Relations Question 16
Question: The range of $ f(x)=co{s^{-1}}( \frac{1+x^{2}}{2x} )+\sqrt{2-x^{2}} $ is
Options:
A) $ { 0,1+\frac{\pi }{2} } $
B) $ { 0,1+\pi } $
C) $ { 1,1+\frac{\pi }{2} } $
D) $ { 1,1+\pi } $
Answer:
Correct Answer: C
Solution:
- [c] $ {{\cos }^{-1}}( \frac{1+x^{2}}{2x} ) $ is define if $ | \frac{1+x^{2}}{2x} |\le 1 $ and $ x\ne 0 $ or $ 1+x^{2}-2| x |\le 0 $ or $ {{(| x |-1)}^{2}}\le 0 $ or $ x=1,-1 $
Thus, the domain of $ f(x) $ is {1, -1}. Hence, the range is $ {1,1+\pi } $ .
BETA
