A) 3
B) 4
C) 1
D) 2
Correct Answer: A
Solution :
| [a] Let \[y={{\left( 1+\frac{1}{x} \right)}^{x}}\] |
| Taking logarithm of both sides, we get |
| \[\log y=x\left[ \log \left( 1+\frac{1}{x} \right) \right]\] |
| \[\Rightarrow \frac{1}{y}{{y}_{1}}(x)=\frac{{{x}^{2}}}{x+1}\left( -\frac{1}{{{x}^{2}}} \right)+\log \left( 1+\frac{1}{x} \right)\] |
| \[=-\frac{1}{x+1}+\log \left( 1+\frac{1}{x} \right)...(1)\] |
| Since, \[y(2)={{(1+1/2)}^{2}}=9/4\] |
| So, \[{{y}_{1}}(2)=(9/4)\left( -\frac{1}{3}+\log \frac{3}{2} \right)\] |
| Again differentiate eq. (1) w.r.t (x), we get |
| \[\frac{y(x){{y}_{2}}(x)-{{[{{y}_{1}}(x)]}^{2}}}{{{(y(x))}^{2}}}=\frac{1}{{{(1+x)}^{2}}}-\frac{1}{x(x+1)}\] |
| By putting \[x=2\], we get |
| \[\frac{y(2){{y}_{2}}(2)-{{({{y}_{1}}(2))}^{2}}}{{{(y(2))}^{2}}}=\frac{-1}{18}\] |
| Now, put value of y(2) and \[{{y}_{1}}(2)\] |
| \[\Rightarrow {{y}_{2}}(2)=\left( \frac{9}{4} \right){{\left( -\frac{1}{3}+\log \frac{3}{2} \right)}^{2}}-\frac{1}{8}\] |
| \[{{\left( {{y}_{2}}(2)+\frac{1}{8} \right)}^{4}}=9{{\left( \log \frac{3}{2}-\frac{1}{3} \right)}^{2}}\] |
| \[\Rightarrow \] Required expression = 3 |
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