and $d$ are non-zero real constants. Then :


Let $f(x)=\frac{x}{\sqrt{\mathrm{a}^{2}+x^{2}}}-\frac{\mathrm{d}-x}{\sqrt{\mathrm{b}^{2}+(\mathrm{d}-x)^{2}}}, x \in \mathbf{R}$ where $\mathrm{a}, \mathrm{b}$

and $d$ are non-zero real constants. Then :

  1. (1) $f$ is an increasing function of $x$

  2. (2) $f$ is a decreasing function of $x$

  3. (3) $f^{\prime}$ is not a continuous function of $x$

  4. (4) $f$ is neither increasing nor decreasing function of $x$

Correct Option: 1, 2




$f^{\prime}(x)=\frac{\sqrt{a^{2}+x^{2}}-\frac{x(2 x)}{2 \sqrt{a^{2}+x^{2}}}}{\left(a^{2}+x^{2}\right)}$

$+\frac{\sqrt{b^{2}+(x-d)^{2}}-\frac{(x-d) 2(x-d)}{2 \sqrt{b^{2}+(x-d)^{2}}}}{\left(b^{2}+(x-d)^{2}\right)}$

$=\frac{a^{2}+x^{2}-x^{2}}{\left(a^{2}+x^{2}\right)^{3 / 2}}+\frac{b^{2}+(x-d)^{2}-(x-d)^{2}}{\left(b^{2}+(x-d)^{2}\right)^{3 / 2}}$

$=\frac{a^{2}}{\left(a^{2}+x^{2}\right)^{3 / 2}}+\frac{b^{2}}{\left(b^{2}+(x-d)^{2}\right)^{3 / 2}}>0$

$\Rightarrow f^{\prime}(x)>0, \forall x \in R$

$\Rightarrow f(x)$ is increasing function.

Hence, $f(x)$ is increasing function.

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