JEE Main 2022 — Mathematics Algebra

Given,

$$f(x)=0\Rightarrow {(x-p)}^{2}-q=0$$f(x)=0⇒(x−p)2−q=0

So, roots are $p+\sqrt{q},p-\sqrt{q}$p+q​,p−q​

Now absolute difference between roots will be $2\sqrt{q}$2q​.

Now given ${a}_{1},{a}_{2},{a}_{3},{a}_{4}$a1​,a2​,a3​,a4​ are in $A.P$A.P and its mean is $p$p

Now let ${a}_{1},{a}_{2},{a}_{3},{a}_{4}$a1​,a2​,a3​,a4​ be {a}_{1}=p-3d,{a}_{2}=p-d,{a}_{3}=p+d&{a}_{4}=p+3d

Now given $|f({a}_{i})|=500$∣f(ai​)∣=500

So, $|f({a}_{4})|=500$∣f(a4​)∣=500

$$\Rightarrow |{({a}_{4}-p)}^{2}-q|=500$$⇒∣(a4​−p)2−q∣=500

$$\Rightarrow {({a}_{4}-p)}^{2}-q=500$$⇒(a4​−p)2−q=500

$$\Rightarrow 9{d}^{2}-q=500....(1)$$⇒9d2−q=500....(1)

And using $|f({a}_{i})|=500\forall i=1,2,3,4$∣f(ai​)∣=500∀i=1,2,3,4

We get ${|f({a}_{4})|}^{2}={|f({a}_{3})|}^{2}$∣f(a4​)∣2=∣f(a3​)∣2

$$\Rightarrow {({({a}_{4}-p)}^{2}-q)}^{2}={({({a}_{3}-p)}^{2}-q)}^{2}$$⇒((a4​−p)2−q)2=((a3​−p)2−q)2

$$\Rightarrow 9{d}^{2}-q+{d}^{2}-q=0$$⇒9d2−q+d2−q=0

So, $2q=10{d}^{2}\Rightarrow q=5{d}^{2}$2q=10d2⇒q=5d2

$$\Rightarrow {d}^{2}=\frac{q}{5}$$⇒d2=5q​

From equation $(1)$(1) we get,

$$9(\frac{q}{5})-q=500$$9(5q​)−q=500

$$\Rightarrow \frac{4q}{5}=500$$⇒54q​=500

$$\Rightarrow q=\frac{500\times 5}{4}$$⇒q=4500×5​

Now absolute difference is $2\sqrt{q}=2\times \sqrt{\frac{500\times 5}{4}}=2\times \frac{50}{2}=50$2q​=2×4500×5​​=2×250​=50