Let 𝒇(𝒙) = 𝒙² − 𝟐𝒙 + 𝟐 be a continuous function defined on 𝒙 ∈ [𝟏, 𝟑]. The point 𝒙 at which the tangent of 𝒇(𝒙) becomes parallel to the straight line joining 𝒇(𝟏) and 𝒇(𝟑) is

The correct answer is 2.

To find the point $x$ at which the tangent to the curve $f\left(x\right)=x^2-2x+2$ is parallel to the secant line (straight line) joining the points $(1,f(1\left)\right)$ and $(3,f(3\left)\right)$, we apply Lagrange's Mean Value Theorem.

According to Lagrange's Mean Value Theorem, if a function is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there exists a point $x=c$ in the interval $(a,b)$ such that the derivative $f^{\prime }\left(c\right)$ is equal to the average rate of change of the function over the interval:
$$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$$

Let us calculate the required components step-by-step:

Step 1: Find the function values at the endpoints of the interval $[1,3]$.
For $a=1$:
$f\left(1\right)=1^2-2\left(1\right)+2=1-2+2=1$
For $b=3$:
$f\left(3\right)=3^2-2\left(3\right)+2=9-6+2=5$

Step 2: Find the slope of the line joining $f\left(1\right)$ and $f\left(3\right)$.
The slope of the straight line joining $(1,1)$ and $(3,5)$ is:
$\text{Slope}=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{5-1}{3-1}=\frac{4}{2}=2$

Step 3: Find the derivative of the function $f\left(x\right)$.
The derivative representing the slope of the tangent at any point $x$ is:
$f^{\prime }\left(x\right)=\frac{d}{dx}(x^2-2x+2)=2x-2$

Step 4: Equate the slope of the tangent to the slope of the secant line.
We set $f^{\prime }\left(x\right)=2$ and solve for $x$:
$2x-2=2$
$2x=4$
$x=2$

Since $x=2$ lies in the open interval $(1,3)$, the point at which the tangent of the function becomes parallel to the line joining the endpoints is indeed $2$.