A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
| We have, |
| \[{{\sec }^{2}}(a+2)x+{{a}^{2}}-1=0\]\[\Rightarrow \]\[{{\tan }^{2}}(a+2)x+{{a}^{2}}=0\] |
| \[\therefore \]\[\tan (a+2)x=0\]and \[a=0\]\[\Rightarrow \]\[\tan \,(2x)=0\]and \[a=0\]\[\Rightarrow \]\[2x=nx\]and \[a=0\]\[\Rightarrow \]\[x=\frac{nx}{2}\]and \[a=0\] |
| \[x\in \,(-\pi ,\pi )\] |
| \[\therefore \]\[x=0,\frac{\pi }{2},-\frac{\pi }{2}\] |
| Number of ordered pair of \[(a,x)\] is 3. i.e. \[(0,0)\left( 0,\frac{\pi }{2} \right)\] and \[\left( 0,-\frac{\pi }{2} \right).\] |
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