A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
Given, \[4{{y}^{2}}+2{{\cos }^{2}}x=4y-{{\sin }^{2}}x\] \[\Rightarrow \,4{{y}^{2}}-4y+1+{{\cos }^{2}}x=0\] \[\Rightarrow \,{{(2y-1)}^{2}}+{{\cos }^{2}}x=0\] \[\therefore \,\,y=\frac{1}{2}\,\,and\,\,x=0\Rightarrow \,x=\frac{\pi }{2},\,\frac{3\pi }{2}\] So, two ordered pairs are possible i.e., \[\left( \frac{\pi }{2},\,\frac{1}{2} \right)\] and \[\left( \frac{3\pi }{2},\,\frac{1}{2} \right)\]You need to login to perform this action.
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