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Let f :(0, \infty )\rightarrow (0,\infty ) be a differentiable function such that f(1)=e and \lim_{t\rightarrow x}\frac{t^{2}f^{2}(x)-x^{2}f^{2}(t)}{t-x}=0. If f(x)=1, then x is equal to:
Option: 1 \frac{1}{e}
Option: 2 2e
Option: 3 \frac{1}{2e}
Option: 4 e

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\begin{aligned} &L=\operatorname{Lim}_{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}\\ &\text { using L H'opital. rule }\\ &\mathrm{L}=\operatorname{Lim}_{\mathrm{t} \rightarrow \mathrm{x}} \frac{2 \mathrm{tf}^{2}(\mathrm{x})-\mathrm{x}^{2} \cdot 2 \mathrm{f}^{\prime}(\mathrm{t}) \cdot \mathrm{f}(\mathrm{t})}{1}\\ &\Rightarrow \mathrm{L}=2 \mathrm{xf}(\mathrm{x})(\mathrm{f}(\mathrm{x})-\mathrm{xf}(\mathrm{x}))=0(\text { given }) \end{aligned}

\\\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{xf}^{\prime}(\mathrm{x}) \Rightarrow \int \frac{\mathrm{f}^{\prime}(\mathrm{x}) \mathrm{dx}}{\mathrm{f}(\mathrm{x})}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\ \Rightarrow \ell \mathrm{n} \operatorname{lf}(\mathrm{x})|=\ell \mathrm{n}| \mathrm{x} \mid+\mathrm{C} \\ \because \mathrm{f}(1)=\mathrm{e}, \mathrm{x}>0, \mathrm{f}(\mathrm{x})>0 \\ \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{ex}, \text { if } \mathrm{f}(\mathrm{x})=1 \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}

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himanshu.meshram

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