If a, b 0,then the minimum value of y = fra | vedclass

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(A) Given the function $y = \frac{b^2}{a-x} + \frac{a^2}{x}$.To find the minimum,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{b^2}{(a-x

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$\frac{(a+b)^2}{a}$

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(A) Given the function $y = \frac{b^2}{a-x} + \frac{a^2}{x}$.To find the minimum,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{b^2}{(a-x)^2} - \frac{a^2}{x^2}$.Setting $\frac{dy}{dx} = 0$,we get $\frac{b^2}{(a-x)^2} = \frac{a^2}{x^2}$,which implies $\frac{b}{a-x} = \pm \frac{a}{x}$.Since $0  0$,we take the positive root: $\frac{b}{a-x} = \frac{a}{x} \Rightarrow bx = a^2 - ax \Rightarrow x(a+b) = a^2 \Rightarrow x = \frac{a^2}{a+b}$.Using the second derivative test: $\frac{d^2y}{dx^2} = \frac{2b^2}{(a-x)^3} + \frac{2a^2}{x^3}$.Since $x = \frac{a^2}{a+b}$ lies in $(0, a)$,both terms are positive,so $\frac{d^2y}{dx^2} > 0$,confirming a minimum.Substituting $x = \frac{a^2}{a+b}$ into $y$:$y_{\min} = \frac{b^2}{a - \frac{a^2}{a+b}} + \frac{a^2}{\frac{a^2}{a+b}} = \frac{b^2}{\frac{a^2+ab-a^2}{a+b}} + (a+b) = \frac{b^2(a+b)}{ab} + (a+b) = \frac{b(a+b)}{a} + (a+b) = (a+b)(\frac{b}{a} + 1) = (a+b)(\frac{a+b}{a}) = \frac{(a+b)^2}{a}$.

If $a, b > 0$,then the minimum value of $y = \frac{b^2}{a-x} + \frac{a^2}{x}$ for $0 < x < a$ is

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