Let f be a twice differentiable function defined on R such that f0=1,f'0=2and f1x≠0for all x∈R . If fxf'xf'xf"x=0 for all x∈R, then the value of f1 lies in the interval

fxf"x=f1x2⇒ f1x2−fxf11xf1x2=0 ⇒ ddxfxf1x=0⇒ fxf1x=k ⇒ k=12fx=±ke2x f0=1⇒f1=e2=7.38

Let f be a twice differentiable function defined on R such that f0=1,f'0=2and f1x≠0for all x∈R . If fxf'xf'xf"x=0 for all x∈R, then the value of f1 lies in the interval

fxf"x=f1x2⇒ f1x2−fxf11xf1x2=0 ⇒ ddxfxf1x=0⇒ fxf1x=k ⇒ k=12fx=±ke2x f0=1⇒f1=e2=7.38

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Let $f$ be a twice differentiable function defined on R such that $f (0) = 1, f^{'} (0) = 2$ and $f^{1} (x) \neq 0$ for all $x \in R$ . If $|\begin{matrix} f (x) & f^{'} (x) \\ f^{'} (x) & f^{"} (x) \end{matrix}| = 0$ for all $x \in R$ , then the value of $f (1)$ lies in the interval

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$(0, 3)$

$(9, 12)$

$(3, 6)$

$(6, 9)$

answer is A.

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$f (x) f^{"} (x) = {(f^{1} (x))}^{2} \Rightarrow \frac{{(f^{1} (x))}^{2} - f (x) f^{11} (x)}{{(f^{1} (x))}^{2}} = 0 \Rightarrow \frac{d}{d x} (\frac{f (x)}{f^{1} (x)}) = 0$ $\Rightarrow \frac{f (x)}{f^{1} (x)} = k \Rightarrow k = \frac{1}{2}$

$f (x) = \pm k e^{2 x} f (0) = 1 \Rightarrow f (1) = e^{2} = 7.38$

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Let f be a twice differentiable function defined on R such that f0=1,f'0=2and f1x≠0for all x∈R . If fxf'xf'xf"x=0 for all x∈R, then the value of f1 lies in the interval

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Let $f$ be a twice differentiable function defined on R such that $f (0) = 1, f^{'} (0) = 2$ and $f^{1} (x) \neq 0$ for all $x \in R$ . If $|\begin{matrix} f (x) & f^{'} (x) \\ f^{'} (x) & f^{"} (x) \end{matrix}| = 0$ for all $x \in R$ , then the value of $f (1)$ lies in the interval