JEE Main · 2005 · 08 May 2005 · Offline · Inverse Trigonometric Functions

Let

f : (- 1, 1) \to B

, be a function defined by

f (x) = tan^{- 1} \frac{2 x}{1 - x ^{2}}

, then

f

is both one-one and onto when

B

is the interval

Answer: (D)

Solution

On the interval

(- 1, 1)

the identity

tan^{- 1} \frac{2 x}{1 - x ^{2}} = 2 tan^{- 1} x

holds, so

f (x) = 2 tan^{- 1} x

. As

x

runs over

(- 1, 1)

tan^{- 1} x

runs over

(- \frac{π}{4}, \frac{π}{4})

, so

f

takes exactly the values in

(- \frac{π}{2}, \frac{π}{2})

, and being strictly increasing it is one-one. For

f

to be onto as well, the codomain

B

must equal this range, namely

(- \frac{π}{2}, \frac{π}{2})

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