Mathematics · 2019
JEE Main · 11 January 2019, Shift 2 · Q72
Let f(x)=x/(√(a^2+x^2))-(d -x)/(√(b^2+( d -x)^2)), x ∈ R, where a, b and d are non-zero real constants. Then:
Let $\displaystyle f(x)=\frac{x}{\sqrt{\mathrm{a}^2+x^2}}-\frac{\mathrm{d}-x}{\sqrt{\mathrm{~b}^2+(\mathrm{d}-x)^2}}, x \in \mathbf{R}$,
where $\displaystyle \mathrm{a}, \mathrm{b}$ and d are non-zero real constants. Then :
Official answer
(1)
$\displaystyle f$ is an increasing function of $\displaystyle x$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.