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At enotes, the educators help students work towards their problems, rather than solving their complete assignments for them. So, I will solve some of your problems and you can use them as examples and solve the rest yourself.
a) i) f(3) = `sin(2tan^(-1)(3/3)) = sin (2tan^(-1)(1)) = sin (2 pi/4) = sin(pi/2) = 1`
(ii) `f(-sqrt(3)) = sin(2tan^(-1)(-sqrt3 /3)) = sin(2tan^(-1) (-1/sqrt(3))) `
`= sin(2 * -pi/6) = sin(-pi/3) = -sqrt(3)/2`
b) since the domain of sin function is restricted to [-pi/2, pi/2], the domain of f(x) will also be `[-pi/2, pi/2]`
c) Range of f(x): In the given domain, sin function has a range of [-1,1].
thus the implied range of f(x) is also [-1,1].
f(2) can be calculated in the same way as f(3) was calculated.
`f^(-1) (x)` can be calculated by solving for x and then replacing x by y.
if we solve the function, f(x) = y (say),
we get, `x = 3 tan ((sin^(-1)y)/2)`
and then we replace x by y and vice versa, we get
`f^(-1) (x) = 3 tan((sin^(-1)x)/2)`
substituting x = -3/5 in the inverse function, we get
`f^(-1) (-3/5) = -1`
Try doing rest of the parts on your own, using the above as examples.
hope this helps.
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