For which of the values of x, the rate of increase of the function y=3x2-2x+7 is 4 times the rate of increase of x?

The correct option is 1.

To find the value of $x$ for which the rate of increase of the function $y$ is 4 times the rate of increase of $x$, we need to set up a relation using derivatives with respect to a variable, say time $t$.

Let the rate of increase of $x$ with respect to time $t$ be denoted by:
$\frac{dx}{dt}$
And the rate of increase of $y$ with respect to time $t$ be denoted by:
$\frac{dy}{dt}$

According to the problem description, the rate of increase of the function $y$ is 4 times the rate of increase of $x$. Therefore, we can write the relationship as:
$\frac{dy}{dt}=4\frac{dx}{dt}$

We are given the function:
$y=3x^2-2x+7$

Differentiating both sides of the function with respect to $t$ using the chain rule, we obtain:
$\frac{dy}{dt}=\frac{d}{dt}\left(3x^2-2x+7\right)$

Applying the differentiation rules:
$\frac{dy}{dt}=\left(6x-2\right)\frac{dx}{dt}$

Now, we substitute the relationship $\frac{dy}{dt}=4\frac{dx}{dt}$ into our differentiated equation:
$4\frac{dx}{dt}=\left(6x-2\right)\frac{dx}{dt}$

Assuming that the rate of change of $x$ is non-zero (i.e., $\frac{dx}{dt}\ne 0$), we can divide both sides by $\frac{dx}{dt}$:
$4=6x-2$

Solving for $x$:
Add 2 to both sides of the equation:
$4+2=6x$
$6=6x$
Divide by 6:
$x=1$

Thus, the rate of increase of the function is 4 times the rate of increase of $x$ when $x=1$.