KVPY SB SX Exam 2011 Question Paper

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Question : 86 of 120

Marks: +1, -0

Let

f(x)=x^3+ax^2+bx+c

, where a, b, c are real numbers. If

f(x)

has a local minimum at

x = 1

and a local maximum at

x = -1/3

and

f(2)=0

, then

∫↖{1}↙{-1}f(x)dx

equals

$14/3$
$-14/3$
$7/3$
$-7/3$

Solution:

f'(x)=3(x^2-2/3x-1/3)

=3x^2-2x-1

f(x) = x^3 - x^2 - x + λ

f(2) = 8 – 4 – 2 + λ = 0 ⇒ λ = – 2

f(x) = x^3 - x^2 - x - 2

∫↖{1}↙{-1}f(x)dx=-2∫↖{1}↙{0}(x^2+2)dx

=-2(1/3+2)=-14/3

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