Let f be a twice differentiable function defined on R such that f 0 = 1 ,   f ' 0 = 2 and f ' x…

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

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

$\Rightarrow \frac{f^{\text{'}\text{'}}\left(x\right)}{f^{\text{'}}\left(x\right)}=\frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}$

On integrating both side, we get

$\ln \left(f^{\text{'}}\left(x\right)\right)=\ln f\left(x\right)+\ln c$

$\Rightarrow f^{\text{'}}\left(x\right)=\mathit{cf}\left(x\right)$

$\Rightarrow \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}=c$

Again integrating, we get

$\ln f\left(x\right)=cx+k_1$

$\Rightarrow f\left(x\right)=k{e}^{cx}$

Since, $f\left(0\right)=1=k$

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

Now, $f\left(x\right)=e^{2x}$

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