the potential of the 2 plates will be different initially

so a current will flow from higher potential to lower until a state of eqlm is reached when no current flows through the resistor.

when the potential of both the plates will be same then there will be no electric field  as there is no electric field for equipotential surface.

tell me if i m correct????????

some1 plz reply

rates assured for trying also

which of  these field can be circular in shape?

a)electrostatic field                                       b)magnetic field

c)gravitational field                                         d)none

as we can see that this function is continous as well as diffrentiable foor all real no so applying rolles theorem we get that ther ther is only one root of the eqn

the thing on the rhs will be become zero sometime if we go on diffrentiating  but the thing on lhs will not become zero unless a=0 which is not possible .

so it has only one root

hope u got it.............

hey let a,b,c,d be its roots

then a+b+c+d=1............................(1)

 ab+bc+....=-12........................(2)

abc+bcd+.....=-12.............(3)

abcd=132..........(4)

we have 4 eqn and 4 unknown solve for a,b,c,d

thats the only way i see a way of getting the ans

hope u got it...........

radius vector is defined as the distance of the curve from the origin

so the radius vector will be =(a²+b²)^1/2

tell me if i m correct?

yes u r correct man and the ans will be 2c

absolutely correct.

hey friend in which interval u r talking abt?

we get f^'(x)= 1+cos x which if >= 0 will be only increasing and if only >0 then it will be strictly increasing

hope u got it..........

hey i think its not that difficult

we can rearrange the given eqn as

lim x tends to pi/2    (a^{cot x - cos x}  - 1)a^{cos x} /cot x- cos x

now it is in a standard form so we get (after putting the limit x=pi/2 in the eqn)

ans = log_ea

hope u got it..........

ok then let us analyse this ques carefully but in the 1st look x,y=1 or 0 came to my mind so i just posted the ans without thinking any other thing.

yes we will get the desired result if we diffrentiate the function and then put x=e

what was the problem in this ques.

hey i have tried to solve it plz see my soln in the diffrential calculus forum and tell me if i m correct

hey it can be proved by mathematical induction

put  p(1) it satisfies the function where p(x) is the function

then assume that p(m)is true and using this prove that p(m+1)is true.

i have checked p(1) it is true

hope u got it........

my soln is not as perfect as hsbhatt sir's but it can b useful for mcq

by the binomial sereis

(1+x)ⁿ=1+nx+n(n-1)x²/2 + ........

now taking 2^{root 3} (1+1)^{root 3} we get that in the 2nd term of the expansion we get a multiple of root 3 and we will get it in many other terms and they will surely not cancel each other

also we know that rational +irrational=irrational so we it will be irrational.

hope u got it........

i m not getting ur question can u plz write it in a more clear way plz.