Let f be twice differentiable on R, f(0)=1, f'(0)=2, f''(x)0.

Question

Let $f$ be twice differentiable on $\mathbb{R}$, $f(0)=1$, $f'(0)=2$, $f''(x)\ne0$. If
$$\begin{vmatrix}f(x)&f'(x)\f'(x)&f''(x)\end{vmatrix}=0\quad\forall x,$$
then $f(1)$ lies in:

Accepted Answer

$f(x)f''(x)=(f'(x))^2 \implies \dfrac{f''}{f'}=\dfrac{f'}{f} \implies f'(x)=cf(x)$.

At $x=0$: $2=c \implies c=2$. Then $f(x)=e^{2x}$ (using $f(0)=1$), so $f(1)=e^2\in(6,9)$.

Solutions · 1