A) - 3
B) - 2
C) 1
D) None of these
Correct Answer: D
Solution :
\[{{\sin }^{2}}\theta \le 1\] \\[\frac{{{x}^{2}}+{{y}^{2}}+1}{2x}\le 1\] \[{{x}^{2}}+{{y}^{2}}-2x+1\le 0\]. \[{{(x-1)}^{2}}+{{y}^{2}}\le 0\] It is possible, iff \[x=1\] and \[y=0\], i.e., It also depends on value of y. Hence, option (d) is correct.
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