AIEEE 2005 | Functions Question 161 | Mathematics

Let $$f:( - 1,1) \to B$$, be a function defined by
$$f\left( x \right) = {\tan ^{ - 1}}{{2x} \over {1 - {x^2}}}$$,
then $$f$$ is both one-one and onto when B is the interval

$$\left( {0,{\pi \over 2}} \right)$$

$$\left[ {0,{\pi \over 2}} \right)$$

$$\left[ { - {\pi \over 2},{\pi \over 2}} \right]$$

$$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$

AIEEE 2005

MCQ (Single Correct Answer)

-1

A real valued function f(x) satisfies the functional equation

f(x - y) = f(x)f(y) - f(a - x)f(a + y)

where a is given constant and f(0) = 1, f(2a - x) is equal to

- f(x)

f(x)

f(a) + f(a - x)

f(- x)

AIEEE 2004

MCQ (Single Correct Answer)

-1

The graph of the function y = f(x) is symmetrical about the line x = 2, then

$$f\left( x \right) = - f\left( { - x} \right)$$

$$f\left( {2 + x} \right) = f\left( {2 - x} \right)$$

$$f\left( x \right) = f\left( { - x} \right)$$

$$f\left( {x + 2} \right) = f\left( {x - 2} \right)$$