JEE Advanced 2022 Paper 1 · Mathematics · Quadratic Equations · hard

Let x,y,z be real numbers with x+y+z=4 and x^2+y^2+z^2=6. Let M and m be the greatest and the least possible values of the product xyz. Find M-m.

Numerical answer

Let

x,y,z

be real numbers with

x+y+z=4

and

x^2+y^2+z^2=6

. Let

M

and

m

be the greatest and the least possible values of the product

xyz

. Find

M-m

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 0.15

From the two conditions,

xy+yz+zx=\frac{16-6}{2}=5

. Hence

x,y,z

are the roots of

t^3-4t^2+5t-p=0

where

p=xyz

. All three roots are real exactly when the horizontal line

s=p

meets the graph of

g(t)=t^3-4t^2+5t

in three points (counting multiplicity). Now

g'(t)=3t^2-8t+5=(3t-5)(t-1)

, so

g

has a local maximum

g(1)=2

and a local minimum

g(5/3)=\frac{50}{27}

. Three real roots occur iff

\frac{50}{27}\le p\le2

. Thus

M=2

(roots

1,1,2

) and

m=\frac{50}{27}

(roots

\frac53,\frac53,\frac23

), and

M-m=\frac{4}{27}\approx0.148

, i.e.

0.15

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