IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - functions prob based on limits and integration

aditi_g

let a function y = f(x) satisfy the following conditions..

dy = y + _[0]

^{[1] y dx}

^-----

^dx

^{and f(0)= 1}

^{thn a) f '' (0) =}

^{b ) f ( 1) =}

^c)

_{[ x}

_{[ 0] f (x) -1}

_-----------

_x

_{d) 1/2 f ' ( ln ( 3 - e)) = ??}

_{find the values of all the options...........}

_{rates guaranteed}

aditi_g

hey all u genius guys out thr plz plz plz solve this one........and do post the method too

abhujabal

dy/dx=y + _[0]

^[1] ydx

This can be written as dy/dx = y+k where k is the constant representing the value of the integral......

Now this is variable separable form....so solving this differential equation we get

log_e(y+k) = x

y+k = e^x

f(0)=1, substituting the value we get y+e=e^x

Now this equation can be used to answer all the other questions.....

abhujabal

Now f(x)=e^x-e

So f'(x)=e^x and f''(x)=e^x also

Now f(1)=e-e=0

f''(0)=e⁰ =1

1/2 f'(ln(3-e))= (3-e)/2

_{[x ]}

_{[ 0]} (f(x)-1)/x = _{[ x]}

_{[ 0]} (e^x-e-1)/x = 1

I hope all these answers are correct......

Please rate ......