If the angle between vectors $x\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-2x\hat{j}+\hat{k}$ is acute, the $x$ lies in the interval of

The correct option is C $(-\infty ,1]$
Let the angle between given vectors be $\theta$ , given that it is accute

We have $cos\theta =\frac{(xi+j+k).(i-2xj+k)}{|xi+j+k||i-2xj+k|}=\frac{-x+1}{\sqrt{x^2+2}\sqrt{4x^2+2}}$

Since $\theta$ is accute , the value of $cos\theta$ must be positive

$\Rightarrow \frac{-x+1}{\sqrt{x^2+2}\sqrt{4x^2+2}}>0$

$\Rightarrow x<1$

Therefore option $C$ is correct