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 :
Correct Option: 1, 2
$f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{(d-x)}{\sqrt{b^{2}+(d-x)^{2}}}$
$=\frac{x}{\sqrt{a^{2}+x^{2}}}+\frac{(x-d)}{\sqrt{b^{2}+(x-d)^{2}}}$
$f^{\prime}(x)=\frac{\sqrt{a^{2}+x^{2}}-\frac{x(2 x)}{2 \sqrt{a^{2}+x^{2}}}}{\left(a^{2}+x^{2}\right)}$
$+\frac{\sqrt{b^{2}+(x-d)^{2}}-\frac{(x-d) 2(x-d)}{2 \sqrt{b^{2}+(x-d)^{2}}}}{\left(b^{2}+(x-d)^{2}\right)}$
$=\frac{a^{2}+x^{2}-x^{2}}{\left(a^{2}+x^{2}\right)^{3 / 2}}+\frac{b^{2}+(x-d)^{2}-(x-d)^{2}}{\left(b^{2}+(x-d)^{2}\right)^{3 / 2}}$
$=\frac{a^{2}}{\left(a^{2}+x^{2}\right)^{3 / 2}}+\frac{b^{2}}{\left(b^{2}+(x-d)^{2}\right)^{3 / 2}}>0$
$\Rightarrow f^{\prime}(x)>0, \forall x \in R$
$\Rightarrow f(x)$ is increasing function.
Hence, $f(x)$ is increasing function.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.