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Question

Mathematics Question on range

Range of f(x)=sin−1x+tan−1x+sec−1xf(x) = sin^{-1} x + tan^{-1} x + sec^{-1} x is

A

(π4,3π4)\left(\frac{\pi}{4}, \frac{3\pi }{4}\right)

B

[π4,3π4]\left[\frac{\pi}{4}, \frac{3\pi }{4}\right]

C

\left\\{\frac{\pi}{4}, \frac{3\pi }{4}\right\\}

D

None of these

Answer

\left\\{\frac{\pi}{4}, \frac{3\pi }{4}\right\\}

Explanation

Solution

f(x)=sin−1x+tan−1x+sec−1xf(x ) = sin^{-1}x + tan^{-1}x + sec^{-1}x Domain of sin−1x=[−1,1]sin^{-1}x = [-1,1] Domain of tan−1x=(−∞,∞)tan^{-1}x = ( -\infty, \infty) Domain of sec−1=(−∞,∞)−(−1,1)sec^{-1} = ( -\infty, \infty) - (-1,1) Domain of f(x)=[−1,1]∩(−∞,∞)∩[(−∞,∞)−(−1,1)]f(x) = [ - 1, 1] \cap (-\infty, \infty) \cap [(-\infty, \infty) - (-1, 1)] =−1,1 = \\{-1,1\\} Now f(−1)=sin−1+tan−1(−1)+sec−1(−1)f(-1 ) = sin^{-1} + tan^{-1}( - 1 ) + sec^{-1}( -1 ) =−π2−π4+π=π4= -\frac{\pi}{2} - \frac{\pi }{4} +\pi = \frac{\pi }{4} and f(1)=sin−1(1)+tan−1(1)+sec−1(1)f\left(1\right) = sin^{-1}\left(1\right) + tan^{-1}\left(1\right) + sec^{-1}\left(1\right) =π2+π4+0= \frac{\pi }{2} + \frac{\pi }{4} + 0 =3π4= \frac{3\pi }{4} Range of f\left(x\right) = \left\\{\frac{\pi }{4}, \frac{3\pi }{4}\right\\}