Question

Let $\underset{h\to 0}{lim}\frac{h^2}{f(x+2h)-2f(x+h)+f\left(x\right)}=\frac{x^{1-x}}{1+x(1+\ln x{)}^2}.$ If $\underset{x\to 0^{+}}{lim}f\left(x\right)=1$ and $f\left(1\right)=2,$ then the value of $\frac{f\left(3\right)}{f\left(2\right)}$ is

Answer 1

$L=\underset{h\to 0}{lim}\frac{h^2}{f(x+2h)-2f(x+h)+f\left(x\right)}$
Using L-Hospital’s rule, we get
$L=\underset{h\to 0}{lim}\frac{2h}{2f^{\prime }(x+2h)-2f^{\prime }(x+h)}$
Again, using L-Hospital’s rule, we get
$L=\underset{h\to 0}{lim}\frac{2}{4f^{\prime \prime }(x+2h)-2f^{\prime \prime }(x+h)}$
$\Rightarrow \frac{1}{f^{\prime \prime }\left(x\right)}=\frac{x^{1-x}}{1+x(1+\ln x{)}^2}$
$\Rightarrow f^{\prime \prime }\left(x\right)=\frac{x(1+\ln x{)}^2+1}{x^{1-x}}$
$\Rightarrow f^{\prime \prime }\left(x\right)=x^{x-1}\left(x\right(1+\ln x{)}^2+1)$
$\Rightarrow f^{\prime \prime }\left(x\right)=x^x(1+\ln x{)}^2+x^{x-1}$
$\Rightarrow \int f^{\prime \prime }\left(x\right)dx=\int \underset{II}{\underset{}{x^x(1+\ln x)}}\cdot \underset{I}{\underset{}{(1+\ln x})}dx+\int x^{x-1}dx$
$\Rightarrow f^{\prime }\left(x\right)=x^x(1+\ln x)+C_1$
$\Rightarrow \int f^{\prime }\left(x\right)dx=\int x^x(1+\ln x)dx+\int C_{1}dx$
$\Rightarrow f\left(x\right)=x^x+C_{1}x+C_2$

Given, $\underset{x\to 0^{+}}{lim}f\left(x\right)=1$
$\Rightarrow \underset{x\to 0^{+}}{lim}(x^x+C_{1}x+C_2)=1$
$\Rightarrow 1+0+C_2=1\Rightarrow C_2=0$
$\therefore f\left(x\right)=x^x+C_{1}x$
Also, $f\left(1\right)=2$
$\Rightarrow f\left(1\right)=1+C_1=2\Rightarrow C_1=1$
$\therefore f\left(x\right)=x^x+x$
Hence, $\frac{f\left(3\right)}{f\left(2\right)}=\frac{30}{6}=5$