Medium INTEGER +4 / -1 PYQ · JEE Mains 2022

For $\mathrm{p}, \mathrm{q} \in \mathbf{R}$, consider the real valued function $f(x)=(x-\mathrm{p})^{2}-\mathrm{q}, x \in \mathbf{R}$ and $\mathrm{q}>0$. Let $\mathrm{a}_{1}$, $\mathrm{a}_{2^{\prime}}$ $\mathrm{a}_{3}$ and $\mathrm{a}_{4}$ be in an arithmetic progression with mean $\mathrm{p}$ and positive common difference. If $\left|f\left(\mathrm{a}_{i}\right)\right|=500$ for all $i=1,2,3,4$, then the absolute difference between the roots of $f(x)=0$ is ___________.

Answer (integer) 50

Solution

<p>$\because$ ${a_1},{a_2},{a_3},{a_4}$</p> <p>$\therefore$ ${a_2} = p - 3d,\,{a_2} = p - d,\,{a_3} = p + d$ and ${a_4} = p + 3d$</p> <p>Where $d > 0$</p> <p>$\because$ $\left| {f({a_i})} \right| = 500$</p> <p>$\Rightarrow |9{d^2} - q| = 500$</p> <p>and $|{d^2} - q| = 500$ ..... (i)</p> <p>either $9{d^2} - q = {d^2} - q$</p> <p>$\Rightarrow d = 0$ not acceptable</p> <p>$\therefore$ $9{d^2} - q = q - {d^2}$</p> <p>$\therefore$ $5{d^2} - q = 0$ ..... (ii)</p> <p>Roots of $f(x) = 0$ are $p + \sqrt q$ and $p - \sqrt q$</p> <p>$\therefore$ absolute difference between roots $= \left| {2\sqrt q } \right| = 50$</p>

About this question

Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations

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