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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)
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