The range of the function f(x)=sin^-1[x²+1/2]+cos^-1[x²-1/2],where [.] denotes the greatest integer function,is

1.{π/2,π}

2.{0,π/2}

3.{π}

4.(0,π/2)

Hi Menka,

As x^2 is always positive.

So x^2 + 1/2 is always greater than or equal to 1/2.

So [x^2 + 1/2] can be 0 or 1 (note that inverse of sin is not defined when it's domain is greater than 1)

similarly [x^2 - 1/2] can be -1, 0, or 1.

Please note for the combined [x^2 + 1/2] and [x^2 - 1/2]........ x^2 must be less than 3/2 for inv(sin) to be defined.

So [x^2 - 1/2] can take values only -1 or 0.

So we can see that when [x^2 + 1/2] = 0, ie when x^2 < 1/2, we have [x^2 - 1/2] = -1.

And hence f(x) = invsin(0) + invcos(-1) = 0 + pi = pi

Similarly, when [x^2 + 1/2] = 1, we have 1/2 <= x^2 < 3/2, and hence [x^2 - 1/2] = 0

And hence f(x) = invsin(1) + invcos(0) = pi/2 + pi/2 = pi.

As we have exhausted the entire domain of the function, there can be no other value that f(x) can take.

Hence in both the cases, f(x) = pi,

And hence the range of the function is pi.

Which is option (3).

Hope that helps.

All the best,

Regards,

Ashwin (IIT Madras)