Question

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 :-

• A. $f$ is a decreasing function of $x$
• B. $f$ is neither increasing nor decreasing function of $x$
• C. $f^{\prime }$ is not continuous function of $x$
• D. $f$ is an increasing function of $x$

Answer 1

The correct option is D $f$ is an increasing function of $x$

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

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

Hence $f\left(x\right)$ is increasing.