NIMCET 2008 Question Paper with solutions

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Question : 17 of 120

Marks: +1, -0

If a, b, c are the roots of the equation

x^{3}-3 p x^{2}+3 q x-1=0,

then the centroid of the triangle with vertices

(a, {1}/{a}),(b, {1}/{b})

and

(c, {1}/{c})

is at the point

$(p, q)$
$( {p}/{3}, {q}/{3})$
$(p+q, p-q)$
$(3 p, 3 q)$

Solution:

a.

b

and

c

are roots of

x^{3}-3 p x^{2}+3 a x-1=0

⇒ \;\; a+b+c=3 p ; a b+b c+c a=3 q

and

a b c=1

⇒ \;\; \;{1}/{a}+\;{1}/{b}+\;{1}/{c}=\;{3 q}/{a b c}

⇒ \;\; \;{1}/{a}+\;{1}/{b}+\;{1}/{c}=3 q

Now, centroid of triangle with vertices

(a, \;{1}/{a}) ⋅(b, \;{1}/{b}),(c, \;{1}/{c})

is

(\;{a+b+c}/{3} ⋅ \;{\;{1}/{a}+{1}/{b}+{1}/{c}}/{3})=(p, q)

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