Continuity And Differentiability Question 5
Question 5 - 2024 (29 Jan Shift 2)
Let $f(x)=\sqrt{\lim _{r \rightarrow x}{\frac{2 r^{2}\left[(f(r))^{2}-f(x) f(r)\right]}{r^{2}-x^{2}}-r^{3} e^{\frac{f(r)}{r}} }}$ 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
$f(1)=1, f(a)=0$
$f^{2}(x)=\operatorname{Lim} _{r \rightarrow x}\left(\frac{2 r^{2}\left(f^{2}(r)-f(x) f(r)\right)}{r^{2}-x^{2}}-r^{3} e^{\frac{f(r)}{r}}\right)$
$=\operatorname{Lim} _{r \rightarrow x}\left(\frac{2 r^{2} f(r)}{r+x} \frac{(f(r)-f(x))}{r-x}-r^{3} e^{\frac{f(r)}{r}}\right)$
$f^{2}(x)=\frac{2 x^{2} f(x)}{2 x} f^{\prime}(x)-x^{3} e^{\frac{f(x)}{x}}$
$y^{2}=x y \frac{d y}{d x}-x^{3} e^{\frac{y}{x}}$
$\frac{y}{x}=\frac{d y}{d x}-\frac{x^{2}}{y} e^{\frac{y}{x}}$
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$v=v+x \frac{dv}{dx}-\frac{x}{v} e^{v}$
$\frac{d v}{d x}=\frac{e^{v}}{v} \Rightarrow e^{-v} v d v=d x$
Integrating both side
$e^{v}(x+c)+1+v=0$
$f(1)=1 \Rightarrow x=1, y=1$
$\Rightarrow c=-1-\frac{2}{e}$
$e^{v}\left(-1-\frac{2}{e}+x\right)+1+v=0$
$e^{\frac{y}{x}}\left(-1-\frac{2}{e}+x\right)+1+\frac{y}{x}=0$
$x=a, y=0 \Rightarrow a=\frac{2}{e}$
$ae=2$
