1take x=tanθ
then y=2θ = 2 tan-1x
=> dy/dx = 2/(1+x2)
this derivative doesnot exist for x2=-1 but this is never possible ... so, ans should be (d).
The derivative of the fn y=sin-1[(2x)/(1+x2)] doesn't exist for
a)all values of x satisfying |x|<1
b)x=1,-1
c)all values of x for which |x|>1
d)none
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12 Answers
11y????
if we take x=tanθ then [(2x)/(1+x^2)]= sin2θ
then sin^-1([(2x)/(1+x2)] ) = sin^-1(sin2θ) = 2θ =2 tan^-1 x
whr am i wrong???
and y should dat at all matter .... whr theta is varying....
in the ultimate expression we have to deal with x...
62yes.. my wrong..
ur mistake is that it should have been
not 2θ
sin(sin-1x)=x
and
sin-1(sinx)=x
are both not correct!
which one is?
62no 1st one is correct!
the second one is not always true..
try with x=2pi!
1if -Π/2 ≤ 2θ ≤ Π/2
then -Π/4 ≤ θ ≤ Π/4
or -Π/4 ≤tan-1x ≤ Π/4
or tan -Π/4 ≤ x ≤ tan Π/4
or -1 ≤ x ≤ 1
u r assuming that x lies b/w -1 nd 1
62dear the assumption is
x=tanθ
That is all we are assuming!
Right.. now why u are getting confused is not strange but bound to happen.!
In Sky's method.. uptil here
"if we take x=tanθ then [(2x)/(1+x^2)]= sin2θ "
I dont see any mistake..
The mistake comes in the next step...
I dont know if ur message is for me or for sky! but i wud like to discuss this more with u :)
1so shdu we take cases as such
wen
-pi/2<x<pi/2 then y=2tan^(-1)x .
so dy/dx=2/(1+x^2)
if pi/2<x<pi......
if pi<x<3pi/2.............
and soo on bt evn if we do like this i m gettin option d as correct
since x term is alwys cumin wid a squre sign
correct me ........if m wrng :S
62@integrations:
-pi/2<x<pi/2 then y=2tan^(-1)x .
so dy/dx=2/(1+x^2)
if pi/2<x<pi......
if pi<x<3pi/2.............
Here our domain for x is only -pi/2<x<pi/2
bcos u have chosen it that way..
so where is the question of taking cases : if pi/2<x<pi......
if pi<x<3pi/2.............????
we already know that x is not in these limits!!
