IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

srichaitanya

domain of root of (sin^-1(2x) +pi/ 6)

rphy

is it x belongs to [-1/4, 1/2] ?

srichaitanya

please solve

rphy

dude firstly the term underneath the root has to be greater than or equal to 0.
we get x is greater than or equal to -1/4. ...... (1)

we also know that sin^-1x is defined for x in [-1,1].

-1<= 2x <= 1
therefore -1/2<= x <= 1/2 ..... (2)

from (1) & (2)

x belongs to [-1/4, 1/2]

pls rate if correct

elessar_iitkgp

For the square root to be defined,
sin^-1 2x +

/6 >= 0
sin^-1 2x >= -

/6
2x>sin(-

/6)
x> -1/4 -----------(1)

For sin^-1 2x to be defined
-1<= 2x <=1

the two inequalities yield
x

[-1/4, 1/2]

magiclko

(sin^-1 2x +

/6 )

this is defined only if

sin^-1 2x +

/6

0
=> sin^-1 2x

-

/6
=> 2x

-1/2
=> x

-1/4

also sin^-1 2x is defined only for

-1

2x

1

=>-1/2

x

1/2

thus x

[-1/4, 1/2]

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