A) \[\left\{ 0,\,\,1+\frac{\pi }{2} \right\}\]
B) \[\left\{ 0,\,\,1+\pi \right\}\]
C) \[\left\{ 1,\,\,1+\frac{\pi }{2} \right\}\]
D) \[\left\{ 1,\,\,1+\pi \right\}\]
Correct Answer: C
Solution :
[c] \[{{\cos }^{-1}}\left( \frac{1+{{x}^{2}}}{2x} \right)\] is define if \[\left| \frac{1+{{x}^{2}}}{2x} \right|\le 1\] and \[x\ne 0\] or \[1+{{x}^{2}}-2\left| x \right|\le 0\] or \[{{(\left| x \right|-1)}^{2}}\le 0\] or \[x=1,-1\] Thus, the domain of \[f(x)\]is {1, -1}. Hence, the range is\[\{1,1+\pi \}\].
You need to login to perform this action.
You will be redirected in
3 sec
