Continuity And Differentiability Question 5
Question 5 - 2024 (29 Jan Shift 2)
Let $f(x)=\sqrt{\lim _{r \rightarrow x}\left{\frac{2 r^{2}\left[(f(r))^{2}-f(x) f(r)\right]}{r^{2}-x^{2}}-r^{3} e^{\frac{f(r)}{r}}\right}}$ be differentiable in $(-\infty, 0) \cup(0, \infty)$ and $f(1)=1$.
Then the value of ea, such that $f(a)=0$, is equal to
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Answer (2)
Solution
$\mathrm{f}(1)=1, \mathrm{f}(\mathrm{a})=0$
$f^{2}(x)=\operatorname{Lim}_{r \rightarrow x}\left(\frac{2 r^{2}\left(f^{2}(r)-f(x) f(r)\right)}{r^{2}-x^{2}}-r^{3} e^{\frac{f(r)}{r}}\right)$
$=\operatorname{Lim}_{r \rightarrow x}\left(\frac{2 r^{2} f(r)}{r+x} \frac{(f(r)-f(x))}{r-x}-r^{3} e^{\frac{f(r)}{r}}\right)$
$f^{2}(x)=\frac{2 x^{2} f(x)}{2 x} f^{\prime}(x)-x^{3} e^{\frac{f(x)}{x}}$
$y^{2}=x y \frac{d y}{d x}-x^{3} e^{\frac{y}{x}}$
$\frac{y}{x}=\frac{d y}{d x}-\frac{x^{2}}{y} e^{\frac{y}{x}}$
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}-\frac{\mathrm{x}}{\mathrm{v}} \mathrm{e}^{\mathrm{v}}$
$\frac{d v}{d x}=\frac{e^{v}}{v} \Rightarrow e^{-v} v d v=d x$
Integrating both side
$\mathrm{e}^{\mathrm{v}}(\mathrm{x}+\mathrm{c})+1+\mathrm{v}=0$
$\mathrm{f}(1)=1 \Rightarrow \mathrm{x}=1, \mathrm{y}=1$
$\Rightarrow \mathrm{c}=-1-\frac{2}{\mathrm{e}}$
$\mathrm{e}^{\mathrm{v}}\left(-1-\frac{2}{\mathrm{e}}+\mathrm{x}\right)+1+\mathrm{v}=0$
$\mathrm{e}^{\frac{y}{x}}\left(-1-\frac{2}{\mathrm{e}}+\mathrm{x}\right)+1+\frac{\mathrm{y}}{\mathrm{x}}=0$
$\mathrm{x}=\mathrm{a}, \mathrm{y}=0 \Rightarrow \mathrm{a}=\frac{2}{\mathrm{e}}$
$\mathrm{ae}=2$
