Let f be a twice differentiable function defined on R such that...

We have, \left|\begin{cases}f\left(x\right) & {f}^{'}\left(x\right) \\ {f}^{'}\left(x\right) & {f}^{' '}\left(x\right)\end{cases}\right| = 0

⇒ f\left(x\right){f}^{' '}\left(x\right) - {\left({f}^{'}\left(x\right)\right)}^{2} = 0

⇒ \dfrac{{f}^{' '}\left(x\right)}{{f}^{'}\left(x\right)} = \dfrac{{f}^{'}\left(x\right)}{f\left(x\right)}

On integrating both side, we get

$\ln \left({f}^{'}\left(x\right)\right) = lnf\left(x\right) + lnc$

$⇒ {f}^{'}\left(x\right) = cf \left(x\right)$

$⇒ \dfrac{{f}^{'}\left(x\right)}{f\left(x\right)} = c$

Again integrating, we get

$lnf\left(x\right) = cx + {k}_{1}$

$⇒ f\left(x\right) = k{e}^{cx}$

Since, $f\left(0\right) = 1 = k$

Therefore, ${f}^{'}\left(0\right) = c = 2$

Now, $f\left(x\right) = {e}^{2x}$

Hence, $f\left(1\right) = {e}^{2} \in \left(6, 9\right)$