A) (a) 2
B) (b) \[\frac{1}{2}\]
C) (c) \[log3\]
D) (d) 1
Correct Answer: B
Solution :
[b] \[y=1-\log 2+\frac{\log 2}{\left| \!{\nderline {\, 2 \,}} \right. }-\frac{{{\left( \log \right)}^{3}}}{\left| \!{\nderline {\, 3 \,}} \right. }+......\infty ={{e}^{-\log 2}}\] \[\left[ \because {{e}^{-x}}=1.\frac{x}{\left| \!{\nderline {\, 1 \,}} \right. }+\frac{{{x}^{2}}}{\left| \!{\nderline {\, 2 \,}} \right. }-\frac{{{x}^{3}}}{\left| \!{\nderline {\, 3 \,}} \right. }+.....\infty \right]={{e}^{\log \left( \frac{1}{2} \right)}}=\frac{1}{2}\] Option [b] is correct.
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