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Solution

Let $\displaystyle y = \left(\dfrac{1}{x}\right)^x = e^{x\ln(1/x)} = e^{-x\ln x}$ Take $\ln$ on both sides: $\ln y = -x\ln x$ Differentiate w.r.t $x$: $\dfrac{1}{y}\dfrac{dy}{dx} = -(\ln x + 1)$ $\Rightarrow \dfrac{dy}{dx} = -y(\ln x + 1)$ For maximum or minimum, set $\dfrac{dy}{dx}=0$: $\ln x + 1 = 0 \Rightarrow x = \dfrac{1}{e}$ Now, $y_{\max} = \left(\dfrac{1}{1/e}\right)^{1/e} = e^{1/e}$

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Solution

**Solution:** Let $y = a^{x} + \dfrac{a}{a^{a x}} = a^{x} + a^{1 - a x}$ Let $t = a^{x}$, $t > 0$. Then $y = t + \dfrac{a}{t^{a}}$ Differentiate: \[ \dfrac{dy}{dt} = 1 - a^2 t^{-a-1} = 0 \Rightarrow t^{a+1} = a^2 \Rightarrow t = a^{\tfrac{2}{a+1}}. \] Substitute back: \[ y_{\min} = a^{\tfrac{2}{a+1}} + a^{1 - a\tfrac{2}{a+1}} = 2\sqrt{a}. \] $\boxed{\text{Answer: (A) }2\sqrt{a}}$

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Solution

$y' = 3x^2 - 3 = 0 \Rightarrow x = 1$ Now, $y(0) = 2$, $y(1) = 0$, $y(2) = 8 - 6 + 2 = 4$ Hence, maximum value = 4.

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Solution

$f'(x) = e^{x^2}(1 - x^2)$. Setting $f'(x) = 0 \Rightarrow 1 - x^2 = 0 \Rightarrow x = \pm 1$.

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Solution

From $y'=a/x + 2bx + 1=0$ → solving gives $a=2,\ b=-\tfrac{1}{2}$.