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Q.
If $y=\left(1+1/x\right)^{x},$ then $\frac{2\sqrt{y_{2}\left(2\right)+1/8}}{\left(log\,3/2-1/3\right)}$ is equal to -
Continuity and Differentiability
Solution:
Let $y=\left(1+\frac{1}{x}\right)^{x}$
Taking logarithm of both sides, we get
$log \,y=x\left[log\left(1+\frac{1}{x}\right)\right]$
$\Rightarrow \frac{1}{y}y_{1}\left(x\right)=\frac{x^{2}}{x+1}\left(-\frac{1}{x^{2}}\right)+log\left(1+\frac{1}{x}\right)=-\frac{1}{x+1}+log\left(1+\frac{1}{x}\right)\,.......\left(1\right)$
Since, $y \left(2\right) = \left(1 + 1/2\right)^{2} = 9/4$
so, $y_{1}\left(2\right)=\left(9/4\right)\left(-\frac{1}{3}+loh \frac{3}{2}\right)$
Again differentiate eq $\left(1\right)$ w.r.t $\left(x\right),$ we get
$\frac{y\left(x\right)y_{2}\left(x\right)-\left[y_{1}\left(x\right)\right]^{2}}{\left(y\left(x\right)\right)^{2}}=\frac{1}{\left(1+x\right)^{2}}-\frac{1}{x\left(x+1\right)}$
By putting $x = 2$, we get
$\frac{y\left(2\right)y_{2}-\left(y_{1}\left(2\right)\right)^{2}}{\left(y\left(2\right)\right)^{2}}=\frac{-1}{18}$
Now, put value of $y \left(2\right)$ and $y_{1}\left(2\right)$
$\Rightarrow y_{2}\left(2\right)=\left(\frac{9}{4}\right)\left(-\frac{1}{3}+log \frac{3}{2}\right)^{2}-\frac{1}{8}$
$4\left(y_{2}\left(2\right)+\frac{1}{8}\right)=9\left(log \frac{3}{2}-\frac{1}{3}\right)^{2}$
$\Rightarrow $ Required expression $= 3$