}

[ ] 3cosx

ln(t)=3sinx +c

t=e^{(3sinx + c)}

f(x)=e^{(3sinx + c)}

f(0)=e^{(3sin 0).}e^c =1

c=0

f(x)=e^3sinx

see its me afterall i won

must be me

Hi Tejasvi

(a) A gets a number larger than B
   except for the case where B gets n, A can always get greater number than B
   Let the number that B selects be r.
   then the number of cards from which a can choose is (n-r),

therefore the probability is=(n-r)/n =1-(r/n).

(b)A gets a number one greater than B
    excepting the case where B gets n A can always try and get a number one greater than B's
   therefore the number of favourable case is (n-1)
   and therefor P(B)=(n-1)/n

(c)A gets an even number
    If n is an odd number then
         number of favourable cases is (n-1)/2
    therfore P(C1)=(n-1)/2n

    If n is an even number then
          number of favourable cases is n/2
    therfore P(C2)=n/2n=1/2 always

Hope this solution is correct and clear enough

george

Please make the question more clear.because three sprigs attached can be two on one side and one on the other or something like that

In a city every person has at least 1 tooth and at most 32 teeth.Also, no two persons have an identical set of teeth.Disregarding the shape and size and considering only the positioning of the teeth, find the maximum population of the city.

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