A) 1.648
B) 1.547
C) 1.447
D) 1.348
Correct Answer: A
Solution :
\[\because \,\,\,\,{{e}^{x}}=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+....\] Putting \[x=\frac{1}{2}\]on both the sides \[\sqrt{e}={{e}^{1/2}}\] = \[1+\frac{1}{2}+\frac{{{\left( \frac{1}{2} \right)}^{2}}}{2!}+\frac{{{\left( \frac{1}{2} \right)}^{3}}}{3!}+\frac{{{\left( \frac{1}{2} \right)}^{4}}}{4!}+....\] = \[1+\frac{1}{2}+\frac{1}{{{2}^{2}}2}+\frac{1}{{{2}^{3}}6}+\frac{1}{{{2}^{4}}.24}+....\] \[1+0.5+0.1250+0.0208+0.0026\] \[=1.648\], (approximately).
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