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

We have the real-valued function

$$f(x)=\dfrac{x}{\sqrt{a^{2}+x^{2}}}-\dfrac{\,d-x\,}{\sqrt{\,b^{2}+(d-x)^{2}}},\qquad x\in\mathbb R,$$

where the given constants $$a,\;b,\;d$$ are all non-zero. To decide whether $$f(x)$$ is increasing or decreasing, we differentiate it and study the sign of the derivative.

First, recall the differentiation rule

$$\dfrac{d}{dx}\Bigl\{u(x)\,[v(x)]^{n}\Bigr\}=u'(x)\,[v(x)]^{n}+u(x)\,n\,[v(x)]^{\,n-1}\,v'(x).$$

Let us treat the two terms of $$f(x)$$ one by one.

Derivative of the first term. Write

$$g(x)=\dfrac{x}{\sqrt{a^{2}+x^{2}}}=x\,(a^{2}+x^{2})^{-1/2}.$$

Applying the rule, we get

$$g'(x)=1\cdot(a^{2}+x^{2})^{-1/2}+x\left(-\tfrac12\right)(a^{2}+x^{2})^{-3/2}\,(2x).$$

Simplifying step by step,

$$g'(x)=(a^{2}+x^{2})^{-1/2}-x^{2}(a^{2}+x^{2})^{-3/2} =\dfrac{a^{2}+x^{2}-x^{2}}{(a^{2}+x^{2})^{3/2}} =\dfrac{a^{2}}{(a^{2}+x^{2})^{3/2}}.$$

Because $$a^{2}>0$$ and the denominator is positive, we have

$$g'(x)>0\quad\text{for every }x\in\mathbb R.$$

Derivative of the second term. Put

$$u=d-x,\qquad h(x)=\dfrac{u}{\sqrt{\,b^{2}+u^{2}}}=u\,(b^{2}+u^{2})^{-1/2}.$$

First differentiate with respect to $$u$$:

$$\frac{d}{du}h(u)=1\cdot(b^{2}+u^{2})^{-1/2}+u\left(-\tfrac12\right)(b^{2}+u^{2})^{-3/2}\,(2u) =(b^{2}+u^{2})^{-1/2}-u^{2}(b^{2}+u^{2})^{-3/2} =\dfrac{b^{2}}{(b^{2}+u^{2})^{3/2}}.$$

Now, since $$u=d-x$$, we have $$du/dx=-1$$, so by the chain rule

$$h'(x)=\dfrac{d}{dx}h(u)=\dfrac{b^{2}}{(b^{2}+u^{2})^{3/2}}\cdot(-1) =-\,\dfrac{b^{2}}{(b^{2}+(d-x)^{2})^{3/2}}.$$

Again, $$b^{2}>0$$ and the denominator is positive, hence

$$h'(x)<0\quad\text{for every }x\in\mathbb R.$$

Derivative of the whole function. Remember that

$$f(x)=g(x)-h(x).$$

Therefore

$$f'(x)=g'(x)-h'(x) =\dfrac{a^{2}}{(a^{2}+x^{2})^{3/2}}-\!\left(-\,\dfrac{b^{2}}{(b^{2}+(d-x)^{2})^{3/2}}\right) =\dfrac{a^{2}}{(a^{2}+x^{2})^{3/2}}+\dfrac{b^{2}}{(b^{2}+(d-x)^{2})^{3/2}}.$$

Both numerators, $$a^{2}$$ and $$b^{2},$$ are strictly positive, and both denominators are positive for every real $$x.$$ Hence

$$f'(x)>0\quad\text{for all }x\in\mathbb R.$$

Because the derivative is strictly positive everywhere, the function $$f(x)$$ is strictly increasing on the entire real line. Moreover, each part of $$f'(x)$$ is a quotient of continuous functions whose denominators never vanish, so $$f'(x)$$ itself is continuous; thus option C is incorrect.

Therefore $$f(x)$$ is an increasing function, matching option A, while options B, C and D are false.

Hence, the correct answer is Option A.