Statement -1: The function f(x) = x^2(e^x + e | vedclass

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(C) Let $f(x) = x^2e^x + x^2e^{-x}.$For $x > 0,$ $x^2e^x$ is clearly increasing because its derivative $2xe^x + x^2e^x = xe^x(2+x) > 0$ for $x > 0.$No

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Statement $-1$ is true; Statement $-2$ is false.

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(C) Let $f(x) = x^2e^x + x^2e^{-x}.$For $x > 0,$ $x^2e^x$ is clearly increasing because its derivative $2xe^x + x^2e^x = xe^x(2+x) > 0$ for $x > 0.$Now consider $h(x) = x^2e^{-x}.$$h'(x) = 2xe^{-x} - x^2e^{-x} = xe^{-x}(2-x).$For $0  0,$ but for $x > 2,$ $h'(x) Thus,$h(x)$ is not increasing for all $x > 0,$ so Statement $-2$ is false.Now,$f'(x) = \frac{d}{dx}(x^2e^x) + \frac{d}{dx}(x^2e^{-x}) = (2x+x^2)e^x + (2x-x^2)e^{-x} = 2x(e^x+e^{-x}) + x^2(e^x-e^{-x}).$Since $e^x > e^{-x}$ for all $x > 0,$ $e^x - e^{-x} > 0.$Thus,$f'(x) > 0$ for all $x > 0,$ meaning $f(x)$ is increasing for all $x > 0.$Therefore,Statement $-1$ is true and Statement $-2$ is false.

Statement $-1:$ The function $f(x) = x^2(e^x + e^{-x})$ is increasing for all $x > 0.$
Statement $-2:$ The functions $g(x) = x^2e^x$ and $h(x) = x^2e^{-x}$ are increasing for all $x > 0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b).$

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Statement $-1:$ The function $f(x) = x^2(e^x + e^{-x})$ is increasing for all $x > 0.$Statement $-2:$ The functions $g(x) = x^2e^x$ and $h(x) = x^2e^{-x}$ are increasing for all $x > 0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b).$