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Let f be a twice differentiable function defined on R such that f\left ( 0 \right )=1,f'\left ( 0 \right )=2 and f'\left ( x \right )\neq 0 for all x\; \epsilon \; R. If \begin{vmatrix} f(x) & f'(x)\\ f'(x) & f''(x) \end{vmatrix}=0 , for all x\; \epsilon \; R, then the value of f\left ( 1 \right ) lies in the interval :
 
Option: 1 \left ( 0,3 \right )
Option: 2 \left ( 9,12 \right )
Option: 3 \left ( 6,9 \right )
Option: 4 \left ( 3,6 \right )

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\left|\begin{array}{ll} f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x) \end{array}\right|=0

\\\mathrm{f}(\mathrm{x}) \mathrm{f}^{\prime \prime}(\mathrm{x})-\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^{2}=0 \\ \\\frac{\mathrm{f}^{\prime \prime}(\mathrm{x})}{\mathrm{f}^{\prime}(\mathrm{x})}=\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ \\\ln \left(\mathrm{f}^{\prime}(\mathrm{x})\right)=\ln \mathrm{f}(\mathrm{x})+\ln \mathrm{c}

\\f^{\prime}(x)=\operatorname{cf}(x) \\ \\\frac{f^{\prime}(x)}{f(x)}=c \\ \\\ln f(x)=c x+k_1

\\f(x)=k e^{c x} \\ \\f(0)=1=k \\ \\f^{\prime}(0)=c=2 \\ \\f(x)=e^{2 x} \\ \\f(1)=e^{2} \in(6,9)

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Suraj Bhandari

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