JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Quadratic Equations and Inequalities

For

p, q \in R

, consider the real valued function

f (x) = (x - p)^{2} - q, x \in R

and

q > 0

. Let

a_{1}, a_{2}, a_{3}

and

a_{4}

be in an arithmetic progression with mean p and positive common difference. If

∣ f (a_{i}) ∣ = 500

for all

i = 1, 2, 3, 4

, then the absolute difference between the roots of

f (x) = 0

Answer: 50

Solution

The roots of

f (x) = 0

are

p \pm q

, so the absolute difference is

2 q

.

Let the AP be

a_{1} = p - \frac{3 d}{2}, a_{2} = p - \frac{d}{2}, a_{3} = p + \frac{d}{2}, a_{4} = p + \frac{3 d}{2}

(mean is

p

).

From

∣ f (a_{4}) ∣ = 500

(a_{4} - p)^{2} - q = 500 \Rightarrow \frac{9 d ^{2}}{4} - q = 500

... (1)

From

∣ f (a_{1}) ∣ = ∣ f (a_{2}) ∣

(a_{1} - p)^{2} - q = - (a_{2} - p)^{2} + q

(since they have opposite signs).

This gives

\frac{9 d ^{2}}{4} - q + \frac{d ^{2}}{4} - q = 0 \Rightarrow q = \frac{5 d ^{2}}{4}

.

Substituting in (1):

\frac{9 d ^{2}}{4} - \frac{5 d ^{2}}{4} = 500 \Rightarrow d^{2} = 500 \Rightarrow q = 625

.

Therefore,

2 q = 2 \times 25 = 50

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer