If a, b, c, d and p are distinct real numbers such that (a^2+b^2+c^2) p^2-2 p(a b+b c+c d)+(b^2+. .c^2+d^2) ≤ 0 , then

Correct answer: D. a, b, c, d are in G.P. The given relation can be written as (a^2 p^2-2 a b p+b^2)+(b^2 p^2+c^2-2 b p c)+ (c^2 p^2+d^2-2 p c d) ≤ 0 or (a p-b)^2+(b p-c)^2+(c p-d)^2 ≤ 0 ...(1) Since a, b, c, d and p are all real, the inequality (1) is possible only when each of factor is zero. i.e., a p-b=0, b p-c=0 and c p-d=0 or p=(b)/(a)=(c)/(b)=(d)/(c) or a, b, c, d are in G.P.

JEE Main 201212 MayOnline

Exam: JEE Main
Year: 2012
Date: 12 May
Mode: Online
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2012 · 12 May · Online · Mathematics · Quadratic Equation

If a, b, c, d and p are distinct real numbers such that...

a, b, c, d and p

are distinct real numbers such that \left(a^2+b^2+c^2\right) p^2-2 p(a b+b c+c d)+\left(b^2+\right. \left.c^2+d^2\right) \leq 0$, then

Options

A $a, b, c, d$ are in A.P.
B $a b=c d$
C $a c=b d$
D $a, b, c, d$ are in G.P.

Correct answer

D. $a, b, c, d$ are in G.P.

Step-by-step solution

The given relation can be written as

\left(a^2 p^2-2 a b p+b^2\right)+\left(b^2 p^2+c^2-2 b p c\right)+ \left(c^2 p^2+d^2-2 p c d\right) \leq 0

\quad(a p-b)^2+(b p-c)^2+(c p-d)^2 \leq 0

...(1) Since

a, b, \mathrm{c}, d

and p are all real, the inequality (1) is possible only when each of factor is zero. i.e., a p-b=0, b p-c=0 and

c p-d=0 or \quad p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c} or a, b, c, d

are in G.P.

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