KVPY SX SB Exam 2015 Question Paper

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Question : 86 of 120

Marks: +1, -0

Let

\f(x) = α x^2- 2 +1/x

where

α

is a real constant. The smallest

α

for which f(x) ≥ 0 for all x > 0 is -

$2^2/3^3$
$2^3/3^3$
$2^4/3^3$
$2^5/3^3$

Solution:

\f(x) = α x^2- 2 +1/x

\f(x) = {α x^3- 2x +1}/x

∀x(0,∞)

so

ϕ(x) =αx^ 3 - 2x + 1

should be positive

ϕ(x) =αx^ 3 - 2x + 1

ϕ'(x) = 3αx ^2 - 2 = 0

x=±√{2/{3α}}

Clearly

x=√{2/{3α}}

point of minima

ϕ(√{2/{3α}})≥0

√{2/{3α}}\{α.2/{3α}-2\}+1≥0

√{2/{3α}} ({ -4/3})+1≥0

√{2/{3α}} ({ 4/3})≤1

√{2/{3α}} ≤3/4

2/{3α} ≤3^2/4^2

α≥32/27

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