A) 0
B) 2
C) 4
D) \[\infty \]
Correct Answer: B
Solution :
Since,\[{{\sin }^{-1}}\sqrt{x},\,\,{{\cos }^{-1}}\sqrt{x}\]are defined for\[x\le 1\]and\[x\ge 0\]. \[\therefore \] \[0\le \sqrt{{{x}^{2}}-x+1}\le 1\] and \[0\le \sqrt{{{x}^{2}}-x}\le 1\] \[\Rightarrow \] \[-1\le {{x}^{2}}-x\le 0\] and \[0\le {{x}^{2}}-x\le 0\] \[\Rightarrow \] \[{{x}^{2}}-x=0\] \[\Rightarrow \] \[x=0,\,\,1\] Hence, number of solutions are two.You need to login to perform this action.
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