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 is

Question

For p,q in R, consider the real valued functionn f(x) = (x - p)^2 - q,x in R and q > 0. Letn a_1,a_2,a_3 and a_4 be in an arithmetic progressionn with mean p and positive common difference. Ifn | f( a_i ) | = 500 for all i = 1,2,3,n 4 , then the absolute difference between the roots of f(x) = 0 is

DivineJEE · Accepted Answer

We have f(x)=(x-p)^2-q, x in R, q>0.nnnn n f(x)=0 n => (x-p)^2-q=0 => x=p+sqrt(q), p-sqrt(q)nn nnn nLet a_1=a, a_2=a+d, a_3=a+2 d and a_4=a+3 dnnnnTherefore p=(4 a+6 d / 4) => a=(p-(3 / 2) d)nnn nNow, |f(a_i)|=500nnnn n |f(a_1)|=500 n [\(p-(3 / 2) d)-p\^2-q]=500 => (9 / 4) d^2-q=500 n |(a_1-p)^2-q|=500nn nnn nNow, |f(a_1)|^2=|f(a_2)|^2nnnn n => [(a_1-p)^2-q]^2=[(a_2-p)^2-q]^2 n => (a_1-p)^4+q^2-2 q(a_1-p)^2=(a_2-p)^4+q^2-2 q(a_2-p)^2 n => (a_1-p)^4-2 q(a_1-p)^2=(a_2-p)^4-2 q(a_2-p)^2 n => ((-3 / 2) d)^4-2 q((-3 / 2) d)^2=((-3 / 2) d+d)^4-2 q((-3 / 2) d+d)^2 n => d^2=(4 q / 5)nn nnn nFrom (i), we have (9 / 4) x (4 q / 5)-q=500 => q=625nn∴ Absolute difference between the roots isnnnn=2 sqrt(q)=2 x 25=50nnn

Explanation

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