Let $f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}, x \in \mathbb{R}$ wherea, b and d…
Let $f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}, x \in \mathbb{R}$ wherea, b
and d are non-zero real constants. Then :
- $f$ is an increasing function of $x$
- $f$ is a decreasing function of $x$
- $f^{\prime}$ is not a continuous function of $x$
- $f$ is neither increasing nor decreasing function of $x$
Solution
$f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}-\frac{(d-x)}{\sqrt{b^{2}+(d-x)^{2}}}$
$=\frac{x}{\sqrt{a^{2}+x^{2}}}+\frac{(x-d)}{\sqrt{b^{2}+(x-d)^{2}}}$
$\begin{array}{l}
f^{\prime}(x)=\frac{\sqrt{a^{2}+x^{2}}-\frac{x(2 x)}{2 \sqrt{a^{2}+x^{2}}}}{\left(a^{2}+x^{2}\right)} \\
=\frac{a^{2}+x^{2}-x^{2}}{\left(a^{2}+x^{2}\right)^{3 / 2}}+\frac{b^{2}+(x-d)^{2}-(x-d)^{2}}{\left(b^{2}+(x-d)^{2}\right)^{3 / 2}} \\
+\frac{\sqrt{b^{2}+(x-d)^{2}}-\frac{(x-d) 2(x-d)}{2 \sqrt{b^{2}+(x-d)^{2}}}}{\left(a^{2}+(x-d)^{2}\right)} \\
\Rightarrow f^{\prime}(x)>0, \square x \in R
\end{array}$
$\Rightarrow f(x)$ is increasing function.
Hence, $f(x)$ is increasing function
Asked in: JEE Main 2019 (11 Jan Shift 2)
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