A) \[{{x}^{2}}-y=1\]
B) \[{{x}^{2}}+y=1\]
C) \[{{x}^{2}}-y+1=0\]
D) \[{{x}^{2}}-y=2\]
Correct Answer: D
Solution :
[d] \[{{t}^{2}}+\frac{1}{{{t}^{2}}}={{\left( t+\frac{1}{t} \right)}^{2}}-2\] \[\Rightarrow \,\,y={{x}^{2}}-2\,\,or\,{{x}^{2}}-y=2.\]You need to login to perform this action.
You will be redirected in
3 sec