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

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"$$\left| {\matrix{ {f(x)} & {f'(x)} \cr {f'(x)} & {f''(x)} \cr } } \right| = 0$$

$\Rightarrow f(x).f''(x) - {\left( {f'(x)} \right)^2} = 0$

Dividing by ${\left( {f(x)} \right)^2}$, we get

$$ \Rightarrow {{f(x).f''(x) - {{\left( {f'(x)} \right)}^2}} \over {{{\left( {f(x)} \right)}^2}}} = 0$$

$\Rightarrow {d \over {dx}}\left( {{{f'(x)} \over {f(x)}}} \right) = 0$

Integrating both side,

${{f'(x)} \over {f(x)}} = c$ (constant)

At, $x = 0$, ${{f'(0)} \over {f(0)}} = c$

$\Rightarrow {2 \over 1} = c$

$\Rightarrow c = 2$

$\therefore$ ${{f'(x)} \over {f(x)}} = 2$

$\Rightarrow \int {{{f'(x)} \over {f(x)}}} dx = 2\int {dx}$

$\Rightarrow \ln |f(x)|\, = 2x + c'$

at x = 0,

$ln|f(0)|\, = 0 + c'$

$\Rightarrow 0 = 0 + c'$

$\Rightarrow c' = 0$

$\therefore$ $n|f(x)| = 2x$

$\Rightarrow f(x) = {e^{2x}}$

$f(1) = {e^2} = {(2.71)^2} = 7.34$

So it lie between (6, 9)."

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

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