IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

DON007

LOOK , U HAVE LEARNED SIN inverse x + COS inverse x = pie/2

& TAN inv x + cot INV X = pie/2

f(x) = cot inverse of(( 3x-x^3)/(1- 3x^2)) and g(x) = sin inverse of ((1-x^2)/(1+x^2))

therefore,

f(x)= cot inverse of(( 3x-x^3)/(1- 3x^2) ) = pie/2 - tan inverse of(( 3x-x^3)/(1- 3x^2))

= pie/2 - 3 tan inverse x

g(x)= sin inverse of ((1-x^2)/(1+x^2)) = pie/2 - cos inverse of ((1-x^2)/(1+x^2))

= pie/2 - 2 taninverse x

now putting values in the limit...........

then lim (f(x)- f(t)) / (g(x) - g(t)) , we get

( pie/2 - 3 tan inverse x ) - ( pie/2 - 3 tan inverse t )

lim -----------------------------------------------------------------------------------

( pie/2 - 2 tan inverse x ) - ( pie/2 - 2 tan inverse t)

3[ tan inverse t - tan inverse x]

lim ---------------------------------------------------------

2 [ tan inverse t - tan inverse x]

cancellin out .......................

WE GET ANS =3/2.

PLEASE I TRULY DESERVE A SALUTE.............

( KIDDING)

THANKU