Let f(x) = frac{x}{sqrt{a^2 + x^2}} - frac{d | vedclass

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(A) Given $f(x) = \frac{x}{\sqrt{a^2 + x^2}} - \frac{d - x}{\sqrt{b^2 + (d - x)^2}}$.To determine the nature of the function,we find the derivative $f

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$f$ is an increasing function of $x$

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(A) Given $f(x) = \frac{x}{\sqrt{a^2 + x^2}} - \frac{d - x}{\sqrt{b^2 + (d - x)^2}}$.To determine the nature of the function,we find the derivative $f'(x)$ with respect to $x$.Using the quotient rule or chain rule,let $u = x$ and $v = \sqrt{a^2 + x^2}$. Then $\frac{d}{dx}(\frac{u}{v}) = \frac{v(1) - u(\frac{x}{\sqrt{a^2+x^2}})}{a^2+x^2} = \frac{a^2+x^2-x^2}{(a^2+x^2)^{3/2}} = \frac{a^2}{(a^2+x^2)^{3/2}}$.Similarly,for the second term,let $g(x) = \frac{d-x}{\sqrt{b^2+(d-x)^2}}$. Let $u = d-x$,so $du = -dx$. The derivative of $\frac{u}{\sqrt{b^2+u^2}}$ is $\frac{b^2}{(b^2+u^2)^{3/2}}$. Since there is a negative sign in front of the term,we have $\frac{d}{dx}(-\frac{d-x}{\sqrt{b^2+(d-x)^2}}) = -(\frac{d}{dx} \frac{d-x}{\sqrt{b^2+(d-x)^2}}) = -(\frac{b^2}{(b^2+(d-x)^2)^{3/2}} \cdot (-1)) = \frac{b^2}{(b^2+(d-x)^2)^{3/2}}$.Thus,$f'(x) = \frac{a^2}{(a^2+x^2)^{3/2}} + \frac{b^2}{(b^2+(d-x)^2)^{3/2}}$.Since $a^2, b^2 > 0$ and the denominators are always positive,$f'(x) > 0$ for all $x \in R$.Therefore,$f$ is an increasing function of $x$.

Let $f(x) = \frac{x}{\sqrt{a^2 + x^2}} - \frac{d - x}{\sqrt{b^2 + (d - x)^2}}$,$x \in R$,where $a, b$ and $d$ are non-zero real constants. Then:

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