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

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(A) Given $f(x) = e^x - x$ and $g(x) = x^2 - x$. $h(x) = (fog)(x) = f(g(x)) = e^{x^2-x} - (x^2-x) = e^{x^2-x} - x^2 + x$. Now,differentiate $h(x)$ wit

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

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(A) Given $f(x) = e^x - x$ and $g(x) = x^2 - x$. $h(x) = (fog)(x) = f(g(x)) = e^{x^2-x} - (x^2-x) = e^{x^2-x} - x^2 + x$. Now,differentiate $h(x)$ with respect to $x$: $h'(x) = e^{x^2-x}(2x-1) - 2x + 1$. $h'(x) = (2x-1)(e^{x^2-x} - 1)$. For $h(x)$ to be increasing,we require $h'(x) \geq 0$. $(2x-1)(e^{x^2-x} - 1) \geq 0$. Let $u = x^2-x$. Since $e^u - 1$ has the same sign as $u$,we have $(2x-1)(x^2-x) \geq 0$. $(2x-1)x(x-1) \geq 0$. Using the sign scheme for the critical points $0, \frac{1}{2}, 1$: For $x \in [0, \frac{1}{2}]$,$(2x-1) \leq 0$ and $x(x-1) \leq 0$,so the product is $\geq 0$. For $x \in [1, \infty)$,$(2x-1) > 0$ and $x(x-1) \geq 0$,so the product is $\geq 0$. Thus,$h(x)$ is increasing for $x \in [0, \frac{1}{2}] \cup [1, \infty)$.

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 $h(x)=(fog)(x)$ is increasing is

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