Statement-1: The equation $$x\log x = 2 - x$$ is satisfied by at least one value of $$x$$ lying between 1 and 2.
Statement-2: The function $$f(x) = x\log x$$ is an increasing function in $$[1, 2]$$ and $$g(x)=2-x$$ is a decreasing function in $$[1, 2]$$ and the graphs represented by these functions intersect at a point in $$[1, 2]$$.

$$h(x) = x \log x + x - 2$$

At $$x = 1$$: $$h(1) = 1 \cdot \log(1) + 1 - 2 = 0 + 1 - 2 = -1$$.

At $$x = 2$$: $$h(2) = 2 \cdot \log(2) + 2 - 2 = 2 \log(2)$$. Since $$\log(2)$$ is positive, $$h(2) > 0$$.

Because $$h(x)$$ is a continuous function that changes sign from negative to positive between $$x=1$$ and $$x=2$$, the Intermediate Value Theorem (IVT) guarantees there is at least one value $$c \in (1, 2)$$ such that $$h(c) = 0$$. Thus, Statement-1 is true.

For $$f(x) = x \log x$$: The derivative is $$f'(x) = \log x + 1$$. In the interval $$[1, 2]$$, $$\log x$$ is $$\ge 0$$, so $$f'(x) > 0$$. This confirms $$f(x)$$ is increasing.

For $$g(x) = 2 - x$$: The derivative is $$g'(x) = -1$$. Since the derivative is negative, $$g(x)$$ is decreasing.

If one function is strictly increasing and the other is strictly decreasing, they can intersect at most once. Since we already proved via IVT that they must intersect at least once, the behavior described in Statement-2 explains exactly why and how that intersection point exists in the given interval.