Let f(x) = e^x - x and g(x) = x^2 - x,forall | vedclass

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(A) $h(x) = f(g(x))$$h'(x) = f'(g(x)) \cdot g'(x)$Given $f(x) = e^x - x$,so $f'(x) = e^x - 1$.Given $g(x) = x^2 - x$,so $g'(x) = 2x - 1$.Thus,$h'(x) =

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$\left[ 0, \frac{1}{2} \right] \cup [1, \infty)$

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(A) $h(x) = f(g(x))$$h'(x) = f'(g(x)) \cdot g'(x)$Given $f(x) = e^x - x$,so $f'(x) = e^x - 1$.Given $g(x) = x^2 - x$,so $g'(x) = 2x - 1$.Thus,$h'(x) = (e^{g(x)} - 1) \cdot g'(x) = (e^{x^2 - x} - 1)(2x - 1)$.For $h(x)$ to be increasing,we require $h'(x) \geq 0$,which means $(e^{x^2 - x} - 1)(2x - 1) \geq 0$.Case $1$: Both factors are non-negative.$e^{x^2 - x} - 1 \geq 0 \Rightarrow e^{x^2 - x} \geq 1 \Rightarrow x^2 - x \geq 0 \Rightarrow x(x - 1) \geq 0 \Rightarrow x \in (-\infty, 0] \cup [1, \infty)$.$2x - 1 \geq 0 \Rightarrow x \geq \frac{1}{2}$.Intersection: $x \in [1, \infty)$.Case $2$: Both factors are non-positive.$e^{x^2 - x} - 1 \leq 0 \Rightarrow x^2 - x \leq 0 \Rightarrow x \in [0, 1]$.$2x - 1 \leq 0 \Rightarrow x \leq \frac{1}{2}$.Intersection: $x \in [0, \frac{1}{2}]$.Combining both cases,the set of $x$ is $[0, \frac{1}{2}] \cup [1, \infty)$.

Let $f(x) = e^x - x$ and $g(x) = x^2 - x$,$\forall x \in R$. Then the set of all $x \in R$,where the function $h(x) = (f \circ g)(x)$ is increasing is

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