Statement-1: The equation x log x = 2 - x is | vedclass

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(A) Let $h(x) = f(x) - g(x) = x \log x - (2 - x) = x \log x + x - 2$.We evaluate the function $h(x)$ at the endpoints of the interval $[1, 2]$:$h(1) =

Vedclass · Accepted Answer

Statement-$1$ is true; Statement-$2$ is true; Statement-$2$ is a correct explanation for Statement-$1$.

Vedclass · Answer

(A) Let $h(x) = f(x) - g(x) = x \log x - (2 - x) = x \log x + x - 2$.We evaluate the function $h(x)$ at the endpoints of the interval $[1, 2]$:$h(1) = 1 \cdot \log(1) + 1 - 2 = 0 + 1 - 2 = -1$.$h(2) = 2 \cdot \log(2) + 2 - 2 = 2 \log(2) = \log(4)$.Since $\log(4) > 0$ (as $4 > 1$),we have $h(1)  0$.Since $h(x)$ is a continuous function on $[1, 2]$,by the Intermediate Value Theorem,there exists at least one $c \in (1, 2)$ such that $h(c) = 0$,which means $f(c) = g(c)$. Thus,Statement-$1$ is true.For Statement-$2$,$f'(x) = \log x + x(1/x) = \log x + 1$. For $x \in [1, 2]$,$f'(x) \geq 1 > 0$,so $f(x)$ is increasing. $g'(x) = -1  g(2)$,the graphs must intersect in $(1, 2)$. Thus,Statement-$2$ is true and provides a correct explanation for Statement-$1$.

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