Question

If $y={\left(1+1/x\right)}^x,$ then $\frac{2\sqrt{y_2\left(2\right)+1/8}}{\left(log3/2-1/3\right)}$ is equal to -

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

Let $y={\left(1+\frac{1}{x}\right)}^x$
Taking logarithm of both sides, we get
$logy=x\left[log\left(1+\frac{1}{x}\right)\right]$
$\Rightarrow \frac{1}{y}{y}_1\left(x\right)=\frac{x^2}{x+1}\left(-\frac{1}{x^2}\right)+log\left(1+\frac{1}{x}\right)=-\frac{1}{x+1}+log\left(1+\frac{1}{x}\right).......\left(1\right)$
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 $=3$