IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

rohitrocks

Please fnd the range of

(x2+14x+9) / (x2+2x+3)

tell detailed solution

Q.2> the inverse of the function (10 raise to power x - 10 raise to power -x)/ ( 10 raise to the power x +10 raise to the power -x) is ??

plzz tell detailed solution...!!

thanks,

need urgently

plzz hlp!

rohitrocks

plzz some 1 help!!!!!!!!!!!!

Arjun

hey rohith is it 2x....or x raised to 2.........

plzzz reply so that i can continue........

rohitrocks

x square

akku

Hi rohit

1)RANGE : X2+14X=9/X2+2X+3

(-5,4)

2)Inverse y=1/2log(base 10) (1+x)/(1-x)

Pls tell if its correct i'll post the detailed soln

DO RATE ME !

Arjun

for da second question.........

let

y=10^x-10^-x/10^x+10^-x

takin 10^-xcommon.......frm both

y=10^-x(10^2x-1)/10^-x(10^2x+1)

y=(10^2x-1)/(10^2x+1)

(10^2x-1)=y10^2x+y

10^2x-10^2xy =1+y

10^2x(1-y)=(1+y)

10^2x=1+y/1-y

takin log to base 10 on both sides........

2x=log(1+y/1-y)

x=1/2log(1+y/1-y)

or

f^-1(y)=1/2log(1+x/1-x)

hope u got it......

thank u......

rohitrocks

u r right

gr88

plzz give me detaled solution!!

rohitrocks

thankss a lotttt

Arjun!

First ques.. plzz!!

Thanks

akku

1) y=x2+14x+9/x2+2x+3
x2+14x+9=yx2+2xy+3y
x2(y-1)+x(2y-14)+3y-9=0
Since x is real D=B2-4AC>0
4(y-7)^2-4(y-1)(3y-9)>0
-2y2-2y+40>0
=>y^2+y-20<0
(y+5)(y-4)<0
=> y

(-5,4)
DONT FORGET TO RATE

Arjun

akku has answered it.......gr88888

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