IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ruhi

hi
plz help me in this question:
find the values of a,b,c so that _{[ x]}

_{[ 0]} (ae^x -bcosx+ce^-x )/xsinx = 2.

puneet

Open the series of cosx e^xand e^-xby power series method .....

Now use two concepts .. these are to be used in general while solving such questions

First the limit has to exist so the constant and coefficeint with x have to be zero since we have xsinx in the denominator ...

Second the coefficent of x² term should give the limit .....

I hope this helps .. reply back to this post if things are not clear ..

cheers !!!

cvramana

Lim x -> 0 (ae^x -bcosx+ce^-x )/xsinx = 2.

On applying the limit we get a - b + c / 0 = 2, which is not possible.

Therefore

a - b + c = 0 ---------------------(1)

Since the function is 0 / 0 form

We apply L? Hospital rule, that is, we differentiate both Numerator and Denominator separately and apply the limit once again.

We get

Lim x -> 0 a e^x + b sin x - c e^-x / sinx + x cos x = a - c / 0 =2,

Which is not possible. Therefore

A - c = 0 --------------------(2)

Repeating the process we get

Lim x -> 0 a e^x + b cos x + c e^-x / cos x + cos x - x sin x = a + b + c / 2 = 2

Therefore

a + b + c = 4 -------------------------(3)

solving 1,2, and 3 we get

a =1, b= 2, c=1