JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 21)
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
50
Explanation
$$\because$$ $${a_1},{a_2},{a_3},{a_4}$$
$$\therefore$$ $${a_2} = p - 3d,\,{a_2} = p - d,\,{a_3} = p + d$$ and $${a_4} = p + 3d$$
Where $$d > 0$$
$$\because$$ $$\left| {f({a_i})} \right| = 500$$
$$ \Rightarrow |9{d^2} - q| = 500$$
and $$|{d^2} - q| = 500$$ ..... (i)
either $$9{d^2} - q = {d^2} - q$$
$$ \Rightarrow d = 0$$ not acceptable
$$\therefore$$ $$9{d^2} - q = q - {d^2}$$
$$\therefore$$ $$5{d^2} - q = 0$$ ..... (ii)
Roots of $$f(x) = 0$$ are $$p + \sqrt q $$ and $$p - \sqrt q $$
$$\therefore$$ absolute difference between roots $$ = \left| {2\sqrt q } \right| = 50$$
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