Explanation
Given that $$f (x)$$is a differentiable function of $$ x$$ and that $$f(x)$$ . $$f (y)$$ = $$f (x) $$+ $$f (y)$$ + $$f (xy) -2$$ and that$$f (2) =5$$.
Then $$f (3)$$ is equal to?
$$\displaystyle\mathrm {f'(x)}=\lim_{h \rightarrow 0}\frac{ \mathrm{f (x+h) - f(x)} }{h}$$ $$\displaystyle =\lim_{h\to 0}\frac{\mathrm {f(x) f(h)- f(x)}}{h}$$ $$\displaystyle = \mathrm{f(x)}\lim_{h\to 0}\frac{\mathrm {f(h)-1}}{h}$$ $$\displaystyle =\mathrm{ f(x)}\lim_{h\to 0}\mathrm{ g(h)}$$, $$( \because \mathrm{f(x)}= 1+ \mathrm{x g(x)}\; )$$
$$= \mathrm{f(x)} $$ ($$\displaystyle \because \lim_{x \rightarrow 0} \mathrm{g(x)} =1$$) $$\Rightarrow \displaystyle \mathrm{ f'(x)}= \mathrm {f(x)}$$
If $$e^y+xy=$$e then
$$e^y+xy=e$$
On substituting $$x=$$ 0, we get $$e^y=e$$
$$y = 1$$ when $$x = 0$$
On differentiating the relation (i) we get
$$e^y\dfrac{dy}{dx}+1.y+x.\dfrac{dy}{dx}=0$$
On substituting, $$x = 0$$, $$y = 1$$ we get
$$e^{y}\left( \dfrac{dy}{dx} \right)^{2}+e^{y}\dfrac{d^{2}y}{dx^{2}}+\dfrac{dy}{dx}+\dfrac{dy}{dx}+x\dfrac{d^{2}y}{dx^{2}}=0$$
On substituting$$x = 0$$, $$y = 1, \dfrac{dy}{dx}=\dfrac{-1}{e} we get$$
$$=\dfrac{d^2y}{dx^2}=\dfrac{1}{e^2}\rightarrow \lambda=2$$
Hence, option 'A' is correct.
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