A) \[\frac{1}{5}\]
B) \[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-....\]
C) \[{{e}^{-2}}\]
D) \[{{e}^{-1}}\]
Correct Answer: A
Solution :
| [a] As v of charged particle is remaining constant, it means, force acting on charged particle is zero. |
| So, \[{{\tan }^{-1}}\frac{bc}{a(c-a)}\] |
| \[{{\tan }^{-1}}\frac{bc}{a}\]\[{{y}^{2}}=8x\] |
| \[y=x+2\]\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x-12y-2\text{ }z+20=0,\] |
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